RBSE Solutions Class 11 Chemistry Chapter 12 Organic Chemistry Some Basic Principles and Techniques

Rajasthan Board RBSE Class 11 Chemistry Chapter 12 Organic Chemistry: Some Basic Principles and Techniques

RBSE Class 11 Chemistry Chapter 12 Text Book Questions

RBSE Class 11 Chemistry Chapter 12 Multiple Choice Questions

 

Question 1. Which of the following method is used for estimation of nitrogen?
(a) Liebig's method
(b) Lassaigne's method
(c) Kjeldahl's method
(d) Carius method
Answer: (c) Kjeldahl's method
In simple words: The Kjeldahl method is specifically used to measure the amount of nitrogen present in a chemical sample.

🎯 Exam Tip: Remember specific named tests for element estimation: Kjeldahl for nitrogen, Carius for halogens, Liebig for carbon/hydrogen, and Lassaigne for general element detection.

 

Question 3. IUPAC name of iso-butane is :
(a) 2-methyl butane
(b) 2-methyl propane
(c) 2-ethyl butane
(d) 2-butyne
Answer: (b) 2-methyl propane
In simple words: The correct scientific name for the compound called iso-butane is 2-methyl propane.

🎯 Exam Tip: Always draw the structure of the common name (like iso-butane) first, then apply IUPAC naming rules to find the correct systematic name.

 

Question 4. Prefix used for -COOH functional group is :
(a) Carbomayl
(b) Carbonyl
(c) Carboxy
(d) Alkoxycarbonyl
Answer: (c) Carboxy
In simple words: When the -COOH group is part of a larger, more complex molecule, the prefix "carboxy" is used in its name.

🎯 Exam Tip: Distinguish between prefixes and suffixes for functional groups, as this depends on whether the group is the main functional group or a substituent.

 

Question 5. Which of the following substituent do not show +I effect ?
(a) Br
(b) R-OH
(c) FeCl3
(d) -CHR2
Answer: (a) Br
In simple words: The bromine atom (Br) does not push electrons away from itself; instead, it pulls them, so it does not show a positive inductive (+I) effect.

🎯 Exam Tip: Remember that halogens like Br generally exhibit a -I effect (electron-withdrawing), while alkyl groups and electron-donating groups show a +I effect.

 

Question 7. Which of the following is most stable carbocation?
(a) \( \ce{+CH3} \)
(b) \( \ce{+CH2CH3} \)
(c) \( \ce{(CH3)2+CH} \)
(d) \( \ce{(CH3)3C+} \)
Answer: (d) \( \ce{(CH3)3C+} \)
In simple words: The carbocation with three methyl groups attached to the positively charged carbon is the most stable because these groups help spread out the positive charge, making it less reactive.

🎯 Exam Tip: The stability of carbocations increases with the number of alkyl groups attached to the positively charged carbon, due to the +I effect and hyperconjugation from the alkyl groups.

 

Question 8. Homolytic fission results in formation of:
(a) Carbocation
(b) Carbene
(c) Nitrene
(d) Free radical
Answer: (d) Free radical
In simple words: When a chemical bond breaks evenly, with each atom taking one electron from the bond, it forms two very reactive particles called free radicals.

🎯 Exam Tip: Understand the difference between homolytic (forms radicals) and heterolytic (forms ions) fission, as they dictate the type of reactive intermediates formed in a reaction.

 

Question 9. Which of the following group do not give addition reactions?
(a) \( \ce{C=C} \)
(b) \( \ce{C=C} \)
(c) \( \ce{C=O} \)
(d) \( \ce{CH3-CH3} \)
Answer: (d) \( \ce{CH3-CH3} \)
In simple words: A single bond between two carbon atoms, like in \( \ce{CH3-CH3} \), does not participate in addition reactions because it is already saturated and stable.

🎯 Exam Tip: Addition reactions typically occur across multiple bonds (double or triple bonds) in unsaturated compounds like alkenes, alkynes, and carbonyls, where new atoms add without losing any existing atoms.

 

Question 11. Write the name of two neutral and two negatively charged nucleophiles.
Answer:
Neutral nucleophiles: Ammonia \( \ce{(NH3)} \), Water \( \ce{(H2O)} \)
Negatively Charged nucleophiles: Cyanide ion \( \ce{(CN-)} \), Chloride ion \( \ce{(Cl-)} \)
In simple words: Neutral nucleophiles have extra electrons but no overall charge, while negatively charged nucleophiles have extra electrons and carry a negative charge.

🎯 Exam Tip: Nucleophiles are electron-rich species that donate electrons to an electron-deficient center. They can be neutral with lone pairs or carry a negative charge.

 

Question 12. Write the name of the method used for estimation of halogens.
Answer: Carius method is used for estimation of Halogens.
In simple words: The Carius method is a special test done to find out how much of a halogen (like chlorine, bromine, or iodine) is present in a chemical sample.

🎯 Exam Tip: Relate the Carius method specifically to halogens; this helps in quickly recalling the correct test for each element.

 

Question 13. Write the formula for percentage of nitrogen in Kjeldhal's method.
Answer:
% of Nitrogen \( = \frac{1.4 \times \text{Normality of acid} \times \text{Volume of acid used for complete neutralization of NH}_3}{\text{weight of the compound taken}} \)
In simple words: To find the percentage of nitrogen, you multiply 1.4 by the acid's strength and the amount of acid used to neutralize ammonia, then divide by the sample's weight.

🎯 Exam Tip: Ensure correct values for normality and volume are used, and that the constant 1.4 is remembered for nitrogen estimation in Kjeldahl's method.

 

Question 14. Which method is used for estimation of carbon and hydrogen?
Answer: Liebig method is used for estimation of carbon and hydrogen.
In simple words: To measure how much carbon and hydrogen are in a substance, chemists use a test called the Liebig method.

🎯 Exam Tip: Associate the Liebig method directly with the quantitative estimation of carbon and hydrogen in organic compounds.

 

Question 16. What is the hybridisation of carbon atom in carbanion?
Answer: \( \text{sp}^3 \).
In simple words: In a carbanion, the carbon atom uses \( \text{sp}^3 \) hybridization, which means its orbitals mix in a specific way.

🎯 Exam Tip: Remember that a carbanion has a negatively charged carbon with a lone pair and three bonds, leading to \( \text{sp}^3 \) hybridization and a pyramidal geometry.

 

Question 17. Given the order of stability of primary, secondary and tertiary carbocation.
Answer:
Increasing order of stability : Primary Carbocation \( < \) Secondary Carbocation \( < \) Tertiary Carbocation
(Least Stable) (Most stable)
In simple words: Tertiary carbocations are the most stable, followed by secondary, and then primary carbocations, which are the least stable.

🎯 Exam Tip: The stability of carbocations increases with increasing alkyl substitution due to the electron-donating inductive effect and hyperconjugation.

 

Question 18. Define Carbene.
Answer: Carbene is defined as a neutral species with two valence electrons and two unshared pair electrons; it is a very reactive intermediate.

In simple words: A carbene is a molecule that has a carbon atom with only two bonds and two electrons that are not shared with other atoms, making it highly reactive.

🎯 Exam Tip: Focus on the key characteristics: neutral, divalent carbon, and having two unshared electrons. Drawing the structure helps illustrate the concept.

 

Question 19. Write the IUPAC name of trimethyl methane.
Answer: 2-Methyl propane.
In simple words: Trimethyl methane is also known by its chemical name 2-methyl propane.

🎯 Exam Tip: Convert the common name into its structure first, then apply IUPAC rules to derive the systematic name.

 

Question 20. Write the IUPAC name of OHC-CH2-CH2-COOH
Answer: 4-Formyl-1-Butanoic Acid
In simple words: The chemical compound OHC-CH2-CH2-COOH is scientifically named 4-Formyl-1-Butanoic Acid.

🎯 Exam Tip: When multiple functional groups are present, identify the highest priority functional group (like -COOH here) to determine the parent chain and then use prefixes for other groups.

 

Question 21. Write the prefix and suffix used for -OH group.
Answer:
Prefix of -OH group = hydroxy
Suffix of -OH group = - ol
In simple words: When -OH is a main part of the name, it ends with "-ol", but if it's a side part, it starts with "hydroxy-".

🎯 Exam Tip: Clarify when to use "hydroxy" (as a substituent prefix) versus "-ol" (as a principal functional group suffix).

 

Question 22. What is the difference between the mass of two successive members of homologous series ?
Answer: The difference between the mass of two successive members of homologous series is 14. This is because the molecular formula of consecutive members differs by a \( \ce{-CH2} \) group, where carbon has an atomic mass of 12 and hydrogen has an atomic mass of 1, so \( 12 + (2 \times 1) = 14 \).
In simple words: Each step up in a homologous series adds a \( \ce{CH2} \) unit, which means the mass increases by 14 units.

🎯 Exam Tip: Remember that a homologous series is characterized by a constant \( \ce{CH2} \) difference between successive members, resulting in a molecular mass difference of 14 amu.

 

Question 23. Write the name and structure of two aromatic heterocyclic compounds.
Answer: Two examples of aromatic heterocyclic compounds are Furan and Pyrrole.
In simple words: Aromatic heterocyclic compounds are ring-shaped molecules that include at least one atom other than carbon in the ring, such as oxygen or nitrogen, and have special stability due to their electron arrangement.

🎯 Exam Tip: When asked for structures, ensure you correctly represent the atoms and bonds, including lone pairs or double bonds that contribute to aromaticity in heterocyclic rings.

RBSE Class 11 Chemistry Chapter 12 Short Answer Type Questions

 

Question 24. Which method is used to test nitrogen, sulphur and halogens?
Answer: The Lassaigne test method is used to test nitrogen, sulphur, and halogens. This test involves two main steps:

• Preparation of sodium fusion extract
• Detection of elements using sodium fusion extract.

In simple words: To check if a substance contains nitrogen, sulfur, or halogens, you perform the Lassaigne test, which involves first preparing a special extract and then testing it.

🎯 Exam Tip: Remember the two key stages of the Lassaigne test: converting covalent organic compounds into ionic inorganic salts (fusion extract) and then identifying the elements present.

 

Question 25. Write the structural formula and IUPAC name of the following compounds:
(i) Formic acid
(ii) Ethyl acetate
(iii) Ethyl methyl ether
Answer:

In simple words: Knowing the structural formula helps you derive the IUPAC name, which is the official scientific name for each compound.

🎯 Exam Tip: Practice drawing structures and applying IUPAC rules for various functional groups. Pay attention to numbering and prioritizing functional groups.

 

Question 27. Write two examples of molecules showing -I effect.
Answer: Two examples of molecules showing the -I (negative inductive) effect are Nitrobenzene and Chlorobenzene.

In these molecules, \( \ce{-NO2} \) and \( \ce{-Cl} \) atoms or groups exhibit the -I effect. This is because they are electron-withdrawing groups or atoms, meaning they pull electrons towards themselves through the single bonds in the molecule.
In simple words: Nitrobenzene and Chlorobenzene show the -I effect because the nitro group (\( \ce{NO2} \)) and the chlorine atom (\( \ce{Cl} \)) are greedy for electrons and pull them away from the rest of the molecule.

🎯 Exam Tip: Remember that electron-withdrawing groups like halogens, nitro groups, and carboxylic acids display the -I effect, decreasing electron density in the rest of the molecule.

 

Question 28. Write four examples of neutral electrophiles.
Answer: Four examples of neutral electrophiles are \( \ce{BF3} \), \( \ce{AlCl3} \), \( \ce{SO3} \), and \( \ce{ZnCl2} \).
In simple words: Neutral electrophiles are molecules that love electrons but don't have a charge themselves. They have empty spaces for electrons to fit into.

🎯 Exam Tip: Neutral electrophiles are often Lewis acids, which are electron-deficient compounds that can accept a pair of electrons. Look for central atoms with incomplete octets.

 

Question 29. Write the following in decreasing order of +I effect:
\( \ce{(CH3)2CH-} \), \( \ce{CH3-} \), \( \ce{(CH3)3C-} \), \( \ce{-CH2CH3} \)
Answer: The decreasing order of +I effect is:
\( \ce{(CH3)3C-} > \ce{(CH3)2CH-} > \ce{-CH2CH3} > \ce{CH3-} \)
In simple words: Groups with more branches and more carbon atoms, like \( \ce{(CH3)3C-} \), push electrons away the most, making them have the strongest positive inductive effect.

🎯 Exam Tip: The +I effect increases with the number of alkyl groups attached to the carbon atom, as alkyl groups are electron-donating. Tertiary alkyl groups have the strongest +I effect.

 

Question 31. Why alkanes are called paraffins ?
Answer: Alkanes are called paraffins because the word "paraffins" comes from Latin, meaning "little affinity" (parum = little and affinis = reactivity). Alkanes have very low reactivity towards general chemical reagents, which is why they were given this name.
In simple words: Alkanes are called paraffins because they don't react easily with other chemicals.

🎯 Exam Tip: Connect the term "paraffin" to the low reactivity of alkanes due to their stable single bonds and nonpolar nature.

 

Question 32. Write the IUPAC name of chloroform, formic acid and iso-pentane
Answer:

In simple words: The IUPAC names provide a clear, standard way to identify these common chemical compounds based on their structure.

🎯 Exam Tip: For common compounds, practice both their common names and their systematic IUPAC names, especially for isomers like iso-pentane.

 

Question 33. Write the formula of isobutyl alcohol and butyl chloride.
Answer:

In simple words: The structural formulas show how the atoms are connected in isobutyl alcohol and butyl chloride.

🎯 Exam Tip: Distinguish between different isomers (like isobutyl vs. n-butyl) when drawing structural formulas to ensure accuracy.

 

Question 35. In which compounds, electrophilic addition reactions are generally found? Explain.
Answer: Electrophilic addition reactions are generally found in organic compounds containing multiple bonds, such as alkenes and alkynes. For example, ethene reacts with bromine by an electrophilic addition reaction. In this reaction, all parts of the adding reagent appear in the product.
\( \ce{C=C + E+ -> E-C-C+ -> E-C-C-N} \)
In simple words: Electrophilic addition reactions happen a lot in compounds with double or triple bonds, like alkenes, where a positive-seeking part of a molecule adds itself across the bond.

🎯 Exam Tip: Remember that electrophilic addition reactions are characteristic of unsaturated compounds, where the electron-rich multiple bond attacks an electrophile.

 

Question 36. Explain rearrangement reactions with examples.
Answer: In a rearrangement reaction, a molecule undergoes a reorganization of its constituent parts, leading to a new isomeric compound. For example, an alkene, when heated with a strong acid, can form another isomeric alkene. This often happens to form a more stable intermediate or product.
\( \ce{H3C-CH=CH-CH3 + H+ (acid) -> H3C-C+(CH3)-CH3 -> H3C-CH(CH3)-CH2-CH3} \)
In simple words: Rearrangement reactions are like molecules changing their internal structure to become a different but related molecule, often to become more stable.

🎯 Exam Tip: Focus on the movement of atoms or groups within the same molecule to form an isomer, often driven by the formation of more stable intermediates like carbocations.

 

Question 37. Write the difference between homolytic and heterolytic cleavage.
Answer:

In simple words: Homolytic cleavage means a bond breaks evenly, with each atom getting one electron, forming free radicals. Heterolytic cleavage means one atom takes both electrons, forming ions.

🎯 Exam Tip: Clearly differentiate the products formed (free radicals vs. ions) and the electron movement (one electron to each atom vs. both electrons to one atom) for each type of bond cleavage.

 

Question 38. Explain the free radical reactions.
Answer: Free radicals are chemical species that contain a singly occupied orbital, meaning they have one unpaired electron. They are neutral and tend to be highly reactive. These are formed when covalent bonds break evenly, a process called homolytic cleavage. Free radicals have a strong tendency to pair their unpaired electron and readily combine with each other or with other molecules. In a free radical reaction, free radicals are formed as reaction intermediates. These reactions mainly occur in the presence of UV light or heat. For example, methyl radicals can combine to form ethane:
\( \ce{CH3· + CH3· -> CH3-CH3} \) (Ethane)
In simple words: Free radical reactions involve very reactive particles called free radicals, which have one unpaired electron and try to pair up, often starting with light or heat.

🎯 Exam Tip: When explaining free radical reactions, mention their formation (homolytic cleavage), their high reactivity due to the unpaired electron, and common conditions like UV light or heat.

 

Question 39. Write the IUPAC name of following:
\( \ce{H2C=CH-C(=O)-CH3} \) and \( \ce{CH2CH2CH2CH3} \) (1-cyclopropyl butane)
Answer:
For \( \ce{H2C=CH-C(=O)-CH3} \): But-3-ene-2-one
For \( \ce{CH2CH2CH2CH3} \) (1-cyclopropyl butane): This is actually a structure with a cyclopropane ring. Assuming the intent of the question for "1 cyclopropyl butane" refers to a cyclopropane ring attached to a butyl chain, its IUPAC name would be 1-cyclopropylbutane. The provided structure text \( \ce{CH2CH2CH2CH3} \) is a linear butane chain, which doesn't match "1 cyclopropyl butane". If the question meant a cyclopropyl group attached to a butane chain, the IUPAC name would be 1-cyclopropylbutane for a straight chain, or more complex if branched. Given the provided "1 cyclopropyl butane", this implies a cyclopropane ring attached to a butane chain.
In simple words: The first compound is a ketone with a double bond. The second compound is a four-carbon chain with a three-carbon ring attached to its first carbon.

🎯 Exam Tip: Pay close attention to functional group priorities, double/triple bond positions, and numbering for correct IUPAC naming, especially for cyclic and branched compounds.

 

Question 40. Give the structural formula of 2-methoxy- 2-methyl propane.
Answer: The structural formula of 2-methoxy-2-methyl propane is:
\( \ce{(CH3)3C-OCH3} \)
(which is a tertiary butyl methyl ether).
In simple words: This compound has a central carbon with three methyl groups and one methoxy group (\( \ce{-OCH3} \)).

🎯 Exam Tip: For ethers, correctly identify the alkoxy group (-OR) and the main alkane chain, ensuring proper numbering for substituents.

 

Question 41. Write the odd term in Thiophene and Pyridine.
Answer:
In simple words: Thiophene has a sulfur atom in its ring, while pyridine has a nitrogen atom. They are both odd terms in their respective categories.

🎯 Exam Tip: Understand that heterocyclic compounds contain atoms other than carbon in their ring structure. Being able to draw and identify common heterocycles like thiophene and pyridine is essential.

 

Question 42. Which of the following will have more hyperconjugation ? Explain with reason
\( \ce{(CH3)2CH+} \), \( \ce{CH2CH3+} \)
Answer: \( \ce{(CH3)2CH+} \) carbocation will have more hyperconjugation than \( \ce{CH2CH3+} \). This is because \( \ce{(CH3)2CH+} \) is a secondary carbocation, which has two methyl groups directly attached to the positively charged carbon. In contrast, \( \ce{CH2CH3+} \) (ethyl carbocation) is a primary carbocation with only one ethyl group. A greater number of alkyl groups in a carbocation means more alpha-hydrogens are available for hyperconjugation. Therefore, more hyperconjugation leads to greater stability of the carbocation.
In simple words: The carbocation with two methyl groups attached (\( \ce{(CH3)2CH+} \)) has more hyperconjugation than the one with only one ethyl group (\( \ce{CH2CH3+} \)). More alkyl groups means more ways to stabilize the positive charge through electron sharing.

🎯 Exam Tip: Hyperconjugation increases with the number of alpha-hydrogens (hydrogens on carbon atoms adjacent to the positively charged carbon). Tertiary carbocations have the most alpha-hydrogens, hence the greatest stability.

 

Question 43. Classify the following reactions :
(a) \( \ce{CH3CH2Br + HS- -> CH3CH2SH + Br-} \)
(b) \( \ce{(CH3)2C=CH2 + HCl -> (CH3)2-CCl-CH3} \)
(c) \( \ce{CH3CH2Br + HO- -> CH2=CH2 + H2O + Br-} \)
(d) \( \ce{(CH3)3C-CH2OH + HBr -> (CH3)3C-CH2Br + H2O} \)
Answer:
(a) \( \ce{CH3CH2Br + HS- -> CH3CH2SH + Br-} \)
This reaction is called nucleophilic substitution reaction.
(b) \( \ce{(CH3)2C=CH2 + HCl -> (CH3)2C(Cl)CH3} \)
This reaction is called electrophilic addition reaction.
(c) \( \ce{CH3CH2Br + HO- -> CH2=CH2 + H2O + Br-} \)
This reaction is called elimination reaction.
(d) \( \ce{(CH3)3C-CH2OH + HBr -> (CH3)3C-CH2Br + H2O} \)
This reaction is called nucleophilic substitution reaction.
In simple words: We classify these reactions based on how the atoms change: substitution means one atom replaces another, addition means new atoms join without losing any, and elimination means atoms are removed to form a multiple bond.

🎯 Exam Tip: To classify reactions, identify the key changes: is an atom/group replaced (substitution), are atoms added across a multiple bond (addition), or are atoms removed to form a multiple bond (elimination)?

 

Question 44. Explain substitution reactions with examples.
Answer: Substitution reactions are those reactions in which one atom or a group of atoms in a molecule is replaced by another atom or group of atoms from another molecule. In this process, the overall structure of the main molecule often stays the same, except for the changed part.
For example, \( \ce{CH3Cl + NaOH (aq.) -> CH3OH + NaCl} \)
In this reaction, the chlorine atom in chloromethane (\( \ce{CH3Cl} \)) is replaced by the hydroxyl group (\( \ce{OH-} \)) from sodium hydroxide (\( \ce{NaOH} \)), forming methanol (\( \ce{CH3OH} \)) and sodium chloride (\( \ce{NaCl} \)).
Another example is the chlorination of methane under UV light:
\( \ce{CH4 + Cl2 (UV light) -> CH3Cl + HCl} \)
Here, a hydrogen atom in methane is substituted by a chlorine atom.
In simple words: A substitution reaction is like swapping parts, where one atom or group in a molecule is replaced by another, while the rest of the molecule stays intact.

🎯 Exam Tip: When explaining substitution reactions, clearly identify the atom or group being replaced and the one doing the replacing, and provide simple, illustrative examples.

 

Question 45. Explain Beldstein test.
Answer: The Beilstein test is a qualitative chemical test used to detect the presence of halogens (chlorine, bromine, and iodine) in an organic compound. To perform the test, the tip of a copper wire is first heated in a burner flame until it shows no further color. Once cooled slightly, the wire is dipped into the unknown solid or liquid sample and then heated in the flame again. A green flash or persistent green flame indicates the presence of chlorine, bromine, or iodine. Fluorine is not detected by this test because copper fluoride is not volatile and does not impart color to the flame.
In simple words: The Beilstein test uses a copper wire and flame to check for halogens in a substance; if it turns green, halogens are likely present.

🎯 Exam Tip: Remember the specific steps of the Beilstein test: cleaning the wire, introducing the sample, heating, and observing the characteristic green flame color for halogens.

 

Question 46. Write the IUPAC name of glycerol and crotonic acid.
Answer:
For Glycerol:
For Crotonic acid:
In simple words: Glycerol is a simple alcohol with three hydroxyl groups, while crotonic acid is a carboxylic acid with a double bond.

🎯 Exam Tip: Pay attention to the number of carbon atoms in the main chain and the positions of functional groups and double bonds for correct IUPAC naming.

RBSE Class 11 Chemistry Chapter 12 Long Answer Type Questions

 

Question 47. Explain the following fundamental concepts in organic chemistry:
(a) Properties of Carbon atom.
(b) Van't Hoff and Le Bel's Principle.
(c) Homologous Series.
Answer:
(a) Properties of Carbon atom:

• Carbon atoms show tetravalency, meaning their valency is four.
• Carbon atoms combine with other carbon atoms to form various open-chained and closed-chained compounds.
• Carbon can form single bonds, double bonds, and triple bonds with other atoms, allowing for a wide range of molecular structures.

(b) Van't Hoff and Le Bel's Principle: According to this principle, in an organic compound, the four valencies of carbon are directed towards the corners of a regular tetrahedron, with the carbon atom at the center. The four covalent bonds of carbon are oriented at an angle of 109°28' to each other, and all four valencies are considered equal. This theory shows that the four valencies of carbon are not in the same plane. This principle helps in understanding the three-dimensional shape of carbon compounds.
(c) Homologous Series: A homologous series is a family of organic compounds where all members have the same functional group and similar chemical properties. Each successive member in the series differs from the previous one by a \( \ce{-CH2-} \) group in its molecular formula. This results in a systematic increase in molecular mass by 14 atomic mass units for each subsequent member. Members of a homologous series are prepared by similar methods and show a gradual change in their physical properties like boiling point and melting point.
In simple words: (a) Carbon can form four bonds and connect in many ways, forming long chains and rings. (b) The Van't Hoff and Le Bel principle explains that carbon's four bonds point to the corners of a 3D shape, making molecules spread out in space. (c) A homologous series is a group of similar chemicals where each one is just a little bit bigger than the last by adding a \( \ce{CH2} \) unit.

🎯 Exam Tip: For fundamental concepts, ensure you define the term clearly and provide key characteristics. For Van't Hoff and Le Bel's principle, emphasize the tetrahedral geometry and bond angle.

 

Question 48. Explain the following :
(a) Inductive effect.
(b) Electromeric effect.
(c) Hyperconjugation.
Answer:
(a) Inductive effect: The inductive effect is the permanent displacement of sigma (\( \sigma \)) electrons along a saturated carbon chain due to the presence of an electron-withdrawing or electron-donating group at one end of the chain. This effect is permanent but decreases rapidly with distance along the carbon chain. It is of two types:
(i) -I effect: The electron-withdrawing nature of groups or atoms is known as the -I effect. These groups pull electron density towards themselves. Examples include \( \ce{-NO2} \), \( \ce{-CN} \), \( \ce{-SO3H} \), \( \ce{-F} \), \( \ce{-Cl} \).
(ii) +I effect: The electron-donating nature of groups or atoms is known as the +I effect. These groups push electron density away from themselves. Examples include alkyl groups like \( \ce{(CH3)3C-} \), \( \ce{(CH3)2CH-} \), \( \ce{CH3CH2-} \), \( \ce{CH3-} \).
(b) Electromeric Effect: The electromeric effect is defined as the complete transfer of a shared pair of pi (\( \pi \))-electrons to one of the atoms joined by a multiple bond on the demand of an attacking reagent. It is a temporary effect, meaning it only occurs in the presence of an attacking reagent. It is represented by 'E' and the shifting of electrons is shown by a curved arrow. It is of two types:
(i) +E effect: When the transfer of electrons takes place towards the attacking reagent, it is called the +E effect. This typically occurs in addition reactions where the pi electrons move towards the more positive part of the reagent. For example, in the addition of acid to an alkene:
\( \ce{C=C + H+ -> C-C+} \)
(ii) -E effect: When the transfer of electrons takes place away from the attacking reagent, it is called the -E effect. This occurs when the pi electrons move away from the attacking reagent. For example, in the addition of cyanide ion to a carbonyl compound:
\( \ce{C=O + CN- -> C-O-} \)
(c) Hyperconjugation: Hyperconjugation is the delocalization of sigma (\( \sigma \)) electrons or lone pair of electrons into an adjacent pi (\( \pi \))-orbital or p-orbital. It is also known as "no bond resonance" or the "Baker-Nathan effect". It occurs due to the overlap of a \( \sigma \)-bonding orbital (from a C-H bond) with an empty or partially filled p-orbital or \( \pi \)-orbital on an adjacent carbon atom.
Conditions for Hyperconjugation: There must be an alpha-CH group (a carbon with hydrogen atoms attached to it, next to an \( \text{sp}^2 \) hybridized carbon) or a lone pair on an atom adjacent to an \( \text{sp}^2 \) hybridized carbon or other atoms like nitrogen, oxygen, etc.
In simple words: (a) The inductive effect is a slight, permanent push or pull of electrons through a single bond, making parts of a molecule slightly charged. (b) The electromeric effect is a temporary, complete shift of electrons in a double or triple bond that happens only when another chemical tries to react with it. (c) Hyperconjugation is when electrons from a single bond spread out into an empty or nearby double bond area, helping to stabilize the molecule.

🎯 Exam Tip: For each effect, define it, state if it's permanent or temporary, explain the direction of electron displacement, and provide typical examples of groups or conditions that cause it. Remember hyperconjugation involves sigma electrons delocalizing into pi or empty p orbitals.

 

Question 49. Write the chemistry for qualitative analysis of nitrogen, sulphur and bromine.
Answer: Lassaigne's test is a common method used to find out if nitrogen, sulfur, or bromine are present in an organic compound. The test involves these main steps:
Step 1: Preparation of Sodium Fusion Extract
A small amount of the organic substance is heated strongly (fused) with a tiny piece of sodium metal in a special glass tube (fusion tube). Once it becomes red hot, the tube is quickly put into distilled water. The mixture is then boiled for a few minutes, cooled, and filtered. The clear liquid obtained is known as the sodium fusion extract (or Lassaigne extract), and it is usually alkaline.
Step 2: Detection of Elements using Sodium Fusion Extract
During the fusion reaction, the elements in the organic compound react with sodium as follows:
If nitrogen is present: \( \text{Na} + \text{C} + \text{N} \rightarrow \text{NaCN} \)
If sulfur is present: \( 2\text{Na} + \text{S} \rightarrow \text{Na}_2\text{S} \)
If both nitrogen and sulfur are present: \( \text{Na} + \text{S} + \text{C} + \text{N} \rightarrow \text{NaSCN} \)
If bromine is present: \( \text{Na} + \text{Br} \rightarrow \text{NaBr} \)

(i) Test for Nitrogen: To test for nitrogen, the sodium fusion extract is boiled with iron (II) sulfate and then made acidic by adding concentrated sulfuric acid (\( \text{H}_2\text{SO}_4 \)). If nitrogen is present, a distinct Prussian blue color will form, confirming its presence.
The chemical reactions involved are:
\( 6\text{NaCN} + \text{FeSO}_4 \rightarrow \text{Na}_4[\text{Fe(CN)}_6] + \text{Na}_2\text{SO}_4 \) (Sodium ferrocyanide is formed)
\( 3\text{Na}_4[\text{Fe(CN)}_6] + 4\text{FeCl}_3 \rightarrow \text{Fe}_4[\text{Fe(CN)}_6]_3 + 12\text{NaCl} \) (Ferric ferrocyanide, which is Prussian blue)

(iii) Test for both Nitrogen and Sulfur: If both nitrogen and sulfur are present in the compound, when the sodium fusion extract is treated with \( \text{Fe}^{3+} \) ions, a characteristic blood-red color appears. This color confirms the presence of both nitrogen and sulfur.
The reaction is:
\( \text{Fe}^{3+} + \text{SCN}^- \rightarrow [\text{Fe(SCN)}]^{2+} \) (A blood-red complex is formed)

(iv) Test for Bromine: To test for bromine, the sodium fusion extract is first made acidic using nitric acid. After this, silver nitrate solution is added. If bromine is present, a pale yellow solid (precipitate) will form. This pale yellow precipitate is partially soluble in ammonium hydroxide (\( \text{NH}_4\text{OH} \)), which helps confirm the presence of bromine.
The reaction is:
\( \text{NaBr} + \text{AgNO}_3 \rightarrow \text{AgBr} \downarrow + \text{NaNO}_3 \) (Silver bromide is a pale yellow precipitate)
In simple words: Lassaigne's test is a way to find nitrogen, sulfur, and bromine in organic substances. You burn the substance with sodium, then test the liquid (extract) with different chemicals to see specific color changes or solids that show if these elements are there.

🎯 Exam Tip: Remember the specific color changes and solubility patterns for each element's confirmation test, especially the Prussian blue for nitrogen and the pale yellow precipitate (partially soluble in \( \text{NH}_4\text{OH} \)) for bromine. These are key for scoring marks.

 

Question 50. Explain Kjeldahl and Duma's method. Derive the expression for percentage of nitrogen.
Answer:
Kjeldahl's Method:
This method is used to determine the amount of nitrogen in organic compounds. It works for compounds that convert completely into ammonium sulfate when heated strongly with concentrated sulfuric acid. However, it cannot be used for organic compounds where nitrogen is part of a ring structure (like in pyridine) or for groups such as \( \text{-NO}_2 \) and \( \text{-N=N-} \).
The method involves the following steps:
(i) Digestion: A known mass (about 0.5 g) of the organic compound is heated with concentrated sulfuric acid (\( \text{H}_2\text{SO}_4 \)) in a Kjeldahl's flask. Small amounts of potassium sulfate (\( \text{K}_2\text{SO}_4 \)) and copper sulfate (\( \text{CuSO}_4 \)) are added. Potassium sulfate helps increase the boiling point of the sulfuric acid, and copper sulfate acts as a catalyst for the reaction. After 3 to 4 hours, the organic compound completely breaks down to form ammonium sulfate.
The reaction is:
Organic compound (containing nitrogen) \( + \text{H}_2\text{SO}_4 \xrightarrow{\text{Cu}^{2+}, \text{heat}} (\text{NH}_4)_2\text{SO}_4 \)

(ii) Distillation: The ammonium sulfate obtained from the digestion step is then heated with an excess amount of a strong base, such as sodium hydroxide (\( \text{NaOH} \)). This process releases ammonia gas (\( \text{NH}_3 \)). The ammonia gas produced is then carefully absorbed into a known excess quantity of a standard acid, like sulfuric acid.
The reactions are:
\( (\text{NH}_4)_2\text{SO}_4 + 2\text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2\text{NH}_3 + 2\text{H}_2\text{O} \)
\( 2\text{NH}_3 + \text{H}_2\text{SO}_4 \text{ (excess acid)} \rightarrow (\text{NH}_4)_2\text{SO}_4 \)
The unreacted acid is then measured by titrating it against a standard alkali. By knowing how much acid reacted with the ammonia, the amount of ammonia and, consequently, the mass of nitrogen can be calculated.

Expression for Percentage of Nitrogen (Kjeldahl's Method):
Let,
Weight of the organic compound \( = \text{W g} \)
Normality of the standard acid solution \( = \text{N} \)
Volume of acid used up by ammonia \( = \text{V mL} \)
The mass of nitrogen in \( \text{V mL} \) of standard acid is:
Mass of nitrogen \( = \frac{14 \times \text{N} \times \text{V}}{1000} \text{ g} \)
Therefore, the percentage of nitrogen in the organic sample is:
Percentage of nitrogen \( = \frac{\text{Mass of nitrogen}}{\text{Mass of organic compound}} \times 100 \)
\( \implies \) Percentage of nitrogen \( = \frac{14 \times \text{N} \times \text{V}}{1000} \times \frac{100}{\text{W}} \)
\( \implies \) Percentage of nitrogen \( = \frac{1.4 \times \text{N} \times \text{V}}{\text{W}} \)

Duma's Method:
Duma's method can be used to determine the percentage of nitrogen in all types of organic compounds. In this method, a known mass of the organic compound is heated with copper oxide in an atmosphere of carbon dioxide. The carbon and hydrogen present in the compound are converted to carbon dioxide and water, respectively, while nitrogen gas is released. Any nitrogen oxides formed during the process are reduced back to free nitrogen by passing them over heated copper gauze.
The resulting mixture of gases (carbon dioxide, water vapor, and nitrogen) is collected over an aqueous solution of potassium hydroxide. The potassium hydroxide solution absorbs all gases except nitrogen. Thus, the volume of nitrogen gas collected is measured. From this measured volume, the percentage of nitrogen in the compound can be calculated.

Expression for Percentage of Nitrogen (Duma's Method):
Let,
Mass of the organic compound \( = \text{W g} \)
Volume of nitrogen gas collected \( = \text{V}_1 \text{ mL} \)
Atmospheric pressure \( = \text{P mm Hg} \)
Vapor pressure of water at the collection temperature \( = \text{p mm Hg} \)
Temperature at which gas is collected \( = \text{T}_1 \text{ K} \)

First, calculate the actual pressure of nitrogen gas:
Pressure of \( \text{N}_2 \) gas, \( \text{P}_1 = (\text{P} - \text{p}) \text{ mm Hg} \)

Next, convert the volume of nitrogen to Standard Temperature and Pressure (STP):
Volume of nitrogen at STP, \( \text{V} = \frac{\text{P}_1 \times \text{V}_1 \times 273}{760 \times \text{T}_1} \text{ mL} \)

Then, calculate the mass of this volume of nitrogen at STP. (Remember that 1 mole of \( \text{N}_2 = 28 \text{ g} = 22400 \text{ mL} \) at STP).
Mass of \( \text{V mL} \) of nitrogen at STP \( = \frac{28 \times \text{V}}{22400} \text{ g} \)

Finally, calculate the percentage of nitrogen in the compound:
Percentage of nitrogen \( = \frac{\text{Mass of nitrogen}}{\text{Mass of organic compound}} \times 100 \)
\( \implies \) Percentage of nitrogen \( = \frac{28 \times \text{V}}{22400} \times \frac{100}{\text{W}} \)
In simple words: Kjeldahl's method finds nitrogen by turning it into ammonia, which is then measured with acid. Duma's method directly measures nitrogen gas released when a substance is burned. Both methods use calculations to find the percentage of nitrogen in the original sample.

🎯 Exam Tip: Pay close attention to the conditions and limitations of each method (e.g., Kjeldahl's method doesn't work for ring nitrogens). Ensure you can write out both the reaction equations and the final percentage calculation formulas accurately.

 

Question 51. Explain the formation, stability and geometry of the following reaction intermediates : carbanion, carbene and free radicals.
Answer:
Here's an explanation of the formation, stability, and geometry of the given reaction intermediates:

1. Carbanion:
A carbanion is a negatively charged ion where a carbon atom carries an unshared pair of electrons, giving it a negative charge.
Formation: Carbanions usually form when a bond to a carbon atom breaks unevenly (heterolytically), and the carbon atom takes both electrons from the broken bond. This leaves the carbon with a negative charge and an extra pair of electrons.
Stability: There are three main types of carbanions: primary, secondary, and tertiary. The stability of carbanions decreases as the number of electron-donating alkyl groups attached to the negatively charged carbon increases. This is because electron-donating groups push more electron density onto the already electron-rich carbon, making it less stable.
The general order of stability for carbanions is:
Primary > Secondary > Tertiary
(Most stable)                       (Least stable)
For example, a methyl carbanion (\( \text{CH}_3^- \)) is more stable than a primary carbanion, which is more stable than a secondary, and a tertiary carbanion (\( (\text{CH}_3)_3\text{C}^- \)) is the least stable.
Geometry: The carbon atom in a carbanion is \( \text{sp}^3 \) hybridized. Its geometry is pyramidal, similar to that of an ammonia molecule. The lone pair of electrons occupies one of the \( \text{sp}^3 \) hybrid orbitals, pushing the other three bonds into a pyramid shape.

2. Carbene:
A carbene is a neutral molecule containing a carbon atom with two bonds and two unshared valence electrons.
Formation: Carbenes are typically generated by the breakdown (photolysis or thermolysis) of aliphatic diazo compounds or ketenes.
For example: \( \text{CH}_2\text{N}_2 \text{ (diazomethane)} \xrightarrow{\text{h}\nu} :\text{CH}_2 \text{ (carbene)} + \text{N}_2 \)
Stability: Carbenes exist in two electronic states: singlet and triplet. Both forms are highly reactive intermediates. Generally, the triplet state is considered more stable than the singlet state because of Hund's rule, which favors maximum spin multiplicity.
Geometry:
* In the **singlet state**, the carbon atom is \( \text{sp}^2 \) hybridized. It has a bent (V-shaped) geometry with a bond angle of about \( 120^\circ \), and the two unshared electrons are paired in one \( \text{sp}^2 \) orbital.
* In the **triplet state**, the carbon atom is \( \text{sp} \) hybridized. It has a linear geometry with a bond angle of \( 180^\circ \), and the two unshared electrons are unpaired, each occupying a separate \( \text{p} \) orbital.

3. Free Radicals:
Free radicals are chemical species (atoms or molecules) that contain at least one unpaired electron in their outermost shell. They are usually neutral but are highly reactive.
Formation: Free radicals are formed through homolytic fission, where a covalent bond breaks evenly, with each atom receiving one electron from the shared pair. This process often occurs under UV light or high temperatures.
Stability: Free radicals are classified as primary, secondary, and tertiary based on the number of alkyl groups attached to the carbon atom with the unpaired electron. The stability of free radicals increases with the number of electron-donating alkyl groups (due to inductive effect and hyperconjugation), which help delocalize the unpaired electron.
The general order of stability for free radicals is:
Tertiary > Secondary > Primary > Methyl
(Most stable)                                   (Least stable)
For instance, a tertiary butyl radical \( ((\text{CH}_3)_3\text{C}^\cdot) \) is more stable than a methyl radical \( (\text{CH}_3^\cdot) \).
Geometry: The carbon atom in alkyl free radicals is typically \( \text{sp}^2 \) hybridized. Its geometry is trigonal planar, with a bond angle of approximately \( 120^\circ \). The unpaired electron resides in an unhybridized \( \text{p} \) orbital.
In simple words: These are temporary, very reactive chemical parts that form during reactions. Carbanions have a negative carbon, are shaped like a pyramid, and are more stable if fewer groups push electrons onto them. Carbenes have a carbon with two extra electrons, can be flat or straight, and are usually made with light. Free radicals have an unpaired electron, are flat, and are more stable if more groups help spread out that electron.

🎯 Exam Tip: When describing reaction intermediates, always include details about their charge (if any), how they are formed, their relative stability (with an order), and their geometric shape and hybridization. Examples of compounds for each intermediate type can also be very helpful.

 

Question 52. (a) Write short notes on reactions and stability of free radials.
(b) Write the method for preparation and stability of carbocation.
Answer:
(a) Free Radicals:
Free radicals are chemical species that have a single unpaired electron in their outermost orbital. They are electrically neutral and are very reactive, always looking for another electron to form a pair. They combine easily with each other or with other molecules to achieve this stable electron pair.

**Reactions of Free Radicals:**
**(i) Recombination into Hydrocarbons:** Free radicals can combine with each other to form larger, stable hydrocarbon molecules.
Example: \( \text{CH}_3^\cdot + \text{CH}_3^\cdot \rightarrow \text{CH}_3\text{-CH}_3 \text{ (Ethane)} \)

**Stability of Free Radicals:** Free radicals are categorized as primary, secondary, or tertiary. The stability of free radicals increases with the number of alkyl groups attached to the carbon atom that holds the unpaired electron. This is because alkyl groups are electron-donating, helping to spread out the electron density of the unpaired electron through inductive effect and hyperconjugation. So, tertiary free radicals are the most stable, while primary and methyl radicals are less stable.
Order of stability: Tertiary > Secondary > Primary > Methyl

(b) Carbocation:
A carbocation is an ion where a carbon atom has a positive charge, forming only three bonds, and is therefore electron-deficient. Carbocations are highly unstable and reactive intermediates. They are generally formed when a bond breaks unevenly (heterolytic fission), with one atom taking both electrons from the bond.

**Methods for Preparation of Carbocations:**
**(i) Direct Ionization:** Compounds like alkyl halides can undergo direct ionization, where a halide ion leaves, resulting in the formation of a carbocation.
Example: \( (\text{CH}_3)_3\text{C-Cl} \rightarrow (\text{CH}_3)_3\text{C}^+ + \text{Cl}^- \)
**(ii) Protonation of Unsaturated Compounds:** Unsaturated compounds (those with double or triple bonds) can react with a proton (\( \text{H}^+ \)) from an acid. The proton adds to one carbon of the multiple bond, leaving the other carbon with a positive charge, forming a carbocation.
Example: \( \text{-CH=CH-} + \text{H}^+ \rightarrow \text{-CH-CH}_2^+ \)
\( \implies \text{C=O} + \text{H}^+ \rightarrow \text{C-OH}^+ \)

**Stability of Carbocations:** Alkyl groups attached to the positively charged carbon help to stabilize carbocations. This stabilization occurs through both the inductive effect (where alkyl groups donate electrons) and hyperconjugation. The more alkyl groups that are connected to the carbon bearing the positive charge, the more stable the carbocation will be.
The relative order of stability for carbocations is:
Tertiary > Secondary > Primary > Methyl
(Most stable)                                   (Least stable)
In simple words: Free radicals are super reactive atoms with one lonely electron; they react by finding another electron. Tertiary free radicals are the most stable. Carbocations are positively charged carbons that form when bonds break unevenly or with acid. They are more stable when surrounded by more groups that push electrons towards them.

🎯 Exam Tip: For both free radicals and carbocations, always discuss their formation, their stability order with a brief reason (like inductive effect or hyperconjugation), and provide a relevant chemical example or reaction for each.

 

Question 54. Explain IUPAC system of nomenclature with examples.
Answer: The **IUPAC System of Nomenclature** provides a standardized, worldwide method for naming chemical compounds. IUPAC stands for the International Union of Pure and Applied Chemistry. This system was first introduced in 1947 and has been updated over the years to keep up with new chemical discoveries. Its most detailed rules were published in 1979 and revised in 1993.

A full IUPAC name for an organic compound typically consists of four main parts:
1.  **Root Word:** Indicates the number of carbon atoms in the longest continuous carbon chain.
2.  **Suffix(es):** Tells about the nature of carbon-carbon bonds (single, double, triple) and the main functional group.
3.  **Prefix(es):** Describes any side chains, substituents, or other functional groups that are not the highest priority.
4.  **Infix:** Used for special structural features like cyclic, spiro, or bicyclic arrangements.
The complete systematic IUPAC name can generally be represented as:
**Prefix + Infix + Root word + Primary Suffix + Secondary Suffix**

Let's look at each part in more detail:

**1. Root Word:**
The root word specifies the number of carbon atoms in the principal (longest continuous) carbon chain of the compound.

**2. Suffix(es):** These are divided into two categories:
**(i) Primary Suffix:** This suffix is added right after the root word. It shows the presence of saturation (single bonds) or unsaturation (double or triple bonds) within the main carbon chain.

**(ii) Secondary Suffix:** This suffix is added after the primary suffix. It indicates the primary functional group present in the organic compound. If there are multiple functional groups, the one with the highest priority determines the secondary suffix, while others are named using prefixes.

**3. Prefix(es):**
Prefixes are used to indicate side chains, other substituents, and functional groups that are considered less important than the main functional group (i.e., lower priority). They are added before the root word or infix.

It's important to remember that alkyl groups, halogens, nitro, and alkoxy groups are considered to have lower priority compared to double and triple bonds.

**4. Infix:**
Infixes are special terms placed between the prefix and the root word to describe unique structural features of the main carbon chain. They are sometimes also referred to as primary prefixes.
* The "**Cyclo**" infix is used when the parent carbon chain forms a ring structure (cyclic compound).
* The "**Spiro**" infix is used for spiro compounds, which are molecules where two rings share only one common carbon atom.
* The "**Bicyclo**" infix is used for bicyclic compounds, where two rings share two or more carbon atoms.

**Steps for Writing an IUPAC Name:**
To name an organic compound systematically, follow these steps:
1.  **Select the Parent Chain and Root Word:** Find the longest continuous carbon chain that contains the main functional group and/or multiple bonds. This chain determines the root word.
2.  **Add Primary Suffix:** Attach the appropriate primary suffix (-ane, -ene, -yne) to the root word based on the presence and number of single, double, or triple bonds in the parent chain.
3.  **Add Secondary Suffix (if applicable):** If the molecule has a functional group, add its secondary suffix after the primary suffix. If there are multiple functional groups, choose the one with the highest priority as the main group.
4.  **Add Infix (if applicable):** If the compound is cyclic, spiro, or bicyclic, insert the corresponding infix (cyclo-, spiro-, or bicyclo-) before the root word.
5.  **Add Prefixes:** Finally, include prefixes for all side chains and other substituents. These prefixes should be listed alphabetically, with their positions indicated by numbers (locants).

Example: The IUPAC name for \( \text{CH}_3\text{-CH(CH}_3\text{)-CH}_2\text{-CH}_2\text{-OH} \) is 3-Methylbutan-1-ol.
In simple words: The IUPAC system gives a standard name to every chemical compound using rules. You build the name from a root word (number of carbons), suffixes (for bonds and main chemical group), prefixes (for side groups), and sometimes an infix (for rings). You follow a set order to put these parts together correctly.

🎯 Exam Tip: Practice naming compounds by breaking them down into root word, suffixes, and prefixes. Always identify the longest carbon chain and the highest priority functional group first. Knowing the common prefixes and suffixes by heart is essential.