Samacheer Kalvi Class 11 Chemistry Solutions Chapter 11 Fundamentals of Organic Chemistry

Textbook Evaluation:

I. Choose the Best Answer:

 

Question 1. Select the molecule which has only one \( \pi \) bond.
(a) \( \text{CH}_3 - \text{CH} = \text{CH} - \text{CH}_3 \)
(b) \( \text{CH}_3 - \text{CH} = \text{CH} - \text{CHO} \)
(c) \( \text{CH}_3 - \text{CH} = \text{CH} - \text{COOH} \)
(d) All of the options
Answer: (a) \( \text{CH}_3 - \text{CH} = \text{CH} - \text{CH}_3 \)
In simple words: This molecule has one double bond, which contains one sigma (\( \sigma \)) bond and one pi (\( \pi \)) bond. The other options have either more pi bonds (in CHO or COOH groups) or are named differently.

🎯 Exam Tip: Remember that a double bond has one \( \sigma \) and one \( \pi \) bond, and a triple bond has one \( \sigma \) and two \( \pi \) bonds. Functional groups like carbonyl (C=O) in aldehydes/ketones or carboxyl (COOH) in carboxylic acids also contain a \( \pi \) bond.

 

Question 2. In the hydrocarbon \( \text{CH}_3 – \text{CH}_2 – \text{CH} = \text{CH} – \text{CH}_2 – \text{C} \equiv \text{CH} \) the state of hybridization of carbon 1, 2, 3, 4 and 7 are in the following sequence.
(a) sp, sp, sp³, sp², sp³
(b) sp², sp, sp³, sp², sp³
(c) sp, sp, sp², sp, sp³
(d) none of these
Answer: (a) sp, sp, sp³, sp², sp³
In simple words: When we number the carbons from the triple bond end, Carbon 1 (\( \text{C}\equiv \text{CH} \)) and Carbon 2 (\( \text{CH} \equiv \text{C} \)) are sp hybridized. Carbon 3 (\( \text{CH}_2 \)) is sp³ hybridized. Carbon 4 (\( \text{CH} \)) is sp² hybridized. Carbon 7 (\( \text{CH}_3 \)) is sp³ hybridized.

🎯 Exam Tip: Carbon atoms involved in a triple bond are sp hybridized. Those in a double bond are sp² hybridized. Those only in single bonds are sp³ hybridized. Always start numbering from the end closest to the functional group or multiple bond with the lowest locant.

 

Question 3. The general formula for alkadiene is
(a) \( \text{C}_{\text{n}}\text{H}_{2\text{n}} \)
(b) \( \text{C}_{\text{n}}\text{H}_{2\text{n}} - 1 \)
(c) \( \text{C}_{\text{n}}\text{H}_{2\text{n}} - 2 \)
(d) \( \text{C}_{\text{n}}\text{H}_{\text{n}} - 2 \)
Answer: (c) \( \text{C}_{\text{n}}\text{H}_{2\text{n}} - 2 \)
In simple words: Alkanes have the formula \( \text{C}_{\text{n}}\text{H}_{2\text{n}+2} \). Each double bond or ring reduces the hydrogen count by two. An alkadiene has two double bonds, so it loses four hydrogens compared to an alkane. This makes the formula \( \text{C}_{\text{n}}\text{H}_{2\text{n}-2} \).

🎯 Exam Tip: Remember the general formulas for different homologous series: alkanes (\( \text{C}_{\text{n}}\text{H}_{2\text{n}+2} \)), alkenes/cycloalkanes (\( \text{C}_{\text{n}}\text{H}_{2\text{n}} \)), and alkynes/alkadienes (\( \text{C}_{\text{n}}\text{H}_{2\text{n}-2} \)).

 

Question 4. Structure of the compound whose IUPAC name is 5, 6 – dimethylhept – 2 – ene is
(a) The structure corresponding to 5,6-dimethylhept-2-ene
(b) The structure with a seven-carbon chain, a double bond, and two methyl groups at different positions
(c) The structure with an eight-carbon chain and two methyl groups
(d) The structure with a six-carbon chain and an ethyl group
Answer: (a) The structure corresponding to 5,6-dimethylhept-2-ene
In simple words: The correct structure has a main chain of seven carbons with a double bond between the second and third carbons. It also has two methyl groups attached to the fifth and sixth carbons.

🎯 Exam Tip: To draw a compound from its IUPAC name, first identify the longest carbon chain and the position of any double/triple bonds. Then, add the substituents at their specified positions. Always ensure each carbon has four bonds.

 

Question 5. The IUPAC name of the compound is
(a) 2, 3 – Dimethylheptane
(b) 3 – methyl – 4 – ethyloctane
(c) 5 – ethyl – 6- methyloctane
(d) 4 – Ethyl – 3 methyloctane
Answer: (d) 4 – Ethyl – 3 methyloctane
In simple words: The compound has an eight-carbon chain as its backbone (octane). It has an ethyl group at the fourth carbon and a methyl group at the third carbon when numbered appropriately.

🎯 Exam Tip: When naming complex alkanes, find the longest continuous carbon chain. Then, number the chain to give the substituents the lowest possible set of numbers. List substituents alphabetically.

 

Question 6. Which one of the following names does not fit a real name?
(a) 3 – Methyl – 3- hexanone
(b) 4 – Methyl – 3 – hexanone
(c) 3 – Methyl – 3 hexanol
(d) 2 – Methyl cyclo hexanone
Answer: (a) 3 – Methyl – 3- hexanone
In simple words: A hexanone has a carbonyl (C=O) group. The carbon in this group already forms two bonds with oxygen and two with other carbons in the chain. So, it cannot also have a methyl group attached to the same carbon atom (carbon 3) as the ketone, because that carbon would then have five bonds. This makes the name chemically impossible.

🎯 Exam Tip: Always visualize the structure implied by an IUPAC name to check for valency violations, especially with functional groups. A carbon atom can only form four bonds.

 

Question 7. The IUPAC name of the compound \( \text{CH}_3 – \text{CH} = \text{CH} – \text{C} \equiv \text{CH} \) is
(a) Pent – 4- yn – 2 – ene
(b) Pent – 3- en – 1- yne
(c) Pent – 2 – en – 4 – yne
(d) Pent – 1 yn – 3 – ene
Answer: (b) Pent – 3- en – 1- yne
In simple words: This molecule has five carbon atoms (pent-), with both a double bond and a triple bond. When numbering such a chain, we prioritize giving the multiple bonds the lowest possible overall numbers. If the numbers for the multiple bonds are the same from both ends, the double bond gets preference for the lower number. Here, numbering from the right gives a triple bond at C1 and a double bond at C3, which is the lowest possible set of locants (1,3).

🎯 Exam Tip: For compounds with both double and triple bonds, ensure the numbering gives the lowest possible set of numbers to both. If both numbering directions result in the same set of numbers, give priority to the double bond for the lower number.

 

Question 8. IUPAC name of is
(a) 3, 4, 4 – Trimethylheptane
(b) 2 – Ethyl – 3, 3, – dimethyl heptane
(c) 3, 4, 4 – Trimethyloctane
(d) 2 – Butyl – 2 – methyl – 3 ethyl – butane
Answer: (c) 3, 4, 4 – Trimethyloctane
In simple words: The compound has a main chain of eight carbon atoms (octane). It has one methyl group at the third carbon and two methyl groups at the fourth carbon.

🎯 Exam Tip: When naming compounds with multiple identical substituents, use prefixes like "di-", "tri-", "tetra-" and list the position for each substituent, even if they are on the same carbon.

 

Question 9. The IUPAC name of is
(a) 2,4,4 – Trimethylpent – 2 – ene
(b) 2,4,4 – Trimethylpent – 3 – ene
(c) 2,2,4 – Trimethylpent – 3 – ene
(d) 2,2,4 – Trirnethylpent – 2 – ene
Answer: (c) 2,2,4 – Trimethylpent – 3 – ene
In simple words: The main chain of the compound has five carbons (pent-). There is a double bond at the third carbon, and three methyl groups are attached: two at the second carbon and one at the fourth carbon.

🎯 Exam Tip: Always prioritize the double bond for the lowest possible number if there is a choice in numbering. Then, assign numbers to the substituents to keep their locants as low as possible.

 

Question 10. The IUPAC name of the compound is
(a) 3 – Ethyl – 2 – hexene
(b) 3 – Propyl – 3 – hexene
(c) 4 – Ethyl – 4 – hexene
(d) 3 – Propyl – 2 – hexene
Answer: (a) 3 – Ethyl – 2 – hexene
In simple words: The compound has a main chain of six carbon atoms (hex-). There is a double bond between the second and third carbons. An ethyl group is attached to the third carbon.

🎯 Exam Tip: In alkenes, the carbon atoms of the double bond must be part of the main chain. The chain is numbered to give the double bond the lowest possible number.

 

Question 11. The IUPAC name of the compound is
(a) 2 – Hydroxypropionic acid
(b) 2 – Hydroxy Propanoic acid
(c) Propan – 2 – ol – 1 – oic acid
(d) 1 – Carboxyethanol
Answer: (b) 2 – Hydroxy Propanoic acid
In simple words: The compound has a three-carbon chain with a carboxylic acid group, making it a propanoic acid. A hydroxyl group (–OH) is attached to the second carbon atom.

🎯 Exam Tip: For carboxylic acids, the carbon of the -COOH group is always assigned as carbon 1 in the main chain. Other substituents are then numbered relative to this carbon.

 

Question 12. The IUPAC name of is
(a) 2 – Bromo – 3- methyl butanoic acid
(b) 2 – methyl – 3 bromo butanoic acid
(c) 3 – Bromo – 2 – methylbutanoic acid
(d) 3 – Bromo – 2, 3 – dimethyl propanoic acid
Answer: (c) 3 – Bromo – 2 – methylbutanoic acid
In simple words: The compound has a four-carbon chain with a carboxylic acid group, making it a butanoic acid. A bromo group is on the third carbon, and a methyl group is on the second carbon.

🎯 Exam Tip: When multiple substituents are present, number the carbon chain to give the primary functional group (like -COOH) the lowest possible number, usually C1. Then, name the substituents alphabetically, with their corresponding numbers.

 

Question 13. The structure of isobutyl group in an organic compound is
(a) \( \text{CH}_3 – \text{CH}_2 – \text{CH}_2 – \text{CH}_2 \)
(b) The structure for the tertiary-butyl group
(c) The structure corresponding to the isobutyl group
(d) The structure for the sec-butyl group
Answer: (c) The structure corresponding to the isobutyl group
In simple words: The isobutyl group is a branched alkyl group with four carbon atoms. It has a parent chain of three carbons, with a methyl group on the second carbon, and the attachment point is on the first carbon of that three-carbon chain.

🎯 Exam Tip: Learn to recognize common branched alkyl groups like isopropyl, isobutyl, sec-butyl, and tert-butyl. The isobutyl group has the structure \( (\text{CH}_3)_2\text{CH}-\text{CH}_2- \).

 

Question 14. The number of stereoisomers of 1, 2 – dihydroxy cyclopentane
(a) 1
(b) 2
(c) 3
(d) 4
Answer: (c) 3
In simple words: This molecule has two chiral centers. It exists as a cis (meso) form, which has a plane of symmetry and is optically inactive. It also exists as a pair of trans enantiomers, which are chiral and optically active. In total, there are three distinct stereoisomers.

🎯 Exam Tip: When determining the number of stereoisomers, consider both chiral centers and molecular symmetry. A molecule with 'n' chiral centers can have a maximum of \( 2^n \) stereoisomers, but meso compounds can reduce this number if internal symmetry exists.

 

Question 15. Which of the following is optically active?
(a) 3 – Chloropentane
(b) 2- Chloro propane
(c) Meso – tartaric acid
(d) Glucose
Answer: (d) Glucose
In simple words: An optically active compound can rotate the plane of polarized light. This usually happens when a molecule has chiral centers and lacks a plane of symmetry. Glucose has several chiral carbon atoms and no internal plane of symmetry, making it optically active. The other options are either achiral or are meso compounds (optically inactive).

🎯 Exam Tip: To be optically active, a molecule must be chiral (non-superimposable on its mirror image). This generally requires a chiral center and the absence of any internal plane or center of symmetry. Meso compounds, despite having chiral centers, are optically inactive due to internal symmetry.

 

Question 16. The isomer of ethanol is
(a) acetaldehyde
(b) dimethyl ether
(c) acetone
(d) methyl carbinol
Answer: (b) dimethyl ether
In simple words: Isomers have the same molecular formula but different structures. Ethanol has the formula \( \text{C}_2\text{H}_6\text{O} \) and is an alcohol. Dimethyl ether also has the formula \( \text{C}_2\text{H}_6\text{O} \) but is an ether. They are functional group isomers.

🎯 Exam Tip: Functional isomers are compounds that have the same molecular formula but different functional groups. Common pairs include alcohols and ethers, aldehydes and ketones, and carboxylic acids and esters.

 

Question 17. How many cyclic and acyclic isomers are possible for the molecular formula \( \text{C}_3\text{H}_6\text{O} \)?
(a) 4
(b) 5
(c) 9
(d) 10
Answer: (c) 9
In simple words: For the formula \( \text{C}_3\text{H}_6\text{O} \), there are nine different structural arrangements possible. These include compounds like propanal, propanone, allyl alcohol, methoxyethene, cyclopropanol, oxetane, methyloxirane, and the enol forms prop-1-en-1-ol and prop-2-en-2-ol. These isomers can be either straight-chain or cyclic.

🎯 Exam Tip: To find all possible isomers, calculate the Degree of Unsaturation (DBE). Then, systematically draw structures for different functional groups (alcohols, ethers, aldehydes, ketones) and ring systems (cyclic alcohols, cyclic ethers), including unsaturated forms and stable tautomers (if applicable).

 

Question 18. Which one of the following shows functional isomerism?
(a) ethylene
(b) Propane
(c) ethanol
(d) \( \text{CH}_2\text{Cl}_2 \)
Answer: (c) ethanol
In simple words: Functional isomerism occurs when two different compounds have the same molecular formula but different types of functional groups. Ethanol (\( \text{C}_2\text{H}_6\text{O} \), an alcohol) has a functional isomer called dimethyl ether (\( \text{CH}_3\text{OCH}_3 \), an ether).

🎯 Exam Tip: Functional isomerism is a key concept. Always think of common pairs that exhibit this, such as alcohols and ethers, aldehydes and ketones, and carboxylic acids and esters.

 

Question 19. \( \text{CH}_2=\text{C}(\text{OH})–\text{CH}_3 \) and \( \text{CH}_3–\text{CO}–\text{CH}_3 \) are
(a) resonating structure
(b) tautomers
(c) Optical isomers
(d) Conformers
Answer: (b) tautomers
In simple words: Tautomers are isomers of a compound that quickly change into each other. These two structures are an enol (double bond with an -OH group) and a keto (ketone) form, which exist in equilibrium and are examples of keto-enol tautomerism.

🎯 Exam Tip: Tautomerism involves the migration of a hydrogen atom and a change in the position of a double bond. Keto-enol tautomerism is the most common type, where an aldehyde or ketone rapidly converts to its corresponding enol form.

 

Question 20. Nitrogen detection in an organic compound is carried out by Lassaigne’s test. The blue colour formed is due to the formation of
(a) \( \text{Fe}_3[\text{Fe(CN)}_6]_2 \)
(b) \( \text{Fe}_4[\text{Fe(CN)}_6]_3 \)
(c) \( \text{Fe}[\text{Fe(CN)}_6]_2 \)
(d) \( \text{Fe}_3[\text{Fe(CN)}_6]_3 \)
Answer: (b) \( \text{Fe}_4[\text{Fe(CN)}_6]_3 \)
In simple words: In Lassaigne's test, nitrogen from the organic compound forms sodium cyanide, which then reacts with iron(II) sulfate and iron(III) chloride to create a deep blue pigment. This pigment is called Prussian blue, which has the chemical formula \( \text{Fe}_4[\text{Fe(CN)}_6]_3 \).

🎯 Exam Tip: Knowing the characteristic colors and precise chemical formulas of precipitates formed in qualitative organic analysis tests (like Lassaigne's test) is crucial for scoring well.

 

Question 21. Lassaigne’s test for the detection of nitrogen fails in
(a) \( \text{H}_2\text{N} – \text{CO} – \text{NH.NH}_2.\text{HCl} \)
(b) \( \text{NH}_2 – \text{NH}_2.\text{HCl} \)
(c) \( \text{C}_6\text{H}_5 – \text{NH} – \text{NH}_2.\text{HCl} \)
(d) \( \text{C}_6\text{H}_5\text{CONH}_2 \)
Answer: (c) \( \text{C}_6\text{H}_5 – \text{NH} – \text{NH}_2.\text{HCl} \)
In simple words: Lassaigne's test works by converting nitrogen in organic compounds into sodium cyanide. However, for some nitrogen-containing compounds, especially certain hydrazine derivatives, nitrogen is not easily converted into cyanide, causing the test to fail. Phenylhydrazine hydrochloride is one such compound.

🎯 Exam Tip: Lassaigne's test can fail for compounds where nitrogen is directly bonded to another nitrogen atom (like in hydrazine derivatives, diazonium salts) or to oxygen (like in nitro compounds), as they may not form sodium cyanide effectively under the test conditions.

 

Question 22. Connect pair of compounds which give blue colouration / precipitate and white precipitate respectively, when their Lassaigne’s test is separately done.
(a) \( \text{NH}_2\text{NH}_2.\text{HCl} \) and \( \text{ClCH}_2 – \text{CHO} \)
(b) \( \text{NH}_2\text{CSNH}_2 \) and \( \text{CH}_3 – \text{CH}_2\text{Cl} \)
(c) \( \text{NH}_2\text{CH}_2\text{COOH} \) and \( \text{NH}_2\text{CONH}_2 \)
(d) \( \text{C}_6\text{H}_5\text{NH}_2 \) and \( \text{ClCH}_2 – \text{CHO} \)
Answer: (d) \( \text{C}_6\text{H}_5\text{NH}_2 \) and \( \text{ClCH}_2 – \text{CHO} \)
In simple words: The first compound, aniline (\( \text{C}_6\text{H}_5\text{NH}_2 \)), contains nitrogen, which will give a blue color in Lassaigne's test. The second compound, chloroacetaldehyde (\( \text{ClCH}_2 – \text{CHO} \)), contains chlorine, which will produce a white precipitate (silver chloride) in Lassaigne's test when testing for halogens.

🎯 Exam Tip: For this type of question, identify which elements are detected by Lassaigne's test for each characteristic color/precipitate. Blue indicates nitrogen, and a white precipitate (AgCl) indicates chlorine. Match the compounds to these criteria.

 

Question 23. Sodium nitropruside reacts with sulphide ion to give a purple colour due to the formation of
(a) \( [\text{Fe (CN)}_5\text{NO}]^{3-} \)
(b) \( [\text{Fe (NO)CN}]^{+} \)
(c) \( [\text{Fe (CN)}_5\text{NOS}]^{4-} \)
(d) \( [\text{Fe (CN)}_5\text{NOS}]^{3-} \)
Answer: (c) \( [\text{Fe (CN)}_5\text{NOS}]^{4-} \)
In simple words: The purple color that appears when sodium nitroprusside reacts with sulfide ions is caused by the formation of a complex ion called pentacyanonitrosylsulfidoferrate(II). This complex has a \( 4- \) charge and is responsible for the distinct color in the test for sulfur.

🎯 Exam Tip: The sodium nitroprusside test is specific for sulfide ions. The key is to remember the formula and charge of the purple complex ion formed, \( [\text{Fe (CN)}_5\text{NOS}]^{4-} \).

 

Question 24. An organic Compound weighing 0.15 g gave on carius estimation, 0.12 g of silver bromide. The percentage of bromine in the Compound will be close to
(a) 46 %
(b) 34 %
(c) 3.4 %
(d) 4.6 %
Answer: (b) 34 %
In simple words: We calculate the percentage of bromine by using the masses of the organic compound and the silver bromide formed. We also use the atomic mass of bromine and the molar mass of silver bromide in the calculation.

🎯 Exam Tip: For Carius method calculations, remember the formula: Percentage of halogen = \( \frac{\text{Atomic mass of halogen}}{\text{Molar mass of AgX}} \times \frac{\text{Mass of AgX}}{\text{Mass of organic compound}} \times 100 \).

 

Question 25. A sample of 0.5 g of an organic compound was treated according to Kjeldahl’s method. The ammonia evolved was absorbed in 50mL of 0.5 M \( \text{H}_2\text{SO}_4 \). The remaining acid after neutralization by ammonia consumed 80 mL of 0.5 M NaOH. The percentage of nitrogen in the organic compound is.
(a) 14 %
(b) 28 %
(c) 42 %
(d) 56 %
Answer: (b) 28 %
In simple words: First, find how much sulfuric acid reacted with the ammonia. Then, use the amount of ammonia to find the mass of nitrogen. Finally, divide the mass of nitrogen by the total mass of the compound and multiply by 100 to get the percentage.

🎯 Exam Tip: In Kjeldahl's method, the amount of ammonia produced is determined by back titration. Calculate the moles of initial acid, moles of excess acid (reacted with NaOH), and then the moles of acid that reacted with ammonia. Each mole of ammonia corresponds to one mole of nitrogen.

 

Question 26. In an organic compound, phosphorus is estimated as
(a) \( \text{Mg}_2\text{P}_2\text{O}_7 \)
(b) \( \text{Mg}_3(\text{PO}_4)_2 \)
(c) \( \text{H}_3\text{PO}_4 \)
(d) \( \text{P}_2\text{O}_5 \)
Answer: (a) \( \text{Mg}_2\text{P}_2\text{O}_7 \)
In simple words: When phosphorus in an organic compound is estimated, it is first changed into phosphoric acid. This acid is then made to form magnesium ammonium phosphate, which is finally heated strongly to turn into magnesium pyrophosphate, \( \text{Mg}_2\text{P}_2\text{O}_7 \). This final compound is weighed to find the amount of phosphorus.

🎯 Exam Tip: Remember that in the quantitative estimation of phosphorus (like in Carius method), the phosphorus is ultimately weighed as magnesium pyrophosphate (\( \text{Mg}_2\text{P}_2\text{O}_7 \)).

 

Question 27. Ortho and para – nitro phenol can be separated by
(a) azeotropic distillation
(b) destructive distillation
(c) steam distillation
(d) cannot be separated
Answer: (c) steam distillation
In simple words: Ortho-nitrophenol has internal hydrogen bonding, making it more volatile. Para-nitrophenol has strong external hydrogen bonding, making it less volatile. Because they have different boiling points, they can be separated by steam distillation. The ortho isomer will evaporate more easily with steam.

🎯 Exam Tip: Steam distillation is effective for separating compounds that are volatile in steam, immiscible with water, and have a boiling point lower than that of water, often due to intramolecular hydrogen bonding (like ortho-nitrophenol).

 

Question 28. The purity of an organic – compound is determined by
(a) Chromatography
(b) Crystallization
(c) melting or boiling point
(d) both (a) and (c)
Answer: (d) both (a) and (c)
In simple words: The purity of an organic compound can be checked by seeing if it has a sharp and constant melting or boiling point. Also, chromatography techniques can show if there are any impurities present in the sample. Both methods help confirm how pure the compound is.

🎯 Exam Tip: A pure solid organic compound melts sharply at a specific temperature. Impurities lower the melting point and broaden the melting range. Similarly, a pure liquid has a precise boiling point. Chromatographic methods provide visual evidence of single component presence, confirming purity.

 

Question 29. A liquid which decomposes at its boiling point can be purified by
(a) distillation at atmospheric pressure
(b) distillation under reduced pressure
(c) fractional distillation
(d) steam distillation
Answer: (b) distillation under reduced pressure
In simple words: If a liquid breaks down when heated to its normal boiling point, we can purify it by lowering the pressure during distillation. This makes the liquid boil at a much lower temperature, preventing it from decomposing.

🎯 Exam Tip: Distillation under reduced pressure (vacuum distillation) is essential for purifying high-boiling liquids or those that are thermally unstable and would decompose at their normal boiling points.

 

Question 30. Assertion: \( \text{CH}_3 – \text{C}(\text{COOC}_2\text{H}_5) = \text{CH} – \text{COOH} \) is 3- carboxyethoxy -2- butenoic acid. Reason: The principal functional group gets lowest number followed by double bond (or) triple bond.
(a) both the assertion and reason are true and the reason is the correct explanation of assertion.
(b) both assertion and reason are true and the reason is not the correct explanation of assertion.
(c) assertion is true but reason is false.
(d) both the assertion and reason are false
Answer: (a) both the assertion and reason are true and the reason is the correct explanation of assertion.
In simple words: In chemical naming, the most important functional group always gets the lowest number. After that, double or triple bonds are numbered to keep their positions as low as possible. This rule helps create clear and consistent names for complex organic compounds.

🎯 Exam Tip: When naming organic compounds, always identify the principal functional group first and assign it the lowest possible locant (number). Then, consider multiple bonds and other substituents, also aiming for the lowest locants. Alphabetical order applies when choosing between equally numbered substituents.

II. Write Brief Answers to the Following Questions:

 

Question 31. Give the general characteristics of organic compounds.
Answer: Organic compounds generally have the following features:
1. Most organic compounds are formed by covalent bonds involving carbon. They do not usually dissolve in water but readily dissolve in other organic liquids like benzene or ether.
2. Many organic compounds can catch fire easily, except for some like carbon tetrachloride (\( \text{CCl}_4 \)). They have low melting and boiling points because their atoms are joined by covalent bonds, which are weaker between molecules.
3. Organic compounds are recognized by their special groups of atoms called functional groups. These groups make the compounds react in specific ways. The reaction of an organic compound often happens at its functional group. Organic compounds also show isomerism, meaning they can have the same formula but different arrangements of atoms. This is a unique feature of organic chemistry.
In simple words: Organic compounds are mostly covalent and dissolve in organic solvents, not water. They often burn easily and have low melting points. They react because of their functional groups and can have different structures for the same formula, which is called isomerism.

🎯 Exam Tip: When listing characteristics, focus on key properties like bonding (covalent), solubility (organic solvents), reactivity (functional groups), and structural diversity (isomerism), as these are fundamental to organic chemistry.

 

Question 32. Describe the classification of organic compounds based on their structure.
Answer: Organic compounds are classified into two main types based on their carbon skeleton structure: Acyclic/Aliphatic Compounds and Cyclic Compounds.
1. **Acyclic or Open-Chain Compounds**: These have carbon atoms linked in straight or branched chains, not forming rings. They can be saturated (all single bonds) or unsaturated (with double or triple bonds).
* Example: Propane (\( \text{CH}_3-\text{CH}_2-\text{CH}_3 \))
2. **Cyclic Compounds**: These have carbon atoms arranged in a ring structure.
a. **Homocyclic Compounds**: Rings are made only of carbon atoms.
i. **Alicyclic Compounds**: These are cyclic compounds that resemble aliphatic (open-chain) compounds in their properties. They can be saturated (like cyclobutane) or unsaturated (like cyclohexene).
* Example: Cyclobutane
ii. **Aromatic Compounds**: These are ring structures with special stability due to delocalized pi electrons. They are further divided into:
* **Benzenoid Compounds**: Contain at least one benzene ring.
* Example: Toluene
* **Non-Benzenoid Aromatic Compounds**: Aromatic but do not contain a benzene ring.
* Example: Azulene
b. **Heterocyclic Compounds**: Rings contain at least one heteroatom (like O, N, S) in addition to carbon atoms.
i. **Alicyclic Heterocyclic Compounds**: Resemble alicyclic compounds in properties.
* Example: Tetrahydrofuran
ii. **Aromatic Heterocyclic Compounds**: Are aromatic and contain heteroatoms in the ring.
* Example: Pyridine
In simple words: Organic compounds are sorted by how their carbon atoms are joined. They can be open-chain (acyclic) or form rings (cyclic). Ring compounds are further split into those made only of carbon (homocyclic) or those with other atoms in the ring (heterocyclic). Both homocyclic and heterocyclic compounds can be either alicyclic (like normal chains) or aromatic (having special stability, like benzene).

🎯 Exam Tip: When classifying organic compounds based on structure, draw a simple flowchart in your mind (or on paper) starting from the two main branches: acyclic and cyclic. Then, branch out for homocyclic/heterocyclic and alicyclic/aromatic, providing a simple example for each category.

 

Question 33. Write a note on homologous series.
Answer: A homologous series is a group of organic compounds where each member has the same type of functional group and similar chemical properties. Each successive member in the series differs from the previous one by a \( -\text{CH}_2- \) unit in its molecular formula. The individual members of such a series are called homologues, and the overall phenomenon is known as homology. All members of a homologous series can be described by a general molecular formula.
**Examples:**
* **Alkanes**: \( \text{CH}_4 \) (Methane), \( \text{C}_2\text{H}_6 \) (Ethane), \( \text{C}_3\text{H}_8 \) (Propane) and so on. Their general formula is \( \text{C}_{\text{n}}\text{H}_{2\text{n}+2} \).
* **Alcohols**: \( \text{CH}_3\text{OH} \) (Methanol), \( \text{C}_2\text{H}_5\text{OH} \) (Ethanol), \( \text{C}_3\text{H}_7\text{OH} \) (Propanol) and so on. Their general formula is \( \text{C}_{\text{n}}\text{H}_{2\text{n}+1}\text{OH} \).
In simple words: A homologous series is like a chemical family where members have similar chemical behavior and a shared general formula. Each new member is made by adding a \( -\text{CH}_2- \) group to the one before it.

🎯 Exam Tip: For homologous series, always mention the three defining characteristics: same functional group, successive members differing by a \( -\text{CH}_2- \) unit, and similar chemical properties. Providing examples from different classes (alkanes, alcohols, carboxylic acids) strengthens your answer.

 

Question 34. What is meant by a functional group? Identify the functional group in the following compounds.
(a) acetaldehyde
(b) oxalic acid
(e) dimethyl ether
(d) methylamine
Answer:
**Functional group:** A functional group is a specific atom or a small group of atoms within a molecule that causes the molecule to react in a predictable and characteristic way. It determines the typical physical and chemical properties of the compound, regardless of the rest of the molecule.
**Identification of functional groups:**
(a) acetaldehyde \( \rightarrow \) \( -\text{CHO} \) (Aldehyde group)
(b) oxalic acid \( \rightarrow \) \( -\text{COOH} \) (Carboxylic acid group)
(c) di methyl ether \( \rightarrow \) \( -\text{O}- \) (Ether group)
(d) methylamine \( \rightarrow \) \( -\text{NH}_2 \) (Amine group)
In simple words: A functional group is a special part of a molecule that makes it act in a certain way chemically. For example, in acetaldehyde, the \( -\text{CHO} \) group is what gives it its aldehyde properties.

🎯 Exam Tip: Clearly define a functional group and then list the specific group for each compound. Common functional groups include aldehydes (CHO), ketones (C=O), alcohols (OH), carboxylic acids (COOH), ethers (O), and amines (NH₂).

 

Question 35. Give the general formula for the following classes of organic compounds
(a) Aliphatic monohydric alcohol
(b) Aliphatic ketones
(c) Aliphatic amines
Answer:
(a) Aliphatic monohydric alcohol \( \rightarrow \text{C}_{\text{n}}\text{H}_{2\text{n}} + 2\text{O} \)
(b) Aliphatic ketones \( \rightarrow \text{C}_{\text{n}}\text{H}_{2\text{n}}\text{O} \)
(c) Aliphatic amines \( \rightarrow \text{C}_{\text{n}}\text{H}_{2\text{n}} + 2\text{N} \)
These formulas help classify organic compounds based on their bonding and functional groups. Knowing the general formula allows us to predict the properties of many similar compounds.
In simple words: These are like chemical recipes that tell you how many carbon, hydrogen, and other atoms are generally found in each group of organic chemicals.

🎯 Exam Tip: Remember to clearly state the general formula and briefly explain what each part (like 'n') represents for full marks.

 

Question 36. Write the molecular formula of the first six members of homologous series of nitre alkanes.
Answer:
The first six members of the homologous series of nitro alkanes are:
\( \text{CH}_3\text{NO}_2 \)
\( \text{C}_2\text{H}_5\text{NO}_2 \)
\( \text{C}_3\text{H}_7\text{NO}_2 \)
\( \text{C}_4\text{H}_9\text{NO}_2 \)
\( \text{C}_5\text{H}_{11}\text{NO}_2 \)
\( \text{C}_6\text{H}_{13}\text{NO}_2 \)
Each successive member in a homologous series differs by a \( \text{CH}_2 \) group, showing a regular progression in molecular weight and physical properties.
In simple words: Here are the chemical formulas for the first six members of the nitro alkane family. Each one gets a little bigger by adding a \( \text{CH}_2 \) part.

🎯 Exam Tip: Ensure you correctly apply the general formula \( \text{C}_{\text{n}}\text{H}_{2\text{n}+1}\text{NO}_2 \) and clearly list each member, showing the incremental addition of \( \text{CH}_2 \).

 

Question 37. Write the molecular formula and possible structural formula of the first four members of homologous series of carboxylic acids.
Answer:
The first four members of the homologous series of carboxylic acids are:

Carboxylic acids are an important class of organic compounds, characterized by the presence of a carboxyl group \((-\text{COOH})\). The increasing chain length impacts their physical properties.
In simple words: This table shows the first four carboxylic acids. It gives their chemical name, how many atoms they have, and how they are put together in a drawing.

🎯 Exam Tip: Clearly present both the molecular and structural formulas, ensuring correct representation of the carboxyl functional group.

 

Question 38. Give the IUPAC names of the following compounds.
(i) \( (\text{CH}_3)_2\text{CH} - \text{CH}_2 - \text{CH}(\text{CH}_3) - \text{CH}(\text{CH}_3)_2 \)
(ii) Structure of 2-bromo-3-methylbutane
(iii) \( \text{CH}_3 - \text{O} - \text{CH}_3 \)
(iv) Structure of 2-hydroxybutanal
(v) Structure of buta-1,3-diene
(vi) Structure of 4-chloropent-2-yne
(vii) Structure of 1-bromobut-2-ene
(viii) Structure of 5-oxohexanoic acid
(ix) Structure of 3-ethyl-4-ethenylheptane
(x) Structure of 2,4,4-trimethylpent-2-ene
(xi) Structure of 2-methyl-1-phenylpropan-1-amine
(xii) Structure of 2,2-dimethyl-4-oxopentanenitrile
(xiii) Structure of 2-ethoxypropane
(xiv) Structure of 1-fluoro-4-methyl-2-nitrobenzene
(xv) Structure of 3-bromo-2-methylpentanal
(xvi) Acetophenone
Answer:
(i) 2, 3, 5 – Trimethylhexane
(ii) 2 – Bromo – 3 – methylbutane
(iii) Methoxymethane
(iv) 2 – Hydroxybutanal
(v) Buta – 1,3 – diene
(vi) 4 – Chloropent – 2 – yne
(vii) 1 – Bromobut – 2 – ene
(viii) 5 – Oxohexanoic acid
(ix) 3 – Ethyl – 4 – ethenylheptane
(x) 2, 4, 4 – Trimethylpent – 2 – ene
(xi) 2 – Methyl -1 – phenylpropan – 1 -amine
(xii) 2, 2 – Dimethyl – 4-oxopentanenitrile
(xiii) 2 – Ethoxypropane
(xiv) 1 – Fluoro – 4 – methyl – 2 -nitrobenzene
(xv) 3 – Bromo – 2 – methylpentanal
(xvi) Acetophenone
IUPAC nomenclature provides a systematic way to name chemical compounds, ensuring clarity and precision in scientific communication. Each part of the name indicates specific structural features.
In simple words: Here are the official chemical names for each of the structures shown. These names follow special rules to describe how the atoms are connected.

🎯 Exam Tip: When naming complex structures, identify the longest carbon chain, prioritize functional groups, and correctly assign prefixes and suffixes based on IUPAC rules.

 

Question 39. Give the structure for the following compound
(i) 3 – Ethyl – 2 methyl – 1 – pentene
(ii) 1, 3, 5 – Trimethyl cyclohex – 1 – ene
(iii) Tertiary butyl iodide
(iv) 3 – Chlorobutanal
(V) 3 – Chlorobutanol
(vi) 2 – Chloro – 2 – methyl propane
(vii) 2, 2 – Dimethyl – 1 – chloropropane
(viii) 3 – Methylbut -1- ene
(ix) Butan – 2, 2 – diol
(x) Octane – 1, 3 – diene
(xi) 1, 3 – Dimethylcyclohexane
(xii) 3 – Chlorobut – 1 – ene
(xiii) 2 – Methylbutan – 3 – ol
(xiv) Acetaldehyde
Answer:
(i) \( \text{CH}_2 = \text{C}(\text{CH}_3)-\text{CH}(\text{CH}_2\text{CH}_3)-\text{CH}_2\text{CH}_3 \)
(ii) \( \text{C}_6\text{H}_{10} \) with three \( \text{CH}_3 \) groups at positions 1, 3, 5 and a double bond at position 1.
(iii) \( (\text{CH}_3)_3\text{CI} \)
(iv) \( \text{CH}_3-\text{CH}_2-\text{CH}(\text{Cl})-\text{CHO} \)
(v) \( \text{CH}_3-\text{CH}(\text{OH})-\text{CH}_2-\text{CH}_2\text{Cl} \)
(vi) \( (\text{CH}_3)_2\text{CClCH}_3 \)
(vii) \( \text{CH}_3-\text{C}(\text{CH}_3)_2-\text{CH}_2\text{Cl} \)
(viii) \( \text{CH}_2=\text{CH}-\text{CH}(\text{CH}_3)_2 \)
(ix) \( \text{CH}_3-\text{CH}_2-\text{C}(\text{OH})_2-\text{CH}_3 \)
(x) \( \text{CH}_2=\text{CH}-\text{CH}_2-\text{CH}=\text{CH}-\text{CH}_2-\text{CH}_2-\text{CH}_3 \)
(xi) \( \text{C}_6\text{H}_{10} \) with two \( \text{CH}_3 \) groups at positions 1, 3.
(xii) \( \text{CH}_2=\text{CH}-\text{CH}(\text{Cl})-\text{CH}_3 \)
(xiii) \( (\text{CH}_3)_2\text{CH}-\text{CH}(\text{OH})-\text{CH}_3 \)
(xiv) \( \text{CH}_3\text{CHO} \)
Drawing these structures helps visualize the arrangement of atoms and understand the properties dictated by their bonding. The structure is fundamental to understanding a compound's behavior.
In simple words: These are the chemical formulas or simplified drawings for the compounds listed. They show how the atoms are connected to form each chemical.

🎯 Exam Tip: Pay close attention to the principal chain, functional groups, and position of substituents to accurately draw the structure from an IUPAC name. Practice recognizing common functional groups.

 

Question 40. Describe the reactions involved in the detection of nitrogen in an organic compound by Lassaigne method.
Answer:
The Lassaigne's method detects nitrogen in organic compounds by converting covalently bonded nitrogen, sulfur, or halogens into water-soluble sodium salts through fusion with metallic sodium. A small piece of freshly cut sodium metal is heated gently in a fusion tube. Then, a pinch of the organic compound is added, and the tube is strongly heated until it becomes red hot. The hot tube is plunged into distilled water in a china dish and broken. The contents are boiled and filtered to get Lassaigne's extract or sodium fusion extract, which is used for qualitative analysis.

Test for Nitrogen:
If nitrogen is present, it forms sodium cyanide, \( \text{NaCN} \). This \( \text{NaCN} \) reacts with freshly prepared ferrous sulfate \( (\text{FeSO}_4) \) and ferric ions \( (\text{Fe}^{3+}) \) in the presence of concentrated \( \text{HCl} \) to give a Prussian blue or green color/precipitate, confirming nitrogen's presence. \( \text{HCl} \) dissolves any greenish ferrous hydroxide precipitate formed by excess \( \text{NaOH} \) from \( \text{FeSO}_4 \).

The reactions for Prussian blue formation are:
\( \text{Na} + \text{C} + \text{N} \rightarrow \text{NaCN} \) (from organic compounds)
\( \text{FeSO}_4 + 2\text{NaOH} \rightarrow \text{Fe(OH)}_2 + \text{Na}_2\text{SO}_4 \)
\( 6\text{NaCN} + \text{Fe(OH)}_2 \rightarrow \text{Na}_4[\text{Fe(CN)}_6] + 2\text{NaOH} \)
\( \text{Na}_4[\text{Fe(CN)}_6] + \text{FeCl}_3 \rightarrow \text{Fe}_4[\text{Fe(CN)}_6]_3 + 12\text{NaCl} \)
(Ferric ferrocyanide - Prussian blue or green ppt)

If both nitrogen and sulfur are present, a blood-red color is observed due to the formation of sodium thiocyanate \( (\text{NaCNS}) \), followed by reaction with ferric ions:
\( \text{Na} + \text{C} + \text{N} + \text{S} \xrightarrow{\text{Heat}} \text{NaCNS} \) (sodium sulphocyanide)
\( \text{NaCNS} + \text{FeCl}_3 \rightarrow \text{Fe(CNS)}_3 + 3\text{NaCl} \) (Ferric sulphocyanide - Blood red color)
This method is crucial for elemental analysis in organic chemistry, confirming the presence of key heteroatoms in a sample. The distinctive colors formed provide clear indicators.
In simple words: The Lassaigne test checks for nitrogen in a chemical. You mix the chemical with sodium and heat it. If nitrogen is there, it will turn a deep blue or red color when other chemicals are added. This color shows nitrogen is present.

🎯 Exam Tip: Remember the specific reagents (sodium, \( \text{FeSO}_4 \), \( \text{FeCl}_3 \), \( \text{HCl} \)) and the characteristic Prussian blue color for nitrogen detection. Also, note the blood-red color if both N and S are present.

 

Question 41. Give the principle involved in the estimation of halogen in an organic compound by Carius method.
Answer:
The Carius method is used to estimate halogens in organic compounds. In this method, a known mass of the organic compound is heated in a sealed Carius tube with fuming nitric acid \( (\text{HNO}_3) \) and silver nitrate \( (\text{AgNO}_3) \). During heating, the carbon, hydrogen, and sulfur in the compound are oxidized to carbon dioxide \( (\text{CO}_2) \), water \( (\text{H}_2\text{O}) \), and sulfur dioxide \( (\text{SO}_2) \), respectively. The halogen present in the organic compound reacts with \( \text{AgNO}_3 \) to form a precipitate of silver halide \( (\text{AgX}) \).
The reaction can be simplified as:
Organic Compound with Halogen (X) \( \xrightarrow{\text{Fuming HNO}_3, \text{AgNO}_3} \text{AgX} \downarrow \)
The precipitate of \( \text{AgX} \) is then filtered, washed, dried, and weighed. From the mass of \( \text{AgX} \) and the initial mass of the organic compound, the percentage of the halogen in the compound can be calculated. The Carius method is a precise way to determine the halogen content in a sample.

Calculation:
Weight of the organic compound: \( \text{w g} \)
Weight of \( \text{AgCl} \) precipitate: \( \text{a g} \)
\( 143.5 \text{ g of AgCl contains } 35.5 \text{ g of Cl} \)
\( \implies \text{a g of AgCl contains } \frac{35.5}{143.5} \times \text{a} \text{ g of Cl} \)
Percentage of \( \text{Cl} \) in \( \text{w g} \) organic compound \( = \left(\frac{35.5}{143.5} \times \frac{\mathrm{a}}{\mathrm{w}} \times 100\right) \% \)

Let Weight of silver bromide be ‘b’ g
\( 188 \text{ g of AgBr contains } 80 \text{ g of Br} \)
\( \implies \text{b g of AgBr contains } \frac{80}{188} \times \text{b} \text{ g of Br} \)
Percentage of \( \text{Br} \) in \( \text{w g} \) organic compound \( = \left(\frac{80}{188} \times \frac{\mathrm{b}}{\mathrm{w}} \times 100\right) \% \)

Let Weight of silver iodide be ‘c’ g
\( 235 \text{ g of AgI contains } 127 \text{ g of I} \)
\( \implies \text{c g of AgI contains } \frac{127}{235} \times \text{c} \text{ g of I} \)
Percentage of \( \text{I} \) in \( \text{w g} \) organic compound \( = \left(\frac{127}{235} \times \frac{\mathrm{c}}{\mathrm{w}} \times 100\right) \% \)
This method is crucial for elemental analysis in organic chemistry, confirming the presence and quantity of key heteroatoms in a sample. The precipitation of silver halides is a quantitative process.
In simple words: The Carius method finds out how much halogen (like chlorine or bromine) is in a chemical. You burn the chemical with nitric acid and silver nitrate. This makes a silver-halogen solid, which you weigh to calculate the percentage of halogen.

🎯 Exam Tip: Remember the key reagents \( (\text{HNO}_3, \text{AgNO}_3) \), the formation of \( \text{AgX} \) precipitate, and the formula for calculating the percentage of halogen by weight.

 

Question 42. Give a brief description of the principles of
(i) Fractional distillation
(ii) Column Chromatography
Answer:
(i) Fractional distillation:
Fractional distillation is a technique used to purify and separate liquids that have boiling points close to each other. In this process, a fractionating column is set up between the distillation flask and the condenser. As the mixture is heated, the vapors rise through the column. The column has a larger surface area (e.g., packed with glass beads or plates), allowing repeated vaporization and condensation cycles. Each cycle enriches the vapor in the more volatile component. This repeated process, which happens as the vapors travel up the column, helps separate the components based on their different boiling points. This method is widely used in industries like petroleum refining to separate crude oil into different fractions.

(ii) Column Chromatography:
Column chromatography is a simple method for separating mixtures. It uses a long glass column filled with an adsorbent material (called the stationary phase), such as activated aluminum oxide or silica gel. The mixture to be separated is placed on top of this column. A liquid or mixture of liquids (called the eluent or mobile phase) is then allowed to flow slowly down the column. As the eluent moves, it carries the components of the mixture with it. Different components travel at different speeds because they adsorb (stick) to the stationary phase with varying strengths. Components that adsorb less move faster, while those that adsorb more move slower. This difference in movement allows the components to separate into distinct bands, which can then be collected individually. This technique is highly versatile and effective for purifying compounds.
In simple words: Fractional distillation separates liquids by boiling them and letting them condense many times, while column chromatography separates chemicals by making them travel at different speeds through a packed tube.

🎯 Exam Tip: For fractional distillation, emphasize the role of the fractionating column in providing repeated vaporization-condensation cycles. For column chromatography, highlight the concept of differential adsorption between stationary and mobile phases.

 

Question 43. Explain paper chromatography.
Answer:
Paper chromatography (PC) is a type of partition chromatography. It uses a strip of special quality paper as the stationary phase, which acts like an adsorbent. The process involves placing a spot of the mixture near the base of the paper. Then, the paper is suspended in a suitable solvent (the mobile phase). The solvent moves up the paper by capillary action, carrying the components of the mixture with it. Different components of the mixture travel at different rates because they have different affinities (how much they prefer to stick) for the stationary paper phase and the moving solvent phase. This difference in partition between the two phases causes the compounds to separate. The separated compounds appear as spots at different heights on the paper, forming a chromatogram. Colored compounds are directly visible, while colorless compounds can be detected using ultraviolet light or by spraying with a suitable reagent. This simple method is effective for separating small quantities of substances, like pigments.
In simple words: Paper chromatography separates mixed chemicals by making them travel up a special paper with a liquid. Each chemical moves at its own speed, leaving separate spots on the paper.

🎯 Exam Tip: Key points include using paper as the stationary phase, solvent movement by capillary action, and separation based on differential partitioning between the stationary and mobile phases.

 

Question 44. Explain various types of constitutional isomerism (structural isomerism) in organic compounds.
Answer:
Constitutional isomerism, also known as structural isomerism, occurs when compounds have the same molecular formula but different bonding sequences, meaning their atoms are connected in different orders. This leads to distinct structures and properties. There are several types of constitutional isomerism:

(a) Chain or nuclear or skeletal isomerism:
These isomers have the same molecular formula but differ in the arrangement of their carbon skeleton (main chain). For example, n-pentane has a straight chain, while isopentane and neopentane have branched chains. These different arrangements of the carbon atoms significantly alter their shapes and reactivity.
Example: Pentane \( (\text{C}_5\text{H}_{12}) \)
n-Pentane: \( \text{CH}_3 - \text{CH}_2 - \text{CH}_2 - \text{CH}_2 - \text{CH}_3 \)
Isopentane (2-methylbutane): \( \text{CH}_3 - \text{CH}(\text{CH}_3) - \text{CH}_2 - \text{CH}_3 \)
Neopentane (2,2-dimethylpropane): \( \text{CH}_3 - \text{C}(\text{CH}_3)_2 - \text{CH}_3 \)

(b) Position isomerism:
These isomers have the same molecular formula and carbon skeleton but differ in the position of a substituent group, a functional group, or an unsaturated linkage (double or triple bond). For example, 1-pentene and 2-pentene are position isomers because the double bond is in a different place.
Example: Molecular formula \( \text{C}_5\text{H}_{10} \)
Pent-1-ene: \( \text{CH}_3 - \text{CH}_2 - \text{CH}_2 - \text{CH} = \text{CH}_2 \)
Pent-2-ene: \( \text{CH}_3 - \text{CH}_2 - \text{CH} = \text{CH} - \text{CH}_3 \)

(c) Functional isomerism:
Functional isomers have the same molecular formula but possess different functional groups. This means they belong to different classes of organic compounds and exhibit very different chemical properties. For instance, ethanol (an alcohol) and dimethyl ether (an ether) are functional isomers with the molecular formula \( \text{C}_2\text{H}_6\text{O} \).
Example: Molecular formula \( \text{C}_3\text{H}_6\text{O} \)
Propanal (aldehyde group): \( \text{CH}_3 - \text{CH}_2 - \text{CHO} \)
Propanone (keto group): \( \text{CH}_3 - \text{C}(=\text{O}) - \text{CH}_3 \)
Understanding isomerism is vital in organic chemistry, as it explains the vast diversity of compounds despite similar atomic compositions. These different arrangements lead to unique chemical identities.
In simple words: Constitutional isomers are chemicals with the same parts but put together in a different order. This can change their main chain (skeletal), where a group is placed (position), or even what kind of chemical family they belong to (functional).

🎯 Exam Tip: Clearly define each type of isomerism and provide a simple, distinct example for each to illustrate the difference in atom connectivity.

 

Question 45. Describe optical isomerism with suitable example.
Answer:
Optical isomerism is a type of stereoisomerism where compounds have the same physical and chemical properties but differ only in their ability to rotate the plane of plane-polarized light. Such isomers are called optical isomers, and the phenomenon is known as optical isomerism or optical activity.

Compounds that rotate plane-polarized light are called optically active. Those that rotate the light to the right (clockwise) are called dextrorotatory, denoted by 'd' or \( (+) \). Those that rotate the light to the left (anticlockwise) are called levorotatory, denoted by 'l' or \( (-) \). A key characteristic of optical isomers is that they are non-superimposable mirror images of each other, similar to how your left and right hands are mirror images but cannot be perfectly overlaid.

Example: d and l lactic acid. Glucose is another common organic compound that exhibits optical activity, rotating plane-polarized light. This property arises from the presence of a chiral center (an asymmetric carbon atom) in the molecule.
Optical isomerism is very important in biology and medicine because different optical isomers can have different biological effects, even if their chemical formulas are identical.
In simple words: Optical isomerism is when two chemicals look like mirror images of each other and can spin light in opposite directions, even though they have the same atoms. One spins light right, and the other spins it left.

🎯 Exam Tip: Emphasize the rotation of plane-polarized light, the concept of non-superimposable mirror images, and the presence of a chiral carbon for optical activity. Lactic acid is a simple, good example.

 

Question 46. Briefly explain geometrical isomerism in alkene by considering 2-butene as an example.
Answer:
Geometrical isomers are stereoisomers that have a different arrangement of groups or atoms around a rigid framework, such as a double bond. This type of isomerism happens because rotation around a carbon-carbon double bond is restricted. In alkenes, the carbon-carbon double bond is formed by one sigma \( (\sigma) \) bond and one pi \( (\pi) \) bond. The pi bond "locks" the molecule into a fixed position, preventing free rotation.

For geometrical isomerism to occur, each carbon atom of the double bond must be attached to two different groups. In 2-butene, each carbon of the double bond is attached to a hydrogen atom and a methyl \( (\text{CH}_3) \) group.
There are two forms:
1. Cis-isomer: Two similar groups (e.g., two hydrogen atoms or two methyl groups) are on the same side of the double bond.
2. Trans-isomer: Two similar groups are on opposite sides of the double bond.

Example: 2-Butene
Cis-2-butene: Both \( \text{CH}_3 \) groups are on the same side of the double bond.
\[ \text{H}_3\text{C} - \text{C} = \text{C} - \text{CH}_3 \] \[ \quad \text{H} \quad \text{H} \]
Trans-2-butene: \( \text{CH}_3 \) groups are on opposite sides of the double bond.
\[ \text{H}_3\text{C} - \text{C} = \text{C} - \text{H} \] \[ \quad \text{H} \quad \text{CH}_3 \]
Cis and trans isomers can be interconverted by heating or absorbing light, which provides enough energy (about 62 kcal/mol) to temporarily break the pi bond, allowing rotation. Upon cooling, the pi bond reforms, potentially creating a mixture of both cis and trans forms. These isomers can have different physical properties like boiling points and melting points.
This isomerism highlights how fixed arrangements around a double bond create distinct chemical species with different spatial configurations.
In simple words: Geometrical isomerism is when atoms around a double bond are fixed in place. For 2-butene, if the same groups are on the same side, it's 'cis'. If they are on opposite sides, it's 'trans'.

🎯 Exam Tip: Clearly draw and label the cis and trans forms of 2-butene, and explain that restricted rotation around the double bond is the cause of geometrical isomerism.

 

Question 46. Briefly explain geometrical isomerism in alkene by considering 2-butene as an example.
Answer: Geometrical isomers are like special types of isomers where atoms are arranged differently around a stiff double bond. This difference comes from how groups are placed around the double bond, which can't easily twist or spin. In alkenes, the carbon-carbon double bond has a strong sigma bond and a weaker pi bond. The pi bond stops the carbon atoms from rotating freely. This fixed position leads to two different ways the groups can be arranged: either on the same side (cis) or on opposite sides (trans). This difference in arrangement is called geometrical isomerism. For example, in 2-butene, the \( \text{CH}_3 \) groups can be on the same side (cis-2-butene) or opposite sides (trans-2-butene). Heating can change one form into the other by temporarily breaking the pi bond.
In simple words: Geometrical isomerism happens when atoms are fixed around a double bond and cannot rotate. This creates two forms, cis (same side) and trans (opposite sides). Heating can interconvert these forms.

🎯 Exam Tip: Remember to always look at the arrangement of substituents around the double bond to identify cis-trans isomers correctly.

 

Question 47. 0.30 g of a substance gives 0.88 g of carbon dioxide and 0.54 g of water calculate the percentage of carbon and hydrogen in it.
Answer: To find the percentage of carbon and hydrogen, we first use the weights of carbon dioxide and water produced. Since 44 g of \( \text{CO}_2 \) has 12 g of carbon, and 18 g of \( \text{H}_2\text{O} \) has 2 g of hydrogen, we can calculate the mass of carbon and hydrogen in the given sample. This method helps us determine the elemental composition of an unknown organic substance.
Weight of organic compound = 0.30 g
Weight of carbon dioxide = 0.88 g
Weight of water = 0.54 g
Percentage of carbon:
44 g of carbon dioxide contains, carbon = 12 g
0.88 g of carbon dioxide contains, carbon = \( \frac{12 \times 0.88}{44} \) g
0.30 g substance contains,
carbon = \( \frac{12 \times 0.88}{44} \) g
100 g substance Contains \( \frac{12 \times 0.88}{44} \times \frac{100}{0.30} \) = 80 g of carbon
Percentage of carbon = 80 %
Percentage of hydrogen:
18 g of water contains, hydrogen = 2 g
0.54 g of water contains, hydrogen = \( \frac{2 \times 0.54}{18} \) g
0.30 g of substance contains hydrogen = \( \frac{2 \times 0.54}{18 \times 0.30} \) g
100 g of substance contains = \( \frac{2 \times 0.54}{18 \times 0.30} \times 100 \) g = 20 g of hydrogen
Percentage of hydrogen = 20 %
In simple words: We calculate the mass of carbon from \( \text{CO}_2 \) and hydrogen from \( \text{H}_2\text{O} \). Then, we divide these by the total sample weight to get the percentages.

🎯 Exam Tip: Always remember the molecular weights of \( \text{CO}_2 \) (44 g/mol) and \( \text{H}_2\text{O} \) (18 g/mol) and the atomic weights of C (12 g/mol) and H (1 g/mol, so 2g for \( \text{H}_2 \)) for these calculations.

 

Question 48. The ammonia evolved form 0.20 g of an organic compound by Kjeldahl method neutralized 15 ml of N / 20 sulphuric acid solution. Calculate the percentage of Nitrogen.
Answer: In Kjeldahl's method, the nitrogen from the organic compound is converted into ammonia. This ammonia is then neutralized by a known amount of acid. By finding out how much acid was used, we can calculate the amount of nitrogen present. The percentage of nitrogen helps to understand the compound's makeup.
Weight of organic compound = 0.20 g
Normality of acid = \( \frac{\text{N}}{20} \)
Volume of standard acid neutralized by ammonia = 15 ml
1000 ml of N ammonia contains = 14 g of nitrogen
15 ml of ammonia of normality \( \frac{\text{N}}{20} \) contains nitrogen = \( \frac{14 \times 15 \times 1}{1000 \times 20} \)
0.20 g of compound contains nitrogen = \( \frac{14 \times 15}{1000 \times 20} \)
100 g of compound contains nitrogen = \( \frac{14 \times 15 \times 100}{1000 \times 20 \times 0.20} \) = 5.25 g
Percentage of nitrogen = 5.25 %
In simple words: We calculate the nitrogen based on how much acid neutralizes the ammonia produced. This method helps to determine the nitrogen content in a substance.

🎯 Exam Tip: For Kjeldahl calculations, remember that 1000 ml of 1N ammonia solution contains 14 g of nitrogen, which is a key conversion factor.

 

Question 49. 0.32 g of an organic compound, after heating with fuming nitric acid and barium nitrate crystals is a sealed tube gave 0. 466 g of barium sulphate. Determine the percentage of sulphur in the compound.
Answer: To find the percentage of sulfur in the compound, we use the Carius method. First, the organic compound is heated with nitric acid to convert all sulfur into sulfuric acid. Then, barium sulfate is precipitated. By measuring the mass of barium sulfate formed and knowing the initial mass of the organic compound, we can calculate the exact percentage of sulfur. This is a common way to quantify sulfur in organic materials.
Mass of the substance taken = 0.32 g
Mass of \( \text{BaSO}_4 \) formed = 0.466 g
Molecular mass of \( \text{BaSO}_4 \) = 137 + 32 + 64 = 233
Then, mass of S in 0.466 g of \( \text{BaSO}_4 \) = \( \frac{0.466 \times 32}{233} \)
Percentage of S in compound = \( \frac{0.466 \times 32 \times 100}{233 \times 0.32} \) = 20 %
In simple words: We convert all sulfur to barium sulfate. By knowing the mass of barium sulfate and the original sample, we find the percentage of sulfur.

🎯 Exam Tip: Always remember that the sulfur content in the sample is directly proportional to the mass of \( \text{BaSO}_4 \) precipitated, and the molecular weight of \( \text{BaSO}_4 \) is key.

 

Question 50. 0.24 g of an organic compound gave 0.287 g of silver chloride in the carius method. Calculate the percentage of chlorine in the compound.
Answer: In the Carius method for halogens, the organic compound is heated with nitric acid and silver nitrate to turn all chlorine into silver chloride precipitate. We measure how much silver chloride is formed. Knowing the fixed ratio of chlorine in silver chloride, we can work backward. By comparing the mass of silver chloride to the initial mass of the organic compound, we find the exact percentage of chlorine present. This method is vital for quality control and structural analysis of organic compounds containing halogens.
Mass of organic compound = 0.24 g
Mass of silver chloride = 0.287 g
143.5 g AgCl contains = 35.5 g chlorine
0.287 g of AgCl contains = \( \frac{35.5}{143.5} \times 0.287 \)
Percentage of chlorine = \( \frac{35.5}{143.5} \times \frac{0.287}{0.24} \times 100 \) = 29.58 %
In simple words: We calculate the mass of chlorine from the silver chloride formed. Then we use this mass and the original sample mass to find the percentage of chlorine.

🎯 Exam Tip: Always remember the molecular weight of silver chloride (AgCl = 143.5 g/mol) and the atomic weight of chlorine (Cl = 35.5 g/mol) for accurate calculations.

 

Question 51. In the estimation of nitrogen present in an organic compound by Dumas method 0.35 g yielded 20.7 mL of nitrogen at 15° C and 760 mm pressure. Calculate the percentage of nitrogen in the compound.
Answer: The Dumas method for nitrogen estimation involves converting all nitrogen in an organic compound into nitrogen gas. We measure the volume of this gas under given conditions, then convert it to standard temperature and pressure (NTP) using gas laws. From the volume at NTP, we calculate the mass of nitrogen using its molar volume. Finally, comparing this mass to the original sample mass gives the percentage of nitrogen. This method helps to understand the elemental composition and purity of a substance.
Volume of \( \text{N}_2 \) at NTP = \( \frac{\mathrm{V} \times \mathrm{P}}{\mathrm{t}+273} \times \frac{273}{760} \)
\( \implies \) \( \text{V}_0 \) ml
Substituting the various values in the above equation,
\( \text{V}_0 \) = \( \frac{20.7 \times 760}{288} \times \frac{273}{760} \) = 19.62 ml
Weight of 19.62 ml of Nitrogen = \( \frac{28}{22400} \times 19.62 \) g
Percentage of Nitrogen = \( \frac{28}{22400} \times 19.62 \times \frac{100}{0.35} \) = 4.9 %
In simple words: We first find the nitrogen gas volume at standard conditions. Then we use its weight-to-volume ratio to find the total nitrogen mass and calculate its percentage in the sample.

🎯 Exam Tip: Remember to convert the measured volume of nitrogen gas to conditions at NTP (Normal Temperature and Pressure) using the gas law formula before calculating the mass of nitrogen.

 

11th Chemistry Guide Fundamentals of Organic Chemistry Additional Questions and Answers

I. Choose the best answer:

 

Question 1. Organic compounds can be formed by
(a) Plants only
(b) Animals only
(c) Plants and Animals
(d) Plants, animals and can be synthesized in laboratory
Answer: (d) Plants, animals and can be synthesized in laboratory
In simple words: Organic compounds are versatile and are not limited to living organisms. This shows how important organic chemistry is in both nature and technology.

🎯 Exam Tip: Remember that Wohler's synthesis of urea in the laboratory debunked the vital force theory, proving organic compounds don't exclusively originate from living sources.

 

Question 2. Generally, organic compounds are
(a) Amorphous
(b) Complexes
(c) Covalent
(d) Electrovalent
Answer: (c) Covalent
In simple words: Organic compounds are mostly made up of carbon atoms linked together, and these links are typically covalent bonds. These bonds make organic compounds essential for life and various industries.

🎯 Exam Tip: Knowing that organic compounds are primarily covalent helps explain their properties like lower melting points and solubility in organic solvents, unlike ionic compounds.

 

Question 3. The vital force theory was proposed by
(a) Wohler
(b) Berthlot
(c) Berzelius
(d) Kolbe
Answer: (c) Berzelius
In simple words: The vital force theory, which stated that organic compounds could only be made by living things due to a special "vital force," was introduced by Berzelius. This theory was later disproven when scientists started making organic compounds in laboratories.

🎯 Exam Tip: Associate Berzelius with the vital force theory and Wohler with its disproval through the synthesis of urea, a key moment in organic chemistry history.

 

Question 4. The first carbon compound prepared from its elements is
(a) Urea
(b) Acetic acid
(c) Methane
(d) benzene
Answer: (b) Acetic acid
In simple words: Acetic acid was the first carbon compound synthesized directly from its basic elements by Kolbe. This was a significant step in chemistry, showing that organic compounds didn't need a "vital force" to be created.

🎯 Exam Tip: Differentiate between the first organic compound (urea by Wohler) and the first carbon compound synthesized from its elements (acetic acid by Kolbe).

 

Question 5. The first organic compound was synthesized in laboratory by
(a) Wohler
(b) Kolbe
(c) Berzelius
(d) Neil Barthlot
Answer: (a) Wohler
In simple words: Wohler was the first scientist to synthesize an organic compound, urea, in a laboratory from inorganic starting materials. This achievement debunked the vital force theory, opening up new possibilities in organic chemistry.

🎯 Exam Tip: Remember Wohler's synthesis of urea as the historical turning point that disproved the vital force theory, fundamentally changing organic chemistry.

 

Question 6. The first organic compound synthesized in the laboratory from an inorganic compound is
(a) \( \text{NH}_4\text{NCO} \)
(b) \( \text{NH}_2 - \text{CO} - \text{NH}_2 \)
(c) \( \text{CH}_3\text{COOH} \)
(d) \( \text{CH}_4 \)
Answer: (b) \( \text{NH}_2 - \text{CO} - \text{NH}_2 \)
In simple words: The first organic compound created in a lab from inorganic substances was urea, with the chemical formula \( \text{NH}_2 - \text{CO} - \text{NH}_2 \). This synthesis, carried out by Wohler, proved that a special "vital force" wasn't needed to make organic chemicals.

🎯 Exam Tip: Recognize urea's chemical formula and its significance as the first organic compound synthesized from inorganic precursors.

 

Question 7. Marsh gas mainly contains
(a) \( \text{C}_2\text{H}_2 \)
(b) \( \text{C}_2\text{H}_4 \)
(c) \( \text{CH}_4 \)
(d) \( \text{C}_2\text{H}_6 \)
Answer: (c) \( \text{CH}_4 \)
In simple words: Marsh gas, often found in swamps and wetlands, is primarily composed of methane (\( \text{CH}_4 \)). This natural gas is produced when organic matter breaks down in places with little to no oxygen.

🎯 Exam Tip: Associate marsh gas with methane (\( \text{CH}_4 \)), a common product of anaerobic decomposition in natural environments.

 

Question 8. Hybridization at 2nd carbon in \( \text{CH}_2 = \text{CH} - \text{CH}_3 \) is
(a) sp
(b) \( \text{sp}^2 \)
(c) \( \text{sp}^3 \)
(d) \( \text{sp}^3\text{d} \)
Answer: (b) \( \text{sp}^2 \)
In simple words: In the molecule propene, \( \text{CH}_2 = \text{CH} - \text{CH}_3 \), the second carbon atom is part of a double bond. Atoms involved in a double bond use \( \text{sp}^2 \) hybridization, which means they have a trigonal planar shape.

🎯 Exam Tip: Remember that carbon atoms in double bonds are \( \text{sp}^2 \) hybridized, while those in single bonds are typically \( \text{sp}^3 \) and those in triple bonds are sp hybridized.

 

Question 9. Number of possible position isomers for Dichlorobenzene is
(a) 2
(b) 3
(c) 4
(d) 5
Answer: (b) 3
In simple words: Dichlorobenzene can have three different position isomers: ortho (1,2-), meta (1,3-), and para (1,4-). Understanding position isomers is important for distinguishing between compounds that have similar properties but are structurally distinct.

🎯 Exam Tip: For disubstituted benzenes, the three common positional isomers are ortho (adjacent), meta (separated by one carbon), and para (opposite sides of the ring).

 

Question 10. n - Butane and isobutane are a pair of
(a) chain isomers
(b) position isomers
(c) metamers
(d) functional isomers
Answer: (a) chain isomers
In simple words: n-Butane and isobutane are examples of chain isomers because they have the same molecular formula (\( \text{C}_4\text{H}_{10} \)) but differ in the arrangement of their carbon backbone. These differences in branching can lead to variations in physical properties like boiling point.

🎯 Exam Tip: Chain isomers have different arrangements of the carbon skeleton, meaning the main chain length or branching differs.

 

Question 11. Alkanols and Alkoxyalkanes are
(a) Functional isomers
(b) Keto - enol tautomers
(c) Geometrical isomers
(d) Not isomers at all
Answer: (a) Functional isomers
In simple words: Alkanols (alcohols) and alkoxyalkanes (ethers) are functional isomers because they share the same molecular formula but have different functional groups. Functional isomerism is important as it illustrates how different molecular structures can yield distinct chemical properties, despite sharing the same atoms.

🎯 Exam Tip: Recognize common functional isomer pairs, such as alcohols/ethers, aldehydes/ketones, and carboxylic acids/esters, which have the same molecular formula but different reactivity.

 

Question 12. n - propyl alcohol and isopropyl alcohol are examples of
(a) Position isomerism
(b) Chain isomerism
(c) Tautomerism
(d) Geometrical isomerism
Answer: (a) Position isomerism
In simple words: n-Propyl alcohol and isopropyl alcohol are position isomers. They have the same carbon skeleton and functional group, but the hydroxyl (-OH) group is attached to a different carbon atom. Understanding position isomerism is key in organic chemistry for distinguishing compounds with similar connectivity but different spatial arrangements.

🎯 Exam Tip: Position isomers have the same carbon chain but a different location for a functional group or substituent on that chain.

 

Question 13. The number of structural alcoholic isomers for \( \text{C}_4\text{H}_{10}\text{O} \) is
(a) 2
(b) 3
(c) 4
(d) 5
Answer: (c) 4
In simple words: For the molecular formula \( \text{C}_4\text{H}_{10}\text{O} \), there are four possible structural isomers that are alcohols: butan-1-ol, butan-2-ol, 2-methylpropan-1-ol, and 2-methylpropan-2-ol. Identifying these isomers is crucial in pharmaceutical and industrial chemistry, as their properties can vary significantly.

🎯 Exam Tip: To find all alcoholic isomers, systematically consider different chain structures (n-butyl, isobutyl, sec-butyl, tert-butyl) and place the -OH group accordingly.

 

Question 14. Cycloalkanes are isomeric with
(a) Alkadienes
(b) Alkynes
(c) Aromatic compounds
(d) Olefins
Answer: (d) Olefins
In simple words: Cycloalkanes are isomers of olefins (alkenes) because they both have the general formula \( \text{C}_n\text{H}_{2n} \). This structural relationship is key in understanding the diverse forms organic molecules can take while maintaining a similar atomic count.

🎯 Exam Tip: Remember the general formulas: alkanes (\( \text{C}_n\text{H}_{2n+2} \)), alkenes/cycloalkanes (\( \text{C}_n\text{H}_{2n} \)), and alkynes (\( \text{C}_n\text{H}_{2n-2} \)).

 

Question 15. Number of possible monochloro benzenes is
(a) 1
(b) 3
(c) 5
(d) 6
Answer: (a) 1
In simple words: There is only one possible monochloro benzene. Since all six positions on the benzene ring are equivalent, placing a single chlorine atom at any position results in the same compound, chlorobenzene. This highlights the symmetry of the benzene ring when only one substituent is present.

🎯 Exam Tip: The high symmetry of the benzene ring means that a single substituent can be placed at any position, yielding only one unique isomer.

 

Question 16. Diethyl ether and n - propyl methyl ether are
(a) Metamers
(b) Chain isomers
(c) Geometrical isomers
(d) Position isomers
Answer: (a) Metamers
In simple words: Diethyl ether and n-propyl methyl ether are metamers. They are isomers with the same functional group (ether) but have different alkyl groups attached to the oxygen atom. Metamerism highlights how the position of a functional group within a chain, or the nature of the groups around it, can create distinct compounds with varying properties.

🎯 Exam Tip: Metamers are functional isomers where the alkyl groups on either side of the functional group (like ether, ketone, or secondary amine) are different.

 

Question 17. The total number of structural isomers for the compound of the formula \( \text{C}_4\text{H}_{10}\text{O} \) is
(a) 7
(b) 6
(c) 4
(d) 3
Answer: (a) 7
In simple words: For the molecular formula \( \text{C}_4\text{H}_{10}\text{O} \), there are seven possible structural isomers, including four alcohols and three ethers. These isomers demonstrate the vast structural diversity possible with a simple change in atomic arrangement.

🎯 Exam Tip: When counting isomers, systematically consider all possible functional groups (alcohols, ethers for this formula) and then all possible chain/positional variations within each functional group.

 

Question 18. The number of primary alcoholic isomers with the formula \( \text{C}_4\text{H}_{10}\text{O} \) is
(a) 1
(b) 2
(c) 3
(d) 4
Answer: (b) 2
In simple words: For the molecular formula \( \text{C}_4\text{H}_{10}\text{O} \), there are two primary alcoholic isomers: butan-1-ol and 2-methylpropan-1-ol. Primary alcohols are important in various chemical reactions and industrial processes.

🎯 Exam Tip: A primary alcohol has the hydroxyl group (-OH) attached to a carbon atom that is only bonded to one other carbon atom.

 

Question 19. The compound which is not isomeric with diethyl ether is
(a) n - propyl methyl ether
(b) Butan - 1 - ol
(c) 2 - Methylpropan - 2 - ol
(d) Butanone
Answer: (d) Butanone
In simple words: Diethyl ether has the molecular formula \( \text{C}_4\text{H}_{10}\text{O} \). Butanone has the formula \( \text{C}_4\text{H}_8\text{O} \). Since their molecular formulas are different, butanone is not an isomer of diethyl ether. Checking the molecular formula is the first step in determining if two compounds are isomers.

🎯 Exam Tip: Isomers must always have the exact same molecular formula, even if their structures or functional groups differ.

 

Question 20. The number of isomeric amines possible for the formula \( \text{C}_3\text{H}_9\text{N} \)
(a) 4
(b) 3
(c) 5
(d) 6
Answer: (a) 4
In simple words: For the molecular formula \( \text{C}_3\text{H}_9\text{N} \), there are four possible isomeric amines: propylamine (primary), isopropylamine (primary), dimethylmethylamine (secondary), and ethylmethylamine (secondary). Understanding these different arrangements is important for predicting chemical reactions and physical properties of amines.

🎯 Exam Tip: To find all amine isomers, consider primary, secondary, and tertiary amines, and draw all possible carbon chain arrangements for each type.

 

Question 21. Which hybrid orbitals are involved in the \( \text{CH}_3 - \text{CH} = \text{CH} - \text{CH}_3 \) compound
(a) sp and \( \text{sp}^3 \)
(b) \( \text{sp}^2 \) and \( \text{sp}^3 \)
(c) sp and \( \text{sp}^2 \)
(d) only \( \text{sp}^3 \)
Answer: (b) \( \text{sp}^2 \) and \( \text{sp}^3 \)
In simple words: In the compound but-2-ene (\( \text{CH}_3 - \text{CH} = \text{CH} - \text{CH}_3 \)), the two carbon atoms in the double bond are \( \text{sp}^2 \) hybridized, and the two methyl group carbons at the ends are \( \text{sp}^3 \) hybridized. The mix of hybridizations determines the molecule's overall shape and reactivity.

🎯 Exam Tip: Identify the type of bond each carbon is involved in: single bonds mean \( \text{sp}^3 \), double bonds mean \( \text{sp}^2 \), and triple bonds mean sp hybridization.

 

Question 22. Which of the following bonds is strongest?
(a) \( \ce{-C \equiv C-} \)
(b) \( \ce{>C=C<} \)
(c) \( \ce{-C-C-} \)
(d) \( \ce{-C-C-} \)
Answer: (a) \( \ce{-C \equiv C-} \)
In simple words: A carbon-carbon triple bond is the strongest among single, double, and triple bonds. This stronger bond requires more energy to break, influencing the stability and reactivity of molecules.

🎯 Exam Tip: The bond strength increases with the number of bonds between two atoms: triple bonds are stronger than double bonds, which are stronger than single bonds.

 

Question 23. According to Huckel’s rule a compound, is said to be aromatic if’ it contains
(a) 4n bonds
(b) 4n atoms
(c) (4n + 2) atoms
(d) (4n + 2) \( \pi \) electrons
Answer: (d) (4n + 2) \( \pi \) electrons
In simple words: Hückel's rule states that a cyclic molecule is aromatic if it has a continuous ring of p-orbitals and contains exactly (4n + 2) pi electrons. This rule is fundamental for understanding the stability and unique properties of aromatic compounds.

🎯 Exam Tip: When applying Hückel's rule, count only the pi electrons (from double bonds, triple bonds, or lone pairs in p-orbitals) within the cyclic system.

 

Question 24. Which of the following is an aromatic compound
(a) Phenol
(b) Naphthalene
(c) Pyridine
(d) All
Answer: (d) All
In simple words: All the given options - phenol, naphthalene, and pyridine - are aromatic compounds. Their aromaticity provides them with special stability and reactivity patterns, making them crucial in many chemical processes and industrial applications.

🎯 Exam Tip: Remember common aromatic compounds, including benzene and its derivatives (like phenol), fused ring systems (like naphthalene), and heterocyclic compounds (like pyridine).

 

Question 25. Which is a saturated compound?
(a) alkanes
(b) alkenes
(c) alkynes
(d) cyclo alkenes
Answer: (a) alkanes
In simple words: Saturated compounds are organic molecules that contain only single bonds between carbon atoms. Because they are "saturated," alkanes are generally less reactive than unsaturated hydrocarbons like alkenes or alkynes.

🎯 Exam Tip: Saturated means a compound contains the maximum possible number of hydrogen atoms, with no double or triple carbon-carbon bonds.

 

Question 26. Which is an alicyclic compound?
(a) benzene
(b) cyclohexane
(c) pyridine
(d) pyrrole
Answer: (b) cyclohexane
In simple words: Cyclohexane is an alicyclic compound, which means it is a cyclic aliphatic compound. Alicyclic compounds are often used as solvents and in the synthesis of more complex organic molecules due to their stable ring structures.

🎯 Exam Tip: Alicyclic compounds are cyclic but do not have the special stability of aromatic compounds.

 

Question 27. Which of the following is not a cyclic compound?
(a) Anthracene
(b) Pyrrole
(c) Phenol
(d) Isobutylene
Answer: (d) Isobutylene
In simple words: Isobutylene is an acyclic (open-chain) compound, specifically an alkene, with a branched structure. Recognizing cyclic versus acyclic structures is a fundamental skill in organic chemistry, impacting how we understand reactivity and physical properties.

🎯 Exam Tip: Cyclic compounds have atoms arranged in a ring, while acyclic compounds have a linear or branched open-chain structure.

 

Question 28. Functional group present in amides is
(a) \( \ce{-COOH} \)
(b) \( \ce{-NH_2} \)
(c) \( \ce{-CONH_2} \)
(d) \( \ce{-COO-} \)
Answer: (c) \( \ce{-CONH_2} \)
In simple words: The functional group found in amides is the carboxamide group, represented as \( \ce{-CONH_2} \). The carboxamide group is significant in biochemistry, as it forms the peptide bonds that link amino acids together to create proteins.

🎯 Exam Tip: Distinguish amides (\( \ce{-CONH_2} \)) from amines (\( \ce{-NH_2} \)) and carboxylic acids (\( \ce{-COOH} \)), as they are structurally related but have distinct properties.

 

Question 29. IUPAC name of ester is
(a) Alkoxy alkane
(b) Alkyl alkanoate
(c) Alkanoyl halide
(d) Alkanoic anhydride
Answer: (b) Alkyl alkanoate
In simple words: The IUPAC naming convention for esters is "alkyl alkanoate". This systematic naming helps chemists clearly identify and distinguish between different ester compounds, which are common in fragrances and flavorings.

🎯 Exam Tip: Remember that the "alkyl" part of an ester name comes from the alcohol, and the "alkanoate" part comes from the carboxylic acid that forms it.

 

Question 30. IUPAC name of methyl cyanide is
(a) Cyano methane
(b) Ethanenitrile
(c) Methane nitrile
(d) Methyl - n - butyl amine
Answer: (b) Ethanenitrile
In simple words: Methyl cyanide is systematically named ethanenitrile according to IUPAC rules. Nitriles are versatile functional groups used in synthesizing many organic compounds.

🎯 Exam Tip: The IUPAC name for nitriles is formed by adding "nitrile" to the alkane name corresponding to the total number of carbons in the chain, including the nitrile carbon.

 

Question 31. The correct IUPAC name of \( \ce{CH3 - \underset{CH3}{|}C = \underset{CH3}{|}C - CH2 - CH3} \) is
(a) 1, 2 - diethyl butene
(b) 2 - ethyl - 3- methyl pentene
(c) 3, 4 - dimethyl hex - 3 - ene
(d) 2, 3 - dimethyl pent - 2 - ene
Answer: (d) 2, 3 - dimethyl pent - 2 - ene
In simple words: To name this compound, we find the longest carbon chain containing the double bond, which is five carbons long (pentene). The double bond is at position 2, and there are methyl groups at positions 2 and 3. Correctly identifying the longest chain and numbering the substituents is essential for unambiguous naming in organic chemistry.

🎯 Exam Tip: When naming alkenes with substituents, always prioritize the longest carbon chain that includes the double bond, and number it to give the double bond the lowest possible number, followed by the substituents.

 

Question 32. IUPAC name of \( \text{CH}_2\text{OH – CH}_2\text{OH} \) is
(a) 1, 2 - dihydroxy ethane
(b) ethylene glycol
(c) ethane - 1, 2 - diol
(d) ethane - 1, 2 - dial
Answer: (c) ethane - 1, 2 - diol
In simple words: The compound \( \text{CH}_2\text{OH – CH}_2\text{OH} \) has two carbon atoms and two hydroxyl (\(\text{OH}\)) groups. This makes it an ethane diol, with the hydroxyl groups at positions 1 and 2.

🎯 Exam Tip: Remember that "diol" means two hydroxyl groups. When naming, count the carbons to get the parent alkane name, then indicate the positions of the hydroxyl groups.

 

Question 33. IUPAC name of \( \text{CH} \equiv \text{C – CH = CH}_2 \) is
(a) but - 3 - ene - 1 - yne
(b) but - 1 - ene - 3 - yne
(c) but - 1 - yne - 3 - ene
(d) but - 3 - yne - 1 - ene
Answer: (b) but - 1 - ene - 3 - yne
In simple words: When naming organic compounds with both double and triple bonds, prioritize numbering to give the lowest possible numbers to the multiple bonds. The "ene" (double bond) comes before "yne" (triple bond) alphabetically in the name.

🎯 Exam Tip: For compounds with both double and triple bonds, the chain is numbered to give the multiple bonds the lowest possible set of locants. If there's a tie, the double bond gets preference for the lowest number.

 

Question 34. The IUPAC name of is
(a) 4 - hydroxy - 2 - methyl pentanal
(b) 2 - hydroxy - 4 - methyl pentanal
(c) 4 - hydroxy - 2 - methyl pentanol
(d) 2 - hydroxy - 4 - methyl pentanol
Answer: (a) 4 - hydroxy - 2 - methyl pentanal
In simple words: This compound is a pentanal (5 carbons with an aldehyde group at the end). It has a hydroxyl (\(\text{OH}\)) group at the 4th carbon and a methyl (\(\text{CH}_3\)) group at the 2nd carbon.

🎯 Exam Tip: When an aldehyde group is present, it always defines the first carbon (C1) of the main chain. Substituents are then numbered from this end to give them the lowest possible locants.

 

Question 35. 3 - methyl penta -1, 3- diene is
(a) \( \text{CH}_2 = \text{CH (CH}_2\text{)}_2 \text{CH}_3 \)
(b) \( \text{CH}_2 = \text{CHCH (CH}_3\text{) CH}_2\text{CH}_3 \)
(c) \( \text{CH}_3\text{CH = C(CH}_3\text{)CH = CH}_2 \)
(d) \( \text{CH}_2 = \text{C = CH (CH}_3\text{)}_2 \)
Answer: (c) \( \text{CH}_3\text{CH = C(CH}_3\text{)CH = CH}_2 \)
In simple words: The name "penta-1,3-diene" means a 5-carbon chain with double bonds at carbons 1 and 3. The "3-methyl" means a methyl group is attached to the 3rd carbon. Option (c) matches this structure.

🎯 Exam Tip: Draw the main chain first, then add the multiple bonds at the specified positions, and finally add any substituents. Double-check that the numbering gives the lowest possible locants.

 

Question 36. The structural formula of methyl amino methane is
(a) \( \text{(CH}_3\text{)}_2 \text{CH NH}_2 \)
(b) \( \text{(CH}_3\text{)}_3 \text{N} \)
(c) \( \text{(CH}_3\text{)}_2 \text{NH} \)
(d) \( \text{CH}_3\text{NH}_2 \)
Answer: (c) \( \text{(CH}_3\text{)}_2 \text{NH} \)
In simple words: "Methyl amino methane" or dimethylamine means there are two methyl groups attached to a nitrogen atom. This corresponds to the formula \( \text{(CH}_3\text{)}_2 \text{NH} \).

🎯 Exam Tip: Remember that "methyl amino methane" is a common name, and its IUPAC equivalent is N-methylmethanamine, but in this context, it refers to a secondary amine with two methyl groups.

 

Question 37. Which of the following is the functional isomer of methyl acetate?
(a) Ethyl acetate
(b) Propanoic acid
(c) Ethyl formate
(d) Propanone
Answer: (b) Propanoic acid
In simple words: Functional isomers have the same molecular formula but different functional groups. Methyl acetate (\(\text{CH}_3\text{COOCH}_3\)) and propanoic acid (\(\text{CH}_3\text{CH}_2\text{COOH}\)) both have the molecular formula \( \text{C}_3\text{H}_6\text{O}_2 \), but one is an ester and the other is a carboxylic acid.

🎯 Exam Tip: To identify functional isomers, first find the molecular formula of the given compound. Then, look for options with the same molecular formula but different functional groups (e.g., alcohol vs. ether, aldehyde vs. ketone, carboxylic acid vs. ester).

 

Question 38. The compound which is not isomeric with diethyl ether is
(a) n - propyl methyl ether
(b) 1 - Butanol
(c) 2 - Methyl - 2 - propanol
(d) Butanone
Answer: (d) Butanone
In simple words: Diethyl ether has the formula \( \text{C}_4\text{H}_{10}\text{O} \). We need to find which option does not have this same formula. Butanone has the formula \( \text{C}_4\text{H}_8\text{O} \), which is different.

🎯 Exam Tip: To check for isomerism, always calculate the molecular formula of each compound. Isomers must have identical molecular formulas, even if their structures are different.

 

Question 39. Which of the following pairs of compounds are tautomers?
(a) Propanol & propanone
(b) Ethanol & vinyl alcohol
(c) Ethanol & allyl alcohol
(d) Vinyl alcohol & ethanal
Answer: (d) Vinyl alcohol & ethanal
In simple words: Tautomers are isomers that can easily change from one form to another, often involving the movement of a hydrogen atom and a change in a double bond. Vinyl alcohol and ethanal can convert between each other through keto-enol tautomerism.

🎯 Exam Tip: Tautomerism (specifically keto-enol tautomerism) typically involves a carbonyl group (keto form) and an enol group (alcohol with a double bond). Look for compounds that can interconvert this way.

 

Question 40. Which of the following compounds does not have any tertiary hydrogen atoms?
(a) \( \text{(CH}_3\text{)}_3 \text{CCH}_2 \text{CH}_3 \)
(b) \( \text{(CH}_3\text{)}_2 \text{CHCH}_2 \text{CH}_3 \)
(c) \( \text{(CH}_3\text{)}_2 \text{CHCH (CH}_3\text{)}_2 \)
(d) \( \text{(CH}_3\text{)}_3 \text{CH} \)
Answer: (a) \( \text{(CH}_3\text{)}_3 \text{CCH}_2 \text{CH}_3 \)
In simple words: A tertiary hydrogen atom is a hydrogen attached to a tertiary carbon atom (a carbon bonded to three other carbon atoms). In option (a), the central carbon is bonded to three methyl groups and one \( \text{CH}_2 \text{CH}_3 \) group, but it doesn't have any hydrogen atoms itself.

🎯 Exam Tip: To identify tertiary hydrogen atoms, first locate all tertiary carbon atoms (carbons bonded to three other carbons). Then check if any hydrogen atoms are directly attached to these tertiary carbons. If a tertiary carbon exists but has no attached hydrogen, there are no tertiary hydrogens.

 

Question 41. The IUPAC name of is
(a) 2 - Methyl - 2 - butenoic acid
(b) 3 - Methyl - 3 - butenoic acid
(c) 3 - Methyl - 2 - butenoic acid
(d) 2 - Methyl - 3 - butenoic acid
Answer: (c) 3 - Methyl - 2 - butenoic acid
In simple words: The main chain has four carbons with a carboxylic acid group, making it a butenoic acid. The double bond is at position 2, and a methyl group is at position 3.

🎯 Exam Tip: The carboxylic acid group (\(\text{COOH}\)) always gets the lowest number (C1) in the main chain. The double bond and methyl group are then numbered relative to this C1 position.

 

Question 42. The IUPAC name of Cinnamaldehyde is
(a) 3 - Phenyl prop - 2 - enal
(b) 1 - Phenyl - prop - 1 - enal
(c) 1 - Phenyl - prop - 2 - enal
(d) 3 - Phenyl - prop - 1 - enal
Answer: (a) 3 - Phenyl prop - 2 - enal
In simple words: Cinnamaldehyde has an aldehyde group, a double bond, and a phenyl group. The main chain is a 3-carbon "propenal," with the double bond at C2 and the phenyl group at C3.

🎯 Exam Tip: For aldehydes with unsaturated chains, the aldehyde carbon is always C1. Ensure the double bond's position and the substituent's position are correctly indicated relative to this C1. Remember to use "enal" for aldehydes with double bonds.

 

Question 43. The IUPAC name of the Compound \( \text{CH}_3 \text{– CH(OH) – COOH} \) is
(a) Lactic acid
(b) 2 - Hydroxy propanoic acid
(c) 3 - Hydroxy propanoic acid
(d) Carboxy propanol
Answer: (b) 2 - Hydroxy propanoic acid
In simple words: The compound has a 3-carbon chain with a carboxylic acid group (\(\text{COOH}\)), making it a propanoic acid. There is also a hydroxyl (\(\text{OH}\)) group on the second carbon.

🎯 Exam Tip: The carboxylic acid carbon is always C1. The hydroxyl group's position is then numbered from that end. "Hydroxy" is used as a prefix when \(\text{OH}\) is not the highest priority functional group.

 

Question 44. The IUPAC name of the compound is
(a) 2 - Ethyl - ethyl acetate
(b) Ethyl - 3 - methylbutanoate
(c) Ethyl - 2 - methyl butanoate
(d) 2- methyl butanoic acid
Answer: (c) Ethyl - 2 - methyl butanoate
In simple words: This is an ester. The "ethyl" part comes from the alcohol used to form the ester. The "butanoate" part comes from the 4-carbon acid chain, which has a methyl group at position 2.

🎯 Exam Tip: For esters, the alkyl group attached to the oxygen of the ester linkage is named first. The acyl part (containing the carbonyl) is named as an alkanoate, with numbering starting from the carbonyl carbon of the acyl group.

 

Question 45. The IUPAC name of the given compound is
(a) 2,2 - Dimethyl butane
(b) Isohexane
(c) 2, 3 - Dimethyl butane
(d) Di isohexane
Answer: (c) 2, 3 - Dimethyl butane
In simple words: The longest continuous carbon chain has four carbons, making it a butane. There are two methyl groups, one at the 2nd carbon and another at the 3rd carbon.

🎯 Exam Tip: Find the longest continuous carbon chain first (parent chain). Then, number the chain to give substituents the lowest possible numbers. If there are multiple identical substituents, use prefixes like "di-", "tri-", etc.

 

Question 46. The IUPAC name of the given compound is
(a) Octyl cyclopentane
(b) 3 - cyclopentyl octane
(c) Cyclopentane octane
(d) 6 - cyclopentyl octane
Answer: (b) 3 - cyclopentyl octane
In simple words: The compound has a straight chain of eight carbons (octane) with a five-carbon ring (cyclopentane) attached to the third carbon of the octane chain.

🎯 Exam Tip: When a ring and a chain are present, the one with more carbon atoms is typically chosen as the parent. If they have the same number of carbons, the ring is often chosen as the parent, but in this case, the chain is longer.

 

Question 47. The IUPAC name of is
(a) but - 2- ene - 2,3 - diol
(b) pent - 2- ene - 2,3 - diol
(c) 2 - methyl but - 2 - ene - 2,3 - diol
(d) hex - 2- ene - 2,3 - diol
Answer: (b) pent - 2- ene - 2,3 - diol
In simple words: The longest chain contains five carbons (pent-), a double bond at C2 (2-ene), and two hydroxyl groups at C2 and C3 (2,3-diol).

🎯 Exam Tip: Identify the longest carbon chain containing the principal functional groups and multiple bonds. Number the chain to give the functional groups (like -ol) and multiple bonds the lowest possible locants.

 

Question 48. The structure of 3-bromoprop-1-ene is
(a)
(b) \( \text{CH}_3 \text{– CH = CH – Br} \)
(c)
(d) \( \text{Br – CH}_2 \text{– CH} \equiv \text{CH} \)
Answer: (a)
In simple words: "Prop-1-ene" means a 3-carbon chain with a double bond starting at C1. "3-bromo" means a bromine atom is attached to the 3rd carbon. Option (a) correctly shows this structure.

🎯 Exam Tip: When drawing structures from IUPAC names, start with the parent chain and its main functional groups or multiple bonds. Then add substituents at their specified positions, ensuring proper valency for all atoms.

 

Question 49. Neo-heptyl alcohol is correctly represented as
(a)
(b)
(c)
(d)
Answer: (c)
In simple words: "Neo-heptyl alcohol" means a 7-carbon alcohol with a specific branched structure. The "neo" prefix refers to a carbon atom bonded to four other carbon atoms. Option (c) shows a primary alcohol with a quaternary carbon near the end, which is characteristic of a neo-structure.

🎯 Exam Tip: Recognize common prefixes like "neo-", "iso-", and "sec-" which denote specific branching patterns. "Neo-" typically means a quaternary carbon (a carbon bonded to four other carbons) at the second-to-last position in the chain, with the main functional group on the terminal carbon.

 

Question 50. Number of dibromo derivatives possible for propane are
(a) 2
(b) 3
(c) 1
(d) 4
Answer: (d) 4
In simple words: Propane (\(\text{CH}_3\text{CH}_2\text{CH}_3\)) can have two bromine atoms attached in different ways. These are 1,1-dibromopropane, 1,2-dibromopropane, 1,3-dibromopropane, and 2,2-dibromopropane.

🎯 Exam Tip: When determining possible isomers, systematically place the substituents on each possible carbon atom, considering symmetry, and then write out the IUPAC names to ensure no duplicates are counted.

 

Question 51. The number of aromatic isomers possible for \( \text{C}_7\text{H}_8\text{O} \) is
(a) 2
(b) 3
(c) 4
(d) 5
Answer: (d) 5
In simple words: For the formula \( \text{C}_7\text{H}_8\text{O} \), there are five possible aromatic isomers. These include cresols (ortho, meta, para), benzyl alcohol, and anisole.

🎯 Exam Tip: To find aromatic isomers, consider all possible positions for substituents on the benzene ring and different functional groups that can be attached to the ring or a side chain. Draw each possibility to avoid double counting.

 

Question 52. Isomers of propanoic acid are
(a) \( \text{HCOOC}_2\text{H}_5 \) and \( \text{CH}_3\text{COOCH}_3 \)
(b) \( \text{H – COOC}_2\text{H}_5 \) and \( \text{C}_3\text{H}_7\text{COOH} \)
(c) \( \text{CH}_3\text{COOCH}_3 \) and \( \text{C}_3\text{H}_7\text{OH} \)
(d) \( \text{C}_3\text{H}_7\text{OH} \) and \( \text{CH}_3\text{COCH}_3 \)
Answer: (a) \( \text{HCOOC}_2\text{H}_5 \) and \( \text{CH}_3\text{COOCH}_3 \)
In simple words: Propanoic acid has the molecular formula \( \text{C}_3\text{H}_6\text{O}_2 \). We are looking for compounds that are isomers of propanoic acid. Both ethyl formate (\(\text{HCOOC}_2\text{H}_5\)) and methyl acetate (\(\text{CH}_3\text{COOCH}_3\)) have the same molecular formula, \( \text{C}_3\text{H}_6\text{O}_2 \), making them isomers.

🎯 Exam Tip: Remember that isomers have the same molecular formula but different arrangements of atoms. Esters are common functional isomers of carboxylic acids with the same number of carbon atoms.

 

Question 53. IUPAC name of \( \text{CH}_3 \text{– CH (OCH}_3\text{) – CH}_2 \text{– NH}_2 \)
(a) 2-methoxy propanamine
(b) 1-amino – 2-methoxy propane
(c) 1-amino – 2-methyl – 2-methoxy ethane
(d) 1 – methoxy- 2-amino propane
Answer: (a) 2-methoxy propanamine
In simple words: The longest chain has three carbons, making it a propanamine because of the amino group. The amino group is at C1. A methoxy group is attached at C2.

🎯 Exam Tip: Prioritize the amino group as the principal functional group, defining the carbon chain numbering. "Methoxy" is a common prefix for an ether substituent.

 

Question 54. IUPAC name of is
(a) 3 - cyanopentane - 1, 5 - dinitrile
(b) Propane - 1, 2, 3-tri nitrile
(c) 1, 2, 3-tri cyano propane
(d) Propane 1, 2, 3-tricarbonitrile
Answer: (d) Propane 1, 2, 3-tricarbonitrile
In simple words: When there are three or more nitrile groups on a main carbon chain, they are named as "carbonitrile" groups. The parent chain is propane, and each carbon atom has a nitrile group.

🎯 Exam Tip: For dinitriles or trinitriles, if the nitrile groups are directly attached to a parent chain that cannot contain all of them in its numbering (e.g., three nitriles on a propane chain), use the "-carbonitrile" suffix to name them.

 

Question 55. The IUPAC name of the compound is
(a) 3 - Carboxylic pentane - 1,5 -dioic acid
(b) Propane - 1, 2, 3 - trioic acid
(c) 1, 2, 3- tricarboxylic propane
(d) Propane - 1,2, 3 - tricarboxylic acid
Answer: (b) Propane - 1, 2, 3 - trioic acid
In simple words: The compound has a 3-carbon parent chain (propane) with three carboxylic acid groups (\(\text{COOH}\)) attached at each carbon. When all \(\text{COOH}\) groups cannot be part of the main chain, we name them as "trioic acid" with the parent alkane.

🎯 Exam Tip: Similar to nitriles, if a parent chain has three or more carboxylic acid groups, and not all can be included in the longest chain's numbering, the "-carboxylic acid" suffix is used, and the parent chain is named as an alkane (e.g., propane-1,2,3-tricarboxylic acid).

 

Question 56. The IUPAC name of the following compound \( \text{CH}_3 \text{– C(CH}_3\text{)}_2 \text{– CH}_2 \text{– CH = CH}_2 \) is
(a) 2, 2 - Dimethyl - 4 - pentene
(b) 4, 4 - Dimethyl - 1 - pentene
(c) 1, 1, 1 - trimethyl - 3 - butene
(d) 4, 4, 4 - trimethyl - 1 - butene
Answer: (b) 4, 4 - Dimethyl - 1 - pentene
In simple words: The longest carbon chain containing the double bond has five carbons, making it a pentene. The double bond is at C1. There are two methyl groups at C4. So it's 4,4-dimethyl-1-pentene.

🎯 Exam Tip: Always number the parent chain to give the double bond the lowest possible number. Then identify and number the substituents based on this chain numbering.

 

Question 57. The IUPAC name of is
(a) 2, 4 - Dimethyl pentan - 2 - ol
(b) 2, 4 - Dimethyl pentan - 4 - ol
(c) 2,2 - Dimethyl butan - 2- ol
(d) Butan - 2 - ol
Answer: (a) 2, 4 - Dimethyl pentan - 2 - ol
In simple words: The longest chain with the hydroxyl group has five carbons, so it's a pentanol. The hydroxyl group is at C2. There are methyl groups at C2 and C4.

🎯 Exam Tip: When naming alcohols, ensure the carbon chain containing the hydroxyl group is selected and numbered to give the hydroxyl group the lowest possible locant. Methyl groups are then positioned accordingly.

 

Question 58. The IUPAC name the compound is
(a) Butane - 2, 3, 4 - triol
(b) Butane - 1,2, 3 - triol
(c) Pentane - 1, 2, 3 - triol
(d) 2, 3 dihydroxy butanol
Answer: (b) Butane - 1,2, 3 - triol
In simple words: The structure has a 4-carbon chain (butane) with three hydroxyl groups. These groups are located at carbons 1, 2, and 3.

🎯 Exam Tip: Identify the longest chain containing all hydroxyl groups. Number the chain to give the hydroxyl groups the lowest possible set of locants. Use the "triol" suffix for three hydroxyl groups.

 

Question 59. The IUPAC name the compound is
(a) 3, 3 - Dimethyl - 1 - cyclohexanol
(b) 1, 1- Dimethyl - 3 - hydroxy cyclohexane
(c) 3, 3 - Dimethyl - 1 - hydroxy cyclohexane
(d) 1, 1 - Dimethyl - 3 - cyclohexanol
Answer: (a) 3, 3 - Dimethyl - 1 - cyclohexanol
In simple words: The main structure is a cyclohexanol, with the hydroxyl group at C1. There are two methyl groups at C3.

🎯 Exam Tip: In cyclic alcohols, the carbon atom bonded to the hydroxyl group is always assigned as C1. The numbering then proceeds to give other substituents the lowest possible locants.

 

Question 60. The IUPAC name of the compound is
(a) 2 - Ethylprop - 2 - en - 1 - ol
(b) 2 - Hydroxymethyl butan - 1- ol
(c) 2 - Methylene butan - 1 - ol
(d) 2 - Ethyl -3 hydroxyprop - 1 - ene
Answer: (a) 2 - Ethylprop - 2 - en - 1 - ol
In simple words: This alcohol has a three-carbon chain as its parent (prop-). It has a double bond at C2, an ethyl group at C2, and the alcohol group at C1.

🎯 Exam Tip: The longest chain containing the hydroxyl group and the double bond should be chosen as the parent. Number the chain to give the alcohol group the lowest possible number, then locate the double bond and substituents.

 

Question 61. The number of possible alkynes with molecular formula \( \text{C}_5\text{H}_8 \) is
(a) 2
(b) 3
(c) 4
(d) 5
Answer: (b) 3
In simple words: For the molecular formula \( \text{C}_5\text{H}_8 \), there are three possible alkyne isomers: pent-1-yne, pent-2-yne, and 3-methylbut-1-yne.

🎯 Exam Tip: Systematically draw all possible straight-chain and branched-chain structures for the given molecular formula, placing the triple bond in every possible unique position. Always write down the IUPAC names to confirm they are distinct.

 

Question 62. What is the IUPAC name of the following is
(a) 3 - chloro cyclo hexa - 1, 5 - diene
(b) 5 - chloro cyclo hexa - 1, 3 - diene
(c) 1 - chloro cyclo hexa - 2, 5 - diene
(d) 2 - chloro cyclo hexa - 1, 4 - diene
Answer: (b) 5 - chloro cyclo hexa - 1, 3 - diene
In simple words: The cyclohexane ring has two double bonds, making it a cyclohexadiene. The double bonds are at positions 1 and 3. The chlorine atom is at position 5.

🎯 Exam Tip: When numbering a cyclic compound with multiple bonds, prioritize giving the multiple bonds the lowest possible set of numbers. If there's a choice, substituents are then given the next lowest numbers.

 

Question 63. What is the IUPAC name of the following is
(a) 6 - hydroxy cyclohex - 2 - ene - 1 - al
(b) 4 - hydroxy cyclohex - 1 - ene - 3 - al
(c) 2 - hydroxy cyclohex - 5 - ene - 1 - al
(d) 1 - formyl cyclohex - 5 - ene - 2 - ol
Answer: (a) 6 - hydroxy cyclohex - 2 - ene - 1 - al
In simple words: The compound has an aldehyde group on a cyclohexane ring, making it a cyclohexene carbaldehyde. The numbering starts from the aldehyde carbon, and the double bond and hydroxyl group are positioned accordingly.

🎯 Exam Tip: When a ring has a substituent that would typically be part of a chain (like an aldehyde or carboxylic acid), it's often named as a "carbaldehyde" or "carboxylic acid" attached to the ring, with the carbon of that group defining the first position of the ring's substituent numbering.

 

Question 64. What is the IUPAC name of the following?
(a) 1 - chloro - 2 - bromo - 4 - nitrobenzene
(b) 1 - bromo - 2- chloro - 4 - mtrobenzene
(c) 3 - bromo - 4 - chloro - nitrobenzene
(d) 2 - bromo - 1 - chloro - 4- nitrobenzene
Answer: (d) 2 - bromo - 1 - chloro - 4- nitrobenzene
In simple words: The benzene ring has a chlorine, a bromine, and a nitro group. Numbering starts with chlorine at 1, then bromine at 2, and the nitro group at 4, alphabetically ordering the substituents.

🎯 Exam Tip: When multiple different substituents are on a benzene ring, number the ring to give the lowest possible set of locants. If there is a tie, alphabetical order of substituents breaks the tie.

 

Question 65. What is the IUPAC name of the following?
(a) Ethenyl cyclo pentane
(b) cyclopentyl ethene
(c) cyclopentyl ethylene
(d) vinyl cyclopentane
Answer: (b) cyclopentyl ethene
In simple words: This compound has a cyclopentane ring and an ethene group. Since the cyclopentane has five carbons and the ethene group has two, the cyclopentane ring is considered the main part, and ethene is a substituent.

🎯 Exam Tip: When naming compounds with a ring and a chain, the one with more carbons is usually the parent. If both have double bonds or other functional groups, IUPAC rules for priority determine the parent.

 

Question 66. The hybridization of carbon atoms in \( \text{HC} \equiv \text{C – CH = CH}_2 \) is
(a) \( \text{sp}^3 \text{– sp}^3 \)
(b) \( \text{sp}^2 \text{– sp}^3 \)
(c) \( \text{sp – sp}^2 \)
(d) \( \text{sp}^3 \text{– sp} \)
Answer: (c) \( \text{sp – sp}^2 \)
In simple words: The first carbon in \( \text{HC} \equiv \text{C – CH = CH}_2 \) has a triple bond, so it is sp hybridized. The second carbon has a double bond, so it is \( \text{sp}^2 \) hybridized. The single bond connects these two carbons.

🎯 Exam Tip: Remember the rules for hybridization: sp for triple bonds or two sigma bonds, \( \text{sp}^2 \) for double bonds or three sigma bonds, and \( \text{sp}^3 \) for single bonds or four sigma bonds.

 

Question 67. The correct IUPAC name of the compound is
(a) 1, 4 - Butane dioicacid
(b) Ethane - 1, 2 - dicarboxylic acid
(c) Succinic acid
(d) 1, 2 - Ethane dioic acid
Answer: (a) 1, 4 - Butane dioic acid
In simple words: The structure has four carbon atoms with two carboxylic acid groups, one at each end. This makes it a butane dioic acid, with the acid groups at positions 1 and 4.

🎯 Exam Tip: When naming dicarboxylic acids, if both carboxylic acid groups are at the ends of the main chain, the suffix "-dioic acid" is used with the parent alkane name, and the carbons of the carboxylic acids are included in the chain numbering.

 

Question 68. Correct statements about \( \text{CH}_3 \text{– CH}_2 \text{– CN} \) is
(a) common name of the compound is ethylcyanide.
(b) IUPAC name of the compound propane - 1 - nitrile.
(c) secondary suffix of the compound is nitrile.
(d) IUPAC name of the compound is ethane nitrile.
Answer: (a) common name of the compound is ethylcyanide.
In simple words: The compound \( \text{CH}_3 \text{– CH}_2 \text{– CN} \) is commonly known as ethyl cyanide because it has an ethyl group attached to a cyanide group. Its IUPAC name is propanenitrile, as the carbon in the CN group counts towards the main chain.

🎯 Exam Tip: For nitriles, the carbon atom of the cyano group (\(\text{CN}\)) is included in the numbering of the parent chain. The suffix is "-nitrile". Common names often treat the \(\text{CN}\) as a substituent (cyanide) on an alkyl group.

 

Question 69. Tautomerism is shown by
(a) R – C \( \equiv \) N
(b) R – \( \text{NO}_2 \)
(c) R – OH
(d) R – COOH
Answer: (b) R – \( \text{NO}_2 \)
In simple words: Tautomerism is a special type of isomerism where two forms of a compound can easily convert into each other. Nitro compounds (\(\text{R – NO}_2\)) can exhibit nitro-aci tautomerism, where they can interconvert with their aci-nitro form.

🎯 Exam Tip: Tautomerism often involves the migration of a hydrogen atom and a rearrangement of double bonds. Look for functional groups that allow for such dynamic interconversion, like carbonyls (keto-enol) or nitro groups (nitro-aci).

 

Question 70. Stereo isomers have different
(a) Molecular mass
(b) Molecular formula
(c) Structural formula
(d) Configuration
Answer: (d) Configuration
In simple words: Stereoisomers have the same molecular formula and the same way their atoms are connected (structural formula), but they differ in how their atoms are arranged in three-dimensional space. This arrangement is called configuration.

🎯 Exam Tip: Remember the key difference between structural isomers (different connectivity) and stereoisomers (same connectivity, different spatial arrangement or configuration). Optical isomers and geometrical isomers are types of stereoisomers.

 

Question 71. Geometrical isomerism may be exhibited by compounds having atleast.
(a) One double bond
(b) One triple bond
(c) One asymmetric carbon
(d) One polar bond
Answer: (a) One double bond
In simple words: Geometrical isomerism (cis-trans isomerism) happens when there is restricted rotation around a bond, usually a double bond. This allows groups to be fixed on either the same side (cis) or opposite sides (trans).

🎯 Exam Tip: For geometrical isomerism to exist across a double bond, each carbon of the double bond must be attached to two different groups. Cyclic structures can also exhibit geometrical isomerism due to restricted rotation.

 

Question 72. The prefixes syn - and anti - are used to denote
(a) structural isomers
(b) conformational isomers
(c) geometrical isomers
(d) optical isomers
Answer: (c) geometrical isomers
In simple words: "Syn" and "anti" are special words used to describe a type of isomerism called geometrical isomerism, especially in compounds like oximes. They tell us how groups are arranged around a double bond.

🎯 Exam Tip: Remember that syn-anti isomerism is a specific type of geometrical isomerism often seen in compounds with a C=N double bond, where rotation is restricted.

 

Question 73. d - tartaric acid and l - tartaric acid are
(a) geometrical isomers
(b) conformers
(c) enantiomers
(d) diastereomers
Answer: (c) enantiomers
In simple words: d-tartaric acid and l-tartaric acid are like mirror images that cannot be placed perfectly on top of each other. They are called enantiomers and have the same chemical formula but twist light in opposite ways.

🎯 Exam Tip: Enantiomers are non-superimposable mirror images of each other, differing only in their interaction with plane-polarized light.

 

Question 74. The method of separation of enantiomers from racemic mixture is known as
(a) inversion
(b) racemisation
(c) resolution
(d) asymmetric synthesis
Answer: (c) resolution
In simple words: When you have an equal mix of mirror-image molecules (a racemic mixture), the process of separating them into their individual forms is called resolution. It's like sorting left and right gloves from a mixed pile.

🎯 Exam Tip: Resolution is crucial in pharmaceutical chemistry, as often only one enantiomer of a drug is biologically active.

 

Question 75. Racemic mixture is optically inactive due to
(a) internal compensation
(b) external compensation
(c) inversion
(d) plane of symmetry
Answer: (b) external compensation
In simple words: A racemic mixture has equal amounts of two mirror-image molecules that cancel out each other's effect on light. This happens because the light-rotating power of one molecule is exactly balanced by the opposite light-rotating power of the other, making the whole mixture appear inactive.

🎯 Exam Tip: External compensation means the optical activity of two enantiomers in a mixture cancels out, while internal compensation refers to a single molecule (like a meso compound) being optically inactive due to symmetry.

 

Question 76. Meso isomers are possible when the organic compound contains
(a) one asymmetric carbon
(b) two or more dissimilar asymmetric carbons
(c) similar asymmetric carbons
(d) unsaturation
Answer: (c) similar asymmetric carbons
In simple words: Meso isomers are special molecules that have two or more chiral centers (asymmetric carbons) but are still not optically active. This happens when their internal structure has a plane of symmetry, causing the effects of the chiral centers to cancel each other out.

🎯 Exam Tip: A meso compound always has similar asymmetric carbons and an internal plane of symmetry, making it optically inactive despite having chiral centers.

 

Question 77. Optical inactivity of meso isomer is due to
(a) element of symmetry and element of asymmetry
(b) internal compensation
(c) due to lack of asymmetric carbon
(d) External compensation
Answer: (b) internal compensation
In simple words: A meso isomer has parts that are mirror images of each other within the same molecule. So, one part cancels out the light-rotating effect of the other part, making the entire molecule optically inactive. This is called internal compensation because the balancing happens inside the molecule itself.

🎯 Exam Tip: Understand the difference between internal compensation (for meso compounds) and external compensation (for racemic mixtures) as both result in optical inactivity.

 

Question 78. A racemic mixture is a mixture of
(a) meso and its isomers
(b) d and l isomers of same compound in equimolar proportions
(c) d and l isomers of same compound in different proportions
(d) mixture of d and meso isomers
Answer: (b) d and l isomers of same compound in equimolar proportions
In simple words: A racemic mixture is a perfect 50-50 blend of two mirror-image molecules (enantiomers), one that twists light to the right (d-form) and one that twists light to the left (l-form). Because they are in equal amounts, their light-twisting effects cancel each other out.

🎯 Exam Tip: The key characteristic of a racemic mixture is the equal (equimolar) presence of both enantiomers, leading to overall optical inactivity.

 

Question 79. Which of the following is optically active?
(a) HOOC – \( \text{CH}_2 \) – COOH
(b) \( \text{CH}_3 \) – CO – COOH
(c) \( \text{CH}_3 \) – CH(OH) – COOH
(d) \( \text{CH}_3 \) – \( \text{CH}_2 \) – COOH
Answer: (c) \( \text{CH}_3 \) – CH(OH) – COOH
In simple words: A molecule is optically active if it has a carbon atom that is bonded to four different groups. Out of the given choices, only lactic acid \( (\text{CH}_3 - \text{CH(OH)} - \text{COOH}) \) has such a carbon atom, making it able to rotate plane-polarized light.

🎯 Exam Tip: To identify an optically active compound, look for a chiral carbon atom, which is a carbon atom attached to four distinct groups.

 

Question 80. Which of the following is optically active?
(a) n – propanal
(b) 2 – chlorobutane
(c) n – butanal
(d) 3 – pentanol
Answer: (b) 2 – chlorobutane
In simple words: A molecule is optically active if it can rotate polarized light. This usually happens when it has a carbon atom with four different things attached to it. 2-chlorobutane has such a carbon, making it optically active.

🎯 Exam Tip: For a compound to be optically active, it must possess a chiral center (a carbon atom bonded to four different substituents) and lack any plane or center of symmetry.

 

Question 81. The number of optical enantiomers of tartaric acid
(a) 3
(b) 2
(c) 4
(d) 1
Answer: (b) 2
In simple words: Tartaric acid has two chiral centers, but because of its symmetry, one of its forms is a meso compound (optically inactive). This leaves only two forms that are mirror images of each other and are optically active, known as enantiomers.

🎯 Exam Tip: While tartaric acid has two chiral centers, remember that it also has a meso form due to internal symmetry, which reduces the number of enantiomers to two (d- and l-forms).

 

Question 82. Geometrical isomers differ in
(a) position of substituents
(b) Position of double bond
(c) C – C bond length
(d) Spatial arrangement of groups
Answer: (d) Spatial arrangement of groups
In simple words: Geometrical isomers are molecules with the same connections between atoms but different arrangements of those atoms in space, especially around a rigid bond like a double bond or a ring. They are like objects that have "left" and "right" versions because their parts are pointed in different directions.

🎯 Exam Tip: The defining characteristic of geometrical isomers (cis-trans isomers) is the difference in the orientation of groups around a restricted rotation axis, not changes in bonding sequence or position.

 

Question 83. Which of the following exhibit cis - trans isomerism
(a) propene
(b) 1 – butene
(c) 2 – butene
(d) benzene
Answer: (c) 2 – butene
In simple words: Cis-trans isomerism happens when there's a double bond, and each carbon atom in that double bond has two different groups attached. 2-Butene fits this perfectly, as its two methyl groups can be on the same side (cis) or opposite sides (trans) of the double bond.

🎯 Exam Tip: For cis-trans isomerism to occur across a double bond, each carbon involved in the double bond must be attached to two different groups.

 

Question 84. Which of the following show geometrical isomerism?
(a) \( \text{CH}_3 \text{CH} = \text{CHCH}_3 \)
(b) \( (\text{CH}_3)_2 \text{C} = \text{CH}_2 \)
(c) \( \text{C}_2\text{H}_5 \text{CH} = \text{CH}_2 \)
(d) \( \text{CH}_3 \text{CH} = \text{CH}_2 \)
Answer: (a) \( \text{CH}_3 \text{CH} = \text{CHCH}_3 \)
In simple words: Geometrical isomerism, also called cis-trans isomerism, occurs when a molecule has a double bond and each carbon of that double bond is connected to two different groups. In option (a), 2-butene, each carbon of the double bond is attached to a hydrogen and a methyl group, allowing for cis and trans forms.

🎯 Exam Tip: Remember the condition for geometrical isomerism: each carbon atom of the double bond must be bonded to two different atoms or groups.

 

Question 85. Which of the following does not show geometrical isomerism?
(a) 1, 2 – dichloro – 1 – pentene
(b) 1, 3 – dichloro – 2 – pentene
(c) 1, 1 – dichloro – 1 – pentene
(d) 1, 4 – dichloro – 2 – pentene
Answer: (c) 1, 1 – dichloro – 1 – pentene
In simple words: A molecule cannot show geometrical isomerism if one of the carbons in the double bond has two identical groups attached to it. In 1,1-dichloro-1-pentene, the first carbon in the double bond has two chlorine atoms, so it cannot have cis-trans forms.

🎯 Exam Tip: The rule of thumb for geometrical isomerism is that for a \( \text{C=C} \) double bond, neither carbon atom should have two identical groups attached.

 

Question 86. Geometrical isomerism is not shown by
(a)
(b)
(c) \( \text{CH}_2 = \text{C(CI)CH}_3 \)
(d) \( \text{CH}_3 - \text{CH} = \text{CH} - \text{CH} = \text{CH}_2 \)
Answer: (c) \( \text{CH}_2 = \text{C(CI)CH}_3 \)
In simple words: Geometrical isomerism needs each carbon in a double bond to have two different groups attached. In option (c), the first carbon has two hydrogen atoms attached, and the second carbon has two chlorine atoms attached to the same carbon, breaking the rule for geometrical isomerism.

🎯 Exam Tip: Always check both carbon atoms of the double bond. If either carbon is bonded to two identical groups, then geometrical isomerism is not possible.

 

Question 87. Which of the following is optically active?
(a) Glycerine
(b) Acetaldehyde
(c) Glyceraldehyde
(d) Acetone
Answer: (c) Glyceraldehyde
In simple words: For a molecule to be optically active, it needs to have at least one carbon atom that is bonded to four different groups. Glyceraldehyde has such a carbon, which allows it to rotate plane-polarized light.

🎯 Exam Tip: The presence of a chiral carbon (one bonded to four distinct groups) is the primary requirement for a compound to be optically active, provided it does not have an internal plane of symmetry.

 

Question 88. The minimum number of C atoms for a hydrocarbon to exhibit optical isomerism
(a) 4
(b) 5
(c) 6
(d) 7
Answer: (d) 7
In simple words: To be optically active, a hydrocarbon needs a carbon atom that is bonded to four different groups. You need at least seven carbon atoms to create such a structure in a simple hydrocarbon.

🎯 Exam Tip: A simple hydrocarbon must be large enough to allow for four different alkyl groups (or a hydrogen and three distinct alkyl groups) to be attached to a chiral carbon, which often requires a minimum of 7 carbon atoms.

 

Question 89. Which of the following can form cis - trans isomers?
(a) \( \text{C}_2\text{H}_5\text{Br} \)
(b) \( (\text{CH})_2(\text{COOH})_2 \)
(c) \( \text{CH}_3\text{CHO} \)
(d) \( (\text{CH}_2)_2\text{COOH} \)
Answer: (b) \( (\text{CH})_2(\text{COOH})_2 \)
In simple words: Cis-trans isomers are special forms of molecules that have a double bond and each carbon in that double bond has two different groups. Fumaric acid, which is \( (\text{CH})_2(\text{COOH})_2 \), has a \( \text{C=C} \) double bond where each carbon has a hydrogen and a carboxylic acid group, allowing for cis and trans forms.

🎯 Exam Tip: Recognize that fumaric acid (trans-butenedioic acid) and maleic acid (cis-butenedioic acid) are classic examples of cis-trans isomerism in dicarboxylic acids.

 

Question 90. No.of geometrical isomers possible for the compound \( \text{CH}_3 - \text{CH} = \text{CH} - \text{CH} = \text{CH} - \text{C}_2\text{H}_5 \)
(a) 2
(b) 3
(c) 4
(d) 5
Answer: (c) 4
In simple words: This molecule has two double bonds, and each can exist in either a cis or trans form. So, for the two double bonds, you can have cis-cis, cis-trans, trans-cis, and trans-trans arrangements, giving a total of four possible geometrical isomers.

🎯 Exam Tip: For compounds with 'n' distinct double bonds that can each exhibit cis-trans isomerism, the maximum number of geometrical isomers is \( 2^n \).

 

Question 91. The number of geometrical isomers of \( \text{CH}_3 - \text{CH} = \text{CH} - \text{CH} = \text{CH} - \text{CH} = \text{CHCl} \) is
(a) 2
(b) 4
(c) 6
(d) 8
Answer: (d) 8
In simple words: This molecule has three different double bonds, and each of these double bonds can be arranged in a "cis" or "trans" way. Since there are three such double bonds, we multiply \( 2 \times 2 \times 2 \), which gives 8 possible shapes.

🎯 Exam Tip: For molecules with 'n' distinct double bonds capable of exhibiting geometrical isomerism, the total number of possible geometrical isomers is \( 2^n \).

 

Question 92. Minimum number of C atoms for an alkene hydrocarbon, that shows geometrical & optical isomerism both
(a) 5
(b) 6
(c) 7
(d) 8
Answer: (c) 7
In simple words: To have both geometrical (cis-trans) and optical isomerism, an alkene needs both a double bond with different groups on each side and a carbon atom with four different groups attached (a chiral center). Building such a molecule requires at least seven carbon atoms.

🎯 Exam Tip: An alkene must have a chiral center to be optically active, and appropriate substituents on the double bond to exhibit geometrical isomerism simultaneously. Visualize 3-methylpent-1-ene as an example for optical, but combine with a C=C to have both.

 

Question 93. Among the following compounds which exhibits optical isomerism?
(a) propanol
(b) 2 – propanol
(c) 1 – butanol
(d) 2 – butanol
Answer: (d) 2 – butanol
In simple words: Optical isomerism occurs when a molecule has a carbon atom bonded to four different groups. 2-Butanol has such a carbon, as it is attached to a hydrogen, a hydroxyl group, a methyl group, and an ethyl group, making it optically active.

🎯 Exam Tip: To quickly identify optical isomerism, draw the structure and look for a chiral carbon center—a carbon bonded to four unique substituents.

 

Question 94. Which of the mesoisomer?
(a) \( \text{CH}_2\text{OHCHOHCHO} \)
(b) \( \text{CH}_2\text{OHCHOHCHOHCHO} \)
(c) HOOCCHOHCHOHCOOH
(d) \( \text{HOH}_2\text{CCHOHCHOHCOOH} \)
Answer: (c) HOOCCHOHCHOHCOOH
In simple words: A meso compound is a molecule that has chiral centers but is not optically active because it has an internal plane of symmetry. Tartaric acid, with the formula \( \text{HOOC-CHOH-CHOH-COOH} \), has two chiral carbons but also a symmetrical structure, which makes it a meso isomer.

🎯 Exam Tip: A key characteristic of a meso isomer is the presence of chiral centers combined with an internal plane of symmetry, rendering the molecule optically inactive.

 

Question 95. d – tartaric acid and l – tartaric acid can be separated by
(a) Salt formation
(b) Fractional distillation
(c) Fractional crystallization
(d) Chromatography
Answer: (a) Salt formation
In simple words: d-tartaric acid and l-tartaric acid are mirror-image molecules that are hard to separate. One way to do it is by reacting them with another special molecule to form new salts that can be separated more easily, often through crystallization.

🎯 Exam Tip: Resolution of enantiomers often involves converting them into diastereomeric salts using a chiral resolving agent, which can then be separated by physical methods like fractional crystallization.

 

Question 96. Paper chromatography is
(a) Adsorption chromatography
(b) partition chromatography
(c) Ion exchange chromatography
(d) all of these
Answer: (b) partition chromatography
In simple words: Paper chromatography separates mixtures based on how much each part dissolves in a stationary liquid phase (water in the paper) and how much it dissolves in a moving liquid phase (the solvent). This distribution between two liquids is called partitioning.

🎯 Exam Tip: Remember that paper chromatography relies primarily on the differential partitioning of components between a stationary liquid phase (water adsorbed on cellulose) and a mobile liquid phase (the developing solvent).

 

Question 97. Simple distillation can be used to separate liquids which differ in their boiling points at least by
(a) 5°C
(b) 10°C
(c) 40 – 50°C
(d) 100°C
Answer: (c) 40 – 50°C
In simple words: Simple distillation works best for separating liquids if their boiling points are quite far apart. If the boiling points are very close, you need a more advanced method like fractional distillation. For simple distillation to be effective, there should be a difference of at least 40-50°C.

🎯 Exam Tip: Simple distillation is suitable for separating liquids with significantly different boiling points, typically greater than 25°C to 50°C, to ensure good separation efficiency.

 

Question 98. In adsorption chromatography mobile phase will be
(a) Only solid
(b) Only liquid
(c) Only gas
(d) Liquid as well as gas
Answer: (d) Liquid as well as gas
In simple words: In adsorption chromatography, the moving part that carries the sample can be either a liquid (like in column chromatography) or a gas (like in gas chromatography). It's the substance that moves through the stationary phase.

🎯 Exam Tip: The mobile phase in chromatography is the medium that moves through the stationary phase, carrying the components of the mixture with it. It can be a liquid or a gas, depending on the specific chromatographic technique.

 

Question 99. Which of the following can be used as adsorbent in adsorption chromatography?
(a) Silica gel
(b) Alumina
(c) Cellulose powder
(d) All of these
Answer: (d) All of these
In simple words: In adsorption chromatography, you need a solid material that can "stick" to the different parts of your mixture to separate them. Silica gel, alumina, and cellulose powder are all common materials used for this purpose because they have surfaces that can temporarily hold onto substances.

🎯 Exam Tip: Adsorbents are stationary phases with active sites that selectively bind components; common examples like silica gel and alumina are used widely due to their porous structure and surface chemistry.

 

Question 100. Two substances when separated out on the basis of their extent of adsorption by one material, the phenomenon is called
(a) Chromatography
(b) Crystallization
(c) Sublimation
(d) Steam distillation
Answer: (a) Chromatography
In simple words: When you separate different substances because they stick to a material at different rates, that whole process is called chromatography. It’s like a race where different compounds move at different speeds depending on how strongly they interact with the stationary material.

🎯 Exam Tip: Chromatography is a powerful separation technique based on differential distribution (adsorption, partition, etc.) of components between a stationary phase and a mobile phase.

 

Question 101. Chromatographic technique is used for the separation of
(a) Camphor
(b) Alcohol & Water
(c) Acetone and Methanol
(d) Plant pigments
Answer: (d) Plant pigments
In simple words: Chromatography is very good at separating complex mixtures, especially delicate ones like the different colored pigments found in plants. It’s a gentle way to separate substances without destroying them.

🎯 Exam Tip: Chromatography is particularly effective for separating complex mixtures of organic compounds, especially those that are heat-sensitive or present in small quantities, like natural pigments.

 

Question 102. In column chromatography stationary phase is
(a) only solid
(b) only liquid
(c) only gas
(d) All of these
Answer: (a) only solid
In simple words: In column chromatography, the material that stays still inside the column and does not move is a solid, usually an adsorbent like silica gel or alumina. This solid helps separate the mixture.

🎯 Exam Tip: In column chromatography, the stationary phase is typically a solid adsorbent packed into a glass tube, through which the mobile phase (liquid solvent) flows.

 

Question 103. Which of the following method is used for the purification of solids?
(a) Distillation under reduced pressure
(b) Distillation
(c) Strain distillation
(d) Sublimation
Answer: (d) Sublimation
In simple words: Sublimation is a special way to purify solids where the solid changes directly into a gas when heated, without becoming a liquid first. Then, when cooled, the gas changes back into a pure solid, leaving impurities behind.

🎯 Exam Tip: Sublimation is an excellent purification method for solids that readily convert from solid to gas (and back) without melting, useful for compounds like benzoic acid or camphor.

 

Question 104. Vacuum distillation is used to purify liquids which
(a) are highly volatile
(b) are explosive in nature
(c) soluble in water
(d) decomposes below their B.P’s
Answer: (d) decomposes below their B.P’s
In simple words: Vacuum distillation is used for liquids that would break down or decompose if heated to their normal boiling point. By lowering the pressure, the liquid can boil and distill at a much lower temperature, preventing it from decomposing.

🎯 Exam Tip: Use vacuum distillation for heat-sensitive compounds, as reducing the pressure lowers the boiling point, allowing distillation to occur without thermal degradation.

 

Question 105. Impure Napthalene is purified by
(a) Fractional crystallization
(b) Fractional distillation
(C) solvent extraction
(d) sublimation
Answer: (d) sublimation
In simple words: Naphthalene, being a solid that can turn directly into a gas when heated (and back to solid when cooled), is often purified using sublimation. This process helps separate it from impurities that do not sublime.

🎯 Exam Tip: Naphthalene is a classic example of a substance purified by sublimation due to its significant vapor pressure below its melting point.

 

Question 106. A very common adsorbent used in chromatography is
(a) Powdered charcoal
(b) Alumina
(c) Chalk
(d) Sodium carbonate
Answer: (b) Alumina
In simple words: In chromatography, alumina is a very common solid material used to separate mixtures. It has a good surface for different substances to stick to (adsorb), helping them separate as they move through the column.

🎯 Exam Tip: Alumina (\( \text{Al}_2\text{O}_3 \)) and silica gel (\( \text{SiO}_2 \)) are the most widely used stationary phases (adsorbents) in adsorption chromatography due to their high surface area and selective adsorption properties.

 

Question 107. Simple distillation of liquids involves simultaneously
(a) Vapourisation and condensation
(b) Condensation and vapourisation
(c) Vapourisation and sublimation
(d) Sublimation and condensation
Answer: (a) Vapourisation and condensation
In simple words: In simple distillation, the liquid is first heated until it turns into a gas (vaporization). Then, this gas is cooled down until it turns back into a liquid (condensation), collecting the purified liquid. These two steps happen one after the other.

🎯 Exam Tip: Distillation is essentially a two-step process: heating to vaporize the liquid and then cooling to condense the vapor back into liquid, separating components based on boiling point differences.

 

Question 108. The latest technique for the purification of organic compounds is
(a) Fractional distillation
(b) Chromatography
(c) Vacuum distillation
(d) Crystallization
Answer: (b) Chromatography
In simple words: Chromatography is considered a very modern and versatile way to purify organic compounds, especially for separating complex mixtures or very small amounts of substances that might be difficult to purify by older methods.

🎯 Exam Tip: Chromatography encompasses a wide range of techniques (e.g., HPLC, GC, TLC) that are highly efficient for separating and purifying organic compounds, often surpassing older methods in sensitivity and resolving power.

 

Question 109. Fixed melting point of an organic compound informs
(a) Purity of an organic compound
(b) Conductivity of compound
(c) Chemical nature of compound
(d) Whether the compound is liquid or gas
Answer: (a) Purity of an organic compound
In simple words: A pure organic compound melts very sharply at a specific temperature. If the melting point is not fixed, or if the substance melts over a wide range of temperatures, it usually means the compound is not pure and has impurities. So, a fixed melting point tells us about its purity.

🎯 Exam Tip: A sharp and consistent melting point is a strong indicator of the purity of a crystalline organic compound, as impurities generally lower and broaden the melting range.

 

Question 110. Lassaigne’s test is used in qualitative analysis to detect
(a) Nitrogen
(b) Sulphur
(c) Chlorine
(d) All of these
Answer: (d) All of these
In simple words: Lassaigne's test is a common way in chemistry to check if a substance contains elements like nitrogen, sulfur, or halogens (like chlorine). It helps identify these elements in organic compounds.

🎯 Exam Tip: Lassaigne's test is a general method for detecting the presence of nitrogen, sulfur, and halogens in organic compounds, as these elements are converted to ionic forms that can be easily tested.

 

Question 111. In Lassaigne’s method organic compound is fused with
(a) Sodium metal
(b) Zinc dust
(c) Sodium carbonate and Zinc dust
(d) Calcium metal
Answer: (a) Sodium metal
In simple words: In Lassaigne's test, the organic compound is heated very strongly with a small piece of sodium metal. This reaction helps turn any nitrogen, sulfur, or halogens in the compound into simple salts that can be easily detected.

🎯 Exam Tip: The fusion with sodium metal in Lassaigne's test is crucial to convert covalent organic compounds into ionic sodium salts, which are water-soluble and amenable to standard inorganic qualitative tests.

 

Question 112. Presence of nitrogen in organic compound in Lassaigne’s extract as
(a) Nitrogen gas
(b) \( \text{NH}_3 \)
(c) NO
(d) \( \text{CN}^- \)
Answer: (d) \( \text{CN}^- \)
In simple words: When an organic compound containing nitrogen is heated with sodium in Lassaigne's test, the nitrogen changes into cyanide ions (\( \text{CN}^- \)). These cyanide ions are then detected to confirm the presence of nitrogen.

🎯 Exam Tip: In Lassaigne's test, nitrogen from the organic compound reacts with sodium to form sodium cyanide (\( \text{NaCN} \)), which dissociates into \( \text{Na}^+ \) and \( \text{CN}^- \) in the aqueous extract.

 

Question 113. Medium of Sodium extract is
(a) Neutral
(b) Basic
(c) Acidic
(d) Depends on organic compound
Answer: (b) Basic
In simple words: The sodium extract, also known as Lassaigne's extract, is made by fusing organic compounds with sodium and then dissolving the mixture in water. The unused sodium metal reacts with water to form sodium hydroxide, making the solution basic.

🎯 Exam Tip: The aqueous sodium fusion extract is typically alkaline (basic) due to the formation of \( \text{NaOH} \) from any unreacted sodium metal reacting with water.

 

Question 114. \( \text{H}_2\text{O} \) vapours on passing through anhydrous \( \text{CuSO}_4 \) turns it to
(a) Green
(b) Blue
(c) Violet
(d) White
Answer: (b) Blue
In simple words: Anhydrous copper sulfate is a white powder. When water vapor passes through it, it absorbs the water and changes color to blue. This color change is a test to show if water is present.

🎯 Exam Tip: Anhydrous copper sulfate is a common chemical test for water; its white crystals turn blue upon hydration due to the formation of \( \text{CuSO}_4 \cdot 5\text{H}_2\text{O} \).

 

Question 115. When a nitrogenous organic compound is fused with sodium, the nitrogen present in the compound is converted into
(a) Sodium Nitrate
(b) Sodium nitrite
(c) Sodamide
(d) Sodium cyanide
Answer: (d) Sodium cyanide
In simple words: During Lassaigne's test, when an organic substance with nitrogen is strongly heated with sodium metal, the nitrogen atoms turn into cyanide ions. These then combine with sodium to form sodium cyanide, which is easy to detect.

🎯 Exam Tip: The formation of sodium cyanide (\( \text{NaCN} \)) is the key step in Lassaigne's test for nitrogen, as the ionic nature of \( \text{NaCN} \) allows for subsequent chemical tests in aqueous solution.

 

Question 116. In the Lassaigne’s test the Sulphur present in the organic compound first changes into
(a) \( \text{Na}_2\text{SO}_3 \)
(b) \( \text{CS}_2 \)
(c) \( \text{Na}_2\text{SO}_4 \)
(d) \( \text{Na}_2\text{S} \)
Answer: (d) \( \text{Na}_2\text{S} \)
In simple words: In Lassaigne's test, if there is sulfur in the organic compound, it reacts with the hot sodium metal. This reaction makes sodium sulfide (\( \text{Na}_2\text{S} \)), which is a simple salt that can then be detected in the solution.

🎯 Exam Tip: During sodium fusion in Lassaigne's test, sulfur converts to sodium sulfide (\( \text{Na}_2\text{S} \)), which is then tested for using lead acetate solution to form black lead sulfide.

 

Question 117. Which of the following elements in an organic compound cannot be detected by Lassaigne’s test?
(a) N
(b) S
(c) Cl
(d) H
Answer: (d) H
In simple words: Lassaigne's test helps find elements like nitrogen, sulfur, and halogens in organic compounds. Hydrogen is part of almost all organic compounds, so it's not specifically tested by Lassaigne's method.

🎯 Exam Tip: Remember that Lassaigne's test is primarily for elements other than carbon and hydrogen, which are already assumed to be present in organic compounds.

 

Question 118. A compound which does not give a positive result in the Lassaigne’s test for nitrogen is
(a) Urea
(b) Hydroxyl amine
(c) Glycine
(d) Phenylhydrazine
Answer: (b) Hydroxyl amine
In simple words: Hydroxyl amine does not give a positive result for nitrogen in Lassaigne's test because its nitrogen atom is directly bonded to oxygen, preventing the formation of sodium cyanide. All other options contain nitrogen that forms sodium cyanide with sodium, leading to a positive test.

🎯 Exam Tip: Compounds where nitrogen is directly attached to oxygen (like nitro compounds or hydroxyl amine) or other elements that prevent cyanide formation often fail Lassaigne's test for nitrogen.

 

Question 119. Lassaigne’s test gives a violet colouration with sodium nitroprusside, it indicates presence of
(a) N
(b) S
(c) O
(d) Cl
Answer: (b) S
In simple words: When you get a violet color with sodium nitroprusside in Lassaigne's test, it means sulfur is present in the compound. The violet color comes from a special complex formed with sulfur.

🎯 Exam Tip: The violet color with sodium nitroprusside is a key indicator for sulfur detection; remember this specific color change for the test.

 

Question 120. The presence of halogen in an organic compound is detected by
(a) Iodoform test
(b) Silver nitrate test
(c) Beilstein’s test
(d) Million’s test
Answer: (c) Beilstein’s test
In simple words: Beilstein's test is a quick way to check if halogens are present in an organic compound. You heat the compound on a copper wire in a flame, and if it glows green or blue, halogens are likely there.

🎯 Exam Tip: Beilstein's test is a preliminary test; a positive result needs to be confirmed by Carius method for quantitative estimation.

 

Question 121. The Beilstein’s test in a rapid test used for organic compound to detect
(a) Phosphorous
(b) Sulphur
(c) Halogens
(d) Nitrogen
Answer: (c) Halogens
In simple words: Beilstein's test is a quick way to find out if there are halogens like chlorine, bromine, or iodine in an organic compound. It works by heating the compound with copper, which creates a colored flame if halogens are present.

🎯 Exam Tip: Understand that Beilstein's test is a qualitative test, meaning it tells you *if* halogens are there, but not how much.

 

Question 122. Liebig’s method is used for the estimation of
(a) Nitrogen
(b) Sulphur
(c) Carbon and hydrogen
(d) Halogens
Answer: (c) Carbon and hydrogen
In simple words: Liebig's method is used to measure how much carbon and hydrogen are in an organic compound. It works by burning the compound and collecting the carbon dioxide and water produced.

🎯 Exam Tip: Remember Liebig's method specifically targets carbon and hydrogen, converting them into \( \text{CO}_2 \) and \( \text{H}_2\text{O} \) for measurement.

 

Question 123. In Kjeldahl’s method of estimation of nitrogen, copper sulphate act as
(a) Oxidizing agent
(b) reducing agent
(c) Catalytic agent
(d) Hydrolysing agent
Answer: (c) Catalytic agent
In simple words: In Kjeldahl's method, copper sulfate is added to help the reaction happen faster. It acts as a catalyst, speeding up the conversion of nitrogen into ammonium sulfate without being used up itself.

🎯 Exam Tip: Identifying the role of reagents like copper sulfate (catalyst) or concentrated sulfuric acid (oxidizing/hydrolyzing agent) is crucial for Kjeldahl's method questions.

 

Question 124. Percentage of carbon in an organic compound is determined by
(a) Duma’s method
(b) Kjeldahl’s method
(c) Carius method
(d) Liebig’s method
Answer: (d) Liebig’s method
In simple words: The amount of carbon in an organic compound is figured out using Liebig's method. This involves burning the compound and measuring the carbon dioxide formed.

🎯 Exam Tip: Connect Liebig's method directly with the estimation of carbon and hydrogen, as it's a fundamental quantitative analysis technique.

 

Question 125. Halogen can be estimated by
(a) Duma’s method
(b) Carius method
(c) Leibig’s method
(d) All of these
Answer: (b) Carius method
In simple words: To measure the amount of halogens (like chlorine, bromine, or iodine) in an organic compound, the Carius method is used. This method converts the halogen into a silver halide, which is then weighed.

🎯 Exam Tip: Remember Carius method is for halogens, phosphorus, and sulfur, while Dumas and Kjeldahl are for nitrogen, and Liebig for carbon and hydrogen.

 

Question 126. In Carius method halogens are estimated
(a) \( \text{X}_2 \)
(b) \( \text{BaX}_2 \)
(c) \( \text{PbX}_2 \)
(d) \( \text{AgX} \)
Answer: (d) AgX
In simple words: In the Carius method, when you want to find out how much halogen is in a substance, it is converted into a silver halide compound, which is written as \( \text{AgX} \). This solid \( \text{AgX} \) is then measured.

🎯 Exam Tip: Always associate the final precipitate in Carius method for halogens as silver halide (\( \text{AgX} \)), where X represents the halogen.

 

Question 127. In Dumas method nitrogen in organic compound is estimated in the form of
(a) \( \text{N}_2 \)
(b) NO
(c) \( \text{NH}_3 \)
(d) \( \text{N}_2\text{O}_5 \)
Answer: (a) N$_2$
In simple words: In the Dumas method, all the nitrogen from the organic compound is turned into nitrogen gas, \( \text{N}_2 \). This gas is then measured to find the amount of nitrogen originally present.

🎯 Exam Tip: The key difference between Dumas and Kjeldahl methods for nitrogen estimation is the form in which nitrogen is measured: \( \text{N}_2 \) gas for Dumas, and ammonia for Kjeldahl.

 

Question 128. In Kjeldahl’s method to estimate nitrogen, compound is heated with conc.\( \text{H}_2\text{SO}_4 \) in presence of
(a) \( \text{CaSO}_4 \)
(b) \( (\text{NH}_4)_2\text{SO}_4 \)
(c) \( \text{CuSO}_4 \)
(d) \( \text{P}_2\text{O}_5 \)
Answer: (c) CuSO$_4$
In simple words: When using Kjeldahl's method to find nitrogen, the compound is heated with strong sulfuric acid. Copper sulfate (\( \text{CuSO}_4 \)) is added to act as a catalyst, helping the reaction to go faster.

🎯 Exam Tip: Remember \( \text{CuSO}_4 \) as the catalyst in Kjeldahl's method; it helps convert nitrogen to ammonium sulfate efficiently.

 

Question 129. In organic compounds, Sulphur is estimated as
(a) \( \text{BaSO}_4 \)
(b) \( \text{BaCl}_2 \)
(c) \( \text{Ba}_3(\text{PO}_4)_2 \)
(d) \( \text{H}_2\text{SO}_4 \)
Answer: (a) BaSO$_4$
In simple words: When we want to measure the amount of sulfur in an organic compound, it's changed into barium sulfate (\( \text{BaSO}_4 \)). This is a solid substance that can be easily weighed.

🎯 Exam Tip: The formation of insoluble \( \text{BaSO}_4 \) is the basis for sulfur estimation, allowing for accurate gravimetric analysis.

 

Question 130. In the Liebig’s method, if ‘w’ is the mass of compound taken and ‘x’ is the amount of \( \text{CO}_2 \) formed then
(a) %C = \( \frac{12 \times x}{16 \times w} \)
(b) %C = \( \frac{12}{44} \times \frac{\mathrm{w}}{\mathrm{x}} \times 100 \)
(c) %C = \( \frac{12}{44} \times \frac{x}{w} \times 100 \)
(d) %C = \( \frac{12}{44} \times \frac{x}{w} \)
Answer: (c) %C = \( \frac{12}{44} \times \frac{x}{w} \times 100 \)
In simple words: To find the percentage of carbon in a substance using Liebig's method, you take the mass of carbon (12 g in 44 g of \( \text{CO}_2 \)), multiply it by the mass of \( \text{CO}_2 \) formed (x), divide by the mass of the original compound (w), and then multiply by 100. This formula helps calculate the carbon content accurately.

🎯 Exam Tip: Remember the atomic weight of carbon (12) and the molecular weight of \( \text{CO}_2 \) (44) are crucial for this calculation. The formula is essentially (mass of C / mass of compound) * 100.

 

Question 131. In Dumas method for estimating nitrogen in organic compound, the gas finally collected is
(a) \( \text{N}_2 \)
(b) NO
(c) \( \text{NH}_3 \)
(d) \( \text{H}_2 \)
Answer: (a) N$_2$
In simple words: In Dumas method, all the nitrogen from the organic compound is changed into nitrogen gas, \( \text{N}_2 \). This pure nitrogen gas is then collected and measured to find out how much nitrogen was in the original sample.

🎯 Exam Tip: The Dumas method converts all nitrogen into free \( \text{N}_2 \) gas, which is then measured volumetrically in a nitrometer.

 

Question 132. In Dumas method, the gas which is collected in Nitrometer is
(a) \( \text{N}_2 \)
(b) NO
(c) \( \text{NH}_3 \)
(d) \( \text{H}_2 \)
Answer: (a) N$_2$
In simple words: In Dumas method, the special measuring device called a nitrometer collects nitrogen gas, which is represented as \( \text{N}_2 \). This collected gas is what helps determine the amount of nitrogen in the compound.

🎯 Exam Tip: The nitrometer is specific to collecting nitrogen gas (\( \text{N}_2 \)) in the Dumas method, crucial for accurate volume measurements.

 

Question 133. In Kjeldahl’s method, the nitrogen presence is estimated as
(a) \( \text{N}_2 \)
(b) \( \text{NH}_3 \)
(c) \( \text{NO}_2 \)
(d) \( \text{N}_2\text{O}_3 \)
Answer: (b) NH$_3$
In simple words: In Kjeldahl's method, the nitrogen from the organic compound is converted into ammonia (\( \text{NH}_3 \)). This ammonia is then measured to determine the amount of nitrogen present.

🎯 Exam Tip: Distinguish between Dumas (elemental \( \text{N}_2 \)) and Kjeldahl (ammonia \( \text{NH}_3 \)) as the measured forms of nitrogen.

 

Question 134. In Kjeldahl’s method, nitrogen present in the organic compound is first converted into
(a) \( \text{NH}_3 \)
(b) \( (\text{NH}_4)_2\text{SO}_4 \)
(c) \( \text{N}_2 \)
(d) NO
Answer: (b) (\( \text{NH}_4)_2\text{SO}_4 \)
In simple words: In Kjeldahl's method, the first step is to change the nitrogen in the compound into ammonium sulfate (\( (\text{NH}_4)_2\text{SO}_4 \)). This is done by heating it with strong sulfuric acid. The ammonium sulfate is then further processed to release ammonia.

🎯 Exam Tip: Recognize ammonium sulfate as the initial stable form into which nitrogen is converted during the digestion step of Kjeldahl's method.

 

Question 135. In Liebig’s method for the estimation of C and H, the combustion tube is passed over
(a) CuO pellets
(b) Copper turnings
(c) Iron fillings
(d) Zinc - copper couple
Answer: (a) CuO pellets
In simple words: In Liebig's method, when we measure carbon and hydrogen, the combustion tube contains copper oxide (CuO) pellets. These pellets help to completely burn the organic compound, turning all carbon into \( \text{CO}_2 \) and all hydrogen into water.

🎯 Exam Tip: Copper oxide acts as an oxidizing agent in Liebig's method, ensuring complete combustion of carbon and hydrogen.

 

Question 136. Which gas is introduced into the combustion tube in Liebig’s method?
(a) Pure and dry \( \text{CO}_2 \)
(b) Pure and dry \( \text{N}_2 \)
(c) Pure and dry \( \text{O}_2 \)
(d) Pure and dry He
Answer: (c) Pure and dry O$_2$
In simple words: In Liebig's method, pure and dry oxygen gas (\( \text{O}_2 \)) is passed into the combustion tube. This is to make sure the organic compound burns completely and all the carbon and hydrogen are converted into \( \text{CO}_2 \) and \( \text{H}_2\text{O} \).

🎯 Exam Tip: Supplying pure oxygen ensures complete oxidation of the organic compound for accurate carbon and hydrogen estimation.

 

Question 137. Chromatographic techniques of purification can be used for
(a) Coloured compounds
(b) Liquids
(c) Solids
(d) All of these
Answer: (d) All of these
In simple words: Chromatographic techniques are very versatile and can be used to purify many different types of substances. This includes colored compounds, as well as both liquids and solids, by separating their components.

🎯 Exam Tip: Chromatography is a powerful separation technique applicable to a wide range of compounds, regardless of their physical state or color.

 

Question 138. Two substances when separated on the basis of partition co - efficient between two liquid phase, then the technique is known as
(a) column chromatography
(b) Paper chromatography
(c) GLC
(d) TLC
Answer: (b) Paper chromatography
In simple words: When two substances are separated because they spread differently between two liquids (one fixed, one moving), the method is called paper chromatography. This separation depends on how well each substance dissolves in or is attracted to each liquid layer.

🎯 Exam Tip: Paper chromatography is a specific type of partition chromatography, relying on the differential partitioning of components between a stationary liquid phase (water in paper) and a mobile liquid phase (solvent).

 

Question 139. Ortho and para nitro phenols can be separated by
(a) crystallization
(b) distillation
(c) sublimation
(d) solvent extraction
Answer: (b) distillation
In simple words: Ortho-nitrophenol and para-nitrophenol can be separated using distillation because they have different boiling points. Ortho-nitrophenol forms intramolecular hydrogen bonds, making its boiling point lower, which allows it to be separated by steam distillation.

🎯 Exam Tip: Steam distillation is particularly effective for separating ortho- and para-nitrophenols due to the significant difference in their volatility caused by hydrogen bonding.

 

Question 140. In steam distillation, the sum of the vapour pressure of the volatile compound and that of water is
(a) Equal to atmospheric pressure
(b) Less than atmospheric pressure
(c) More than atmospheric pressure
(d) Exactly half of the atmospheric pressure
Answer: (a) Equal to atmospheric pressure
In simple words: In steam distillation, the total pressure of the vapors from the compound and the water will reach the surrounding atmospheric pressure. This is why the mixture boils at a lower temperature than either component alone.

🎯 Exam Tip: The principle of steam distillation states that the total vapor pressure of the mixture equals the sum of the partial pressures of the components, and boiling occurs when this sum equals the atmospheric pressure.

 

Question 141. Organic compound is fused with metallic sodium for testing nitrogen, sulphur, and halogens because
(a) To make the solution alkaline
(b) To convert into elemental state of nitrogen, sulphur, and halogens
(c) To convert covalent compound into ionic compound
(d) To decrease fusion temperature
Answer: (c) To convert covalent compound into ionic compound
In simple words: Organic compounds are fused with sodium metal to change the nitrogen, sulfur, and halogens from their original covalent form into ionic salts. These ionic salts are easier to detect in water using simple chemical tests. This step ensures that these elements become water-soluble and reactive for subsequent tests.

🎯 Exam Tip: The formation of water-soluble ionic salts (like NaCN, \( \text{Na}_2\text{S} \), NaX) from covalent organic compounds is the primary purpose of Lassaigne's fusion.

 

Question 142. Sodium extract gives blood red colour when treated with \( \text{FeCl}_3 \), Formation of blood-red colour confirms the presence of
(a) Only nitrogen
(b) Only sulphur
(c) Only halogens
(d) Both Nitrogen and Sulphur
Answer: (d) Both Nitrogen and Sulphur
In simple words: When the sodium extract turns blood red after adding iron(III) chloride, it means both nitrogen and sulfur are present in the organic compound. This red color comes from a special complex formed with both elements.

🎯 Exam Tip: A blood-red color with \( \text{FeCl}_3 \) in Lassaigne's test indicates the presence of both nitrogen and sulfur, forming a ferrothiocyanate complex.

 

Question 143. The compound not formed in the positive test for nitrogen with the Lassaigne’s solution of an organic compound is
(a) \( \text{Fe}_4[\text{Fe}(\text{CN})_6]_3 \)
(b) \( \text{Na}_3[\text{Fe}(\text{CN})_6] \)
(c) \( \text{Fe}(\text{CN})_3 \)
(d) \( \text{Na}_3[\text{Fe}(\text{CN})_5\text{NOS}] \)
Answer: (a) b, c, d
In simple words: In the positive test for nitrogen, a specific blue compound called Prussian blue is formed, which is \( \text{Fe}_4[\text{Fe}(\text{CN})_6]_3 \). The other compounds listed are generally not the final product that gives the characteristic blue color.

🎯 Exam Tip: Be aware that \( \text{Na}_3[\text{Fe}(\text{CN})_6] \) and \( \text{Fe}(\text{CN})_3 \) are intermediate species or incorrect formulas for the final Prussian blue precipitate, which is \( \text{Fe}_4[\text{Fe}(\text{CN})_6]_3 \).

 

Question 144. The Lassaigne’s solution when heated with ferrous sulphate and acidified with sulphuric acid gave intense blue colour indicating the presence of nitrogen. The blue colour is due to the formation of
(a) \( \text{Na}_4[\text{Fe}(\text{CN})_6] \)
(b) \( \text{Fe}_3[\text{Fe}(\text{CN})_6]_2 \)
(c) \( \text{Fe}_2[\text{Fe}(\text{CN})_6] \)
(d) \( \text{Fe}_4[\text{Fe}(\text{CN})_6]_3 \)
Answer: (d) Fe$_4$[Fe(CN)$_6$]$_3$
In simple words: The deep blue color that shows nitrogen is present in Lassaigne's test comes from a complex called Prussian blue. This complex has the chemical formula \( \text{Fe}_4[\text{Fe}(\text{CN})_6]_3 \).

🎯 Exam Tip: Memorize the formula \( \text{Fe}_4[\text{Fe}(\text{CN})_6]_3 \) for Prussian blue; it is the definitive product for a positive nitrogen test in the Lassaigne method.

 

Question 145. Which of the following compounds will answer Lassaigne’s test for nitrogen?
(a) \( \text{NH}_2\text{NH}_2 \)
(b) \( \text{NH}_2\text{OH} \)
(c) NaCN
(d) \( \text{NaNO}_3 \)
Answer: (c) NaCN
In simple words: Lassaigne's test works by turning nitrogen in an organic compound into sodium cyanide (NaCN). So, if a compound *already* is NaCN, it will give a positive test because the essential ingredient for the reaction is present.

🎯 Exam Tip: Lassaigne's test *detects* nitrogen by converting it to NaCN. Therefore, NaCN itself will bypass the fusion step and directly give a positive test for nitrogen.

 

Question 146. In Dumas method 0.5 g of an organic compound containing nitrogen gave 112 ml of nitrogen at S.T.P. The percentage of nitrogen in the given compound is
(a) 28
(b) 38
(c) 18
(d) 48
Answer: (a) 28
In simple words: To find the percentage of nitrogen, we know that 22400 mL of nitrogen at STP weighs 28 g. If 112 mL of nitrogen is obtained from 0.5 g of compound, then the percentage of nitrogen is calculated as \( (\frac{28}{22400} \times \frac{112}{0.5}) \times 100 \), which equals 28%. This calculation helps determine the nitrogen content.

🎯 Exam Tip: Remember that at STP, 22.4 L (or 22400 mL) of any gas contains one mole of that gas. For nitrogen (\( \text{N}_2 \)), one mole weighs 28 g.

 

Question 147. 0.73 g of organic compound on oxidation gave 1.32 g of carbon dioxide. The percentage of carbon in the given compound will be
(a) 49.32
(b) 59.32
(c) 29.32
(d) 98.64
Answer: (a) 49.32
In simple words: To find the percentage of carbon, we use the mass of carbon dioxide produced. Since 44 g of \( \text{CO}_2 \) contains 12 g of carbon, 1.32 g of \( \text{CO}_2 \) contains \( \frac{12}{44} \times 1.32 \) g of carbon. The percentage is then \( (\frac{\text{mass of carbon}}{\text{mass of compound}}) \times 100 \), which gives 49.32%. This calculation helps determine the carbon content.

🎯 Exam Tip: Always use the stoichiometric ratio of carbon in carbon dioxide (12/44) when calculating the percentage of carbon from Liebig's method data.

 

Question 148. In an estimation of S by Carius method 0.217 g of the compound gave 0.5825 g of \( \text{BaSO}_4 \). Percentage of S is
(a) 36.78 %
(b) 35.50 %
(c) 36.48 %
(d) 35.69 %
Answer: (a) 36.78 %
In simple words: To find the percentage of sulfur, we use the mass of barium sulfate formed. Since 233 g of \( \text{BaSO}_4 \) contains 32 g of sulfur, 0.5825 g of \( \text{BaSO}_4 \) contains \( \frac{32}{233} \times 0.5825 \) g of sulfur. The percentage is then \( (\frac{\text{mass of sulfur}}{\text{mass of compound}}) \times 100 \), which is 36.78%. This calculation helps determine the sulfur content.

🎯 Exam Tip: The key values for Carius method sulfur estimation are the atomic weight of sulfur (32) and the molecular weight of barium sulfate (233).

II. Very short question and answers (2 Marks):

 

Question 1. What are Acyclic compounds? Give suitable example.
Answer: Acyclic compounds are organic compounds where the carbon atoms are connected in an open chain, which can be straight or branched. These compounds can have only single bonds (saturated) or include multiple bonds (unsaturated). Such compounds are also known as open-chain or aliphatic compounds. They are a fundamental class of organic molecules.

In simple words: Acyclic compounds have carbon atoms arranged in an open chain, not a ring. They can have single, double, or triple bonds.

🎯 Exam Tip: Remember that 'acyclic' means 'not cyclic', referring to straight or branched chain structures. Providing both saturated and unsaturated examples strengthens your answer.

 

Question 2. What are Alicyclic Compounds? Give suitable example.
Answer: Alicyclic compounds are a type of organic compounds that contain rings of carbon atoms, but they behave like open-chain aliphatic compounds in many ways. These compounds can be either saturated (with only single bonds in the ring) or unsaturated (with double or triple bonds in the ring). Their properties often resemble those of their open-chain counterparts.

In simple words: Alicyclic compounds are ring-shaped organic molecules, but they behave like straight-chain compounds. They can have either all single bonds or some double bonds in their ring.

🎯 Exam Tip: Highlight that alicyclic compounds have cyclic structures but non-aromatic properties, differentiating them from aromatic compounds.

 

Question 3. What are Aromatic heterocyclic Compounds? Give example.
Answer: Aromatic heterocyclic compounds are ring structures that contain at least one atom other than carbon in the ring (like nitrogen, oxygen, or sulfur), and they also show aromatic properties similar to benzene. These compounds often have special stability and a particular way of reacting. They are also sometimes called non-benzenoid aromatic compounds.

In simple words: Aromatic heterocyclic compounds are ring-shaped molecules that have atoms other than carbon (like N, O, S) in their ring and also act like aromatic compounds. They are stable and show specific reactions.

🎯 Exam Tip: For aromatic heterocycles, remember the key features: cyclic, contains a heteroatom (N, O, S), and follows Hückel's rule for aromaticity (4n+2 \( \pi \) electrons).

 

Question 4. What is functional group? Give example.
Answer: A functional group is a specific atom or a group of atoms inside a molecule that gives the molecule its main chemical behavior and properties. It reacts in a consistent way, no matter what other parts of the organic molecule are present. The functional group usually determines how the organic compound will react.
Example:
(i) \( -\text{NH}_2 \) - amines
(ii) \( =\text{NH} \) - Imines
(iii) \( \text{-OH} \) - alcohols
(iv) \( -\text{C} \equiv \text{N} \) - nitriles
(v) \( -\text{COOH} \) - carboxylic acids
(vi) \( -\text{CHO} \) - aldehydes
(vii) \( >\text{C=O} \) - ketones
In simple words: A functional group is a small part of a molecule that makes the molecule react in a special way. It's like a tag that tells you what the molecule can do.

🎯 Exam Tip: When defining a functional group, emphasize its role in determining chemical properties and reactivity, along with giving diverse examples.

 

Question 5. Explain the following terms in IUPAC system of nomenclature of organic compounds.
(i) Root word
(ii) prefix
(iii) suffix
Answer:
(i) Root word:
The root word in IUPAC nomenclature tells us how many carbon atoms are in the longest continuous chain of a molecule. It forms the base name for the organic compound. For example, 'meth' for one carbon, 'eth' for two carbons, 'prop' for three carbons, 'but' for four carbons, and so on. Understanding the root word is the first step in naming any organic compound correctly.

(ii) Prefix:
A prefix is added before the root word in a compound's name. It describes any side groups (substituents) or branching chains that are attached to the main carbon chain. Prefixes also indicate if the compound has rings (cyclic structures) or if multiple identical groups are present. These details help to describe the structure more fully. For instance, 'methyl' for a \( \text{CH}_3 \) group, 'chloro' for a Cl atom, or 'cyclo' for a ring.

(iii) Suffix:
A suffix is placed after the root word in a compound's name. It shows the main functional group present in the molecule. If there are multiple functional groups, the suffix indicates the highest priority functional group. For example, '-ane' for single bonds, '-ene' for double bonds, '-ol' for alcohols, or '-oic acid' for carboxylic acids. The suffix helps to clearly identify the compound's class.
In simple words: In naming chemicals, the 'root word' tells us how many carbons are in the main chain. A 'prefix' comes before the root and tells us about side groups or branches. A 'suffix' comes after the root and tells us about the main special part of the molecule.

🎯 Exam Tip: To master IUPAC nomenclature, practice identifying the longest carbon chain (root word), all substituents (prefixes), and the highest priority functional group (suffix) in various structures.

 

Question 6. What is Metamerism? With suitable examples.
Answer: Metamerism is a type of structural isomerism where compounds have the same molecular formula and the same functional group, but they differ in how the carbon atoms are arranged around the functional group. This means the alkyl groups attached to the functional group are different. Metamerism is often seen in compounds like ethers, ketones, esters, and secondary amines, where the functional group acts as a bridge between two alkyl chains.
Example:
For \( \text{C}_4\text{H}_{10}\text{O} \) (molecular formula)
\( \text{CH}_3 - \text{O} - \text{C}_3\text{H}_7 \) - Methyl propyl ether (1 - methoxypropane)
\( \text{C}_2\text{H}_5 - \text{O} - \text{C}_2\text{H}_5 \) - diethyl ether (ethoxyethane)
These two compounds have the same molecular formula (\( \text{C}_4\text{H}_{10}\text{O} \)) and the same functional group (ether, \( -\text{O}- \)), but the alkyl groups attached to the oxygen atom are different. In methyl propyl ether, they are methyl and propyl groups, while in diethyl ether, they are two ethyl groups.
Another example:
\( \text{CH}_3-\text{CH}_2-\text{CO}-\text{CH}_2-\text{CH}_3 \) (Pentan-3-one)
\( \text{CH}_3-\text{CO}-\text{CH}_2-\text{CH}_2-\text{CH}_3 \) (Pentan-2-one)
Both are ketones with molecular formula \( \text{C}_5\text{H}_{10}\text{O} \). The difference is in the position of the carbonyl functional group or the alkyl groups around it.
In simple words: Metamerism is when compounds have the same overall formula and the same main chemical group, but the carbon chains on either side of that main group are arranged differently.

🎯 Exam Tip: Remember that metamers always belong to the same functional group family (e.g., all ethers, all ketones) and differ only in the length/arrangement of carbon chains around the functional group.

 

Question 7. What is meant by stereochemistry?
Answer: Stereochemistry is a branch of chemistry that studies molecules in three dimensions. It looks at how atoms are arranged in space within a molecule and how this arrangement affects the molecule's properties and reactions. Isomers that have the same bonding order but different spatial arrangements of atoms or groups are called stereoisomers. This field is very important because the three-dimensional shape of molecules plays a big role in their biological activities, how they are made in nature, and in drug design.
In simple words: Stereochemistry is the study of how atoms are arranged in 3D space within molecules. It explains why molecules with the same parts can have different shapes and properties.

🎯 Exam Tip: Emphasize the "three-dimensional arrangement" aspect when defining stereochemistry, as this is its core distinguishing feature from other types of isomerism.

 

Question 8. What is enantiomerism?
Answer: Enantiomerism describes when a substance is optically active and exists in two or more forms. These forms have the same physical and chemical properties. However, they differ in how they rotate plane-polarized light. Such optical isomers rotate the light by the same amount but in opposite directions. These isomers are called enantiomers, and the phenomenon is known as enantiomerism. A good example is d- and l-lactic acid, which are mirror images of each other but cannot be perfectly overlapped.
In simple words: Enantiomerism is when two forms of a chemical look like mirror images but cannot be stacked on top of each other perfectly, and they spin light in opposite directions.

🎯 Exam Tip: To identify enantiomers, look for chiral centers (carbon atoms with four different groups attached) and check if the molecules are non-superimposable mirror images.

 

Question 9. What are the conditions for an organic compound is said to be optically active?
Answer: For an organic compound to be optically active, it must meet two main conditions:
(1) The molecule needs to have at least one chiral or asymmetric carbon atom. This means a carbon atom is bonded to four different groups. This unique arrangement is crucial for optical activity.
(2) The molecule must not be superimposable on its mirror image. If a molecule and its mirror image can be perfectly overlaid, then it is not optically active. This means they are distinct, non-identical forms.
In simple words: A chemical is optically active if it has a special carbon atom with four different things attached to it, and its mirror image cannot be placed exactly on top of it.

🎯 Exam Tip: Remember, the presence of a chiral center is a strong indicator, but non-superimposability with its mirror image is the ultimate test for optical activity. Meso compounds are an exception, having chiral centers but being optically inactive due to internal symmetry.

 

Question 10. How will you detect phosphorus present in the given organic compound?
Answer: To detect phosphorus in an organic compound, follow these steps:
**Test for phosphorus:**
A solid sample of the compound is strongly heated with a mixture of sodium carbonate (Na\(_{2}\)CO\(_{3}\)) and potassium nitrate (KNO\(_{3}\)). This heating process oxidizes any phosphorus in the compound into sodium phosphate. The residue from this reaction is then extracted with water and boiled with concentrated nitric acid (HNO\(_{3}\)). Next, a solution of ammonium molybdate is added to the mixture. If phosphorus is present, a canary yellow coloration or precipitate will form, confirming its presence. This chemical change makes it easy to spot phosphorus even in tiny amounts.
In simple words: To find phosphorus, heat the compound with special salts, then add nitric acid and ammonium molybdate. If it turns yellow, phosphorus is there.

🎯 Exam Tip: The formation of a canary yellow precipitate of ammonium phosphomolybdate is the key indicator for the presence of phosphorus in this test.

 

Question 11. Explain the sublimation process for the purification of organic compounds.
Answer: Sublimation is a process used to purify certain organic compounds. In this method, some substances like benzoic acid, naphthalene, and camphor, when heated, change directly from a solid to a vapor without first melting into a liquid. When these vapors are cooled, they turn back into solid form. This change is called sublimation. It is a helpful way to separate a volatile (easily vaporized) solid from a non-volatile solid impurity. This technique is only useful for a few specific substances that can sublime, making it a specialized purification method.
In simple words: Sublimation is when a solid turns directly into a gas when heated, and then back into a solid when cooled, skipping the liquid phase. This is used to clean certain chemicals.

🎯 Exam Tip: Remember that sublimation is only effective for compounds that have a significant vapor pressure below their melting point and are stable at the sublimation temperature.

 

Question 12. What is the need for purifying an organic compound?
Answer: Organic compounds must be purified for several important reasons. To properly study their structure, understand their physical and chemical properties, and determine their biological activities, these compounds need to be in a pure state. Any impurities can affect the results and lead to incorrect conclusions about the compound. A pure compound ensures reliable data for research and applications.
In simple words: Organic compounds need to be clean and pure so we can correctly study how they look, how they react, and what they do.

🎯 Exam Tip: In chemistry, purity is crucial; impurities can drastically alter observed properties, leading to incorrect characterization and application of a compound.

 

III. Short question and answers (3 Marks):

 

Question 1. What is Homologous Series? Give suitable example.
Answer:
**Homologous Series:**
A homologous series is a group of organic compounds where each member has the same characteristic functional group. Each successive member in the series differs from the previous one by a -CH\(_{2}\) unit in its molecular formula. The individual members of such a series are called homologues, and the concept is known as homology. For example, all members of the alkane series follow the general formula C\(_{n}\)H\(_{2n+2}\). These series simplify the study of organic chemistry by grouping similar compounds.
**Examples:**
**Alkanes:**
Methane (CH\(_{4}\))
Ethane (C\(_{2}\)H\(_{6}\))
Propane (C\(_{3}\)H\(_{8}\))
Butane (C\(_{4}\)H\(_{10}\))
Pentane (C\(_{5}\)H\(_{12}\))
**Alcohols:**
Methanol (CH\(_{3}\)OH)
Ethanol (C\(_{2}\)H\(_{5}\)OH)
Propanol (C\(_{3}\)H\(_{7}\)OH)
In simple words: A homologous series is a family of chemicals that are very similar, where each new family member is just a little bit bigger (by a -CH\(_{2}\) group) than the last. Like methane, ethane, propane are all alkanes.

🎯 Exam Tip: When defining homologous series, always mention the same functional group, the -CH\(_{2}\) difference, and a general formula if applicable, along with clear examples.

 

Question 2. What are Alicycic heterocyclic Compounds? Give example.
Answer: Alicyclic heterocyclic compounds are cyclic organic compounds that behave similarly to aliphatic (open-chain) compounds in many of their properties, but they also contain one or more heteroatoms (atoms other than carbon, like oxygen, nitrogen, or sulfur) within their ring structure. These compounds often have properties that combine aspects of both aliphatic and heterocyclic chemistry, making them versatile in organic synthesis. They are important in various chemical reactions and can be found in natural products.
**Example:**

In simple words: Alicyclic heterocyclic compounds are ring-shaped chemicals that act like straight-chain chemicals, but they also have other atoms like oxygen or nitrogen inside their rings. Pyrrolidine and piperidine are examples.

🎯 Exam Tip: For alicyclic heterocycles, emphasize both the cyclic (ring) structure and the presence of a heteroatom, noting their similarity to aliphatic compounds.

 

Question 3. Write the IUPAC name of the following compounds.
a) \( \text{CH}_{3} - \underset{\text{CH}_{3}}{\overset{\text{CH}_{3}}{\text{C}}} - \text{CH}_{2} - \text{CH}_{3} \)
b) \( \text{CH}_{3} - \text{CH}_{2} - \underset{\text{CH}_{3}}{\text{CH}} - \text{CH}_{2}\text{CH}_{3} \)
c) \( \text{CH}_{3} - \text{CH}_{2} - \underset{\text{CH}_{3}}{\overset{\text{CH}_{3}}{\text{C}}} - \text{CH}_{2} - \text{CH}_{3} \)
d) \( \text{CH}_{3} - \text{CH}_{2} - \text{CH}_{2} - \text{CH} = \text{CH} - \text{COOH} \)
Answer:
a) 2,2 - Dimethylbutane
b) 3-Methylpentane
c) 3,3-Dimethylpentane
d) Pent-3-enoic acid (or Hex-4-enoic acid, if the chain is counted including the carbonyl carbon). This name is commonly used when the carbon chain contains both a double bond and a carboxylic acid group, where the carboxylic acid group determines the suffix '-oic acid' and the double bond is indicated by '-en'. The numbering starts from the carboxylic acid carbon. However, given the structure's visual representation, it refers to 2-propylbut-3-enoic acid as the most likely intended answer in the context of the numbering 1 to 5. The general IUPAC rules require the functional group with higher priority (COOH) to be at the lowest possible number. So the correct answer would be 2-propylbut-3-enoic acid, given the implied 5-carbon chain. With the options given, let's re-evaluate (d) 2-propylbut-3-enoic acid for the molecule shown.
*The provided OCR has the answer as "2 – propyl but – 3 – enoic acid", which corresponds to numbering the chain from the COOH side.*
In simple words: These are the correct IUPAC names for the chemical structures given. You name them by finding the longest carbon chain and then adding numbers for any branches or special groups like double bonds or acid parts.

🎯 Exam Tip: Always prioritize the functional group with the highest priority (e.g., carboxylic acid over double bond) when numbering the carbon chain in IUPAC nomenclature. Longest continuous chain is essential.

 

Question 4. Write the IUPAC name of the following compounds.
a)
b)
c)
d)
Answer:
a) 1-ethyl-2,3-dimethylcyclohexane
b) 2-ethyl-1,1-dimethylcyclopentane
c) Cyclopropylmethylcyclopropane (derivative of cyclopropane)
d) 2-Propylcyclobutane (derivative of cycloalkane)
In simple words: These are the correct scientific names for these ring-shaped chemical structures. You find the main ring, then name the groups attached to it and where they are.

🎯 Exam Tip: For cyclic compounds, identify the parent ring and then number the substituents to give them the lowest possible locants, considering alphabetical order if a tie occurs.

 

Question 5. Write the IUPAC name of the following compounds.
a)
b)
c)
d)
Answer:
a) 2-Cyclobutylpropanal
b) 3-Cyclohexylpentan-2-one
c) 4-(Cyclopent-3-en-1-yl)-3-methylbutane-1-oic acid
d) 3-(3-nitrocyclopentyl)prop-2-en-1-oic acid
In simple words: These are the correct scientific names for the given chemical drawings. Each name tells you about the main chain or ring and all the attached groups.

🎯 Exam Tip: When naming complex cyclic compounds with functional groups in the side chain, identify the principal functional group and the main chain, then treat the cyclic part as a substituent, using proper numbering for both.

 

Question 6. Write the IUPAC name of the following compounds.
a)
b)
c)
d)
Answer:
a) 2-(2-Hydroxypropyl)cyclohexan-1-ol
b) 1,4-Dicyclobutylbutane
c) Cyclopentylbenzene
d) 2-Carbonylcyclobutane-1-carboxylic acid
In simple words: These are the official names for the chemical structures shown. You name them by looking at their main parts and any smaller groups attached to them.

🎯 Exam Tip: For compounds with multiple rings or complex substituents, focus on identifying the main parent structure (which might be a benzene ring or a large cycloalkane) and then name the attached groups systematically.

 

Question 7. Write the IUPAC name of the following compounds.
a)
b)
c)
d)
Answer:
a) 1,3,5-Trimethylbenzene (mesitylene)
b) 1-Chloro-3-methylbenzene
c) Phenyldichloromethane (benzal dichloride)
d) Phenyltrichloromethane (benzotrichloride)
In simple words: These are the proper scientific names for the benzene compounds shown. You name them based on the groups attached to the benzene ring and where they are.

🎯 Exam Tip: For substituted benzene rings, use systematic IUPAC names, numbering substituents for the lowest possible combination, or common names like mesitylene when applicable.

 

Question 8. Classify the following compounds based on the structure.
i) \( \text{CH} \equiv \text{C} - \text{CH}_{2} - \text{C} \equiv \text{CH} \)
ii) \( \text{CH}_{3} - \text{CH}_{2} - \text{CH}_{2} - \text{CH}_{2} - \text{CH}_{3} \)
iii)
iv)
Answer:
i) \( \text{CH} \equiv \text{C} - \text{CH}_{2} - \text{C} \equiv \text{CH} \) is an unsaturated open chain compound. It has triple bonds and no ring structure.
ii) \( \text{CH}_{3} - \text{CH}_{2} - \text{CH}_{2} - \text{CH}_{2} - \text{CH}_{3} \) is a saturated open chain compound. It only has single bonds and no ring.
iii) is an aromatic benzenoid compound. It contains a benzene ring.
iv) is an alicyclic compound. It is a cyclic compound that behaves like an open-chain aliphatic compound.
In simple words: We classify chemicals by their structure: some are straight chains with special bonds (like triple bonds), some are straight chains with only single bonds, some have a benzene ring, and some have other rings that act like straight chains.

🎯 Exam Tip: When classifying organic compounds by structure, remember to look for the presence of rings (cyclic vs. acyclic), benzene rings (aromatic vs. non-aromatic), and multiple bonds (saturated vs. unsaturated).

 

Question 9. Give two examples for each of the following type of organic compounds.
(i) non-benzenoid aromatic
(ii) aromatic heterocyclic
(iii) alicyclic
(iv) aliphatic open chain
Answer:
(i) non-benzenoid aromatic: Azulene, Tropone. These compounds are aromatic but do not contain a benzene ring. Azulene has a distinct blue color and an interesting ring structure.
(ii) aromatic heterocyclic: Pyrrole, Pyridine, Furan, Thiophene, Quinoline. These are aromatic compounds with at least one heteroatom (like N, O, S) in their ring.
(iii) alicyclic: Cyclohexane, Cyclopropane. These are cyclic compounds that resemble aliphatic compounds in their properties.
(iv) aliphatic open chain: Methane (CH\(_{4}\)), Ethane (CH\(_{3}\)CH\(_{3}\)), Propane (CH\(_{3}\)CH\(_{2}\)CH\(_{3}\)), Butane. These are compounds with open, non-cyclic carbon chains. Methane is the simplest alkane, a basic building block in organic chemistry.
In simple words: This lists examples for different types of chemical structures: aromatic rings without benzene (like azulene), aromatic rings with other atoms in them (like pyrrole), simple carbon rings (like cyclohexane), and straight carbon chains (like methane).

🎯 Exam Tip: Ensure your examples clearly illustrate the defining features of each class: non-benzenoid aromatics lack a benzene ring but follow Huckel's rule; aromatic heterocycles have a heteroatom within an aromatic ring; alicyclics are cyclic but behave like aliphatics; and aliphatic open chains are non-cyclic saturated or unsaturated hydrocarbons.

 

Question 10. Explain the copper oxide test for the detection of carbon and hydrogen present in the given organic compound.
Answer:
**Copper Oxide Test:**
This test detects carbon and hydrogen in an organic compound. First, the organic substance is thoroughly mixed with about three times its weight of dry copper oxide (CuO). This mixture is then placed in a hard glass test tube, which has a bent delivery tube. The delivery tube dips into lime water in a separate test tube. The mixture is heated strongly.

During heating, if carbon is present, it is oxidized to carbon dioxide (CO\(_{2}\)), which turns the lime water milky. This indicates the presence of carbon. If hydrogen is present, it is oxidized to water (H\(_{2}\)O), which condenses as small droplets on the cooler walls of the test tube. These water droplets can be collected in a bulb and separated using anhydrous copper sulfate (CuSO\(_{4}\)), which turns blue in the presence of water, confirming hydrogen. If no water droplets are formed, hydrogen is not present. This simple test is a foundational step in elemental analysis.

**Reactions:**
\( \text{C} + 2\text{CuO} \rightarrow \text{CO}_{2} + 2\text{Cu} \)
\( 2\text{H} + \text{CuO} \rightarrow \text{H}_{2}\text{O} + \text{Cu} \)
In simple words: To find carbon and hydrogen in a chemical, heat it with copper oxide. Carbon turns into a gas that makes lime water cloudy, and hydrogen turns into water droplets.

🎯 Exam Tip: Key indicators for the copper oxide test are the lime water turning milky (for carbon) and the formation of water droplets that turn anhydrous copper sulfate blue (for hydrogen).

 

Question 11. In an estimation of sulphur by carius method. 0.2175 g of the substance gave 0.5825 g of BaSO\(_{4}\) calculate the percentage composition of S in the compound.
Answer:
**Solution:**
Weight of organic compound = 0.2175 g
Weight of BaSO\(_{4}\) (Barium sulfate) formed = 0.5825 g
Molar mass of BaSO\(_{4}\) = 137 (Ba) + 32 (S) + 4 * 16 (O) = 233 g/mol
Atomic mass of Sulphur (S) = 32 g/mol

From the molar mass, 233 g of BaSO\(_{4}\) contains 32 g of Sulphur.

So, 0.5825 g of BaSO\(_{4}\) contains Sulphur:
Mass of S = \( \frac{32}{233} \times 0.5825 \text{ g} \)
Mass of S \( \approx 0.0799 \text{ g} \)

Now, calculate the percentage of Sulphur in the organic compound:
Percentage of S = \( \left( \frac{\text{Mass of S}}{\text{Weight of organic compound}} \right) \times 100 \)
Percentage of S = \( \left( \frac{0.0799}{0.2175} \right) \times 100 \)
Percentage of S \( \approx 36.73\% \)

The percentage composition of Sulphur in the compound is approximately 36.73%. This calculation helps determine the elemental makeup of the compound accurately.
In simple words: We find the amount of sulfur in a sample by measuring how much barium sulfate is made. Since we know how much sulfur is in barium sulfate, we can work out the percentage of sulfur in the original sample.

🎯 Exam Tip: In Carius method calculations, ensure you use the correct molecular weights for BaSO\(_{4}\) and the atomic weight for sulfur, and always express the final answer as a percentage.

 

Question 12. 0.16 g of an organic compound was heated in a carius tube and H\(_{2}\)SO\(_{4}\) acid formed was precipitated with BaCl\(_{2}\). The mass of BaSO\(_{4}\) was 0.35 g. Find the percentage of sulphur [30.04]
Answer:
**Solution:**
Weight of organic substance (w) = 0.16 g
Weight of Barium sulphate (x) = 0.35 g
Molar mass of BaSO\(_{4}\) = 233 g/mol
Atomic mass of Sulphur (S) = 32 g/mol

The formula to calculate the percentage of sulfur in the compound is:
Percentage of sulphur = \( \left( \frac{\text{Atomic mass of S}}{\text{Molar mass of BaSO}_4} \right) \times \left( \frac{\text{Mass of BaSO}_4}{\text{Mass of organic compound}} \right) \times 100 \)

Percentage of sulphur = \( \left( \frac{32}{233} \right) \times \left( \frac{0.35}{0.16} \right) \times 100 \)

Percentage of sulphur = \( \left( 0.1373 \right) \times \left( 2.1875 \right) \times 100 \)

Percentage of sulphur \( \approx 30.04\% \)

Thus, the percentage of sulfur in the given organic compound is approximately 30.04%. This calculation confirms the sulfur content through quantitative analysis.
In simple words: We calculated the sulfur percentage by using the weight of the organic compound and the barium sulfate formed. We use a formula that connects these weights to the atomic weight of sulfur.

🎯 Exam Tip: Always double-check the formula for percentage calculation in Carius method, specifically ensuring the mass of the element (S) is divided by the molar mass of its compound (BaSO\(_{4}\)), and multiplied by the ratio of actual precipitate to sample mass, then by 100.

 

Question 13. 0.284 g of an organic substance gave 0.287 g AgCl in a Carius method for the estimation of halogen. Find the percentage of Cl in the compound.
Answer:
**Solution:**
Weight of organic substance = 0.284 g
Weight of AgCl (Silver Chloride) formed = 0.287 g
Molar mass of AgCl = 108 (Ag) + 35.5 (Cl) = 143.5 g/mol
Atomic mass of Chlorine (Cl) = 35.5 g/mol

From the molar mass, 143.5 g of AgCl contains 35.5 g of Chlorine.

So, 0.287 g of AgCl contains Chlorine:
Mass of Cl = \( \frac{35.5}{143.5} \times 0.287 \text{ g} \)
Mass of Cl \( \approx 0.0709 \text{ g} \)

Now, calculate the percentage of Chlorine in the organic compound:
Percentage of Cl = \( \left( \frac{\text{Mass of Cl}}{\text{Weight of organic substance}} \right) \times 100 \)
Percentage of Cl = \( \left( \frac{0.0709}{0.284} \right) \times 100 \)
Percentage of Cl \( \approx 24.96\% \)

The percentage of chlorine in the compound is approximately 24.96%. This demonstrates the quantitative determination of halogen in an organic sample. The original calculation yielded 24.56%, likely due to rounding at intermediate steps, so maintaining precision is important.
In simple words: To find the percentage of chlorine, we used the weight of the organic sample and the silver chloride formed. We calculated how much chlorine was in the silver chloride, and then found its percentage in the original compound.

🎯 Exam Tip: Pay close attention to the atomic and molecular weights used. Small rounding errors in intermediate steps can lead to slight deviations in the final percentage.

 

Question 14. 0.185 g of an organic compound when treated with Conc. HNO\(_{3}\) and silver nitrate gave 0.320 g of silver bromide. Calculate the % of bromine in the compound.
(Ag = 108, Br = 80)
Answer:
**Solution:**
Weight of organic substance (w) = 0.185 g
Weight of silver bromide (x) = 0.320 g
Atomic mass of Silver (Ag) = 108
Atomic mass of Bromine (Br) = 80
Molar mass of AgBr = 108 + 80 = 188 g/mol

The formula to calculate the percentage of bromine in the compound is:
Percentage of bromine = \( \left( \frac{\text{Atomic mass of Br}}{\text{Molar mass of AgBr}} \right) \times \left( \frac{\text{Mass of AgBr}}{\text{Mass of organic compound}} \right) \times 100 \)

Percentage of bromine = \( \left( \frac{80}{188} \right) \times \left( \frac{0.320}{0.185} \right) \times 100 \)

Percentage of bromine = \( \left( 0.4255 \right) \times \left( 1.7297 \right) \times 100 \)

Percentage of bromine \( \approx 73.61\% \)

Therefore, the percentage of bromine in the compound is approximately 73.61%. This calculation confirms the bromine content by using the Carius method principles.
In simple words: We figured out how much bromine was in the compound by seeing how much silver bromide it made. We use a formula that accounts for the weights of silver, bromine, and the sample.

🎯 Exam Tip: Always specify the atomic weights used for calculations, as these are critical for accuracy. Ensure the correct molar mass of the silver halide (AgBr) is derived from the given atomic weights.

 

Question 15. 0.40 g of an Iodo – substituted organic compound gave 0.235 g of AgI by carlus method. Calculate the percentage of iodine in the compound. (Ag = 108 I = 127).
Answer:
**Solution:**
Weight of organic substance (w) = 0.40 g
Weight of silver iodide (x) = 0.235 g
Atomic mass of Silver (Ag) = 108
Atomic mass of Iodine (I) = 127
Molar mass of AgI = 108 + 127 = 235 g/mol

The formula to calculate the percentage of iodine in the compound is:
Percentage of iodine = \( \left( \frac{\text{Atomic mass of I}}{\text{Molar mass of AgI}} \right) \times \left( \frac{\text{Mass of AgI}}{\text{Mass of organic compound}} \right) \times 100 \)

Percentage of iodine = \( \left( \frac{127}{235} \right) \times \left( \frac{0.235}{0.40} \right) \times 100 \)

Percentage of iodine = \( \left( 0.5404 \right) \times \left( 0.5875 \right) \times 100 \)

Percentage of iodine \( \approx 31.75\% \)

Thus, the percentage of iodine in the compound is approximately 31.75%. This is a direct application of the Carius method for halogen estimation, ensuring the accuracy of elemental composition. The method is reliable for quantifying halogens in organic samples.
In simple words: We found the amount of iodine in the compound by measuring the silver iodide it produced. We used a formula that accounts for the weights of silver, iodine, and the original sample.

🎯 Exam Tip: Remember to use the correct atomic weight for Iodine (I) and the molecular weight for Silver Iodide (AgI) when calculating percentages in Carius method problems, to ensure precision in your results.

 

Question 16. 0.24 g of organic compound containing phosphorous gave 0.66 g of \(Mg_2P_2O_7\) by the usual analysis. Calculate the percentage of phosphorous in the compound.
Answer:
Weight of organic compound \( (w) = 0.24 \, g \)
Weight of \(Mg_2P_2O_7\) formed \( (x) = 0.66 \, g \)
Molar mass of \(Mg_2P_2O_7 = 222 \, g/mol \)
Mass of phosphorus in \(Mg_2P_2O_7 = 2 \times 31 = 62 \, g/mol \)
Now, we calculate the percentage of phosphorus.
Percentage of Phosphorus \( = \left( \frac{\text{Mass of P in } Mg_2P_2O_7}{\text{Molar mass of } Mg_2P_2O_7} \times \frac{\text{Mass of } Mg_2P_2O_7}{\text{Mass of organic compound}} \times 100 \right) \% \)
\( \implies = \left( \frac{62}{222} \times \frac{0.66}{0.24} \times 100 \right) \% \)
\( \implies = 76.80 \, \% \)
Therefore, the percentage of phosphorus in the compound is 76.80%.
In simple words: We find out how much phosphorus is in the final compound formed and then use that to calculate its percentage in the original sample. This helps us know the amount of phosphorus present.

🎯 Exam Tip: Always remember to use the correct molar mass ratio for the element being estimated (e.g., phosphorus from \(Mg_2P_2O_7\)) and ensure all values are correctly substituted into the formula.

 

Question 17. 0.33 g of an organic compound containing phosphorous gave 0.397 g of \(Mg_2P_2O_7\) by the analysis. Calculate the percentage of P in the compound.
Answer:
Weight of organic substance \( (w) = 0.33 \, g \)
Weight of \(Mg_2P_2O_7\) formed \( (x) = 0.397 \, g \)
Molar mass of \(Mg_2P_2O_7 = 222 \, g/mol \)
Mass of phosphorus in \(Mg_2P_2O_7 = 2 \times 31 = 62 \, g/mol \)
Now, we calculate the percentage of phosphorus.
Percentage of phosphorus \( = \left( \frac{\text{Mass of P in } Mg_2P_2O_7}{\text{Molar mass of } Mg_2P_2O_7} \times \frac{\text{Mass of } Mg_2P_2O_7}{\text{Mass of organic compound}} \times 100 \right) \% \)
\( \implies = \left( \frac{62}{222} \times \frac{0.397}{0.33} \times 100 \right) \% \)
\( \implies = 33.59 \, \% \)
This calculation confirms the amount of phosphorus present in the organic sample.
In simple words: We calculate the phosphorus percentage by comparing the mass of the phosphorus compound formed with the initial organic sample. This shows how much phosphorus was in the original material.

🎯 Exam Tip: Double-check your molar mass calculations for the element and its compound to avoid errors in the final percentage.

 

Question 18. Explain simple distillation process with suitable example.
Answer: Simple distillation is a process used to purify liquids. It works by heating an impure liquid until it turns into vapor. This vapor is then collected and cooled, turning back into a pure liquid in a separate container. This method is effective for liquids that have a large difference in their boiling points (typically around 40 K or more) and do not break down when heated. An example is separating a mixture of nitrobenzene (boiling point 484 K) and benzene (boiling point 354 K). Another example is separating diethyl ether (boiling point 308 K) and ethyl alcohol (boiling point 35 K).
In simple words: Simple distillation purifies liquids by boiling them into vapor and then cooling the vapor back into liquid. This works best when liquids have very different boiling temperatures and don't break apart when heated.

🎯 Exam Tip: Remember that simple distillation is suitable only for mixtures where the components have significantly different boiling points, ideally more than 25°C apart.

 

Question 19. How will you purify an organic compound by differential extraction process?
Answer: Differential extraction is a way to separate a substance from its water solution by shaking it with another liquid called an organic solvent. The organic substance needs to be more soluble in this new solvent than in water. First, the water solution containing the organic substance is put into a separating funnel. Then, a small amount of an organic solvent like ether or chloroform \( (CHCl_3) \) is added. Since the organic solvent does not mix with water, it forms a separate layer. The funnel is gently shaken, which helps the organic substance move from the water layer into the organic solvent layer because it prefers to dissolve there. After shaking, the layers separate again. The bottom layer is then drained out of the separating funnel, leaving the separated substance behind. This process is repeated to get as much of the substance as possible.
In simple words: Differential extraction separates a substance from water by dissolving it in another liquid that doesn't mix with water. The substance moves to the new liquid, which is then separated from the water.

🎯 Exam Tip: The success of differential extraction depends on the organic compound being significantly more soluble in the extracting solvent than in water, and the solvents forming distinct layers.

 

Question 20. Explain the steam distillation process for purifying organic compound.
Answer: Steam distillation is a method for purifying organic compounds that are mixed with non-volatile impurities. This technique works for both solids and liquids. For a compound to be purified by steam distillation, it must not break down at the temperature of steam, it should have a good vapor pressure at 373 K, and it must not dissolve in water. Also, any impurities present should not evaporate easily. In this process, the impure liquid is heated in a round-bottom flask along with a little water. Steam is passed through the mixture, which helps the compound evaporate along with the water vapor. The mixed vapors are then cooled in a condenser, and the pure compound is collected separately from the water.
In simple words: Steam distillation purifies compounds by evaporating them with steam, then cooling the mixed vapors to get the pure compound. It works for compounds that don't mix with water and don't break down when heated.

🎯 Exam Tip: The key principle of steam distillation is that the total vapor pressure of the immiscible liquid mixture equals atmospheric pressure at a temperature lower than the boiling point of either pure component.

 

Question 21. Write a note on azeotropic distillation with suitable example.
Answer: Azeotropic distillation is used to separate liquid mixtures that cannot be separated by normal fractional distillation because they form azeotropes. Azeotropes are mixtures that boil at a constant temperature and distil as a single compound, even though they contain different liquids. A common example is a mixture of ethanol and water, which forms an azeotrope with about 95.87% ethanol and 4.13% water. To separate such mixtures, a third substance is added, which is called a dehydrating agent. This agent, such as benzene \( (C_6H_6) \) or carbon tetrachloride \( (CCl_4) \), helps to lower the vapor pressure of one component. This makes its boiling point higher, allowing the other component to distil over. Glycerol or glycol can also be used; they have high boiling points and reduce the vapor pressure of water more than that of alcohol.
In simple words: Azeotropic distillation separates liquids that boil together as a fixed mixture. It works by adding a third substance to change how they evaporate, allowing one part to boil away from the other.

🎯 Exam Tip: Azeotropic distillation is crucial for separating mixtures like ethanol and water, where traditional distillation methods are ineffective due to the formation of a constant-boiling mixture.

 

Question 22. What is Ring chain isomerism? Give an example.
Answer: Ring-chain isomerism is a type of isomerism where compounds have the same molecular formula but different arrangements of atoms, forming both open-chain and cyclic structures. For example, for the molecular formula \(C_3H_6\), there are two isomers: propene, which is an open-chain unsaturated compound, and cyclopropane, which is a cyclic compound. Both compounds share the same formula but have distinct structural arrangements.
Example:
(i) Propene (\(C_3H_6\)): \(CH_3-CH=CH_2\)
(ii) Cyclopropane (\(C_3H_6\)):

In simple words: Ring-chain isomerism happens when two molecules have the same atoms but one is a straight or branched chain, and the other is a ring. Like propene (a chain) and cyclopropane (a ring) both have the formula \(C_3H_6\).

🎯 Exam Tip: When identifying ring-chain isomers, ensure both structures have the identical molecular formula but one is acyclic (open) and the other is cyclic (ring).

 

Question 23. Briefly explain geometrical isomerism in oximes with suitable example.
Answer: Geometrical isomerism in oximes happens because of restricted rotation around the carbon-nitrogen double bond (\(C=N\)). Instead of "cis" and "trans," we use "syn" and "anti" to describe these isomers. In a syn isomer, the hydrogen atom on the carbon with the double bond and the -OH group on the nitrogen atom are on the same side of the double bond. In contrast, in an anti isomer, these groups are on opposite sides of the double bond. For example, in acetaldoxime, the syn form has the H on carbon and OH on nitrogen on the same side, while the anti form has them on opposite sides. This fixed arrangement creates distinct forms of the molecule.
Example: Acetaldoxime
Syn-Acetaldoxime: When H and OH are on the same side.
Anti-Acetaldoxime: When H and OH are on opposite sides.
In simple words: Geometrical isomerism in oximes is about how hydrogen and the -OH group are arranged around a carbon-nitrogen double bond. If they are on the same side, it's 'syn'; if on opposite sides, it's 'anti'.

🎯 Exam Tip: For oximes, remember that restricted rotation around the \(C=N\) bond leads to syn-anti isomerism, similar to cis-trans isomerism in alkenes.

 

Question 1. Write the structure of the following compounds.
a) Cinnamic acid
b) Lactic acid
c) Phthalic acid
d) Tartaric acid
d) Benzoic acid
Answer:
a) Cinnamic acid: \(C_6H_5-CH=CH-COOH\)
b) Lactic acid: \(CH_3-CH(OH)-COOH\)
c) Phthalic acid: Benzene-1,2-dicarboxylic acid
d) Tartaric acid: \(HOOC-CH(OH)-CH(OH)-COOH\)
e) Benzoic acid: \(C_6H_5-COOH\)
These structures show the specific arrangement of atoms and functional groups that define each compound. For example, cinnamic acid contains a benzene ring, a double bond, and a carboxylic acid group.
In simple words: The structures define what each compound looks like at the atomic level, showing how different parts like rings, double bonds, and acid groups are connected.

🎯 Exam Tip: When writing structures, pay attention to the carbon skeleton, functional groups, and any double or triple bonds. Ensure proper valence for all atoms.

 

Question 2. Explain the methods for the representation of structure of organic compounds with suitable example.
Answer: The structure of organic compounds can be shown in a few different ways to help us understand them better. These methods include:
1. Lewis structure or dot structure: This method shows all the atoms and how they are connected, including all valence electrons as dots or lines. It provides a complete picture of electron distribution.
2. Dash structure or line bond structure: This is a simpler way where a single dash represents a single covalent bond (two shared electrons). Double and triple dashes show double and triple bonds, respectively. Lone pairs of electrons are often not shown in this representation.
3. Condensed structure: This method simplifies dash structures by omitting some or all dashes. It shows groups of atoms together, like \(CH_3\), and indicates multiple identical groups with subscripts, such as \(CH_2CH_2\).
4. Bond line structure: This is the most simplified way, where only lines are used to represent carbon-carbon bonds in a zigzag pattern. Carbon and hydrogen atoms are usually not drawn; carbons are assumed at the ends and bends of lines, and hydrogens are assumed to complete the valency of carbon. Heteroatoms (like oxygen, nitrogen, halogens) are always explicitly written. This method makes it easier to draw complex structures quickly. For example, a zig-zag line represents a carbon chain, and a hexagon represents a benzene ring. These different methods allow chemists to communicate structural information clearly, from detailed electron arrangements to simplified skeletons.
In simple words: We can show how organic compounds are built using different drawings. Some drawings show every electron, some show bonds as lines, some group atoms together, and some just draw zigzag lines for carbon chains. Each way helps us understand the molecule from simple to very detailed.

🎯 Exam Tip: Understand when to use each representation method (Lewis, dash, condensed, bond-line) based on the level of detail required. Bond-line structures are particularly useful for quick representation of complex molecules.

 

Question 3. Explain the molecular model method for the representation of structure of organic compounds.
Answer: Molecular models are physical tools that help us see and understand the three-dimensional shapes of organic molecules. They are usually made from materials like wood, plastic, or metal. These models allow for a better visual understanding of how atoms are arranged in space, which is hard to grasp from flat drawings. There are different types of molecular models:
(i) Frame work model: This model shows only the bonds connecting the atoms, like a skeleton. It highlights the pattern of bonds and the overall shape but does not show the actual size of the atoms.
(ii) Ball and stick model: In this model, atoms are represented by colored balls, and bonds are represented by sticks. This model shows both the atoms and the bonds, giving a clearer idea of bond angles and molecular geometry. For compounds with double or triple bonds, springs can be used instead of sticks to better show the bond characteristics.
(iii) Space filling model: This model emphasizes the actual relative size of each atom based on its van der Waals radius. Atoms are represented by interpenetrating spheres, which gives a realistic view of the molecule's overall size and how much space it takes up. This helps understand how molecules might interact with each other. These models are very helpful in visualizing complex organic structures and their properties.
In simple words: Molecular models are physical toys that show us the 3D shape of molecules. Some models only show the connections (framework), some show balls for atoms and sticks for bonds (ball and stick), and some show the actual size of atoms (space filling). They help us see how molecules are really built in space.

🎯 Exam Tip: Molecular models are excellent for visualizing molecular geometry, bond angles, and steric hindrance, which are crucial for understanding chemical reactivity and properties.

 

Question 4. Explain briefly
(i) Fisher projection
(ii) sawhorse projection
(iii) Newman projection formula with neat example.
Answer:
(i) Fisher projection formula: This method is used to represent three-dimensional molecular structures in a two-dimensional format. In this projection, the chiral atom (or atoms) is placed in the plane of the paper. Horizontal lines represent bonds coming out towards the observer, while vertical lines represent bonds going away from the observer. This makes it easier to visualize and compare stereoisomers. For example, the Fisher projection of tartaric acid clearly shows its two chiral centers.
Example: Fisher projection formula for tartaric acid:
\[ \begin{array}{c} \mathrm{COOH} \\ | \\ \mathrm{H}-\mathrm{C}-\mathrm{OH} \\ | \\ \mathrm{H}-\mathrm{C}-\mathrm{OH} \\ | \\ \mathrm{COOH} \end{array} \]
(ii) Sawhorse projection formula: This projection shows the bond between two carbon atoms diagonally, making it look elongated. The carbon atom on the lower left is considered to be closer to the viewer, and the carbon on the upper right is further away. This method helps to show the spatial relationship between groups attached to adjacent atoms, which is not always clear in Fisher projections. For example, a sawhorse projection can show the eclipsed and staggered conformations of ethane.
(iii) Newman projection formula: In this method, the molecule is viewed directly along the carbon-carbon bond axis. The carbon atom closer to the observer is represented by a point (the "front" carbon), and the carbon atom further away is represented by a larger circle behind the front carbon (the "back" carbon). Bonds from the front carbon are drawn from the center point, while bonds from the back carbon are drawn from the circumference of the circle, usually at 120° angles to each other. This projection is very useful for illustrating different conformations arising from rotation around a single bond. An example is the Newman projection of ethane, showing its staggered and eclipsed forms.
In simple words: Fisher projection flattens a 3D molecule onto paper, showing front and back groups with lines. Sawhorse projection shows a tilted view of a carbon-carbon bond to see groups on adjacent carbons. Newman projection looks straight down a carbon-carbon bond, showing one carbon as a dot and the other as a circle behind it to view rotations.

🎯 Exam Tip: Master the conventions for each projection: horizontal lines forward/vertical lines backward for Fisher, staggered/eclipsed for Sawhorse and Newman. Practice drawing simple molecules in all three.

 

Question 5. What is Tautomerism? Explain different types of Tautomerism with suitable example.
Answer: Tautomerism is a special type of functional isomerism where a single compound exists in two different, easily interconvertible structures called tautomers. These structures differ mainly in the position of at least one atomic nucleus, typically hydrogen, and a double bond. Tautomers are usually in a dynamic equilibrium with each other. The two main types of tautomerism are the dyad system and the triad system.
Dyad system: In this system, a hydrogen atom moves (oscillates) between two directly connected polyvalent atoms.
Example: Hydrogen cyanide and hydrogen isocyanide. \(H-C \equiv N \rightleftharpoons H-N=C\). Here, the hydrogen moves between the carbon and nitrogen atoms.
Triad system: In this system, a hydrogen atom moves (oscillates) between three polyvalent atoms. This involves a 1,3-migration of a hydrogen atom from one polyvalent atom to another within the molecule. The most important type is keto-enol tautomerism. The two tautomers are the keto form (containing a carbonyl group \(C=O\)) and the enol form (containing a hydroxyl group and a carbon-carbon double bond \(-C=C-OH\)). The polyvalent atoms involved are usually one oxygen and two carbon atoms. Enolization is the process where the keto-form changes into the enol form. Both tautomeric forms are usually not equally stable; the less stable form is known as the labile form. This dynamic interconversion is crucial in many organic reactions.
Example: Keto-enol tautomerism in acetone:
\[ \begin{array}{c} \mathrm{H}_3\mathrm{C}-\mathrm{C}-\mathrm{CH}_3 \\ \| \\ \mathrm{O} \\ \text{(Keto form)} \end{array} \rightleftharpoons \begin{array}{c} \mathrm{H}_3\mathrm{C}-\mathrm{C}=\mathrm{CH}_2 \\ | \\ \mathrm{OH} \\ \text{(Enol form)} \end{array} \]
In simple words: Tautomerism is when a molecule quickly switches between two different forms by moving a hydrogen atom and a double bond. In the dyad system, hydrogen moves between two atoms. In the triad system (like keto-enol), it moves across three atoms.

🎯 Exam Tip: For tautomerism, clearly identify the migrating hydrogen and the changing positions of the double bond. Keto-enol tautomerism is a frequently tested example; remember that the keto form is usually more stable.

 

Question 6. How is sulphur detected by Lassaigne test?
Answer: Sulphur can be detected in an organic compound using the Lassaigne's test, which involves a few steps:
a) Sodium Nitroprusside Test: A small amount of the Lassaigne's extract (which contains sodium sulfide if sulphur is present) is treated with a freshly prepared sodium nitroprusside solution. If sulphur is present, a deep violet color appears. This color is due to the formation of sodium thionitroprusside, \(Na_4[Fe(CN)_5NOS]\), from the reaction of sodium sulfide with sodium nitroprusside:
\(Na_2S + Na_2[Fe(CN)_5NO] \rightarrow Na_4[Fe(CN)_5NOS]\)
b) Lead Acetate Test: Another portion of the Lassaigne's extract is acidified with acetic acid and then lead acetate solution is added. If sulphur is present, a black precipitate of lead sulfide \( (PbS) \) is formed. This confirms the presence of sulphur.
\((CH_3COO)_2Pb + Na_2S \rightarrow PbS\downarrow + 2CH_3COONa\)
c) Oxidation Test: For this, the organic substance is heated strongly (fused) with a mixture of potassium nitrate \( (KNO_3) \) and sodium carbonate \( (Na_2CO_3) \). Any sulphur present is oxidized to sodium sulfate \( (Na_2SO_4) \):
\(Na_2CO_3 + S + 3O \rightarrow Na_2SO_4 + CO_2\)
The fused mixture is then dissolved in water and acidified with hydrochloric acid \( (HCl) \). Finally, barium chloride \( (BaCl_2) \) solution is added. A white precipitate of barium sulfate \( (BaSO_4) \) indicates the presence of sulphur.
\(BaCl_2 + Na_2SO_4 \rightarrow BaSO_4\downarrow + 2NaCl\)
These tests provide reliable confirmation of sulphur in an organic compound. The formation of the characteristic colors and precipitates are key indicators.
In simple words: To find sulphur, we use the Lassaigne's test. We can add sodium nitroprusside for a violet color, or lead acetate for a black solid. Another way is to burn the sample to turn sulphur into sulfate, then add barium chloride to see a white solid.

🎯 Exam Tip: Remember the characteristic color changes and precipitates for each sulphur detection method: violet with nitroprusside, black with lead acetate, and white \(BaSO_4\) precipitate after oxidation.

 

Question 7. How is halogen detected by using Lassaigne test?
Answer: Halogens (chlorine, bromine, iodine) are detected using the Lassaigne's test after preparing a sodium fusion extract. To do this, a portion of the Lassaigne's filtrate is gently warmed with dilute nitric acid \( (HNO_3) \) and then silver nitrate \( (AgNO_3) \) solution is added. The type of precipitate formed indicates the specific halogen present:
a) Chlorine: A curdy white precipitate that dissolves easily in ammonia solution indicates chlorine. This precipitate is silver chloride \( (AgCl) \).
b) Bromine: A pale yellow precipitate that is only slightly soluble in ammonia solution indicates bromine. This precipitate is silver bromide \( (AgBr) \).
c) Iodine: A yellow precipitate that does not dissolve in ammonia solution indicates iodine. This precipitate is silver iodide \( (AgI) \).
The initial fusion converts the covalently bonded halogen in the organic compound into an ionic sodium halide \( (NaX) \):
\(Na + \text{Heat} \rightarrow NaX \)
The sodium halide then reacts with silver nitrate to form a silver halide precipitate:
\(NaX + AgNO_3 \rightarrow AgX + NaNO_3\)
It's important to remove any nitrogen or sulphur present, as they can interfere with the test. If nitrogen (forming NaCN) or sulphur (forming \(Na_2S\)) are also present, they must be decomposed by boiling the Lassaigne's extract with dilute \(HNO_3\) before adding \(AgNO_3\):
\(NaCN + HNO_3 \xrightarrow{\Delta} NaNO_3 + HCN \uparrow\)
\(Na_2S + 2HNO_3 \xrightarrow{\Delta} 2NaNO_3 + H_2S\)
If these are not removed, \(AgCN\) (white precipitate) or \(Ag_2S\) (black precipitate) could form and give misleading results. For example, \(AgCN\) could be mistaken for \(AgCl\), and \(Ag_2S\) can obscure other precipitates.
In simple words: To find halogens, we add silver nitrate to the sample. A white solid means chlorine, a pale yellow solid means bromine, and a yellow solid means iodine. We must first remove any nitrogen or sulphur that could confuse the results.

🎯 Exam Tip: Remember the color and solubility of silver halide precipitates to distinguish between chlorine, bromine, and iodine. Always ensure nitrogen and sulfur interfering compounds are removed by boiling with \(HNO_3\).

 

Question 8. How will you estimate carbon and hydrogen present in the given organic compound?
Answer: Carbon and hydrogen in an organic compound are estimated by a combustion method called Liebig's method. A known amount of the organic substance is burned completely in a stream of excess oxygen. During this process, all the carbon is converted into carbon dioxide \( (CO_2) \), and all the hydrogen is converted into water \( (H_2O) \):
\(C_xH_y + (x + \frac{y}{4})O_2 \rightarrow xCO_2 + \frac{y}{2}H_2O\)
The amounts of carbon dioxide and water produced are then carefully measured. From these amounts, the percentages of carbon and hydrogen in the original organic compound can be calculated.
The apparatus for this method consists of three main parts:
1. Oxygen Supply: Oxygen is first purified by removing moisture (bubbled through sulphuric acid) and carbon dioxide (passed through soda lime). This ensures only pure, dry oxygen enters the combustion tube.
2. Combustion Tube: A hard glass tube contains:
* A roll of oxidized copper gauze at the inlet to prevent gases from diffusing backward.
* A porcelain boat holding the weighed organic substance.
* Coarse copper oxide packed along two-thirds of the tube, separated by asbestos plugs, to ensure complete oxidation of the organic compound.
* Another roll of oxidized copper gauze at the outlet to reduce any nitrogen oxides formed back into elemental nitrogen.
3. Absorption Apparatus: This part collects the products of combustion:
* A weighed U-tube packed with pumice soaked in concentrated \(H_2SO_4\) absorbs all the water produced.
* A set of bulbs containing a strong potassium hydroxide \( (KOH) \) solution absorbs all the carbon dioxide.
* A guard tube filled with anhydrous \(CaCl_2\) prevents any outside moisture from entering.
Procedure: The combustion tube is first cleaned and dried by heating it in a current of pure oxygen. After cooling, the boat with the weighed organic sample is placed inside. The tube is then heated strongly for about two hours, ensuring complete combustion. A strong current of oxygen is maintained to sweep all products into the absorption apparatus. Finally, the U-tube and KOH bulbs are re-weighed. The increase in their weights corresponds to the mass of water and carbon dioxide formed, respectively.
Calculation:
Let the weight of the organic substance be \( w \, g \).
Increase in weight of \(H_2O = x \, g \).
Increase in weight of \(CO_2 = y \, g \).
Percentage of Hydrogen \( = \left( \frac{2}{18} \times \frac{x}{w} \times 100 \right) \% \)
Percentage of Carbon \( = \left( \frac{12}{44} \times \frac{y}{w} \times 100 \right) \% \)
These calculations provide the composition of hydrogen and carbon in the given organic compound. This process is highly accurate for elemental analysis.
In simple words: To measure carbon and hydrogen, we burn the organic sample in oxygen. All the carbon turns into carbon dioxide, and all the hydrogen turns into water. We weigh the carbon dioxide and water collected, then use those weights to calculate the percentage of carbon and hydrogen in the original sample.

🎯 Exam Tip: Remember that water is absorbed first (by \(H_2SO_4\)) and then carbon dioxide (by KOH). Always use purified oxygen and ensure all apparatus are weighed accurately before and after the experiment.

 

Question 9. Explain the Carius method for the estimation of sulphur in an organic compound.
Answer: In the Carius method, a known amount of the organic substance is strongly heated with fuming nitric acid \( \left(\mathrm{HNO}_{3}\right) \). During this process, carbon \( (\mathrm{C}) \) and hydrogen \( (\mathrm{H}) \) in the compound are oxidized to carbon dioxide \( \left(\mathrm{CO}_{2}\right) \) and water \( \left(\mathrm{H}_{2} \mathrm{O}\right) \), respectively. Any sulphur present in the compound is oxidized to sulphuric acid \( \left(\mathrm{H}_{2} \mathrm{SO}_{4}\right) \). This method ensures that sulphur is completely converted into a form that can be easily measured.
The following reaction takes place:
\( \mathrm{C} \xrightarrow{\text{fum. } \mathrm{HNO}_{3}} \mathrm{CO}_{2} \)
\( \mathrm{2H} \xrightarrow{\text{fum. } \mathrm{HNO}_{3}} \mathrm{H}_{2} \mathrm{O} \)
\( \mathrm{S} \xrightarrow{\mathrm{O}_{2}+\mathrm{H}_{2} \mathrm{O}} \mathrm{H}_{2} \mathrm{SO}_{4} \)
The sulphuric acid formed is then treated with an excess of barium chloride \( \left(\mathrm{BaCl}_{2}\right) \) solution. This quantitatively converts the sulphuric acid into barium sulphate \( \left(\mathrm{BaSO}_{4}\right) \), which is an insoluble precipitate. By measuring the mass of this barium sulphate precipitate, the mass of sulphur in the original compound can be determined, and from that, the percentage of sulphur can be calculated.

**Procedure:**
A known amount of the organic compound is placed in a clean Carius tube. A few milliliters of fuming nitric acid \( \left(\mathrm{HNO}_{3}\right) \) are added to it. The tube is then sealed and placed inside an iron tube, which is heated for about 5 hours. After heating, the tube is allowed to cool. A small hole is made in the tube to release any gases that have formed inside. The tube is then broken, and its contents are collected in a beaker. Excess barium chloride \( \left(\mathrm{BaCl}_{2}\right) \) is added to the beaker. The sulphuric acid formed from the reaction is converted to barium sulphate \( \left(\mathrm{BaSO}_{4}\right) \). The barium sulphate precipitate is filtered, washed, dried, and weighed. From the mass of barium sulphate, the percentage of sulphur is determined.

**Calculation:**
Let the weight of the organic compound be \( w \) g.
Let the mass of the \( \mathrm{BaSO}_{4} \) formed be \( x \) g.
We know that 233 g of \( \mathrm{BaSO}_{4} \) contains 32 g of sulphur.
Therefore, \( x \) g of \( \mathrm{BaSO}_{4} \) contains \( \frac{32}{233} \times x \) g of sulphur.

The percentage of sulphur in the compound is calculated as:
Percentage of sulphur \( = \left( \frac{32}{233} \times \frac{x}{w} \times 100 \right) \% \)
In simple words: This method burns the organic stuff with strong acid, turning all sulphur into sulphuric acid. Then we add another chemical to make a solid called barium sulphate. By weighing this solid, we can figure out how much sulphur was in the original sample.

🎯 Exam Tip: Remember that Carius method is very effective for quantitative estimation of halogens, sulphur, and phosphorus, as it converts them to easily weighable precipitates.

 

Question 10. 0.26 g of an organic compound gave 0.039 g of water and 0.245 g of carbon dioxide on combustion. Calculate the percentage of C & H.
Answer:
Given:
Weight of organic compound \( w = 0.26 \) g
Weight of water \( = 0.039 \) g
Weight of carbon dioxide \( = 0.245 \) g

**Percentage of hydrogen:**
We know that 18 g of \( \mathrm{H}_{2} \mathrm{O} \) contains 2 g of hydrogen.
So, 0.039 g of water contains \( \frac{2}{18} \times 0.039 \) g of hydrogen.
This amount of hydrogen comes from 0.26 g of the organic compound.
Percentage of hydrogen \( = \left( \frac{2}{18} \times \frac{0.039}{0.26} \times 100 \right) \% = 1.66 \% \)

**Percentage of carbon:**
We know that 44 g of \( \mathrm{CO}_{2} \) contains 12 g of carbon.
So, 0.245 g of \( \mathrm{CO}_{2} \) contains \( \frac{12}{44} \times 0.245 \) g of carbon.
This amount of carbon comes from 0.26 g of the organic compound.
Percentage of carbon \( = \left( \frac{12}{44} \times \frac{0.245}{0.26} \times 100 \right) \% = 25.69 \% \)
In simple words: We burn the compound and measure the water and carbon dioxide. Knowing how much hydrogen is in water and how much carbon is in carbon dioxide, we can calculate the percentage of each in the original compound.

🎯 Exam Tip: Ensure you use the correct molecular masses for water (18 g/mol) and carbon dioxide (44 g/mol) in your calculations to get accurate percentages.

 

Question 11. 0.2346 g of an organic compound yielded C, H & O 0.2754 g of H₂O and 0.4488 g CO₂. Calculate the % composition.
Answer:
Given:
Weight of organic substance \( w = 0.2346 \) g
Weight of water \( x = 0.2754 \) g
Weight of \( \mathrm{CO}_{2} \ y = 0.4488 \) g

**Percentage of carbon:**
Percentage of carbon \( = \left( \frac{12}{44} \times \frac{y}{w} \times 100 \right) \% \)
Percentage of carbon \( = \left( \frac{12}{44} \times \frac{0.4488}{0.2346} \times 100 \right) \% = 52.17 \% \)

**Percentage of hydrogen:**
Percentage of hydrogen \( = \left( \frac{2}{18} \times \frac{x}{w} \times 100 \right) \% \)
Percentage of hydrogen \( = \left( \frac{2}{18} \times \frac{0.2754}{0.2346} \times 100 \right) \% = 13.04 \% \)

**Percentage of oxygen:**
Since the compound contains carbon, hydrogen, and oxygen, the percentage of oxygen can be found by subtracting the percentages of carbon and hydrogen from 100%.
Percentage of oxygen \( = [100 - (52.17 + 13.04)] \% \)
Percentage of oxygen \( = [100 - 65.21] \% = 34.79 \% \)
In simple words: We determine the percentages of carbon and hydrogen by measuring the carbon dioxide and water produced when the compound is burned. The remaining percentage is then assigned to oxygen.

🎯 Exam Tip: When calculating percentage composition, always remember to account for oxygen by difference if it's not directly measured, assuming carbon and hydrogen have been fully oxidized.

 

Question 12. Explain the estimation of phosphorous by Carius method.
Answer:
In the Carius method for estimating phosphorus, a known mass of the organic compound, which contains phosphorus, is heated inside a sealed tube with fuming nitric acid \( \left(\mathrm{HNO}_{3}\right) \). During this heating process, the carbon present in the compound is converted to carbon dioxide \( \left(\mathrm{CO}_{2}\right) \), and hydrogen is converted to water \( \left(\mathrm{H}_{2} \mathrm{O}\right) \). The phosphorus, however, is oxidized to phosphoric acid \( \left(\mathrm{H}_{3} \mathrm{PO}_{4}\right) \).

The phosphoric acid is then precipitated. This can be done in one of two ways:
1. As ammonium phosphomolybdate: This precipitate is formed by heating the solution with concentrated nitric acid \( \left(\mathrm{HNO}_{3}\right) \) and then adding ammonium molybdate.
The reaction is:
\( \mathrm{H}_{3} \mathrm{PO}_{4} + 12(\mathrm{NH}_{4})_{2} \mathrm{MoO}_{4} + 21 \mathrm{HNO}_{3} \xrightarrow{\text{heat}} (\mathrm{NH}_{4})_{3} \mathrm{PO}_{4} \cdot 12 \mathrm{MoO}_{3} + 21 \mathrm{NH}_{4} \mathrm{NO}_{3} + 12 \mathrm{HNO}_{3} \)
This precipitate is filtered, washed, dried, and weighed.

2. As magnesium-ammonium phosphate: In an alternative method, the phosphoric acid is precipitated by adding a magnesia mixture (which is a combination of \( \mathrm{MgCl}_{2} \), \( \mathrm{NH}_{4} \mathrm{Cl} \), and ammonia). This precipitate is then washed, dried, and ignited to produce magnesium pyrophosphate \( \left(\mathrm{Mg}_{2} \mathrm{P}_{2} \mathrm{O}_{7}\right) \), which is then weighed.
The following reactions occur:
\( \mathrm{H}_{3} \mathrm{PO}_{4} + \mathrm{MgCl}_{2} + \mathrm{NH}_{3} \xrightarrow{} \mathrm{MgNH}_{4} \mathrm{PO}_{4} \)
\( \mathrm{2MgNH}_{4} \mathrm{PO}_{4} \xrightarrow{\text{heat}} \mathrm{Mg}_{2} \mathrm{P}_{2} \mathrm{O}_{7} + 2 \mathrm{NH}_{3} + \mathrm{H}_{2} \mathrm{O} \)

By knowing the mass of the organic compound and the mass of the ammonium phosphomolybdate or magnesium pyrophosphate formed, the percentage of phosphorus can be accurately calculated. This allows for precise determination of the phosphorus content in the sample.

**Calculations (general formulas):**
Let the mass of the organic compound be \( w \) g.

**If ammonium phosphomolybdate \( (\mathrm{NH}_{4})_{3} \mathrm{PO}_{4} \cdot 12 \mathrm{MoO}_{3} \) is formed:**
Molar mass of \( (\mathrm{NH}_{4})_{3} \mathrm{PO}_{4} \cdot 12 \mathrm{MoO}_{3} = 1877 \) g/mol.
It contains 31 g of phosphorus.
If \( x \) g of \( (\mathrm{NH}_{4})_{3} \mathrm{PO}_{4} \cdot 12 \mathrm{MoO}_{3} \) is formed from \( w \) g of organic compound, then:
Percentage of phosphorus \( = \left( \frac{31}{1877} \times \frac{x}{w} \times 100 \right) \% \)

**If magnesium pyrophosphate \( (\mathrm{Mg}_{2} \mathrm{P}_{2} \mathrm{O}_{7}) \) is formed:**
Molar mass of \( \mathrm{Mg}_{2} \mathrm{P}_{2} \mathrm{O}_{7} = 222 \) g/mol.
It contains 2 atoms of phosphorus, so \( 2 \times 31 = 62 \) g of phosphorus.
If \( y \) g of \( \mathrm{Mg}_{2} \mathrm{P}_{2} \mathrm{O}_{7} \) is formed from \( w \) g of organic compound, then:
Percentage of phosphorus \( = \left( \frac{62}{222} \times \frac{y}{w} \times 100 \right) \% \)
In simple words: To find phosphorus, we burn the compound with strong acid, which turns phosphorus into phosphoric acid. This acid is then made into a solid which is weighed. From the weight of this solid, we can calculate how much phosphorus was in the original sample.

🎯 Exam Tip: Ensure you use the correct molar mass for the specific phosphorus precipitate (ammonium phosphomolybdate or magnesium pyrophosphate) when calculating the percentage of phosphorus.

 

Question 13. Explain the estimation of nitrogen by Dumas method.
Answer:
The Dumas method is used to estimate nitrogen in organic compounds. It works by heating the nitrogen-containing compound with copper oxide \( (\mathrm{CuO}) \) in an atmosphere of carbon dioxide \( (\mathrm{CO}_{2}) \). This process converts all the nitrogen in the compound into free nitrogen gas \( (\mathrm{N}_{2}) \). Any carbon and hydrogen present are oxidized to \( \mathrm{CO}_{2} \) and \( \mathrm{H}_{2} \mathrm{O} \).

The general reaction is:
\( \mathrm{C}_{\mathrm{x}} \mathrm{H}_{\mathrm{y}} \mathrm{N}_{\mathrm{z}} + (2 \mathrm{x} + \frac{\mathrm{y}}{2}) \mathrm{CuO} \rightarrow \mathrm{xCO}_{2} + \frac{\mathrm{y}}{2} \mathrm{H}_{2} \mathrm{O} + \frac{\mathrm{z}}{2} \mathrm{N}_{2} + (2 \mathrm{x} + \frac{\mathrm{y}}{2}) \mathrm{Cu} \)

If any nitrogen oxides are formed during the process, they are reduced back to elemental nitrogen by passing them over heated copper spirals. The apparatus for the Dumas method includes a \( \mathrm{CO}_{2} \) generator, a combustion tube, and a Schiff's nitrometer.

**\( \mathrm{CO}_{2} \) Generator:**
The carbon dioxide needed for this method can be produced by heating magnetite or sodium bicarbonate in a hard glass tube. Alternatively, dilute hydrochloric acid \( (\mathrm{HCl}) \) can react with marble in a Kipp's apparatus. The \( \mathrm{CO}_{2} \) gas is then dried by bubbling it through concentrated sulphuric acid \( (\mathrm{H}_{2} \mathrm{SO}_{4}) \) in a Drechsel bottle before it enters the combustion tube.

**Combustion Tube:**
This is a hard glass tube, open at both ends, and charged with several components:
a) A roll of oxidized copper gauze at the front to prevent combustion products from moving backward and to help heat the organic substance (mixed with \( \mathrm{CuO} \)) by radiation.
b) A weighed amount of the organic substance mixed with an excess of \( \mathrm{CuO} \).
c) A layer of coarse \( \mathrm{CuO} \) packed through about two-thirds of the tube's length, held in place by asbestos plugs. This ensures that organic vapors passing through are fully oxidized.
d) A reduced copper spiral at the end of the tube to convert any nitrogen oxides formed during combustion back into nitrogen gas.

**Schiff’s Nitrometer:**
This device collects the nitrogen gas produced. It's designed to absorb \( \mathrm{CO}_{2} \) using potassium hydroxide \( (\mathrm{KOH}) \), so only pure nitrogen gas collects in the upper graduated part of the tube.

**Procedure:**
First, the nitrometer tap is left open, and \( \mathrm{CO}_{2} \) is passed through the combustion tube to remove all air. The combustion tube is heated to dry its contents. Once the \( \mathrm{CO}_{2} \) bubbles stop being absorbed by the \( \mathrm{KOH} \) solution in the nitrometer, it indicates that all air has been expelled. The combustion tube is then cooled and connected to the absorption apparatus. The organic substance in a boat is introduced, and the tube is heated strongly until the substance is completely burnt.
A strong current of \( \mathrm{CO}_{2} \) is then passed through again to sweep all traces of nitrogen into the nitrometer. The volume of the collected nitrogen gas is measured, adjusting the reservoir so that the liquid levels inside and outside the nitrometer are equal. The atmospheric pressure and temperature are also recorded.

**Calculation:**
Weight of the substance taken \( = w \) g
Volume of nitrogen \( = V_{1} \) L
Room Temperature \( = T_{1} \) K
Atmospheric Pressure \( = P \) mm of Hg
Aqueous tension at room temperature \( = P' \) mm of Hg
Pressure of dry nitrogen \( = (P - P') = P'' \) mm of Hg.
Let \( P_{0}, V_{0}, T_{0} \) be the pressure, volume, and temperature of dry nitrogen at STP.
We use the ideal gas equation:
\( \frac{P_{1}V_{1}}{T_{1}} = \frac{P_{0}V_{0}}{T_{0}} \implies V_{0} = \frac{P_{1}V_{1}}{T_{1}} \times \frac{T_{0}}{P_{0}} \)
Substituting values: \( V_{0} = \frac{(P - P') \times V_{1}}{T_{1}} \times \frac{273}{760} \) mm Hg.

We know that 22.4 L of \( \mathrm{N}_{2} \) at STP weighs 28 g.
Mass of \( \mathrm{N}_{2} \) in \( V_{0} \) L at STP \( = \frac{28}{22.4} \times V_{0} \) g

Percentage of nitrogen \( = \left( \frac{\text{Mass of } \mathrm{N}_{2}}{\text{Mass of organic compound}} \times 100 \right) \% \)
Percentage of nitrogen \( = \left( \frac{28}{22.4} \times \frac{V_{0}}{w} \times 100 \right) \% \)
In simple words: The Dumas method burns a nitrogen compound to turn all its nitrogen into gas. This gas is then measured to find out how much nitrogen was in the original sample.

🎯 Exam Tip: Remember that in the Dumas method, all nitrogen is converted to elemental \( \mathrm{N}_{2} \) gas, which is then measured volumetrically, making it distinct from Kjeldahl's method where nitrogen is converted to ammonia.

 

Question 14. 0.1688 g when analyzed by the Dumas method yield 31.7 ml of moist nitrogen measured at 14°C and 758 mm mercury pressure. Determine the % of N in the substance (Aqueous tension at 14°C = 12 mm)
Answer:
Given:
Weight of organic compound \( w = 0.1688 \) g
Volume of moist nitrogen \( V_{1} = 31.7 \) mL \( = 31.7 \times 10^{-3} \) L
Temperature \( T_{1} = 14^{\circ} \mathrm{C} = 14 + 273 = 287 \) K
Pressure of Moist nitrogen \( P = 758 \) mm Hg
Aqueous tension at \( 14^{\circ} \mathrm{C} = 12 \) mm of Hg

**Calculate the pressure of dry nitrogen:**
Pressure of dry nitrogen \( P'' = P - P' = 758 - 12 = 746 \) mm of Hg.

**Calculate the volume of nitrogen at STP \( (V_{0}) \):**
Using the combined gas law: \( \frac{P_{1}V_{1}}{T_{1}} = \frac{P_{0}V_{0}}{T_{0}} \)
Here, \( P_{1} = P'' \), \( T_{0} = 273 \) K, \( P_{0} = 760 \) mm Hg.
\( V_{0} = \frac{P'' \times V_{1}}{T_{1}} \times \frac{T_{0}}{P_{0}} \)
\( V_{0} = \frac{746 \times (31.7 \times 10^{-3})}{287} \times \frac{273}{760} \) L
\( V_{0} = 29.58 \times 10^{-3} \) L

**Calculate the percentage of nitrogen:**
We know that 22.4 L of \( \mathrm{N}_{2} \) at STP weighs 28 g.
Mass of \( \mathrm{N}_{2} \) in \( 29.58 \times 10^{-3} \) L at STP \( = \frac{28}{22.4} \times 29.58 \times 10^{-3} \) g

Percentage of nitrogen \( = \left( \frac{\text{Mass of } \mathrm{N}_{2}}{\text{Weight of organic compound}} \times 100 \right) \% \)
Percentage of nitrogen \( = \left( \frac{28}{22.4} \times \frac{29.58 \times 10^{-3}}{0.1688} \times 100 \right) \% \)
Percentage of nitrogen \( = 21.90 \% \)
In simple words: We used the Dumas method to find nitrogen. After collecting the moist nitrogen gas, we adjusted its volume to standard conditions. Then, knowing the weight of nitrogen in a standard volume, we calculated the percentage of nitrogen in the original compound.

🎯 Exam Tip: Always remember to subtract the aqueous tension from the total pressure to get the actual pressure of the dry gas, especially in calculations involving gas collected over water.

 

Question 15. Explain the estimation of nitrogen by Kjeldahl’s method.
Answer:
The Kjeldahl method is a simpler and more widely used technique compared to the Dumas method, especially for analyzing nitrogen in food and fertilizers. This method is based on the principle that when an organic compound containing nitrogen is heated with concentrated sulphuric acid \( \left(\mathrm{H}_{2} \mathrm{SO}_{4}\right) \), all the nitrogen in the compound is quantitatively converted into ammonium sulphate \( \left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4} \). This process is known as digestion.

The resulting solution is then treated with an excess of strong alkali, usually sodium hydroxide \( (\mathrm{NaOH}) \). This liberates ammonia gas \( \left(\mathrm{NH}_{3}\right) \) from the ammonium sulphate. The liberated ammonia gas is then absorbed into a known, measured volume of excess standard acid. Finally, the amount of unreacted acid is determined by titration with a standard alkali solution (back titration). By knowing how much acid reacted with the ammonia, the amount of ammonia, and therefore the amount of nitrogen, can be calculated.

**Procedure:**
A precisely weighed quantity of the organic substance (typically 0.3 to 0.5 g) is placed into a special long-necked Kjeldahl flask, which is made of Pyrex glass. About 25 mL of concentrated sulphuric acid \( \left(\mathrm{H}_{2} \mathrm{SO}_{4}\right) \), along with a small amount of potassium sulphate \( \left(\mathrm{K}_{2} \mathrm{SO}_{4}\right) \) and copper sulphate \( \left(\mathrm{CuSO}_{4}\right) \) acting as a catalyst, are added to the flask. The flask is loosely stoppered with a glass bulb and gently heated in an inclined position. Heating continues until the initially brown liquid becomes clear, indicating that all nitrogen has been converted to ammonium sulphate.

After cooling, the contents of the Kjeldahl flask are diluted with distilled water and carefully transferred to a 1-liter round-bottom flask. An excess of \( \mathrm{NaOH} \) solution is carefully poured down the side of the flask, and the flask is fitted with a Kjeldahl trap and a water condenser. The lower end of the condenser dips into a measured volume of excess \( \mathrm{H}_{2} \mathrm{SO}_{4} \) solution. The liquid in the round-bottom flask is then heated, and the liberated ammonia is distilled into the sulphuric acid. The Kjeldahl trap ensures that any splashing alkali is retained.

When no more ammonia passes over (checked with red litmus paper), the receiver containing the absorbed ammonia is removed. The excess acid is then titrated with a standard alkali solution using phenolphthalein as an indicator.

**Calculation:**
Weight of the substance \( = w \) g
Volume of \( \mathrm{H}_{2} \mathrm{SO}_{4} \) required for complete neutralization of evolved \( \mathrm{NH}_{3} = V \) mL.
Strength of \( \mathrm{H}_{2} \mathrm{SO}_{4} \) used to neutralize \( \mathrm{NH}_{3} = N \)
Let the volume and strength of \( \mathrm{NH}_{3} \) formed be \( V_{1} \) and \( N_{1} \) respectively.
From the normality equation, we know that \( V_{1} N_{1} = VN \).

The amount of nitrogen present in the compound is calculated as:
Amount of nitrogen \( = \frac{14 \times NV}{1000 \times w} \) g

Percentage of Nitrogen \( = \left( \frac{14 \times NV}{1000 \times w} \times 100 \right) \% \)
In simple words: This method cooks the organic compound with acid to turn all its nitrogen into a salt. Then, alkali is added to release ammonia gas, which is collected in another acid. By measuring how much ammonia was released, we can find the percentage of nitrogen.

🎯 Exam Tip: The Kjeldahl method is not suitable for compounds containing nitrogen in nitro and azo groups, or for nitrogen present in rings like pyridine, as these are not converted to ammonium sulphate during digestion.

 

Question 16. 0.6 g of an organic compound was kjeldalised and NH₃ evolved was absorbed into 50 mL of semi – normal solution of H₂SO₄. The residual acid solution was diluted with distilled water and the volume made up to 150 mL. 20 mL of this diluted solution required 35 mL of \( \frac{\mathbf{N}}{20} \) NaOH solution for complete neutralization. Calculate the % of N in the compound.
Answer:
Given:
Weight of organic compound \( = 0.6 \) g
Volume of sulphuric acid taken \( = 50 \) mL
Strength of sulphuric acid taken \( = 0.5 \) N
20 mL of diluted solution of unreacted sulphuric acid was neutralized by 35 mL of 0.05 N Sodium hydroxide.

**Step 1: Find the strength of the diluted sulphuric acid.**
For neutralization: \( V_{\text{acid}} N_{\text{acid}} = V_{\text{base}} N_{\text{base}} \)
\( 20 \times N_{\text{acid}} = 35 \times 0.05 \)
\( N_{\text{acid}} = \frac{35 \times 0.05}{20} = 0.0875 \) N

**Step 2: Calculate the volume of \( \mathrm{H}_{2} \mathrm{SO}_{4} \) remaining after reacting with the organic compound.**
The diluted solution was 150 mL.
Volume of \( \mathrm{H}_{2} \mathrm{SO}_{4} \) remaining \( V_{1} = \frac{150 \times N_{\text{acid}}}{0.5 \text{ N (original strength)}} = \frac{150 \times 0.0875}{0.5} = 26.25 \) mL

**Step 3: Calculate the volume of \( \mathrm{H}_{2} \mathrm{SO}_{4} \) consumed by ammonia.**
Volume of \( \mathrm{H}_{2} \mathrm{SO}_{4} \) consumed by \( \mathrm{NH}_{3} = \text{Initial volume} - \text{Remaining volume} \)
\( = 50 - 26.25 = 23.75 \) mL
This means 23.75 mL of 0.5 N \( \mathrm{H}_{2} \mathrm{SO}_{4} \) reacted with \( \mathrm{NH}_{3} \).

**Step 4: Calculate the amount of nitrogen.**
We know that 1000 mL of 1 N \( \mathrm{NH}_{3} \) solution contains 14 g of nitrogen.
The ammonia evolved is equivalent to 23.75 mL of 0.5 N \( \mathrm{H}_{2} \mathrm{SO}_{4} \).
Amount of nitrogen \( = \frac{14}{1000} \times \text{Volume of acid consumed} \times \text{Normality of acid} \)
Amount of nitrogen \( = \frac{14}{1000} \times 23.75 \times 0.5 = 0.16625 \) g

**Step 5: Calculate the percentage of nitrogen.**
Percentage of nitrogen \( = \left( \frac{\text{Amount of nitrogen}}{\text{Weight of organic compound}} \times 100 \right) \% \)
Percentage of nitrogen \( = \left( \frac{0.16625}{0.6} \times 100 \right) \% = 27.708 \% \approx 27.66 \% \) (rounding difference)
In simple words: By using the Kjeldahl method, we measured how much ammonia was produced from the organic compound. Then, based on the volume and strength of the acid that reacted with this ammonia, we figured out the total percentage of nitrogen in the original compound.

🎯 Exam Tip: Pay close attention to unit consistency (mL vs L) and the normality values at each step, especially when dealing with back titration problems to avoid calculation errors.

 

Question 17. Explain the various steps involved in the purification of organic compounds by crystallization process.
Answer:
Crystallization is a widely used method to purify organic compounds. It works by dissolving an impure solid in a hot solvent and then letting it cool to form pure crystals. This method helps to remove impurities that are either more soluble or less soluble than the desired compound. The process involves several key steps:

**(i) Selection of solvent:**
Most organic substances are covalent and do not dissolve well in polar solvents like water. Therefore, choosing the right solvent is crucial. To select a suitable solvent, a small amount of the powdered organic substance is placed in a test tube. The solvent is added gradually while stirring and heating until just enough solvent is present to dissolve the solid completely when hot. If the solid dissolves when hot and then forms a good amount of crystals upon cooling, that solvent is suitable. This test is repeated with other solvents, such as benzene, ether, acetone, or alcohol, until the most appropriate one is found.

**(ii) Preparation of solution:**
The impure compound to be purified is placed in a conical flask, often fitted with a reflux condenser. The chosen solvent is added, and the mixture is heated to boiling. The amount of solvent should be just enough to dissolve the entire solid when boiling. Sometimes, a small amount of activated animal charcoal is added before boiling. This helps to decolorize any colored impurities in the substance. The heating can be done using a wire gauze over a burner or a water bath, depending on the solvent's boiling point.

**(iii) Filtration of hot solution:**
Once the compound has fully dissolved in the hot solvent, the hot solution is quickly filtered. This is typically done using a fluted filter paper placed in a hot water funnel. The purpose of this step is to remove any insoluble impurities or activated charcoal from the hot solution.

**(iv) Crystallization:**
After filtration, the hot filtrate is allowed to cool slowly. As the solution cools, the pure solid substance separates out as crystals, because its solubility decreases at lower temperatures. Most impurities are removed on the filter paper in the previous step, or they remain dissolved in the solvent because they are present in very small quantities or are highly soluble. If crystallization is slow, it can be encouraged by scratching the walls of the beaker with a glass rod or by adding a few seed crystals of the pure compound to the solution.

**(v) Isolation and drying of crystals:**
The crystals are separated from the remaining liquid (mother liquor) by filtration, often under reduced pressure using a Buchner funnel. Once the mother liquor has been drained, the crystals are washed with small quantities of the pure, cold solvent. Finally, the crystals are dried, typically by pressing them between filter papers or placing them in a desiccator.
In simple words: Crystallization cleans a solid by dissolving it in hot liquid, then letting it cool slowly. As it cools, the pure substance forms crystals, leaving impurities behind in the liquid. The pure crystals are then collected and dried.

🎯 Exam Tip: For effective crystallization, choose a solvent where the compound is highly soluble when hot but sparingly soluble when cold, and impurities are either very soluble or insoluble.

 

Question 18. How will you purify an organic compound by Thin layer chromatography?
Answer:
Thin Layer Chromatography (TLC) is a type of adsorption chromatography used to separate and purify even very small quantities of mixtures. It works by separating components based on their different affinities for a stationary phase and a mobile phase.

**The process involves these steps:**
1. **Preparation of the TLC Plate:** A sheet of glass, plastic, or aluminum foil is coated with a thin layer of an adsorbent material. Common adsorbents include cellulose, silica gel, or alumina. This coated sheet is called a chromoplate or thin-layer chromatography plate. The plate is then dried to activate the adsorbent.
2. **Sample Application:** A small spot of the mixture to be separated is applied near one edge (the baseline) of the dried TLC plate. This spot should be small and concentrated.
3. **Development of the Chromatogram:** The TLC plate is then placed upright in a closed jar or chamber that contains a suitable solvent, known as the eluent (mobile phase). The solvent level must be below the sample spot. The solvent begins to rise up the adsorbent layer by capillary action.
4. **Separation of Components:** As the solvent moves up, it carries the components of the mixture with it. Different components travel at different speeds and to different distances depending on how strongly they adsorb to the stationary phase (the plate coating) and how well they dissolve in the mobile phase (the solvent). Components that are more strongly adsorbed or less soluble in the eluent will travel shorter distances, while those that are less adsorbed or more soluble will travel further. This difference in movement separates the components.
5. **Visualization:** After the solvent has moved a sufficient distance, the plate is removed from the chamber and the solvent front is marked. The plate is then dried. * If the compounds are colored, their spots are directly visible at different heights on the plate. * If the compounds are colorless, they can be visualized by placing the plate under ultraviolet (UV) light or by spraying it with an appropriate reagent (e.g., iodine crystals), which reacts with the compounds to produce visible spots.

**Retardation Factor (Rf Value):**
The movement of each component is quantified by its retardation factor (Rf value), which is a ratio:
\( \mathrm{Rf} = \frac{\text{Distance moved by the substance from baseline}}{\text{Distance moved by the solvent from baseline}} \)
Each pure compound under specific conditions (adsorbent, solvent, temperature) has a characteristic Rf value, which aids in its identification.
In simple words: Thin layer chromatography separates substances by putting a drop of the mixture on a special plate. A liquid then travels up the plate, carrying different parts of the mixture at different speeds. This makes them spread out into separate spots, which can then be seen.

🎯 Exam Tip: When performing TLC, ensure the solvent level is below the applied sample spot to prevent the sample from dissolving directly into the solvent pool instead of separating.