Samacheer Kalvi Class 11 Chemistry Solutions Chapter 13 Hydrocarbons

Textbook Evaluation:

I. Choose the Best Answer:

 

Question 1. The correct statement regarding the comparison of staggered and eclipsed conformations of ethane, is
(a) The eclipsed conformation of ethane is more stable than staggered conformation even though the eclipsed conformation has torsional strain.
(b) The staggered conformation of ethane is more stable than eclipsed conformation, because staggered conformation has no torsional strain.
(c) The staggered conformation of ethane is less stable than eclipsed conformation, because staggered conformation has torsional strain.
(d) The staggered conformation of ethane is less stable than eclipsed conformation, because staggered conformation has no torsional strain.
Answer: (b) The staggered conformation of ethane is more stable than eclipsed conformation, because staggered conformation has no torsional strain.
In simple words: The staggered form of ethane is more stable because its atoms are further apart, reducing stress between them. This makes it a lower energy arrangement.

🎯 Exam Tip: Remember that staggered conformations have minimal torsional strain due to hydrogen atoms being as far apart as possible, making them more stable than eclipsed conformations.

 

Question 2. \( \text{C}_2\text{H}_5\text{Br} + \text{2Na} \xrightarrow{\text{dry ether}} \text{C}_4\text{H}_{10} + \text{2NaBr} \). The above reaction is an example of which of the following
(a) Reimer Tiemann reaction
(b) Wurtz reaction
(c) Aldol condensation
(d) Hoffmann reaction
Answer: (b) Wurtz reaction
In simple words: This reaction is called a Wurtz reaction, where two alkyl halides join together to form a larger alkane with the help of sodium metal in dry ether. It's a way to double the carbon chain.

🎯 Exam Tip: Identify Wurtz reactions by the presence of an alkyl halide reacting with sodium metal in dry ether to form a symmetrical alkane.

 

Question 3. An alkyl bromide (A) reacts with sodium in ether to form 4, 5 – diethyloctane, the compound (A) is
(a) \( \text{CH}_3(\text{CH}_2)_3\text{Br} \)
(b) \( \text{CH}_3(\text{CH}_2)_5\text{Br} \)
(c) \( \text{CH}_3(\text{CH}_2)_3\text{CH(Br)CH}_3 \)
(d) (Image of 3-Bromohexane: \( \text{CH}_3(\text{CH}_2)_2\text{CH(Br)CH}_2\text{CH}_3 \) with a bromine atom on the third carbon)
Answer: (d) (Image of 3-Bromohexane: \( \text{CH}_3(\text{CH}_2)_2\text{CH(Br)CH}_2\text{CH}_3 \) with a bromine atom on the third carbon)
In simple words: To make 4,5-diethyloctane using a Wurtz reaction, you need two identical alkyl bromides that combine. If the final product has 10 carbons and two ethyl groups, the starting bromide must be 3-bromohexane, which is half of the final symmetrical molecule.

🎯 Exam Tip: For Wurtz reactions forming symmetrical alkanes, the alkyl halide must be half of the final carbon chain with the leaving group at the site of new bond formation.

 

Question 4. The C – H bond and C – C bond in ethane are formed by which of the following types of overlap
(a) \( \text{sp}^3 - \text{s} \) and \( \text{sp}^3 - \text{sp}^3 \)
(b) \( \text{sp}^2 - \text{s} \) and \( \text{sp}^2 - \text{sp}^2 \)
(c) \( \text{sp} - \text{sp} \) and \( \text{sp} - \text{sp} \)
(d) \( \text{p} - \text{s} \) and \( \text{p} - \text{p} \)
Answer: (a) \( \text{sp}^3 - \text{s} \) and \( \text{sp}^3 - \text{sp}^3 \)
In simple words: In ethane, carbon atoms are \( \text{sp}^3 \) hybridized. This means the C-H bonds are formed by the overlap of an \( \text{sp}^3 \) orbital from carbon and an s orbital from hydrogen. The C-C bond is formed by the overlap of \( \text{sp}^3 \) orbitals from each carbon.

🎯 Exam Tip: Remember that alkanes have all single bonds, leading to \( \text{sp}^3 \) hybridization for their carbon atoms.

 

Question 5. In the following reaction, (Image of cyclohexane with a \( \text{CH}_3 \) group and a \( \text{Br} \) group on adjacent carbons, reacting with \( \text{Br}_2/\text{h}\nu \) to produce several options.) Answer:
(a) (Image of 1-bromo-1-methylcyclohexane where Br and CH3 are on the same carbon)
(b) (Image of 1-bromo-2-methylcyclohexane, with Br on a different carbon than CH3)
(c) (Image of 1-bromo-3-methylcyclohexane)
(d) (Image of 1-bromo-4-methylcyclohexane)
Answer: (b) (Image of 1-bromo-2-methylcyclohexane, with Br on a different carbon than CH3)
In simple words: When a methylcyclohexane reacts with bromine in the presence of light, the bromine attacks the most stable carbon radical. In this case, it adds to the carbon atom next to the methyl group, leading to 1-bromo-2-methylcyclohexane as a major product.

🎯 Exam Tip: Free radical halogenation reactions like this prefer to form the most stable radical intermediate, typically at a tertiary or secondary carbon position.

 

Question 6. Which of the following is optically active
(a) 2 – methyl pentane
(b) citric acid
(c) Glycerol
(d) none of these
Answer: (a) 2 – methyl pentane
In simple words: A molecule is optically active if it has a chiral center, which is a carbon atom bonded to four different groups. 2-methyl pentane has such a carbon, making it optically active. Citric acid and glycerol do not have a chiral center.

🎯 Exam Tip: To identify an optically active compound, look for a carbon atom bonded to four unique substituents.

 

Question 7. The compound formed at anode in the electrolysis of an aqueous solution of potassium acetate are
(a) \( \text{CH}_4 \) and \( \text{H}_2 \)
(b) \( \text{CH}_4 \) and \( \text{CO}_2 \)
(c) \( \text{C}_2\text{H}_6 \) and \( \text{CO}_2 \)
(d) \( \text{C}_2\text{H}_4 \) and \( \text{Cl}_2 \)
Answer: (c) \( \text{C}_2\text{H}_6 \) and \( \text{CO}_2 \)
In simple words: When potassium acetate is electrolyzed, the acetate ions lose electrons at the anode, combining to form ethane gas and carbon dioxide gas. This process is a common way to make alkanes from carboxylic acid salts.

🎯 Exam Tip: Recall Kolbe's electrolytic method, where the carboxylate ions decarboxylate and couple at the anode to form an alkane and carbon dioxide.

 

Question 8. The general formula for cyclo alkanes
(a) \( \text{C}_\text{n}\text{H}_\text{n} \)
(b) \( \text{C}_\text{n}\text{H}_{2\text{n}} \)
(c) \( \text{C}_\text{n}\text{H}_{2\text{n}-2} \)
(d) \( \text{C}_\text{n}\text{H}_{2\text{n}+2} \)
Answer: (b) \( \text{C}_\text{n}\text{H}_{2\text{n}} \)
In simple words: Cycloalkanes are ring structures made of carbon atoms, and each carbon has two hydrogens attached, giving them the general formula \( \text{C}_\text{n}\text{H}_{2\text{n}} \). This is the same formula as alkenes, but cycloalkanes have all single bonds.

🎯 Exam Tip: Remember that cycloalkanes have two fewer hydrogen atoms than their open-chain alkane counterparts due to the formation of a ring.

 

Question 9. The compound that will react most readily with gaseous bromine has the formula
(a) \( \text{C}_3\text{H}_6 \)
(b) \( \text{C}_2\text{H}_2 \)
(c) \( \text{C}_4\text{H}_{10} \)
(d) \( \text{C}_2\text{H}_4 \)
Answer: (a) \( \text{C}_3\text{H}_6 \)
In simple words: Alkenes, which have double bonds, react very quickly with bromine through an addition reaction, causing the bromine's color to disappear. Propene \( (\text{C}_3\text{H}_6) \) is an alkene and will react readily.

🎯 Exam Tip: Unsaturated hydrocarbons (alkenes and alkynes) undergo addition reactions much more readily than saturated ones (alkanes) with halogens.

 

Question 10. Which of the following compounds shall not produce propene by reaction with HBr followed by elimination (or) only direct elimination reaction
(a) (Image of propan-1-ol: \( \text{CH}_3\text{CH}_2\text{CH}_2\text{OH} \))
(b) \( \text{CH}_3 – \text{CH}_2 – \text{CH}_2 – \text{OH} \)
(c) \( \text{H}_2\text{C} = \text{C} = \text{O} \)
(d) \( \text{CH}_3 – \text{CH}_2 – \text{CH}_2\text{Br} \)
Answer: (c) \( \text{H}_2\text{C} = \text{C} = \text{O} \)
In simple words: To make propene by an elimination reaction, the starting molecule needs to have at least three carbons and a functional group that can be removed. \( \text{H}_2\text{C} = \text{C} = \text{O} \) (ketene) only has two carbons and cannot form propene through elimination.

🎯 Exam Tip: Elimination reactions require a suitable leaving group and adjacent hydrogen atoms to form an alkene; make sure the carbon backbone is appropriate for the desired product.

 

Question 11. Which among the following alkenes on reductive ozonolysis produces only propanone?
(a) 2 – Methyl propene
(b) 2 – Methyl but – 2 – ene
(c) 2, 3 – Dimethyl but – 1- ene
(d) 2, 3 – Dimethyl but – 2 – ene
Answer: (d) 2, 3 – Dimethyl but – 2 – ene
In simple words: Reductive ozonolysis breaks the double bond of an alkene and forms carbonyl compounds. For only propanone to be formed, the original alkene must be 2,3-dimethylbut-2-ene, which breaks into two identical propanone molecules.

🎯 Exam Tip: In reductive ozonolysis, each carbon of the double bond becomes a carbonyl carbon. To yield only one type of ketone (propanone), the double bond must be symmetrically substituted with methyl groups.

 

Question 12. The major product formed when 2 – bromo – 2 – methyl butane is refluxed with ethanolic KOH is
(a) 2 – methylbut – 2- ene
(b) 2 – methyl butan – 1 – ol
(c) 2 – methyl but – 1 – ene
(d) 2 – methyl butan – 2- ol
Answer: (a) 2 – methylbut – 2- ene
In simple words: When 2-bromo-2-methylbutane reacts with alcoholic KOH, it undergoes an elimination reaction, removing a hydrogen and a bromine to form an alkene. According to Saytzeff's rule, the more substituted alkene, 2-methylbut-2-ene, will be the main product.

🎯 Exam Tip: Elimination reactions with alcoholic KOH favor the formation of the more stable (more substituted) alkene, as predicted by Saytzeff's rule.

 

Question 13. Major product of the below mentioned reaction is, \( (\text{CH}_3)_2\text{C} = \text{CH}_2 \xrightarrow{\text{HCl}} \)
(a) 2 – chloro – 1 – iode – 2 – methyl propane
(b) 1 – chloro – 2 – iodo – 2 – methyl propane
(c) 1, 2 – dichloro – 2 – methyl propane
(d) 1, 2 – diiodo – 2 – methyl propane
Answer: (a) 2 – chloro – 1 – iode – 2 – methyl propane
In simple words: When isobutylene \( ((\text{CH}_3)_2\text{C} = \text{CH}_2) \) reacts with HCl, the hydrogen adds to the carbon with more hydrogens, and the chlorine adds to the carbon with fewer hydrogens, following Markovnikov's rule. This forms 2-chloro-2-methylpropane.

🎯 Exam Tip: Remember Markovnikov's rule: in the addition of HX to an unsymmetrical alkene, the hydrogen adds to the carbon with more hydrogens, and the halogen adds to the carbon with fewer hydrogens.

 

Question 14. The IUPAC name of the following compound is (Image of 2-chloro-3-iodo-2-pentene, trans isomer)
(a) trans – 2- chloro – 3- iodo – 2- pentane
(b) cis – 3- iodo – 4 chloro – 3 – pentane
(c) trans – 3 – iodo – 4 – chloro – 3 – pentene
(d) cis – 2 – chloro – 3 iodo – 2 – pentene
Answer: (a) trans – 2- chloro – 3- iodo – 2- pentane
In simple words: The compound shown is a pentene with a double bond at the second carbon. It has a chlorine and iodine atom attached, making it 2-chloro-3-iodo-2-pentene. Because the substituents on the double bond are opposite, it's the trans isomer.

🎯 Exam Tip: When naming alkenes, prioritize the double bond for numbering the parent chain. Determine cis/trans based on the positions of identical or priority groups around the double bond.

 

Question 15. Cis – 2- butene and trans – 2 – butene are
(a) conformational isomers
(b) structural isomers
(c) configurational isomers
(d) optical isomers
Answer: (c) configurational isomers
In simple words: Cis-2-butene and trans-2-butene are types of isomers where the arrangement of atoms can only be changed by breaking and reforming chemical bonds, not by simple rotation. These are known as configurational isomers.

🎯 Exam Tip: Configurational isomers (like cis-trans isomers) can be interconverted only by breaking and reforming bonds, unlike conformational isomers that interconvert by bond rotation.

 

Question 16. Identify the compound (A) in the following reaction (Image of cyclohexene reacting with \( \text{CH}_3\text{CHO} \) and \( \text{O}_3 \), followed by \( \text{Zn}/\text{H}_2\text{O} \) to produce a dicarbonyl compound with 6 carbons, plus Compound A. The dicarbonyl is 1,6-hexanedial, indicating ring opening.)
(a) (Image of an aldehyde with a phenyl group: benzaldehyde)
(b) (Image of an aldehyde with a cyclohexyl group: cyclohexanecarbaldehyde)
(c) (Image of a phenol with a methyl group: cresol)
(d) (Image of an aldehyde with a phenyl group: benzaldehyde)
Answer: (a) (Image of an aldehyde with a phenyl group: benzaldehyde)
In simple words: This is an ozonolysis reaction. Cyclohexene breaks open at the double bond and forms a linear molecule with two aldehyde groups. The byproduct (A) is formaldehyde, which results from the cleavage of the \( \text{CH}_3\text{CHO} \) if it were part of the initial olefin, but the reactant is cyclohexene + CH3CHO, implying that CH3CHO is a reagent. Based on the reaction and expected products, benzaldehyde (a) appears to be an incorrect interpretation in the OCR of the options provided. However, following the OCR's answer, it refers to benzaldehyde. If the question implies addition of \( \text{CH}_3\text{CHO} \) (acetaldehyde) or a similar reagent, then the options are misleading. Assuming the reaction is ozonolysis of cyclohexene forming 1,6-hexanedial, then "Compound A" is not explicitly formed as a distinct organic product unless the question implies ozonolysis of an unknown alkene forming 1,6-hexanedial AND benzaldehyde. Given the specific options provided as images, option (a) is benzaldehyde. Without further context or a clear reaction pathway for "compound A," it's difficult to resolve the discrepancy, so I will stick to the provided answer which points to image (a) (benzaldehyde). For this specific problem, based on the image description, it appears to be a misidentified option or a complex multi-step reaction not fully shown. But for now, as given, (a) is chosen.

🎯 Exam Tip: Ozonolysis is a powerful reaction for breaking carbon-carbon double bonds and forming carbonyl compounds. Pay close attention to the structures of reactants and products.

 

Question 17. (Image of 1,2-dibromoethane reacting to form ethyne: \( \text{CH}_2\text{Br}-\text{CH}_2\text{Br} \xrightarrow{} \text{CH} \equiv \text{CH} \), where A is)
(a) Zn
(b) conc. \( \text{H}_2\text{SO}_4 \)
(c) alc. KOH
(d) dil. \( \text{H}_2\text{SO}_4 \)
Answer: (c) alc. KOH
In simple words: To convert 1,2-dibromoethane into ethyne (acetylene), you need to remove two molecules of HBr. Alcoholic potassium hydroxide (alc. KOH) is used for this type of double dehydrohalogenation, which forms a triple bond.

🎯 Exam Tip: Alcoholic KOH is a strong base commonly used for dehydrohalogenation reactions to form alkenes and alkynes. Remember that a double dehydrohalogenation is needed to go from a dihaloalkane to an alkyne.

 

Question 18. Consider the nitration of benzene using mixed con \( \text{H}_2\text{SO}_4 \) and \( \text{HNO}_3 \) if a large quantity of \( \text{KHSO}_4 \) is added to the mixture, the rate of nitration will be
(a) unchanged
(b) doubled
(c) faster
(d) slower
Answer: (d) slower
In simple words: The nitration of benzene is an acid-catalyzed reaction. Adding a large amount of \( \text{KHSO}_4 \) introduces \( \text{HSO}_4^- \) ions, which consume the \( \text{H}^+ \) ions needed for the reaction. This makes the reaction slower.

🎯 Exam Tip: The nitration of benzene is an electrophilic aromatic substitution reaction where the electrophile \( \text{NO}_2^+ \) is generated by the reaction of nitric acid with sulfuric acid. Any substance that reduces the concentration of the necessary acidic species will slow down the reaction.

 

Question 19. In which of the following molecules, all atoms are co-planar
(a) (Image of Benzene ring)
(b) (Image of Biphenyl, two benzene rings joined)
(c) (Image of Naphthalene, two fused benzene rings)
(d) both (a) and (b)
Answer: (d) both (a) and (b)
In simple words: Benzene itself is a perfectly flat molecule, so all its atoms are in one plane. Biphenyl, which is two benzene rings connected by a single bond, can rotate around that bond. However, in its most stable conformation, biphenyl is also essentially planar, allowing for co-planarity of all atoms, or near-planarity with slight twists. Naphthalene, with two fused rings, is also planar. Given the options, the intent is likely that benzene and biphenyl are considered planar.

🎯 Exam Tip: Benzene's aromaticity requires a planar structure. While biphenyl can have slight twists, it can adopt a planar or near-planar conformation, especially in solutions or crystalline forms, making (d) a plausible answer in this context.

 

Question 20. Propyne on passing through red hot iron tube gives
(a) (Image of 1,3,5-trimethylbenzene, mesitylene)
(b) (Image of 1,2,4-trimethylbenzene)
(c) (Image of 1,2,3-trimethylbenzene)
(d) none of these
Answer: (a) (Image of 1,3,5-trimethylbenzene, mesitylene)
In simple words: When propyne passes through a red-hot iron tube, three molecules of propyne undergo trimerization, combining to form a benzene ring with three methyl groups. This results in 1,3,5-trimethylbenzene, also known as mesitylene.

🎯 Exam Tip: The cyclization of alkynes in a red-hot iron tube is a common method to synthesize benzene and its derivatives. Remember that three molecules of the alkyne are required.

 

Question 21. (Image of allyl chloride, \( \text{CH}_2=\text{CH}-\text{CH}_2\text{Cl} \), reacting with HCl to form compound A) is
(a) (Image of 1,2-dichloropropane, \( \text{CH}_2\text{Cl}-\text{CHCl}-\text{CH}_3 \))
(b) (Image of 1,3-dichloropropane, \( \text{CH}_2\text{Cl}-\text{CH}_2-\text{CH}_2\text{Cl} \))
(c) both (a) and (b)
Answer: (d) (Image of 2-chloropropane with chlorine attached to the second carbon)
In simple words: When allyl chloride reacts with HCl, the chlorine atom adds to the most substituted carbon of the double bond (the middle carbon), and the hydrogen adds to the least substituted carbon (the end carbon). This forms 1,2-dichloropropane, but the options and answer are mixed here. Let me re-evaluate based on the provided answer (d) for the image which is 2-chloropropane. The original image of question 21 is CH2=CH-CH2-Benzene. The question in OCR is CH2=CH-CH2, with HCl, what is (A). If CH2=CH-CH2-Benzene is the reactant, it would be 1-phenylpropene. When 1-phenylpropene reacts with HCl, the chlorine would add to the carbon with fewer hydrogens, giving 2-chloro-1-phenylpropane. However, the image options show simple propane derivatives. Given that the image description for (d) is 2-chloropropane, and the reactant in the OCR image for the question is \( \text{CH}_2=\text{CH}-\text{CH}_2-\text{Cl} \) (allyl chloride) not \( \text{CH}_2=\text{CH}-\text{CH}_2 \). When allyl chloride reacts with HCl, it is an addition reaction. Hydrogen will add to \( \text{CH}_2 \) and Cl will add to \( \text{CH} \), giving \( \text{CH}_3-\text{CHCl}-\text{CH}_2\text{Cl} \) (1,2-dichloropropane). The question and options are inconsistent. I will use the OCR's provided structure for option (d) in the solution which is 1,2-dichloropropane, based on the answer key. So, the question should be \( \text{CH}_2=\text{CH}-\text{CH}_2\text{Cl} \) reacts with HCl. Option (d) in the OCR's answer shows the structure of 1,2-dichloropropane. Therefore, the answer should be (d).

🎯 Exam Tip: Be careful with addition reactions involving compounds that already have a halogen. Markovnikov's rule generally applies, and it's important to correctly identify the major product based on carbocation stability.

 

Question 22. Which one of the following is non aromatic?
(a) (Image of Benzene ring)
(b) (Image of Naphthalene, two fused benzene rings)
(c) (Image of Thiophene, a five-membered ring with one sulfur atom)
(d) (Image of Cyclopentadiene, a five-membered ring with two double bonds)
Answer: (d) (Image of Cyclopentadiene, a five-membered ring with two double bonds)
In simple words: Aromatic compounds follow Huckel's rule (4n+2 pi electrons) and are planar cyclic systems. Benzene, naphthalene, and thiophene are all aromatic. Cyclopentadiene, however, has only 4 pi electrons, so it is not aromatic.

🎯 Exam Tip: To determine if a compound is aromatic, check if it's cyclic, planar, fully conjugated, and contains (4n+2) pi electrons according to Huckel's rule.

 

Question 23. Which of the following compounds will not undergo Friedal – crafts reaction easily?
(a) Nitro benzene
(b) Toluene
(c) Cumene
(d) Xylene
Answer: (a) Nitro benzene
In simple words: Friedel-Crafts reactions are electrophilic aromatic substitutions. Nitrobenzene has a very strong electron-withdrawing nitro group, which deactivates the benzene ring significantly, making it very difficult for it to react in Friedel-Crafts reactions.

🎯 Exam Tip: Strong deactivating groups (like nitro, carbonyl, sulfonyl) on a benzene ring make it unreactive towards Friedel-Crafts alkylation and acylation reactions.

 

Question 24. Some meta-directing substituents in aromatic substitution are given. Which one is most deactivating?
(a) – COOH
(b) – \( \text{NO}_2 \)
(c) -C \( \equiv \) N
(d) -S \( \text{O}_3\text{H} \)
Answer: (b) – \( \text{NO}_2 \)
In simple words: Among the given meta-directing groups, the nitro group (\( \text{NO}_2 \)) is the most powerful electron-withdrawing group. This means it pulls electrons away from the benzene ring more strongly than the others, making the ring least reactive (most deactivating) towards electrophilic attack.

🎯 Exam Tip: Electron-withdrawing groups are deactivating and meta-directing. The strength of deactivation increases with the electron-withdrawing ability of the group.

 

Question 25. Which of the following can be used as the halide component for friedal – crafts reaction?
(a) Chloro benzene
(b) Bromo benzene
(c) Chloro ethene
(d) Isopropyl chloride
Answer: (d) Isopropyl chloride
In simple words: In Friedel-Crafts alkylation, the alkyl halide reacts with a Lewis acid catalyst to form a carbocation, which then attacks the benzene ring. Aryl halides (like chlorobenzene) and vinyl halides (like chloroethene) do not form stable carbocations, so they are generally not used in Friedel-Crafts reactions. Isopropyl chloride is a suitable alkyl halide.

🎯 Exam Tip: For Friedel-Crafts alkylation, the alkyl halide must be capable of forming a stable carbocation intermediate. Aryl and vinyl halides typically do not participate in these reactions.

 

Question 26. An alkane isobtainedbydecarboxylation of sodium propionate. Same alkane can be prepared by
(a) Catalytic hydrogenation of propene
(b) action of sodium metal on iodomethane
(c) reduction of 1 – chloro propane
(d) reduction of bromomethane
Answer: (b) action of sodium metal on iodomethane
In simple words: Decarboxylation of sodium propionate produces ethane \( (\text{CH}_3\text{CH}_3) \). To make the same alkane, ethane, using sodium metal, you can react two molecules of iodomethane with sodium. This is a Wurtz reaction.

🎯 Exam Tip: Decarboxylation of a carboxylic acid salt removes one carbon. Wurtz reaction combines two alkyl halides to form a symmetrical alkane, effectively doubling the carbon chain of the alkyl group.

 

Question 27. Which of the following is aliphatic saturated hydrocarbon
(a) \( \text{C}_8\text{H}_{18} \)
(b) \( \text{C}_9\text{H}_{18} \)
(c) \( \text{C}_8\text{H}_{14} \)
(d) All of the options
Answer: (a) \( \text{C}_8\text{H}_{18} \)
In simple words: Aliphatic saturated hydrocarbons are alkanes, which have the general formula \( \text{C}_\text{n}\text{H}_{2\text{n}+2} \). For n=8, \( \text{C}_8\text{H}_{18} \) fits this formula, indicating it is an alkane. The other options represent an alkene/cycloalkane and an alkyne.

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

 

Question 28. Identify the compound ‘Z’ in the following reaction (Image of \( \text{C}_2\text{H}_5\text{OH} \xrightarrow{\text{Al}_2\text{O}_3, 623 \text{K}} \text{X} \xrightarrow{\text{O}_3} \text{Y} \xrightarrow{\text{Zn}/\text{H}_2\text{O}} \text{Z} \))
(a) Formaldehyde
(b) Acetaldehyde
(c) Formic acid
(d) none of these
Answer: (a) Formaldehyde
In simple words: Starting with ethanol, \( \text{Al}_2\text{O}_3 \) at 623 K dehydrates it to ethene (X). Ozonolysis of ethene (Y) followed by reductive workup with \( \text{Zn}/\text{H}_2\text{O} \) cleaves the double bond, producing two molecules of formaldehyde (Z).

🎯 Exam Tip: This is a classic sequence: dehydration of alcohol to alkene, then ozonolysis of alkene to aldehyde/ketone. Ethene always yields formaldehyde upon ozonolysis.

 

Question 29. Peroxide effect (Kharasch effect) can be studied in case of
(a) Oct – 4 – ene
(b) hex – 3 – ene
(c) pent – 1 – ene
(d) but – 2 – ene
Answer: (c) pent – 1 – ene
In simple words: The peroxide effect (Kharasch effect) is observed only with unsymmetrical alkenes and HBr. Pent-1-ene is an unsymmetrical alkene, so it will show the peroxide effect when reacting with HBr, leading to anti-Markovnikov addition.

🎯 Exam Tip: Remember that the peroxide effect (anti-Markovnikov addition of HBr) only occurs with unsymmetrical alkenes and specifically with HBr, not HCl or HI.

 

Question 30. 2 – butyne on chlorination gives
(a) 1 – chloro butane
(b) 1, 2 – dichloro butane
(c) 1, 1, 2, 2 – tetrachlorobutane
(d) 2, 2, 3, 3 – tetra chloro butane
Answer: (d) 2, 2, 3, 3 – tetra chloro butane
In simple words: When 2-butyne, which has a triple bond, undergoes complete chlorination, each bond of the triple bond adds two chlorine atoms. This results in the addition of a total of four chlorine atoms, forming 2,2,3,3-tetrachlorobutane.

🎯 Exam Tip: Alkynes can undergo addition reactions twice across their triple bond. Complete halogenation of an alkyne results in a tetrahaloalkane.

II. Write Brief Answer to the Following Questions:

 

Question 31. Give IUPAC names for the following compounds
1) \( \text{CH}_3 – \text{CH} = \text{CH} – \text{CH} = \text{CH} – \text{C} \equiv \text{C} – \text{CH}_3 \)
2) (Image of 4, 5, 5 - trimethyl - 2 - heptyne)
3) \( (\text{CH}_3)_3\text{C} – \text{C} \equiv \text{C} – \text{CH}(\text{CH}_3)_2 \)
4) ethyl isopropyl acetylene
5) \( \text{CH} \equiv \text{C} – \text{C} \equiv \text{C} – \text{C} \equiv \text{CH} \)
Answer:
1) 2, 4-octadiene-6-yne. When both double and triple bonds are present, the longest chain containing both is chosen. Numbering gives the lowest locant to the multiple bond that appears first, or to the double bond if there's a tie.
2) 4,5,5-trimethylhept-2-yne. The longest chain with the triple bond is seven carbons. The triple bond is at position 2. Methyl groups are at positions 4, 5, and 5.
3) 2,2,5-trimethylhex-3-yne. The longest chain containing the triple bond is six carbons. The triple bond is at position 3. Methyl groups are at positions 2, 2, and 5.
4) 2,5-dimethylhex-3-yne. Ethyl isopropyl acetylene translates to a 6-carbon chain with a triple bond and two methyl groups.
5) Hexa-1,3,5-triyne. This molecule has six carbons with triple bonds at positions 1, 3, and 5.
In simple words: To name organic compounds, first find the longest carbon chain with the main functional groups like double or triple bonds. Then, number the carbons to give these groups the lowest possible numbers. Finally, list any side groups, making sure to include their positions and the type of bonds. When both double and triple bonds are present, usually the double bond gets preference for numbering if there's a tie in location, but the 'yne' suffix comes last if the double bond is not higher priority.

🎯 Exam Tip: For compounds with multiple functional groups, follow the IUPAC priority rules carefully. The longest chain containing the highest priority functional group determines the parent name.

 

Question 32. Identify the compound A, B, C and D in the following series of reactions
Answer:

The compounds A, B, C, and D are identified through a series of reactions. Starting from ethyl bromide, elimination reaction with alcoholic KOH yields ethylene (A). Ethylene reacts with \( \text{Cl}_2/\text{CCl}_4 \) to form 1,2-dichloroethane (B). Further reaction with \( \text{NaNH}_2 \) gives vinyl chloride (D). Additionally, ozonolysis of ethylene (A) produces formaldehyde (C). This sequence shows common reactions for preparing unsaturated hydrocarbons and their derivatives.
In simple words: We start with ethyl bromide and perform a series of chemical changes. First, we remove an HBr to get ethylene (A). Then, we add chlorine to get a dichloroethane (B). From (B), we remove HCl to get vinyl chloride (D). Also, if we react (A) with ozone, we get formaldehyde (C).

🎯 Exam Tip: When identifying compounds in a reaction series, always consider the reagents used and the type of reaction (e.g., elimination, addition, ozonolysis) at each step to determine the product.

 

Question 33. Write short notes on ortho, para directors in aromatic electrophilic substitution reactions.
Answer: All activating groups in aromatic electrophilic substitution reactions are known as ortho-para directors. Examples include \( -\text{OH} \), \( -\text{NH}_2 \), \( -\text{NHR} \), \( -\text{NHCOCH}_3 \), \( -\text{OCH}_3 \), \( -\text{CH}_2-\text{C}_2\text{H}_5 \), etc. For instance, in a phenolic \( (-\text{OH}) \) group, the lone pair of electrons on the oxygen atom participates in resonance with the benzene ring. This makes the ring more electron-rich, especially at the ortho and para positions, compared to the meta position. As a result, the phenolic group activates the benzene ring, directing incoming electrophiles to the ortho and para positions. A good example to consider is how substituents influence the reactivity of the benzene ring. In aryl halides, halogens have a strong \( -\text{I} \) effect (electron-withdrawing tendency), which generally decreases the electron density of the benzene ring, thus deactivating it towards electrophilic attack. However, the halogens also have lone pairs that can participate in resonance with the \( \pi \) electrons of the benzene ring. This resonance increases electron density at the ortho and para positions. Therefore, halogen groups act as ortho-para directors, even though they are deactivating overall. The balance between inductive and resonance effects determines the overall directing power.
In simple words: Some groups attached to a benzene ring make it easier for new chemicals to join at specific spots, called ortho and para positions. These are called ortho-para directors. They push electrons into the ring, making these spots very attractive. Halogens also direct to ortho-para positions, but they make the whole ring less reactive because they pull electrons away through a different effect.

🎯 Exam Tip: Remember that activating groups are generally ortho-para directors, while deactivating groups (except halogens) are meta directors. Halogens are unique as they are deactivating but still ortho-para directing due to resonance effects.

 

Question 34. How is propyne prepared from an alkene dihalide?
Answer: Propyne can be prepared from an alkene dihalide, specifically propylidene chloride, through a two-step elimination process. First, 1,2-dibromopropane (an alkene dihalide) undergoes dehydrohalogenation with alcoholic KOH, which removes HBr to form 2-methyl-1-propene. Next, treatment with \( \text{NaNH}_2 \) (sodium amide) causes further elimination, resulting in the formation of propyne. This is a common method for preparing alkynes from dihalides.

In simple words: To make propyne from an alkene with two halogens, we first remove one hydrogen and one halogen using strong base, then remove another hydrogen and halogen to create the triple bond.

🎯 Exam Tip: Dehydrohalogenation reactions using strong bases like alcoholic KOH or \( \text{NaNH}_2 \) are crucial for converting dihalides into alkynes.

 

Question 35. An alkylhalide with molecular formula \( \text{C}_6\text{H}_{13}\text{Br} \) on dehydro halogenation gave two isomeric alkenes X and Y with molecular formula \( \text{C}_6\text{H}_{12} \). On reductive ozonolysis, X and Y gave four compounds \( \text{CH}_3\text{COCH}_3 \), \( \text{CH}_3\text{CHO} \), \( \text{CH}_3\text{CH}_2\text{CHO} \) and \( (\text{CH}_3)_2\text{CHCHO} \). Find the alkylhalide.
Answer: The problem asks to identify an alkyl halide \( \text{C}_6\text{H}_{13}\text{Br} \) that undergoes dehydrohalogenation to form two isomeric alkenes, X and Y, with the formula \( \text{C}_6\text{H}_{12} \). When these alkenes are subjected to reductive ozonolysis, they yield four specific carbonyl compounds: \( \text{CH}_3\text{COCH}_3 \) (acetone), \( \text{CH}_3\text{CHO} \) (acetaldehyde), \( \text{CH}_3\text{CH}_2\text{CHO} \) (propanal), and \( (\text{CH}_3)_2\text{CHCHO} \) (isobutanal). To find the alkyl halide, we first need to determine the structures of alkenes X and Y by piecing together the ozonolysis products. Ozonolysis breaks carbon-carbon double bonds, and each fragment gets a carbonyl group. From the ozonolysis products: 1. Acetone (\( \text{CH}_3\text{COCH}_3 \)) means a \( (\text{CH}_3)_2\text{C}= \) fragment. 2. Acetaldehyde (\( \text{CH}_3\text{CHO} \)) means a \( \text{CH}_3\text{CH}= \) fragment. 3. Propanal (\( \text{CH}_3\text{CH}_2\text{CHO} \)) means a \( \text{CH}_3\text{CH}_2\text{CH}= \) fragment. 4. Isobutanal (\( (\text{CH}_3)_2\text{CHCHO} \)) means a \( (\text{CH}_3)_2\text{CHCH}= \) fragment. Since we have four products, it implies that X and Y are different alkenes, and their fragments combine to give these products. If we combine acetone and propanal, we get \( (\text{CH}_3)_2\text{C}=\text{CHCH}_2\text{CH}_3 \) (2-methyl-2-pentene). This would yield acetone and propanal. If we combine isobutanal and acetaldehyde, we get \( (\text{CH}_3)_2\text{CHCH}=\text{CHCH}_3 \) (4-methyl-2-pentene). This would yield isobutanal and acetaldehyde. Let's assume the two alkenes X and Y are: Alkene X: 2-methyl-2-pentene (\( \text{CH}_3-\text{CH}_2-\text{C}(\text{CH}_3)=\text{CH}-\text{CH}_3 \)) Ozonolysis of X: \( \text{CH}_3-\text{CH}_2-\text{C}(\text{CH}_3)=\text{O} \) (Propanone/Acetone) \( \text{CH}_3-\text{CH}=\text{O} \) (Ethanal/Acetaldehyde) - Wait, this is not correct. Ozonolysis of 2-methyl-2-pentene should yield propanone and ethanal. Let's recheck. \( \text{CH}_3-\text{CH}_2-\text{C}(\text{CH}_3)=\text{CH}-\text{CH}_3 \) would give \( \text{CH}_3\text{COCH}_2\text{CH}_3 \) (butanone) and \( \text{CH}_3\text{CHO} \) (acetaldehyde). This doesn't match the list perfectly. Let's reconsider the fragments: - \( (\text{CH}_3)_2\text{C}= \) from \( \text{CH}_3\text{COCH}_3 \) - \( \text{CH}_3\text{CH}= \) from \( \text{CH}_3\text{CHO} \) - \( \text{CH}_3\text{CH}_2\text{CH}= \) from \( \text{CH}_3\text{CH}_2\text{CHO} \) - \( (\text{CH}_3)_2\text{CHCH}= \) from \( (\text{CH}_3)_2\text{CHCHO} \) The sum of carbons in the products for X must be 6, and for Y must be 6. Products: 1. \( \text{CH}_3\text{COCH}_3 \) (3C) 2. \( \text{CH}_3\text{CHO} \) (2C) 3. \( \text{CH}_3\text{CH}_2\text{CHO} \) (3C) 4. \( (\text{CH}_3)_2\text{CHCHO} \) (4C) Let's try to form two \( \text{C}_6\text{H}_{12} \) alkenes from these fragments. Possible alkene X combination: - If X yields \( \text{CH}_3\text{COCH}_3 \) (3C) and \( \text{CH}_3\text{CH}_2\text{CHO} \) (3C), then X is 2-methyl-2-pentene (\( \text{CH}_3-\text{CH}_2-\text{C}(\text{CH}_3)=\text{CH}-\text{CH}_3 \)). This alkene has 6 carbons. \( \text{CH}_3-\text{CH}_2-\text{C}(\text{CH}_3)=\text{CH}-\text{CH}_3 \) on ozonolysis would give \( \text{CH}_3\text{CH}_2\text{COCH}_3 \) (butanone, 4C) and \( \text{CH}_3\text{CHO} \) (acetaldehyde, 2C). This does not match the products provided. Let's try to construct the alkenes by reversing ozonolysis. If alkene X gives \( \text{CH}_3\text{COCH}_3 \) (acetone) and \( \text{CH}_3\text{CHO} \) (acetaldehyde), it must be \( \text{CH}_3-\text{C}(\text{CH}_3)=\text{CH}-\text{CH}_3 \) (2-methyl-2-butene). This alkene has 5 carbons, not 6. So this is not correct. The given products are: Acetone (3C), Acetaldehyde (2C), Propanal (3C), Isobutanal (4C). Since the alkenes X and Y both have \( \text{C}_6\text{H}_{12} \), each alkene must produce two fragments whose carbons sum to 6. So, one alkene (X) must yield a pair of products, and the other alkene (Y) must yield another pair. Possibility 1: Alkene 1 gives Acetone (3C) and Propanal (3C). This means Alkene 1 is \( (\text{CH}_3)_2\text{C}=\text{CHCH}_2\text{CH}_3 \) (2-methyl-2-pentene). Ozonolysis: \( (\text{CH}_3)_2\text{C}=\text{O} \) (acetone) + \( \text{O}=\text{CHCH}_2\text{CH}_3 \) (propanal). This matches. So, X (or Y) is 2-methyl-2-pentene. Alkene 2 gives Acetaldehyde (2C) and Isobutanal (4C). This means Alkene 2 is \( \text{CH}_3\text{CH}=\text{CHCH}(\text{CH}_3)_2 \) (4-methyl-2-pentene). Ozonolysis: \( \text{CH}_3\text{CH}=\text{O} \) (acetaldehyde) + \( \text{O}=\text{CHCH}(\text{CH}_3)_2 \) (isobutanal). This matches. So, Y (or X) is 4-methyl-2-pentene. The alkyl halide \( \text{C}_6\text{H}_{13}\text{Br} \) must be able to form both 2-methyl-2-pentene and 4-methyl-2-pentene upon dehydrohalogenation. This suggests that the bromine atom must be located such that elimination can occur from different beta-hydrogens to yield these two isomers. Let's consider the possible \( \text{C}_6\text{H}_{13}\text{Br} \) structures that could yield these alkenes. 2-methyl-2-pentene: \( \text{CH}_3-\text{CH}_2-\text{C}(\text{CH}_3)=\text{CH}-\text{CH}_3 \) 4-methyl-2-pentene: \( \text{CH}_3-\text{CH}=\text{CH}-\text{CH}(\text{CH}_3)_2 \) For 2-methyl-2-pentene to be formed, the bromine could be on C-2 with a methyl group. \( \text{CH}_3-\text{CH}_2-\text{CBr}(\text{CH}_3)-\text{CH}_2-\text{CH}_3 \) (3-bromo-3-methylpentane) - dehydrohalogenation could give 2-methyl-2-pentene by removing H from C-2 or C-4. If Br is on C-3: \( \text{CH}_3-\text{CH}_2-\text{CHBr}-\text{CH}(\text{CH}_3)-\text{CH}_3 \) (3-bromo-4-methylpentane). Let's try to get the required alkenes from 3-bromo-3-methylpentane: \( \text{CH}_3-\text{CH}_2-\text{CBr}(\text{CH}_3)-\text{CH}_2-\text{CH}_3 \) Elimination of H from C-2: \( \text{CH}_2=\text{C}(\text{CH}_3)-\text{CH}_2-\text{CH}_3 \) (2-methyl-1-pentene). (Not X or Y) Elimination of H from C-4: \( \text{CH}_3-\text{CH}_2-\text{C}(\text{CH}_3)=\text{CH}-\text{CH}_3 \) (2-methyl-2-pentene). (Matches X or Y!) Let's try to get the required alkenes from 3-bromo-2-methylpentane: \( \text{CH}_3-\text{CHBr}-\text{CH}(\text{CH}_3)-\text{CH}_2-\text{CH}_3 \) This is 3-bromo-2-methylpentane. Elimination of H from C-2: \( \text{CH}_2=\text{C}(\text{CH}_3)-\text{CH}_2-\text{CH}_3 \) (2-methyl-1-pentene). Elimination of H from C-4: \( \text{CH}_3-\text{CH}=\text{C}(\text{CH}_3)-\text{CH}_2-\text{CH}_3 \) (2,3-dimethyl-2-butene). Elimination of H from C-1: \( \text{CH}_3-\text{C}(\text{CH}_3)=\text{CH}-\text{CH}_2-\text{CH}_3 \) (2-methyl-2-pentene). (Matches X or Y!) Let's consider 3-bromo-2,3-dimethylpentane, but that would be 7 carbons. We need 6. Consider 3-bromo-2-methylpentane: \( \text{CH}_3-\text{CH}_2-\text{CH}(\text{CH}_3)-\text{CHBr}-\text{CH}_3 \) (this is 3-bromo-2-methylpentane). Dehydrohalogenation of 3-bromo-2-methylpentane: 1. \( \text{CH}_3-\text{CH}_2-\text{CH}(\text{CH}_3)-\text{CH}=\text{CH}_2 \) (4-methyl-1-pentene) 2. \( \text{CH}_3-\text{CH}_2-\text{C}(\text{CH}_3)=\text{CH}-\text{CH}_3 \) (2-methyl-2-pentene) - Matches X or Y! 3. \( \text{CH}_3-\text{CH}=\text{C}(\text{CH}_3)-\text{CH}_2-\text{CH}_3 \) (2-methyl-2-pentene by another name) Let's analyze the given answer's structure and the derived alkenes. The alkyl halide provided in the solution is 3-bromo-2-methylpentane. Structure of 3-bromo-2-methylpentane: \( \text{CH}_3 - \text{CH} (\text{CH}_3) - \text{CHBr} - \text{CH}_2 - \text{CH}_3 \) Dehydrohalogenation will remove Br from C3 and H from an adjacent carbon (C2 or C4). 1. H from C2: \( \text{CH}_3 - \text{C}(\text{CH}_3) = \text{CH} - \text{CH}_2 - \text{CH}_3 \) (2-methyl-2-pentene) - This is one of the target alkenes. 2. H from C4: \( \text{CH}_3 - \text{CH}(\text{CH}_3) - \text{CH} = \text{CH} - \text{CH}_3 \) (4-methyl-2-pentene) - This is the other target alkene. So, 3-bromo-2-methylpentane can indeed produce both 2-methyl-2-pentene and 4-methyl-2-pentene. The ozonolysis products of these two alkenes are: For 2-methyl-2-pentene (\( \text{CH}_3 - \text{C}(\text{CH}_3) = \text{CH} - \text{CH}_2 - \text{CH}_3 \)): - \( \text{CH}_3\text{COCH}_3 \) (Acetone) - \( \text{CH}_3\text{CH}_2\text{CHO} \) (Propanal) These are two of the four given products. For 4-methyl-2-pentene (\( \text{CH}_3 - \text{CH} = \text{CH} - \text{CH}(\text{CH}_3)_2 \)): - \( \text{CH}_3\text{CHO} \) (Acetaldehyde) - \( (\text{CH}_3)_2\text{CHCHO} \) (Isobutanal) These are the other two of the four given products. All four products are accounted for by the ozonolysis of the two alkenes, which can both be formed from 3-bromo-2-methylpentane. Thus, the alkyl halide is 3-bromo-2-methylpentane.

In simple words: We are given an alkyl halide that changes into two different alkenes. When these alkenes are broken down by ozonolysis, they give four specific smaller compounds. By working backward from these smaller compounds, we can figure out the structure of the two alkenes. Then, we find an alkyl halide that can form both of these alkenes when one bromine atom and one hydrogen atom are removed from it. This alkyl halide is 3-bromo-2-methylpentane.

🎯 Exam Tip: Ozonolysis is a powerful tool to determine the position of double bonds in alkenes. Working backward from ozonolysis products is a common strategy in organic chemistry problems to deduce the structure of the parent alkene.

 

Question 36. Describe the mechanism of Nitration of benzene.
Answer: Nitration of benzene is an electrophilic substitution reaction where a nitro group \( (-\text{NO}_2) \) replaces a hydrogen atom on the benzene ring. This reaction is carried out by treating benzene with a nitrating mixture, which is a combination of concentrated nitric acid \( (\text{HNO}_3) \) and concentrated sulfuric acid \( (\text{H}_2\text{SO}_4) \), typically maintained below 333 K. The sulfuric acid acts as a catalyst, helping to generate the electrophile. The mechanism involves three main steps: 1. **Generation of the Electrophile:** The electrophile, the nitronium ion \( (\text{NO}_2^+) \), is formed when concentrated sulfuric acid protonates nitric acid, causing it to lose a water molecule. This positive ion is highly reactive and seeks electrons. \[ \text{H}_2\text{SO}_4 + \text{HONO}_2 \rightarrow \text{H}_2\text{O}-\text{NO}_2^+ + \text{HSO}_4^- \]
\( \implies \) \( \text{H}_2\text{SO}_4 + \text{HNO}_3 \rightarrow \text{NO}_2^+ + \text{H}_3\text{O}^+ + \text{HSO}_4^- \) 2. **Attack of the Electrophile (Formation of Arenium Ion):** The nitronium ion \( (\text{NO}_2^+) \), being electron-deficient, attacks the electron-rich benzene ring. A pair of \( \pi \) electrons from the benzene ring forms a new sigma bond with the nitrogen atom of the nitronium ion. This temporarily disrupts the aromaticity, forming a positively charged intermediate called an arenium ion (or sigma complex), which is resonance-stabilized.

3. **Removal of Proton and Restoration of Aromaticity:** In the final step, a proton \( (\text{H}^+) \) from the carbon bearing the nitro group is removed by a base (typically \( \text{HSO}_4^- \) from sulfuric acid). This allows the electrons from the carbon-hydrogen bond to reform the \( \pi \) system, restoring the aromaticity of the ring and yielding nitrobenzene. This step is usually very fast.

In simple words: To put a nitro group on benzene, we use a mix of strong acids to create a special positive ion called a nitronium ion. This ion then attacks the benzene ring, temporarily making it unstable but also giving it a positive charge. Finally, the ring loses a hydrogen atom, and its stable aromatic structure comes back, now with the nitro group attached. The acid mix helps create the attacking particle and stabilizes the final product.

🎯 Exam Tip: Remember the three key steps: electrophile generation, electrophilic attack (forming the arenium ion), and proton removal to restore aromaticity. The nitronium ion (\( \text{NO}_2^+ \)) is the active electrophile in this reaction.

 

Question 37. How does Huckel rule help to decide the aromatic character of a compound.
Answer: Huckel's rule provides a simple yet effective way to determine if a cyclic, planar molecule exhibits aromatic character. Aromatic compounds are exceptionally stable due to the delocalization of \( \pi \) electrons. Huckel's rule states that a compound is aromatic if it meets the following three criteria: 1. **Cyclic:** The molecule must have a ring structure. 2. **Planar:** All atoms in the ring must lie in the same plane, allowing for continuous overlap of p-orbitals. 3. **Conjugated:** There must be a continuous system of p-orbitals overlapping around the ring, meaning alternating single and double bonds. 4. **Huckel's Number of \( \pi \) Electrons:** The cyclic system must contain \( (4n + 2) \) \( \pi \) electrons, where 'n' is a non-negative integer (n = 0, 1, 2, 3, ...). Let's look at the historical context: In 1865, August Kekulé proposed that benzene has a cyclic planar structure with alternating single and double bonds. However, this model had two issues: * Benzene produces only one type of ortho-disubstituted product, but Kekulé's structure, with fixed single and double bonds, would predict two. * Kekulé's structure failed to explain why benzene primarily undergoes substitution reactions instead of addition reactions, which are typical for alkenes.

To address these objections, Kekulé suggested that benzene exists as a rapidly interconverting mixture of two equivalent forms, leading to the concept of resonance. **Resonance Description of Benzene:** Resonance describes a phenomenon where a substance's actual structure is a hybrid of two or more possible alternative structures. For benzene, Kekulé's structures (I and II) are resonance forms, and the true structure (III) is a resonance hybrid of these forms. The real structure of benzene is not any one of these individual forms but an average, which helps explain its unique stability.

**Spectroscopic Measurements:** Experimental data from spectroscopy supports the resonance hybrid model. All carbon-carbon bonds in benzene have an identical length of 1.40 Å. This value is intermediate between a typical carbon-carbon single bond (1.54 Å) and a carbon-carbon double bond (1.34 Å), confirming that the electrons are delocalized rather than fixed in alternating single and double bonds. **Molecular Orbital Structure:** From a molecular orbital perspective, all six carbon atoms in benzene are \( \text{sp}^2 \) hybridized. Each carbon uses three \( \text{sp}^2 \) orbitals to form sigma bonds: one with a hydrogen atom and two with adjacent carbon atoms. This forms a planar hexagonal ring. The remaining unhybridized p-orbital on each carbon atom (containing one electron each) overlaps laterally above and below the ring, creating a continuous \( \pi \) electron cloud. These six delocalized \( \pi \) electrons are spread over all six carbon atoms, making the molecule highly stable. Because of this strong \( \pi \) delocalization, benzene undergoes substitution reactions (to maintain aromaticity) rather than addition reactions (which would disrupt it). Therefore, Huckel's rule, especially the \( (4n + 2) \) \( \pi \) electron count, helps to define this exceptional stability and reactivity pattern, clearly distinguishing aromatic compounds from non-aromatic and anti-aromatic ones.

In simple words: Huckel's rule helps us know if a ring-shaped chemical is super stable and special, which we call aromatic. For a molecule to be aromatic, it must be a flat ring, have electrons that can move freely all around the ring, and, most importantly, have a specific number of these free-moving electrons that follows the rule of \( (4n + 2) \). If it meets all these points, it's aromatic and very stable.

🎯 Exam Tip: Always check for the four conditions: cyclic, planar, conjugated, and \( (4n + 2) \) \( \pi \) electrons. If any condition is not met, the compound is either non-aromatic or anti-aromatic.

 

Question 38. Suggest the route for the preparation of the following from benzene.
1) 3 – chloro nitrobenzene
2) 4 – chlorotoluene
3) Bromo benzene
4) m – dinitro benzene
Answer:
1) **Preparation of 3-chloro nitrobenzene from benzene:**
To prepare 3-chloro nitrobenzene, we need to add a nitro group and a chloro group to benzene. Since the chloro group is ortho-para directing and the nitro group is meta directing, the order of reactions is important. To get them in meta positions to each other, we first introduce the meta-directing nitro group, then the ortho-para directing chloro group will be meta to the existing nitro group (since the nitro group deactivates ortho and para positions). First, benzene is nitrated to form nitrobenzene. Nitrobenzene is a meta-directing compound, meaning any subsequent electrophilic substitution will occur at the meta position. Then, nitrobenzene is chlorinated in the presence of a Lewis acid catalyst (\( \text{FeCl}_3 \)) to form 3-chloro nitrobenzene.

2) **Preparation of 4-chlorotoluene from benzene:**
To get 4-chlorotoluene, we need to introduce a methyl group and a chloro group. Both are ortho-para directors. To ensure para-substitution, we can either do Friedel-Crafts alkylation first, followed by chlorination, or vice versa. Usually, adding the bulky group first promotes para. First, benzene undergoes Friedel-Crafts alkylation with chloromethane (\( \text{CH}_3\text{Cl} \)) in the presence of anhydrous \( \text{AlCl}_3 \) to form toluene. Toluene's methyl group is an ortho-para director. Then, toluene is chlorinated with \( \text{Cl}_2 \) in the presence of \( \text{FeCl}_3 \). Since the methyl group is an ortho-para director, a mixture of ortho and para products will be formed. 4-chlorotoluene is the major product due to steric hindrance at the ortho position.

3) **Preparation of Bromobenzene from benzene:**
Benzene reacts with bromine (\( \text{Br}_2 \)) in the presence of a Lewis acid catalyst, such as anhydrous ferric bromide (\( \text{FeBr}_3 \)), via electrophilic aromatic substitution to form bromobenzene. The iron catalyst helps to polarize the bromine molecule, making it more electrophilic.

4) **Preparation of m-dinitrobenzene from benzene:**
To prepare m-dinitrobenzene, two nitro groups must be added to the benzene ring. Since the nitro group is a strong meta-director and deactivator, successive nitration under more vigorous conditions will lead to meta-substitution. First, benzene is nitrated with a nitrating mixture to form nitrobenzene. Then, nitrobenzene is nitrated again under harsher conditions (higher temperature and/or stronger nitrating mixture) because the first nitro group deactivates the ring. Since the nitro group is meta-directing, the second nitro group will be added at the meta position, yielding m-dinitrobenzene.

In simple words: To make different benzene compounds, we use specific reactions in a certain order. For example, to make 3-chloro nitrobenzene, we first add a nitro group (which directs new groups to the meta spot) and then a chlorine. To make 4-chlorotoluene, we first add a methyl group and then a chlorine, aiming for the para spot. Bromobenzene is made by adding bromine with an iron catalyst. Finally, to make m-dinitrobenzene, we add two nitro groups, one after another, as the first one directs the second to the meta position.

🎯 Exam Tip: The order of reactions in multi-substituted benzene synthesis is crucial. Always consider the directing nature (ortho-para or meta) and activating/deactivating effects of existing substituents to achieve the desired product.

 

Question 39. Suggest a simple chemical test to distinguish propane and propene.
Answer: A simple chemical test to distinguish between propane (an alkane) and propene (an alkene) involves using bromine water. Propene, being an unsaturated hydrocarbon with a double bond, will react with bromine water. The reddish-brown color of bromine water will disappear as it adds across the double bond to form a colorless dibromo compound. This reaction is a type of addition reaction. Propane, being a saturated hydrocarbon (an alkane) with only single bonds, will not react with bromine water under normal conditions. Thus, the reddish-brown color of the bromine water will persist. This difference in reactivity with bromine water is a classic test for unsaturation.
In simple words: You can tell propane and propene apart using bromine water. Propene, which has a double bond, will make the bromine water lose its reddish-brown color. Propane, with only single bonds, will not change the color of the bromine water.

🎯 Exam Tip: The decolorization of bromine water is a characteristic test for unsaturation (presence of double or triple bonds) in organic compounds. Remember to specify the color change for full marks.

 

Question 40. What happens when Isobutylene is treated with acidified potassium permanganate?
Answer: When isobutylene (2-methylpropene) is treated with acidified potassium permanganate \( (\text{KMnO}_4) \), it undergoes oxidative cleavage. Acidified \( \text{KMnO}_4 \) is a strong oxidizing agent that breaks the carbon-carbon double bond. In this reaction, isobutylene is oxidized to yield acetone and carbon dioxide. The carbon atom of the double bond that is disubstituted (part of the \( (\text{CH}_3)_2\text{C}= \) group) becomes a ketone (acetone), while the carbon atom that is monosubstituted (part of the \( =\text{CH}_2 \) group) is fully oxidized to carbon dioxide because it cannot form a stable carboxylic acid or aldehyde under strong oxidative conditions.

In simple words: When isobutylene reacts with acidified potassium permanganate, its double bond breaks apart. This reaction makes acetone (a type of ketone) and carbon dioxide. This shows how strong oxidizing agents can completely break down alkenes.

🎯 Exam Tip: Strong oxidizing agents like acidified \( \text{KMnO}_4 \) cleave double bonds. Remember that a \( =\text{CH}_2 \) group is oxidized to \( \text{CO}_2 \) and water, a \( =\text{CHR} \) group becomes a carboxylic acid \( (\text{RCOOH}) \), and a \( =\text{CR}_2 \) group forms a ketone \( (\text{R}_2\text{C}=\text{O}) \).

 

Question 41. How will you convert ethyl chloride into
(i) ethane
(ii) n – butane
Answer:
(i) **Conversion of ethyl chloride to ethane:**
Ethyl chloride \( (\text{C}_2\text{H}_5\text{Cl}) \) can be converted to ethane (\( \text{C}_2\text{H}_6 \)) through a reduction reaction. When ethyl chloride is treated with zinc \( (\text{Zn}) \) and hydrochloric acid \( (\text{HCl}) \), the chlorine atom is replaced by a hydrogen atom. This process is a common method for reducing alkyl halides to alkanes.

(ii) **Conversion of ethyl chloride to n-butane:**
Ethyl chloride can be converted to n-butane \( (\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_3) \) via the Wurtz reaction. In this reaction, two molecules of ethyl chloride react with sodium metal \( (\text{Na}) \) in the presence of dry ether. The sodium removes the chlorine atoms, and the two ethyl groups combine to form a longer carbon chain, n-butane. This reaction is useful for synthesizing symmetrical alkanes from alkyl halides.

In simple words: (i) To change ethyl chloride into ethane, we can use zinc and acid to remove the chlorine and add a hydrogen. (ii) To change ethyl chloride into n-butane, we react two ethyl chloride molecules with sodium in dry ether. This joins the two ethyl parts together to make a longer chain.

🎯 Exam Tip: Remember that \( \text{Zn}/\text{HCl} \) is a common reducing agent for alkyl halides to alkanes, and the Wurtz reaction is used for coupling two alkyl halides to form a symmetrical alkane with a longer carbon chain.

 

Question 42. Describe the conformers of n – butane.
Answer: n-Butane \( (\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_3) \) can be thought of as a derivative of ethane where one hydrogen on each carbon is replaced by a methyl group. Due to the rotation around the central carbon-carbon single bond, n-butane exists in various spatial arrangements called conformers. The key conformers are staggered and eclipsed, with different energy levels due to torsional strain and steric strain from groups getting too close. 1. **Eclipsed Conformation:** In the fully eclipsed conformation, the two bulky methyl groups are directly opposite each other, creating maximum repulsion. This arrangement also causes maximum torsional strain due to the hydrogens on adjacent carbons being directly aligned. Consequently, the fully eclipsed conformation is the least stable conformer of n-butane, possessing the highest potential energy. 2. **Staggered Conformation (Anti-periplanar/Anti Form):** The anti (or anti-periplanar) conformation is a type of staggered form where the two methyl groups are as far apart as possible (180° dihedral angle). In this arrangement, the distance between the two methyl groups is maximized, leading to minimum steric repulsion and torsional strain. Because of its minimal strain, the anti form is the most stable conformer of n-butane and has the lowest potential energy. 3. **Gauche Conformation:** The gauche conformation is another staggered form where the two methyl groups are adjacent but not directly aligned, with a dihedral angle of 60°. While still staggered, the methyl groups are closer than in the anti form, leading to some steric repulsion (gauche interaction). This makes the gauche conformation less stable than the anti form but more stable than any eclipsed conformation. 4. **Skew Conformations:** These are all the intermediate conformations between the fully eclipsed and staggered (anti or gauche) forms. They are generally less stable than staggered forms but more stable than fully eclipsed forms. These are constantly interconverting at room temperature. The relative stabilities, from most stable to least stable, are: Anti > Gauche > Eclipsed. The potential energy diagram illustrates these stabilities, with valleys representing stable staggered forms and peaks representing unstable eclipsed forms.

In simple words: n-Butane molecules can twist and turn around their central bond, forming different shapes called conformers. The most stable shape is "anti" where the big groups are furthest apart. The least stable shape is "eclipsed" where the big groups are right on top of each other, causing a lot of pushing. There's also a "gauche" shape which is in-between, where the big groups are still staggered but a bit closer.

🎯 Exam Tip: When explaining conformers, emphasize the role of steric strain (repulsion between bulky groups) and torsional strain (repulsion between electron clouds of aligned bonds) in determining relative stability.

 

Question 43. Write the chemical equations for combustion of propane.
Answer: Combustion is a chemical reaction where a substance reacts rapidly with oxygen, usually producing heat and light. For hydrocarbons like propane, complete combustion typically yields carbon dioxide (\( \text{CO}_2 \)) and water (\( \text{H}_2\text{O} \)). It is an exothermic reaction, meaning it releases energy. The chemical equation for the complete combustion of propane (\( \text{C}_3\text{H}_8 \)) is: \[ \text{C}_3\text{H}_8 (\text{g}) + 5\text{O}_2 (\text{g}) \rightarrow 3\text{CO}_2 (\text{g}) + 4\text{H}_2\text{O} (\text{l}) \] In this reaction, one molecule of gaseous propane reacts with five molecules of gaseous oxygen to produce three molecules of gaseous carbon dioxide and four molecules of liquid water.
In simple words: When propane burns completely, it mixes with oxygen to create carbon dioxide and water. This process also releases a lot of heat, which is why propane is used as fuel.

🎯 Exam Tip: Always ensure the chemical equation is balanced for both atoms and charge on both sides. For complete combustion of hydrocarbons, the products are always carbon dioxide and water.

 

Question 44. Explain Markovnikoff’s rule with suitable example.
Answer: **Markovnikoff’s Rule:** Markovnikoff’s rule is a principle in organic chemistry that predicts the regioselectivity of electrophilic addition reactions of unsymmetrical reagents to unsymmetrical alkenes or alkynes. The rule states: "When an unsymmetrical alkene reacts with a hydrogen halide (\( \text{HX} \), where X is a halogen), the hydrogen atom \( (\text{H}) \) adds to the carbon atom of the double bond that already has a greater number of hydrogen atoms, and the halogen atom \( (\text{X}) \) adds to the carbon atom with fewer hydrogen atoms." This rule can also be stated as: in the addition reaction of an alkene or alkyne, the most electronegative part of the reagent adds to the least hydrogen-attached doubly bonded carbon. The underlying reason for this regioselectivity is the formation of a more stable carbocation intermediate during the reaction. The more substituted carbocation (e.g., tertiary > secondary > primary) is more stable and thus preferentially formed. **Addition of Water (Hydration of Alkenes):** Alkenes typically do not react directly with water under normal conditions. However, in the presence of concentrated sulfuric acid, alkenes undergo hydration to form alcohols. This reaction also follows Markovnikoff’s rule and proceeds via a carbocation mechanism. The sulfuric acid first adds to the alkene according to Markovnikoff's rule, forming an alkyl hydrogen sulfate, which is then hydrolyzed by water to yield the alcohol. **Example:** Let's consider the addition of \( \text{HBr} \) to propene (\( \text{CH}_3-\text{CH}=\text{CH}_2 \)). According to Markovnikoff's rule: - The hydrogen atom from \( \text{HBr} \) will add to the \( \text{CH}_2 \) carbon (which has more hydrogens). - The bromine atom from \( \text{HBr} \) will add to the \( \text{CH} \) carbon (which has fewer hydrogens). Therefore, the major product formed is 2-bromopropane (\( \text{CH}_3-\text{CHBr}-\text{CH}_3 \)), rather than 1-bromopropane.

In simple words: Markovnikoff's rule tells us how two parts of a molecule will add to a double bond. The hydrogen part of the adding molecule goes to the carbon with more hydrogens, and the other part goes to the carbon with fewer hydrogens. For example, when HBr adds to propene, the hydrogen goes to the end carbon, and bromine goes to the middle carbon, making 2-bromopropane the main product.

🎯 Exam Tip: Always identify the "rich get richer" carbon (more hydrogens) and the "poor get poorer" carbon (fewer hydrogens) on the double bond to correctly apply Markovnikoff's rule. The stability of the carbocation intermediate is the underlying principle.

 

Question 45. What happens when ethylene is passed ‘ through cold dilute alkaline potassium permanganate.
Answer: When ethylene \( (\text{CH}_2=\text{CH}_2) \) is passed through cold, dilute, alkaline potassium permanganate solution (Baeyer's reagent), it undergoes a syn-dihydroxylation reaction. This means two hydroxyl \( (-\text{OH}) \) groups are added to the same face of the double bond. The purple color of potassium permanganate disappears, and a brown precipitate of manganese dioxide \( (\text{MnO}_2) \) is formed. This reaction is also known as the Baeyer test and is used to detect the presence of unsaturation (carbon-carbon double or triple bonds) in a compound. In the case of ethylene, the product formed is ethylene glycol (ethane-1,2-diol). \[ \text{CH}_2=\text{CH}_2 + \text{H}_2\text{O} + [\text{O}] \xrightarrow{\text{cold dil alkaline KMnO}_4} \text{CH}_2\text{OH}-\text{CH}_2\text{OH} \] Ethylene Glycol
In simple words: When ethylene gas is bubbled through a cold, weak solution of purple potassium permanganate, the purple color disappears. Two hydroxyl groups join to the ethylene, making ethylene glycol. This is a common way to test for double bonds.

🎯 Exam Tip: Remember that cold, dilute, alkaline \( \text{KMnO}_4 \) (Baeyer's reagent) causes syn-dihydroxylation, forming a diol, and is a visual test for unsaturation. Stronger conditions (hot, concentrated \( \text{KMnO}_4 \)) would cause oxidative cleavage of the double bond.

 

Question 46. Write the structures of following alkanes.
1) 2, 3 – Dimethyl – 6 – (2 – methyl propyl) decane
2) 5 – (2 – Ethyl butyl ) – 3, 3, – dimethyldecane
3) 5 – (1, 2 – Dimethyl propyl) – 2 – methylnonane
Answer:
1) **2, 3 – Dimethyl – 6 – (2 – methyl propyl) decane**
The parent chain is decane, which means it has 10 carbon atoms. - Methyl groups are at positions 2 and 3. - A (2-methyl propyl) group is at position 6.

2) **5 – (2 – Ethyl butyl ) – 3, 3, – dimethyldecane**
The parent chain is decane (10 carbon atoms). - Methyl groups are at positions 3 and 3. - A (2-ethyl butyl) group is at position 5.

3) **5 – (1, 2 – Dimethyl propyl) – 2 – methylnonane**
The parent chain is nonane (9 carbon atoms). - A methyl group is at position 2. - A (1,2-dimethyl propyl) group is at position 5.

In simple words: These are structures of different alkanes, which are long chains of carbon atoms. The names tell us how many carbons are in the main chain and what smaller groups are attached to them, and at which positions. For example, "decane" means a 10-carbon main chain, with "dimethyl" meaning two methyl groups attached.

🎯 Exam Tip: When drawing alkane structures from IUPAC names, always start by drawing the longest continuous carbon chain (the parent alkane). Then, number the chain and attach the substituent groups at their indicated positions, ensuring each carbon has four bonds by adding hydrogen atoms.

 

Question 47. How will you prepare propane from a sodium salt of fatty acid?
Answer: Propane \( (\text{CH}_3\text{CH}_2\text{CH}_3) \) can be prepared from a sodium salt of a fatty acid through a process called decarboxylation using soda lime. Decarboxylation involves the removal of a carboxyl group \( (-\text{COOH}) \) as carbon dioxide. To obtain propane (a 3-carbon alkane), we need to start with a fatty acid that has one more carbon atom, specifically a 4-carbon acid. So, we would use the sodium salt of butyric acid, which is sodium butyrate. When sodium butyrate \( (\text{CH}_3\text{CH}_2\text{CH}_2\text{COONa}) \) is heated with soda lime \( (\text{NaOH} + \text{CaO}) \), the carboxyl group is removed as sodium carbonate \( (\text{Na}_2\text{CO}_3) \), and the remaining alkyl group gains a hydrogen atom to form propane. Calcium oxide \( (\text{CaO}) \) in soda lime helps prevent the fusion of sodium hydroxide glass and acts as a dehydrating agent.

In simple words: To make propane from a sodium salt of a fatty acid, we use sodium butyrate, which has four carbons. We heat it with a mixture called soda lime. This process removes a carbon atom from the salt and leaves us with propane, a three-carbon chain.

🎯 Exam Tip: For decarboxylation to prepare an alkane, always choose a sodium salt of a fatty acid with one more carbon atom than the desired alkane. Soda lime is essential for this reaction.

 

Question 48. Identify A and B

Answer:
A)
b)
The initial reaction of an alcohol with heat in the presence of an acid removes water to form an alkene (A), following Zaitsev's rule for the major product. The subsequent reaction of this alkene with HBr follows Markovnikov's rule, where the bromine atom adds to the carbon with fewer hydrogen atoms, forming the major product (B).
In simple words: When the alcohol is heated with acid, it loses water to become alkene A. When HBr is added to alkene A, it forms alkyl bromide B.

🎯 Exam Tip: Remember to apply Zaitsev's rule for elimination to determine the major alkene product, and Markovnikov's rule for addition to determine the major alkyl halide product.

 

Question 49. Complete the following:
(i) 2 - butyne \( \xrightarrow{\text{Lindlar Catalyst}} \)
(ii) \( \text{CH}_2 = \text{CH}_2 \xrightarrow{\text{I}_2} \)
(iii) \( \text{CH}_3 - \text{CH}_2 \xrightarrow{\text{Zn/C}_2\text{H}_5\text{OH}} \)
\( \text{Br} \)
\( \text{Br} \)
(iv) \( \text{CaC}_2 \xrightarrow{\text{H}_2\text{O}} \)
Answer:
(i) 2 - butyne \( \xrightarrow{\text{Lindlar Catalyst}} \) Cis - 2 - butene

The Lindlar catalyst is a specific type of palladium catalyst that helps hydrogen add to alkynes in a way that creates a cis (same side) double bond.
(ii) \( \text{CH}_2 = \text{CH}_2 \xrightarrow{\text{I}_2} \) 1,2 - diiodoethane

Iodine adds across the double bond of ethene, similar to other halogens, but the product is often less stable and can be formed reversibly.
(iii) \( \text{CH}_3 - \text{CH}_2 \) \( \xrightarrow{\text{Zn/C}_2\text{H}_5\text{OH}} \text{CH}_2 = \text{CH}_2 + \text{ZnBr}_2 \)
\( | \)
\( \text{Br} \)
\( | \)
\( \text{Br} \)

This is a dehalogenation reaction where zinc removes two halogen atoms from adjacent carbons, leading to the formation of an alkene.
(iv) \( \text{CaC}_2 \xrightarrow{\text{H}_2\text{O}} \text{Ca(OH)}_2 + \text{C}_2\text{H}_2 \uparrow \)

Calcium carbide reacts with water in a hydrolysis reaction to produce acetylene gas and calcium hydroxide.
In simple words: These reactions show how alkynes can be reduced to cis alkenes, how halogens add to alkenes, how dihaloalkanes lose halogens to form alkenes, and how calcium carbide reacts with water to produce ethyne.

🎯 Exam Tip: Pay close attention to the reagents and conditions for each conversion, as slight changes can lead to different products (e.g., Lindlar catalyst for cis-alkenes vs. Na/Li in liquid ammonia for trans-alkenes).

 

Question 50. How will distinguish 1 - butyne and 2 - butyne?
Answer:

1 - butyne reacts with ammoniacal \( \text{AgNO}_3 \) solution to form a white precipitate of silver acetylide. This happens because 1-butyne is a terminal alkyne with an acidic hydrogen atom. On the other hand, 2 - butyne does not react with ammoniacal \( \text{AgNO}_3 \) solution because it is an internal alkyne and does not have an acidic hydrogen atom. This difference allows us to distinguish between the two.
In simple words: We can tell 1-butyne and 2-butyne apart by mixing them with a special silver solution. Only 1-butyne will make a white solid because it has a hydrogen atom that can be taken off by the silver solution, while 2-butyne does not.

🎯 Exam Tip: Remember that terminal alkynes (those with a triple bond at the end of the carbon chain) are acidic and can react with metal ions like silver and copper to form precipitates, which is a key distinguishing test.

 

11th Chemistry Guide Hydrocarbons Additional Questions and Answers

 

I. Choose the best answer:

 

Question 1. The IUPAC name of neopentane is
(a) 2 - methyl butane
(b) 2, 2 - dimethyl propane
(c) 2 - methyl propane
(d) 2, 2- dimethyl butane
Answer: (b) 2, 2 - dimethyl propane
In simple words: Neopentane is also known as 2,2-dimethylpropane because it has a propane chain with two methyl groups attached to the second carbon.

🎯 Exam Tip: For IUPAC naming, always find the longest continuous carbon chain (parent chain), number it to give substituents the lowest possible numbers, and list substituents alphabetically.

 

Question 2. Alkanes are also known as
(a) olefins
(b) unsaturated aliphatic hydrocarbons
(c) saturated aromatic hydrocarbon
(d) paraffins
Answer: (d) paraffins
In simple words: Alkanes are also called paraffins, which means they are not very reactive.

🎯 Exam Tip: Knowing common names and classifications for hydrocarbon families helps in quickly identifying their properties and reactions.

 

Question 3. The compressed gas available in cooking gas cylinders is a mixture of:
(a) \( \text{C}_6\text{H}_6 + \text{C}_6\text{H}_5\text{CH}_3 \)
(b) \( \text{C}_2\text{H}_4 + \text{C}_2\text{H}_2 \)
(c) \( \text{C}_2\text{H}_4 + \text{CH}_4 \)
(d) \( \text{C}_4\text{H}_{10} + \text{C}_3\text{H}_8 \)
Answer: (d) \( \text{C}_4\text{H}_{10} + \text{C}_3\text{H}_8 \)
In simple words: Cooking gas, also known as LPG, is mainly a mixture of butane (\( \text{C}_4\text{H}_{10} \)) and propane (\( \text{C}_3\text{H}_8 \)).

🎯 Exam Tip: LPG (Liquefied Petroleum Gas) is primarily composed of propane and butane, making it a common fuel for heating and cooking.

 

Question 4. The gas supplied in cylinders for cooking is
(a) marsh gas
(b) LPG
(c) mixture \( \text{CH}_4 \) and \( \text{C}_2\text{H}_6 \)
(d) mixture of ethane and propane
Answer: (b) LPG
In simple words: The gas in cooking cylinders is called LPG, which stands for Liquefied Petroleum Gas.

🎯 Exam Tip: Be careful not to confuse LPG with natural gas, which is primarily methane (marsh gas).

 

Question 5. Adam’s catalyst is:
(a) platinum metal
(b) palladium
(c) nickel metal
(d) \( \text{PtO}_2 \)
Answer: (d) \( \text{PtO}_2 \)
In simple words: Adam's catalyst is a special platinum oxide (\( \text{PtO}_2 \)) compound used to help chemical reactions, especially to add hydrogen to other chemicals.

🎯 Exam Tip: Adam's catalyst is primarily used for hydrogenation reactions, converting unsaturated compounds to saturated ones, and is highly effective in various organic syntheses.

 

Question 6. Soda lime is
(a) \( \text{NaOH} \)
(b) \( \text{NaOH} + \text{CaO} \)
(c) \( \text{CaO} \)
(d) \( \text{Na}_2\text{CO}_3 \)
Answer: (b) \( \text{NaOH} + \text{CaO} \)
In simple words: Soda lime is a mix of sodium hydroxide (\( \text{NaOH} \)) and calcium oxide (\( \text{CaO} \)) that is used to remove carbon dioxide and water.

🎯 Exam Tip: Soda lime is a useful reagent in decarboxylation reactions, which is often used to prepare alkanes by removing a carboxyl group from a carboxylic acid salt.

 

Question 7. Methyl bromide is converted into ethane by heating it in ether medium with
(a) Al
(b) Mg
(c) Na
(d) Cu
Answer: (c) Na
In simple words: To change methyl bromide into ethane, you heat it with sodium metal in a dry ether solution. This reaction is called the Wurtz reaction.

🎯 Exam Tip: The Wurtz reaction uses sodium metal to couple two alkyl halides, forming a new C-C bond and producing a symmetrical alkane, effectively doubling the carbon chain.

 

Question 8. Hydrocarbon which is liquid at room temperature is
(a) Pentane
(b) Butane
(c) Propane
(d) Ethane
Answer: (a) Pentane
In simple words: Among the choices, pentane is a liquid at room temperature, while butane, propane, and ethane are gases.

🎯 Exam Tip: The physical state of alkanes (gas, liquid, or solid) depends on their molecular weight and intermolecular forces. Generally, alkanes with 1 to 4 carbons are gases, 5 to 17 carbons are liquids, and 18 or more carbons are solids at room temperature.

 

Question 9. Pyrolysis of Methane and respectively are
(a) Exothermic & Endothermic
(b) Endothermic & Exothermic
(c) Both are endothermic
(d) Both are exothermic
Answer: (c) Both are endothermic
In simple words: Pyrolysis of methane, which breaks it down into simpler parts, requires heat, meaning it is an endothermic process.

🎯 Exam Tip: Pyrolysis, also known as cracking, involves breaking chemical bonds through heat in the absence of oxygen, which always requires energy input, making it an endothermic process.

 

Question 10. Select the correct statement about alkanes
(a) they are polar in nature
(b) they are soluble in water
(c) they are non - combustible
(d) their dipole moment is zero
Answer: (d) their dipole moment is zero
In simple words: Alkanes are nonpolar molecules, meaning they have no overall charge difference from one end to the other, so their dipole moment is zero.

🎯 Exam Tip: Alkanes are nonpolar because the electronegativity difference between carbon and hydrogen is very small, and their symmetrical structure often cancels out any slight bond dipoles, resulting in a zero net dipole moment.

 

Question 11. Final products of complete oxidation of hydrocarbon is
(a) Acid
(b) Dihydric alcohol
(c) Aldehyde
(d) \( \text{H}_2\text{O} + \text{CO}_2 \)
Answer: (d) \( \text{H}_2\text{O} + \text{CO}_2 \)
In simple words: When hydrocarbons burn completely with enough oxygen, they always make water and carbon dioxide.

🎯 Exam Tip: Complete combustion of any hydrocarbon always yields carbon dioxide and water, irrespective of the specific hydrocarbon, provided there is sufficient oxygen.

 

Question 12. Isomerisation in alkane can be brought by using
(a) \( \text{Al}_2\text{O}_3 \)
(b) \( \text{Fe}_2\text{O}_3 \)
(c) \( \text{Anh.AlCl}_3 \)/ HCl at 200°C
(d) Conc. \( \text{H}_2\text{SO}_4 \)
Answer: (c) \( \text{Anh.AlCl}_3 \)/ HCl at 200°C
In simple words: To change the shape of an alkane molecule (isomerization), we use anhydrous aluminum chloride with hydrogen chloride gas at 200°C.

🎯 Exam Tip: Anhydrous \( \text{AlCl}_3 \) in the presence of HCl acts as a Lewis acid catalyst for the isomerization of alkanes, converting straight-chain alkanes into branched isomers, which often improves fuel quality.

 

Question 13. In aromatization of n - hexane, the catalyst used is
(a) \( \text{Cr}_2\text{O}_3 \)
(b) \( \text{V}_2\text{O}_5 \)
(c) \( \text{Mo}_2\text{O}_3 \)
(d) All of the options
Answer: (d) All of the options
In simple words: For changing n-hexane into an aromatic compound, catalysts like chromium oxide, vanadium pentoxide, and molybdenum oxide can all be used.

🎯 Exam Tip: Aromatization, or catalytic reforming, converts straight-chain alkanes into aromatic compounds using specific metal oxide catalysts at high temperatures and pressures, forming valuable products like benzene and toluene.

 

Question 14. The most oxidized form of ethane is
(a) \( \text{CO}_2 \)
(b) HCHO
(c) HCOOH
(d) \( \text{CH}_3\text{COOH} \)
Answer: (a) \( \text{CO}_2 \)
In simple words: The most oxidized form of ethane is carbon dioxide, which is what you get when ethane burns completely.

🎯 Exam Tip: Oxidation involves increasing the number of bonds to oxygen or decreasing the number of bonds to hydrogen. Carbon dioxide represents the highest oxidation state of carbon in organic compounds.

 

Question 15. The following substance is used as anti knocking compound
(a) TEL
(b) Lead tetrachioride
(c) Lead acetate
(d) \( \text{C}_2\text{H}_2\text{PbCl} \)
Answer: (a) TEL
In simple words: TEL, or tetraethyl lead, was used in gasoline to stop engine knocking.

🎯 Exam Tip: Tetraethyl lead (TEL) was historically used as an antiknock agent to improve the octane rating of gasoline, but its use has been largely phased out due to environmental concerns about lead pollution.

 

Question 16. Conformational isomers are due to
(a) Free rotation about C - C single bond
(b) Frozen rotation about C - C single bond
(e) Frozen rotation about C - C double bond
(d) Restricted rotation about C - C single bond
Answer: (a) Free rotation about C - C single bond
In simple words: Conformational isomers are different shapes of the same molecule that can change into each other easily by rotating around single carbon-carbon bonds.

🎯 Exam Tip: Free rotation around single bonds allows molecules to adopt various temporary shapes (conformers), which can be visualized using Newman projections or sawhorse representations.

 

Question 17. The most stable conformation of ethane is
(a) Eclipsed
(b) Skew
(c) Staggered
(d) All are equally stable
Answer: (c) Staggered
In simple words: The staggered shape of ethane is the most stable because its hydrogen atoms are as far apart as possible, causing the least pushing against each other.

🎯 Exam Tip: In the staggered conformation, the torsional strain is minimized because the bonds on adjacent carbons are farthest apart, leading to lower energy and greater stability.

 

Question 18. IUPAC name of the following compound

(a) 3 - methyl hexane
(b) 3 - Ethyl pentane
(c) 2, 3 - dimethyl pentane
(d) 2, 2 - dimethyl pentane
Answer: (a) 3 - methyl hexane
In simple words: The chemical name for this compound is 3-methylhexane, meaning it's a hexane chain with a methyl group on the third carbon.

🎯 Exam Tip: When naming, always identify the longest continuous carbon chain first. If there's a tie, choose the chain with more substituents. Number the chain to give the substituents the lowest possible numbers.

 

Question 19. The number of sigma bonds formed in ethane by the overlapping of \( \text{sp}^3 - \text{sp}^3 \) orbitals
(a) 7
(b) 5
(c) 1
(d) 4
Answer: (c) 1
In simple words: In ethane, there is one sigma bond formed by the direct overlap of \( \text{sp}^3 \) orbitals from each of the two carbon atoms.

🎯 Exam Tip: Remember that in alkanes, each carbon-carbon single bond is a sigma bond formed by the head-on overlap of hybridized orbitals (typically \( \text{sp}^3 \) in saturated carbons).

 

Question 20. How many types of carbon atoms are present in 2, 2, 3 - trimethyl pentane
(a) one
(b) Two
(c) Three
(d) Four
Answer: (d) Four
In simple words: In 2,2,3-trimethylpentane, there are four different kinds of carbon atoms based on how many other carbon atoms they are connected to.

🎯 Exam Tip: To determine the number of distinct carbon types, classify each carbon as primary, secondary, tertiary, or quaternary. Carbons in chemically equivalent environments (e.g., all methyl carbons attached to the same quaternary carbon) are considered the same type.

 

Question 21. The correct IUPAC name of the following alkane is

(a) 3, 6 - diethyl - 2- methyloctane
(b) 5 - isopropyl - 3- ethyloctane
(c) 3 - ethyl - 5- isopropyloctane
(d) 3 - isopropyl - 6 - ethyloctane
Answer: (a) 3, 6 - diethyl - 2- methyloctane
In simple words: The correct name for this chemical is 3,6-diethyl-2-methyloctane, which means it has an eight-carbon chain with two ethyl groups and one methyl group attached at specific points.

🎯 Exam Tip: When multiple substituents are present, ensure the parent chain is correctly identified as the longest chain, and substituents are numbered to give the lowest possible numbers collectively, then listed alphabetically.

 

Question 22. In Wurtz reaction, n - hexane is obtained from
(a) n - propyl chloride
(b) n - butyl chloride
(c) Ethyl chloride
(d) isopropyl chloride
Answer: (a) n - propyl chloride
In simple words: In the Wurtz reaction, n-hexane (a six-carbon chain) is made by joining two n-propyl chloride (three-carbon) molecules together.

🎯 Exam Tip: The Wurtz reaction typically joins two identical alkyl halides to form a symmetrical alkane with double the number of carbons in the alkyl group. Thus, for n-hexane (6 carbons), two n-propyl groups (3 carbons each) are needed.

 

Question 23. When sodium acetate is heated with sodalime the reaction is called
(a) Dehydration
(b) Decarboxylation
(c) Dehydrogenation
(d) Dehydrohalogenation
Answer: (b) Decarboxylation
In simple words: When sodium acetate is heated with soda lime, the reaction that removes carbon dioxide is called decarboxylation.

🎯 Exam Tip: Decarboxylation is the process of removing a carboxyl group (as \( \text{CO}_2 \)) from a molecule, often used in organic synthesis to shorten carbon chains or convert carboxylic acid derivatives to hydrocarbons.

 

Question 24. The following substance reacts with water to give ethane
(a) \( \text{CH}_4 \)
(b) \( \text{C}_2\text{H}_5\text{MgBr} \)
(c) \( \text{C}_2\text{H}_4\text{OH} \)
(d) \( \text{C}_2\text{H}_5\text{OC}_2\text{H}_5 \)
Answer: (b) \( \text{C}_2\text{H}_5\text{MgBr} \)
In simple words: Ethylmagnesium bromide reacts with water to produce ethane, which is a two-carbon alkane.

🎯 Exam Tip: Grignard reagents (\( \text{RMgX} \)) are highly reactive and act as strong bases or nucleophiles. They react readily with compounds containing acidic hydrogens (like water) to form alkanes (\( \text{RH} \)).

 

Question 25. Reaction of ROH with R’ MgX produces
(a) RH
(b) R’H
(c) R - R
(d) R’ - R”
Answer: (b) R’H
In simple words: When an alcohol (ROH) reacts with a Grignard reagent (R’MgX), it produces an alkane (R’H) because the alcohol provides an acidic hydrogen.

🎯 Exam Tip: Alcohols have an acidic hydrogen on the oxygen atom. When an alcohol reacts with a Grignard reagent (R’MgX), the R' group of the Grignard reagent acts as a strong base, abstracting this acidic proton to form an alkane (R’H).

 

Question 26. The solvent used in Wurtz reaction is
(a) \( \text{C}_2\text{H}_5\text{OH(aq)} \)
(b) \( \text{CH}_3\text{COOH} \)
(c) \( \text{H}_2\text{O} \)
(d) \( \text{C}_2\text{H}_5\text{OC}_2\text{H}_5 \)(dry)
Answer: (d) \( \text{C}_2\text{H}_5\text{OC}_2\text{H}_5 \)(dry)
In simple words: In the Wurtz reaction, dry diethyl ether is used as the solvent.

🎯 Exam Tip: Dry ether is crucial for the Wurtz reaction because sodium metal reacts vigorously with water or alcohols, which would interfere with the desired coupling reaction.

 

Question 27. Arrange the following in the decreasing order of their boiling points
(i) n - butane
(ii) 2 - methylbutane
(iii) n - pentane
(iv) 2 - methylbutane
(a) i > ii > iii > iv
(b) ii > iii > iv > i
(c) iv > iii > ii > i
(d) iii > ii > iv > i
Answer: (d) iii > ii > iv > i
In simple words: The boiling points decrease as follows: n-pentane, then 2-methylbutane, then the other 2-methylbutane, and finally n-butane. This is because longer, straighter chains have higher boiling points.

🎯 Exam Tip: For isomers, boiling points decrease with increasing branching because branching reduces the surface area available for London dispersion forces, which weakens intermolecular attractions.

 

Question 28. The compound with the highest boiling point is
(a) n - Hexane
(b) n - Pentane
(c) 2, 2 - dimethyl propane
(d) 2 - methyl butane
Answer: (a) n - Hexane
In simple words: Among the options, n-hexane has the longest straight chain, which gives it the highest boiling point.

🎯 Exam Tip: Boiling points generally increase with increasing molecular weight and decrease with increasing branching for compounds with similar molecular weights, due to differences in intermolecular forces.

 

Question 29. The increasing order of reduction of alkyl halides with zinc and dilute HCl is
(a) R - Cl < R - I < R - Br
(b) R - Cl < R - Br < R - I
(c) R - I < R - Br < R - Cl
(d) R - Br < R - I < R - Cl
Answer: (b) R - Cl < R - Br < R - I
In simple words: Alkyl iodides are easiest to reduce, followed by alkyl bromides, and then alkyl chlorides, which are the hardest to reduce.

🎯 Exam Tip: The ease of reduction of alkyl halides follows the trend of bond strength, where the C-I bond is the weakest and easiest to break, while the C-Cl bond is the strongest.

 

Question 30. The volume of oxygen required for the complete combustion of 4 lit of ethane is
(a) 4 lit
(b) 8 lit
(c) 12 lit
(d) 14 lit
Answer: (d) 14 lit
In simple words: To fully burn 4 liters of ethane, you need 14 liters of oxygen. This is based on the chemical reaction ratios between ethane and oxygen.

🎯 Exam Tip: To solve gas volume stoichiometry problems, first write and balance the combustion equation. Then use the stoichiometric coefficients as volume ratios (according to Avogadro's law for gases at constant temperature and pressure).

 

Question 31. The dihedral angle between the hydrogen atoms of 2 methyl groups in staggered conformation of ethane is
(a) 0°
(b) 60°
(c) 120°
(d) 240°
Answer: (b) 60°
In simple words: In the most stable staggered shape of ethane, the hydrogen atoms on the front and back carbons are separated by an angle of 60 degrees.

🎯 Exam Tip: The staggered conformation of ethane is characterized by a dihedral angle of 60° between the hydrogen atoms on the front and back carbons when viewed down the C-C bond (Newman projection), which minimizes torsional strain.

 

Question 32. The distances between the hydrogen nuclei in staggered and eclipsed form in ethane respectively are
(a) 2.55 Å & 2.29 Å
(b) 1.54 Å & 1.34 Å
(c) 3.5 Å & 2.5 Å
(d) 2.29 Å & 2.55 Å
Answer: (a) 2.55 Å & 2.29 Å
In simple words: In the staggered form of ethane, the hydrogens are further apart (2.55 Å), while in the eclipsed form, they are closer (2.29 Å).

🎯 Exam Tip: The greater distance between hydrogen atoms in the staggered conformation (2.55 Å) compared to the eclipsed conformation (2.29 Å) directly relates to its lower energy and higher stability due to reduced steric repulsion.

 

Question 33. Energy barrier between staggered and eclipsed form in ethane is
(a) 0.6 kcal / mole
(b) 2.9 kcal / mole
(c) 12 kcal / mole
(d) 14 cal / mole
Answer: (b) 2.9 kcal / mole
In simple words: It takes about 2.9 kilocalories of energy per mole to change ethane from its stable staggered shape to its less stable eclipsed shape.

🎯 Exam Tip: The energy barrier, known as torsional strain, is the energy difference between the most stable (staggered) and least stable (eclipsed) conformations, representing the energy required for rotation around the C-C bond.

 

Question 34. IUPAC name of the following compound

(a) 3 - Ethyl, 5 - methyl heptane
(b) 5 - Ethyl, 3 - methyl heptane
(c) 2 - Ethyl, 5 - methyl heptane
(d) 4 - Ethyl, 5 - methyl heptane
Answer: (a) 3 - Ethyl, 5 - methyl heptane
In simple words: The correct chemical name for this molecule is 3-ethyl-5-methylheptane, following IUPAC rules for naming carbon chains with branches.

🎯 Exam Tip: Always identify the longest carbon chain first to name the parent alkane. Then, number the chain so that the substituents have the lowest possible position numbers. Finally, list the substituents alphabetically.

 

Question 35. Ethylene is converted to ethane in the presence of Ni at 300°C. In this reaction the hybridization of carbon changes from
(a) \( \text{sp} \) to \( \text{sp}^2 \)
(b) \( \text{sp}^2 \) to \( \text{sp}^3 \)
(c) \( \text{sp}^3 \) to \( \text{sp} \)
(d) \( \text{sp} \) to \( \text{sp}^3 \)
Answer: (b) \( \text{sp}^2 \) to \( \text{sp}^3 \)
In simple words: When ethylene (which has \( \text{sp}^2 \) carbons) turns into ethane (which has \( \text{sp}^3 \) carbons) by adding hydrogen, the way the carbon atoms bond changes from \( \text{sp}^2 \) to \( \text{sp}^3 \).

🎯 Exam Tip: Hydrogenation of alkenes involves the addition of hydrogen across the double bond, converting \( \text{sp}^2 \) hybridized carbons (trigonal planar geometry) into \( \text{sp}^3 \) hybridized carbons (tetrahedral geometry).

 

Question 36. In which of the following reactions in the preparation of ethane a new C - C bond is formed
(a) Sabatier - Senderen’s reaction
(b) Reduction of ethyl iodide
(c) Decarboxylation
(d) Kolbe’s electrolysis
Answer: (d) Kolbe’s electrolysis
In simple words: Kolbe’s electrolysis forms a new carbon-carbon bond when making ethane, unlike other methods that just modify an existing carbon chain.

🎯 Exam Tip: Kolbe's electrolytic method involves the dimerization of carboxylate anions, where two alkyl groups combine to form a new C-C bond, making it suitable for preparing symmetrical alkanes.

 

Question 37. Select the correct statements
(a) eclipsed and staggered ethanes give different products on reaction with chlorine in presence of light
(b) the conformational isomers can be isolated at room temperature
(c) torsional strain is minimum in ethane at dihedral angles 60°, 180° and 300°
(d) steric strain is minimum in gauche form of n - butane
Answer: (c) torsional strain is minimum in ethane at dihedral angles 60°, 180° and 300°
In simple words: In ethane, the pushing forces (torsional strain) are smallest when the hydrogen atoms are spread out at angles of 60°, 180°, and 300°.

🎯 Exam Tip: Torsional strain refers to the repulsion between electron clouds of bonds on adjacent atoms. It is minimized in staggered conformations (dihedral angles of 60°, 180°, 300°), where bonds are maximally separated.

 

Question 38. The fully eclipsed conformation of n - butane is least stable due to the presence of
(a) bond opposition strain only
(b) steric strain only
(c) bond opposition strain as well as steric strain
(d) no strain is present in the molecule
Answer: (c) bond opposition strain as well as steric strain
In simple words: The fully eclipsed form of n-butane is the least stable because its atoms are very close, causing both a pushing-away (steric) strain and an energy barrier from electron clouds overlapping (bond opposition strain). These two types of strain make it uncomfortable and high in energy.

🎯 Exam Tip: Remember that "eclipsed" means atoms are directly aligned, causing maximum repulsion and strain, while "staggered" means they are spread out, leading to less repulsion and more stability.

 

Question 39. Vinyl group among the following is
(a) \( (CH_3)_2 CH - \)
(b) \( HC \equiv C - \)
(c) \( H_2C = CH - CH_2 - \)
(d) \( CH_2 = CH - \)
Answer: (d) \( CH_2 = CH - \)
In simple words: The vinyl group is a special part of a molecule that always has two carbon atoms connected by a double bond, with one hydrogen atom on the first carbon and two on the second, and a free bond to attach to other molecules. It's like a small building block with a double bond.

🎯 Exam Tip: Familiarize yourself with common organic functional groups and their names, like vinyl, allyl, and alkyl, as these are basic building blocks in organic chemistry.

 

Question 40. The alkene that exhibits geometrical isomerism is
(a) propene
(b) 2 - methyl propene
(c) 2 - butene
(d) 2 - methyl - 2 - butene
Answer: (c) 2 - butene
In simple words: Geometrical isomerism happens when atoms around a double bond can be arranged in two different ways (cis or trans). 2-butene can do this because each carbon in the double bond has two different groups attached to it.

🎯 Exam Tip: To exhibit geometrical isomerism, each carbon atom of the double bond must be attached to two different groups. If any carbon has two identical groups, it cannot show geometrical isomerism.

 

Question 41.

(a) 5 - methylocta - 1, 3, 5, 7 - tetraene
(b) 4 - methylocta - 1, 3, 5, 7 - tetraene
(c) 4 - butenylocta -1, 3- diene
(d) octa - 1, 5 - diene
Answer: (b) 4 - methylocta - 1, 3, 5, 7 - tetraene
In simple words: This molecule has a main chain of eight carbons with four double bonds at positions 1, 3, 5, and 7, making it an octatetraene. There is also a methyl group attached at the fourth carbon atom, so its name is 4-methylocta-1,3,5,7-tetraene.

🎯 Exam Tip: When naming complex alkenes, always find the longest continuous carbon chain that includes the maximum number of double bonds, and number it from the end that gives the lowest numbers to the double bonds and substituents.

 

Question 42. \( CH_2 = C (CH_2CH_2CH_3)_2 \)
(a) 2 - Propyl pent - 1 - ene
(b) 2 - Propyl pent - 2- ene
(c) 2 - Propyl pent - 3 - ene
(d) 3 - Propyl pent - 1- ene
Answer: (a) 2 - Propyl pent - 1 - ene
In simple words: In this molecule, the longest chain has five carbon atoms (pentane) with a double bond at the first carbon (pent-1-ene). A propyl group is attached to the second carbon atom, so the full name is 2-propylpent-1-ene.

🎯 Exam Tip: When an alkene has branches, the numbering of the main chain must prioritize giving the double bond the lowest possible number, even if it means substituents get higher numbers.

 

Question 43. The number of sigma (\( \sigma \)) and pi (\( \pi \)) bonds in the following structure are
(a) \( \sigma \) bonds - 33 \( \pi \) bonds - 2
(b) \( \sigma \) bonds - 22 \( \pi \) bonds - 2
(c) \( \sigma \) bonds - 42 \( \pi \) bonds - 2
(d) \( \sigma \) bonds - 40 \( \pi \) bonds - 3
Answer: (a) \( \sigma \) bonds - 33 \( \pi \) bonds - 2
In simple words: Every single bond is a sigma bond, and in double or triple bonds, one is a sigma bond and the others are pi bonds. In this specific molecule, if you count all the bonds, you will find 33 sigma bonds and 2 pi bonds.

🎯 Exam Tip: To correctly count sigma and pi bonds, draw the full Lewis structure. Each single bond is one sigma bond. Each double bond has one sigma and one pi bond. Each triple bond has one sigma and two pi bonds.

 

Question 44. In dehydrohalogenation, hydrogen and halogen are removed from
(a) the same carbon atom
(b) from adjacent carbon atoms
(c) from isolate carbon atoms
(d) from any two carbon atoms
Answer: (b) from adjacent carbon atoms
In simple words: In dehydrohalogenation, a hydrogen atom is taken from one carbon, and a halogen atom is taken from the carbon right next to it. This process creates a double bond between these two carbons.

🎯 Exam Tip: Dehydrohalogenation reactions typically follow Zaitsev's rule, favoring the formation of the more substituted (more stable) alkene product when multiple adjacent hydrogens are available.

 

Question 45. Ethylene readily undergoes the following type of reaction.
(a) Elimination
(b) Addition
(c) Rearrangement
(d) Substitution
Answer: (b) Addition
In simple words: Ethylene has a double bond, which makes it very reactive for addition reactions. In these reactions, other atoms or groups are added across the double bond, breaking it and forming a single bond without losing any existing atoms.

🎯 Exam Tip: Alkenes, with their carbon-carbon double bonds, primarily undergo addition reactions where the pi bond breaks, and new atoms or groups attach to the carbons.

 

Question 46. Baeyer’s reagent is
(a) Aqueous bromine solution
(b) Neutral permanganate solution
(c) Acidified permanganate solution
(d) Alkaline potassium permanganate solution
Answer: (d) Alkaline potassium permanganate solution
In simple words: Baeyer's reagent is a special purple liquid that contains potassium permanganate in an alkaline (basic) solution. It is used to test for the presence of double or triple bonds in organic compounds because it loses its color when it reacts with them.

🎯 Exam Tip: Baeyer's reagent is a cold, dilute, alkaline \( KMnO_4 \) solution. It's a key qualitative test for unsaturation (alkenes and alkynes), as it gets decolorized in the presence of double or triple bonds, and a brown precipitate of \( MnO_2 \) forms.

 

Question 47. Baeyer’s reagent oxidizes ethylene to
(a) Ethylene chlorohydrin
(b) Ethyl alcohol
(c) \( CO_2 \) and \( H_2O \)
(d) Ethane - 1, 2 - diol
Answer: (d) Ethane - 1, 2 - diol
In simple words: When ethylene reacts with Baeyer's reagent, the double bond breaks, and two hydroxyl (-OH) groups are added to the two carbon atoms. This forms a compound called ethane-1,2-diol, also known as ethylene glycol.

🎯 Exam Tip: Baeyer's reagent typically performs a syn-dihydroxylation (adding two -OH groups to the same side of the double bond) on alkenes, forming vicinal diols.

 

Question 48. On reductive ozonolysis ethylene gives
(a) Aldehyde
(b) Ketone
(c) Carboxylic acid
(d) Ether
Answer: (a) Aldehyde
In simple words: Reductive ozonolysis is a reaction that breaks double bonds in alkenes. When ethylene undergoes this reaction, each of its carbon atoms turns into an aldehyde group. Since ethylene is a small molecule, it specifically forms formaldehyde.

🎯 Exam Tip: Reductive ozonolysis splits the alkene at the double bond and forms aldehydes or ketones, depending on the substituents on the double-bonded carbons. For ethylene, both carbons form formaldehyde.

 

Question 49. Polythene is obtained by the polymerization of
(a) Styrene
(b) A mixture of ethylene and styrene
(c) Acetylene
(d) Ethene
Answer: (c) Acetylene
In simple words: Polythene is a type of plastic made by joining many small acetylene molecules together in a long chain. This joining process is called polymerization. Acetylene molecules link up one after another to create the large polythene molecule.

🎯 Exam Tip: Polymerization is a fundamental reaction where many identical small molecules (monomers) link up to form a very large molecule (polymer). Understanding the monomer helps identify the polymer.

 

Question 50. Polytetrafluoroethylene is commercially known as
(a) Teflon
(b) Freon
(c) Lewisite
(d) Westron
Answer: (a) Teflon
In simple words: Polytetrafluoroethylene, a long chemical name, is commonly known by its brand name, Teflon. It's used to make non-stick coatings on pans and other items because of its very slippery properties.

🎯 Exam Tip: Many polymers have common names that are more widely used than their scientific names. Knowing these common names, like Teflon for PTFE, is useful for general knowledge and applications.

 

Question 51. Polythene is
(a) \( (- H_2C = CH_2 -)_n \)
(b) \( (- HC = CH -)_n \)
(c) \( (- H_3C - CH_3 -)_n \)
(d) \( (- H_2C - CH_2 -)_n \)
Answer: (d) \( (- H_2C - CH_2 -)_n \)
In simple words: Polythene is made of many repeating units of \( CH_2 - CH_2 \). These units link up one after another to form a very long chain, and 'n' shows how many times this unit repeats.

🎯 Exam Tip: The repeating unit of a polymer is derived from its monomer. For polythene, the monomer is ethene \( (CH_2=CH_2) \), and when polymerized, the double bond breaks to form a single bond, resulting in the \( -CH_2-CH_2- \) repeating unit.

 

Question 52. When ethanol vapours are passed over alumina heated at 350°C, the main product obtained is
(a) \( C_2H_6 \)
(b) \( C_2H_4 \)
(c) \( C_2H_2 \)
(d) \( C_2H_5OC_2H_5 \)
Answer: (b) \( C_2H_4 \)
In simple words: When you heat ethanol over alumina, it removes water from the ethanol molecule. This process is called dehydration, and it forms ethene (ethylene), which is a gas.

🎯 Exam Tip: Alumina (\( Al_2O_3 \)) at high temperatures acts as a dehydrating agent for alcohols, converting them into alkenes by removing a molecule of water.

 

Question 53. Dehydrohalogenation of ethyl chloride in presence of alc. KOH produces the following
(a) \( HC \equiv CH + KCl + H_2O \)
(b) \( CH_4 + KCl + H_2O \)
(c) \( CH_2 = CH_2 + KCl + H_2O \)
(d) \( C_2H_4 + HCl \)
Answer: (c) \( CH_2 = CH_2 + KCl + H_2O \)
In simple words: When ethyl chloride is heated with alcoholic potassium hydroxide, it loses a hydrogen atom and a chlorine atom. This reaction makes ethene (ethylene), potassium chloride, and water.

🎯 Exam Tip: Alcoholic KOH is a strong base commonly used for dehydrohalogenation (elimination of HX) from alkyl halides, leading to the formation of alkenes.

 

Question 54. Ethylene is prepared by
(a) Dehalogenation of chloroform
(b) Pyrolysis of ethane at 450°C
(c) Dehydration of methanol with \( Al_2O_3 \) / 350°C
(d) Methyl chloride on reduction
Answer: (b) Pyrolysis of ethane at 450°C
In simple words: Ethylene can be made by heating ethane to a very high temperature, around 450°C. This process breaks down the ethane molecule into smaller ones, producing ethylene. This breakdown is called pyrolysis.

🎯 Exam Tip: Pyrolysis (also known as cracking) is a thermal decomposition reaction often used in the industry to break larger hydrocarbons into smaller, more valuable ones like alkenes.

 

Question 55. The peroxide effect involves
(a) Ionic mechanism
(b) Free - radical mechanism
(c) Heterolytic fission of double bond
(d) Homolytic fission of double bond
Answer: (b) Free - radical mechanism
In simple words: The peroxide effect, also called the Kharasch effect, works through a mechanism where highly reactive particles called free radicals are formed and react. These free radicals have an unpaired electron.

🎯 Exam Tip: Remember that the peroxide effect (anti-Markovnikov addition of HBr) is specific to HBr and always proceeds via a free-radical mechanism initiated by peroxides.

 

Question 56. In which of the following will Kharasch effect operate?
(a) \( CH_3 - CH_2 - CH = CH_2 + HCl \)
(b) \( CH_3 - CH_2 - CH = CH_2 + HBr \)
(c) \( CH_3 - CH = CH - CH_3 + HBr \)
(d) \( CH_3 - CH_2 - CH = CH_2 + HI \)
Answer: (b) \( CH_3 - CH_2 - CH = CH_2 + HBr \)
In simple words: The Kharasch effect, which is when HBr adds to an alkene in an unexpected way, only happens with HBr. So, only the reaction involving 1-butene and HBr will show this special effect.

🎯 Exam Tip: The Kharasch effect is observed exclusively with HBr. It does not occur with HCl or HI, as the free-radical addition mechanism is only energetically favorable for HBr.

 

Question 57. Anti Markownikoff addition of HBr is not observed in
(a) Propene
(b) Butene - 1
(c) Butene - 2
(d) pentene - 2
Answer: (c) Butene - 2
In simple words: Anti-Markovnikov addition happens with unsymmetrical alkenes. Butene-2 is a symmetrical alkene, meaning both carbons in its double bond have the same number of hydrogens or attached groups, so this special addition rule does not apply to it.

🎯 Exam Tip: Anti-Markovnikov addition occurs with unsymmetrical alkenes in the presence of peroxides. Symmetrical alkenes, like butene-2, do not show anti-Markovnikov behavior because both carbon atoms of the double bond are equally substituted.

 

Question 58. Conditions used for the formation of ethylene glycol from ethylene
(a) bromine water
(b) cold alkaline \( KMnO_4 \)
(c) dil \( H_2SO_4 \), 60°C
(d) Ag / 200°C
Answer: (b) cold alkaline \( KMnO_4 \)
In simple words: Ethylene glycol is made from ethylene using cold, alkaline potassium permanganate solution. This special reagent helps to add two -OH groups to the double bond, turning the alkene into a diol.

🎯 Exam Tip: The conversion of an alkene to a vicinal diol (ethylene glycol in this case) using cold, dilute, alkaline \( KMnO_4 \) is known as Baeyer's test and is a key reaction for identifying unsaturation.

 

Question 59. The olefin which on ozonolysis gives \( CH_3CH_2CHO \) and \( CH_3CHO \) is
(a) 1 - butene
(b) 2 - butene
(c) 1 - pentene
(d) 2 - pentene
Answer: (d) 2 - pentene
In simple words: Ozonolysis breaks a double bond and adds oxygen atoms to where the bond was. If the products are \( CH_3CH_2CHO \) (propanal) and \( CH_3CHO \) (ethanal), it means the original alkene had a double bond between the carbons that turned into these aldehyde groups. Putting them back together shows the original molecule was 2-pentene.

🎯 Exam Tip: To find the original alkene from ozonolysis products, mentally remove the oxygen atoms from the carbonyl groups and connect the two carbon atoms that were originally double-bonded. This effectively reverses the ozonolysis process.

 

Question 60. In the following reaction, A and B respectively are,

(a) \( C_2H_4 \) and alcoholic KOH / \( \Delta \)
(b) \( C_2H_5Cl \) and aqueous KOH / \( \Delta \)
(c) \( C_2H_5OH \) and aq KOH / \( \Delta \)
(d) \( C_2H_2 \) and \( Br_2 \)
Answer: (a) \( C_2H_4 \) and alcoholic KOH / \( \Delta \)
In simple words: The reaction starts with compound A. It reacts with HBr in the presence of heat and a solvent (like ethyl bromide) to give compound B. Then, compound B reacts with alcoholic KOH and heat to go back to compound A. This shows that A is ethene, which can react with HBr to form ethyl bromide, and then ethyl bromide can react with alcoholic KOH to regenerate ethene.

🎯 Exam Tip: Recognize common reagent pairs used for interconversions. HBr addition to an alkene (A to B) and dehydrohalogenation with alcoholic KOH (B to A) are characteristic reactions for converting alkenes to alkyl halides and back.

 

Question 61. The number of possible alkynes with molecular formula \( C_5H_8 \) is
(a) 3
(b) 4
(c) 5
(d) 6
Answer: (a) 3
In simple words: For the chemical formula \( C_5H_8 \), there are three different alkynes you can draw. These include 1-pentyne, 2-pentyne, and 3-methyl-1-butyne. They all have the same number of carbons and hydrogens but are arranged differently.

🎯 Exam Tip: To find all possible isomers, systematically draw all possible carbon skeletons (straight chain, branched), then place the functional group (triple bond) in all unique positions, checking for duplicates by naming them using IUPAC rules.

 

Question 62. The number of open chain structural isomers possible with molecular formula \( C_5H_8 \) is
(a) 7
(b) 6
(c) 5
(d) 4
Answer: (a) 7
In simple words: For the chemical formula \( C_5H_8 \), you can draw seven different open-chain molecules. These include three alkynes and four dienes, all having the same number of carbon and hydrogen atoms but with different structures.

🎯 Exam Tip: When calculating structural isomers for a given molecular formula, consider all possible types of unsaturation (double bonds, triple bonds, rings) and carbon chain arrangements (straight, branched).

 

Question 63. Alkynes exhibit.
(a) Chain isomerism
(b) Position isomerism
(c) Functional isomerism
(d) All of the options
Answer: (d) All of the options
In simple words: Alkynes can show all types of isomerism mentioned: chain isomerism (different carbon chain shapes), position isomerism (triple bond in different places), and functional isomerism (being isomers of other functional groups like dienes).

🎯 Exam Tip: Alkynes are versatile in isomerism because the triple bond can be at different positions, the carbon chain can be branched, and they can be functional group isomers with dienes, providing a wide range of structural possibilities.

 

Question 64. The IUPAC name of the compound having the formula \( CH \equiv C - CH = CH_2 \) is
(a) Butene - 2 - ye
(b) But - 2- yne - 3 - ene
(c) 3 - butane 1 - ene
(d) But - 1 - ene - 3 - yne
Answer: (d) But - 1 - ene - 3 - yne
In simple words: This molecule has a chain of four carbons (but-). It has a double bond at the first carbon (1-ene) and a triple bond at the third carbon (3-yne). So, its full IUPAC name is but-1-ene-3-yne.

🎯 Exam Tip: When a molecule contains both double and triple bonds, the chain is numbered to give the lowest possible numbers to the multiple bonds. If there's a tie, the double bond gets preference for numbering.

 

Question 65. Acetylene can be obtained by the electrolysis of the following compound
(a) Potassium fumerate
(b) Potassium succinate
(c) Potassium acetate
(d) Potassium formate
Answer: (a) Potassium fumerate
In simple words: Acetylene is formed when you use electricity to break down potassium fumerate, which is a salt of fumaric acid. This process, called electrolysis, helps create the triple bond characteristic of acetylene.

🎯 Exam Tip: Kolbe's electrolytic method can produce alkynes from salts of dicarboxylic acids, such as fumaric acid or maleic acid, by a free-radical mechanism.

 

Question 66. The gas obtained when ethylene chloride reacts with alcoholic potash and sodamide is
(a) \( C_2H_4 \)
(b) \( C_2H_6 \)
(c) \( C_2H_2 \)
(d) \( C_2H_5Cl \)
Answer: (c) \( C_2H_2 \)
In simple words: When ethylene chloride is treated first with alcoholic potash and then with sodamide, it undergoes a double elimination reaction. This removes two hydrogen atoms and two chlorine atoms, resulting in the formation of acetylene gas, which has a triple bond.

🎯 Exam Tip: The reaction of a vicinal dihalide (like ethylene chloride) with strong bases like alcoholic KOH or \( NaNH_2 \) is a classic method for preparing alkynes through double dehydrohalogenation.

 

Question 67. PVC is the polymer of the following
(a) Ethyl chloride
(b) Vinyl Chloride
(c) Allyl Chloride
(d) Ethynyl chloride
Answer: (b) Vinyl Chloride
In simple words: PVC, which stands for Polyvinyl Chloride, is a type of plastic. It is made by joining together many small units of vinyl chloride molecules. These small units link up to form the long polymer chains that make up PVC.

🎯 Exam Tip: The name of a polymer often gives a clue to its monomer. "Poly-" means many, so Polyvinyl Chloride comes from many Vinyl Chloride units.

 

Question 68. Which of the following possess acidic hydrogen
(a) \( C_2H_6 \)
(b) \( C_2H_4 \)
(c) \( C_2H_2 \)
(d) \( CH_4 \)
Answer: (c) \( C_2H_2 \)
In simple words: Acetylene (\( C_2H_2 \)) has hydrogen atoms attached to carbons that are part of a triple bond. These hydrogens are slightly acidic because of the unique electron arrangement in the triple bond, allowing them to be removed by strong bases.

🎯 Exam Tip: Terminal alkynes (alkynes with a triple bond at the end of the chain, like acetylene) have acidic hydrogens, which can be identified by their reaction with strong bases or ammoniacal silver nitrate/cuprous chloride solutions.

 

Question 69. Hydrocarbon which gives oxyacetylene flame
(a) ethane
(b) ethene
(c) ethyne
(d) ethanol
Answer: (c) ethyne
In simple words: Ethyne is another name for acetylene. When acetylene is burned with oxygen, it produces a very hot flame called an oxyacetylene flame. This flame is used for welding and cutting metals because of its high temperature.

🎯 Exam Tip: The oxyacetylene torch relies on the high heat released during the combustion of ethyne (acetylene) in oxygen, making it an industrial application of alkyne chemistry.

 

Question 70. The isomer of propyne
(a) Allene
(b) Propene
(c) Cyclo propane
(d) Propane
Answer: (a) Allene
In simple words: Allene is an isomer of propyne, meaning both molecules have the same chemical formula (\( C_3H_4 \)) but different arrangements of atoms. Propyne has a triple bond, while allene has two double bonds next to each other.

🎯 Exam Tip: Isomers have the same molecular formula but different structural arrangements. Propyne and allene are functional group isomers, with one being an alkyne and the other a cumulated diene.

 

Question 71. Bond angle between C - C in alkyne
(a) 109°.28'
(b) 120°
(c) 180°
(d) 60°
Answer: (c) 180°
In simple words: The carbon atoms involved in a triple bond in an alkyne have a linear shape. This linear arrangement causes the bond angle between the carbon atoms to be 180 degrees.

🎯 Exam Tip: Triple-bonded carbons are sp hybridized, which results in a linear geometry around the triple bond, leading to a 180° bond angle.

 

Question 72. The molecule having linear structure is
(a) Methane
(b) Ethylene
(c) Acetylene
(d) Water
Answer: (c) Acetylene
In simple words: Acetylene, also known as ethyne, has a straight-line structure because of its carbon-carbon triple bond. This linear arrangement makes its bond angles 180 degrees.

🎯 Exam Tip: A linear molecular geometry (180° bond angle) is characteristic of sp-hybridized central atoms, such as the carbon atoms in a triple bond.

 

Question 73. The C - C bond length is shortest in
(a) \( C_2H_6 \)
(b) \( C_2H_2 \)
(c) \( C_6H_6 \)
(d) \( C_2H_4 \)
Answer: (b) \( C_2H_2 \)
In simple words: The carbon-carbon bond in acetylene (\( C_2H_2 \)) is a triple bond. Triple bonds are stronger and shorter than double bonds (like in \( C_2H_4 \)) and single bonds (like in \( C_2H_6 \)), making it the shortest among the options. Benzene (\( C_6H_6 \)) has intermediate bond lengths due to resonance.

🎯 Exam Tip: Bond length decreases as bond order increases: triple bonds are shorter than double bonds, which are shorter than single bonds, due to stronger attraction between the atoms.

 

Question 74. The hydrolysis of \( Mg_2C_3 \) produces
(a) acetylene
(b) propyne
(c) butyne
(d) ethylene
Answer: (b) propyne
In simple words: When magnesium carbide (\( Mg_2C_3 \)) reacts with water, it breaks down to form propyne, a hydrocarbon with a triple bond. This reaction is a specific way to make propyne.

🎯 Exam Tip: Certain metal carbides react with water (hydrolysis) to produce specific hydrocarbons, depending on the structure of the carbide anion. \( Mg_2C_3 \) contains the propynyl anion \( C_3^{4-} \), which yields propyne upon hydrolysis.

 

Question 75. 1 - pentyne and 2 - pentyne can be distinguished by
(a) Silver mirror test
(b) Iodoform test
(c) Addition of \( H_2 \)
(d) Baeyers test
Answer: (a) Silver mirror test
In simple words: 1-pentyne has a triple bond at the end of its carbon chain, which gives it an acidic hydrogen. This acidic hydrogen reacts with Tollen's reagent (ammoniacal silver nitrate) to form a white precipitate, creating a "silver mirror." 2-pentyne, however, has an internal triple bond and no acidic hydrogen, so it will not react.

🎯 Exam Tip: The silver mirror test (Tollen's reagent) specifically detects terminal alkynes due to their acidic C-H bond, which is not present in internal alkynes. This is a crucial distinction in alkyne chemistry.

 

Question 76. Acetylene on reaction with silver nitrate shows
(a) Oxidizing property
(b) Reducing property
(c) Basic nature
(d) Acidic nature
Answer: (d) Acidic nature
In simple words: Acetylene reacts with silver nitrate because its hydrogen atoms, attached to the triple-bonded carbons, are slightly acidic. This allows them to be replaced by silver ions, showing that acetylene behaves as a weak acid.

🎯 Exam Tip: The reaction of terminal alkynes with heavy metal ions (like \( Ag^+ \) or \( Cu^+ \)) indicates their acidic nature, as the hydrogen is replaced by the metal to form insoluble metal acetylides.

 

Question 77. The acidic nature of hydrogens in acetylene cannot be explained by the reaction with
(a) Sodium metal
(b) Ammonical cuprous chloride solution
(c) Ammonical silver nitrate solution
(d) HCN
Answer: (d) HCN
In simple words: The acidic nature of acetylene's hydrogen atoms can be shown by reactions with strong bases or metal salts. However, its reaction with hydrogen cyanide (HCN) is a different type of reaction and does not show its acidic hydrogen.

🎯 Exam Tip: Acidic properties are demonstrated by reactions with bases or species that can accept a proton. HCN typically undergoes addition reactions with alkynes, not reactions that highlight their acidic hydrogen character.

 

Question 78. Westron is the solvent obtained by the reaction of chlorine with
(a) Ethylene
(b) Ethyne
(c) Ethane
(d) Methane
Answer: (b) Ethyne
In simple words: Westron is a solvent that is made by reacting chlorine with ethyne, also known as acetylene. This reaction leads to the formation of 1,1,2,2-tetrachloroethane, which is called Westron.

🎯 Exam Tip: Knowing the industrial names for common organic compounds and their synthetic routes is important for applied chemistry questions.

 

Question 79.

(a) \( CH_2 = CH - CH = CH_2 \)
(b) \( HC \equiv C - CH_3 \)
(c) \( CH_2 = CH - CH_3 \)
(d) \( CH_3 - CH_2 - CH_3 \)
Answer: (b) \( HC \equiv C - CH_3 \)
In simple words: This reaction starts with acetylene, which reacts with sodium amide to form sodium acetylide. Then, this sodium acetylide reacts with methyl bromide. This whole process results in the formation of propyne, where a methyl group replaces one of the acetylene hydrogens.

🎯 Exam Tip: The reaction of a terminal alkyne with a strong base (like \( NaNH_2 \)) followed by an alkyl halide is a classic method for forming new C-C bonds and lengthening the carbon chain of the alkyne.

 

Question 80. Hydration of ethyne to ethanal takes place through the formation of
(a) \( CH_3CH(OH)_2 \)
(b) \( CH_2 = CHOH \)
(c) \( CH_2 = CHO^{-} \)
(d) \( CH \equiv C^{-} \)
Answer: (b) \( CH_2 = CHOH \)
In simple words: When ethyne (acetylene) reacts with water, it first forms an unstable molecule called vinyl alcohol. This vinyl alcohol then quickly changes into ethanal (acetaldehyde) through a process called tautomerism.

🎯 Exam Tip: The hydration of alkynes (especially ethyne) follows Markovnikov's rule and typically proceeds via an enol intermediate, which immediately tautomerizes to a more stable aldehyde or ketone.

 

Question 81.

(a) Acetylene
(b) Acetaldehyde
(c) Acetone
(d) Acetic acid
Answer: (b) Acetaldehyde
In simple words: The reaction starts with calcium carbide, which reacts with water (hydrolysis) to produce acetylene (A). Then, acetylene reacts with mercury sulfate and dilute sulfuric acid (hydration) to form acetaldehyde (B). Acetaldehyde is the final product.

🎯 Exam Tip: Remember the two key steps: calcium carbide with water yields acetylene, and then acetylene hydration using mercury salt as a catalyst produces acetaldehyde.

 

Question 82.

(a) Ethyl chloride
(b) 1, 2 dichloro ethene
(c) Vinyl chloride
(d) Ethylidine chloride
Answer: (c) Vinyl chloride
In simple words: To get vinyl chloride, a process called dehydrohalogenation is often used. This involves removing a hydrogen atom and a halogen atom from an alkyl dihalide using a strong base like alcoholic potassium hydroxide. The resulting compound, vinyl chloride, is an important monomer.

🎯 Exam Tip: Recognize that dehydrohalogenation with alcoholic KOH is a common method to introduce unsaturation. Vinyl chloride is a key intermediate in polymer production.

 

Question 83. \( CH_2 = CH_2 \xrightarrow{\text{Polymerisation}} B \). The Polymer ‘B’ is
(a) orlon
(b) PVC
(c) nylon
(d) Teflon
Answer: (b) PVC
In simple words: When vinyl chloride (\( CH_2=CHCl \)) undergoes polymerization, it forms Polyvinyl Chloride (PVC), which is polymer 'B'. Orlon is polyacrylonitrile, nylon is a polyamide, and Teflon is polytetrafluoroethylene.

🎯 Exam Tip: Remember the monomers for common polymers, as their polymerization reactions are frequently tested.

 

Question 84. Benzene is ______ molecule.
(a) Tetrahedral
(b) Planar
(c) Trigonal
(d) Square planar
Answer: (b) Planar
In simple words: Benzene has a flat, ring-like structure where all its atoms lie in the same plane. This flat shape is important for its special chemical properties.

🎯 Exam Tip: Aromatic compounds like benzene are always planar due to the delocalization of pi electrons, which is key for their stability.

 

Question 85. Bond length of C – C in benzene.
(a) 1.34 Å
(b) 1.39 Å
(c) 1.54 Å
(d) 1.20 Å
Answer: (b) 1.39 Å
In simple words: In benzene, the bond between any two carbon atoms is not a single bond or a double bond, but something in between, making its length 1.39 Å. This special length shows that electrons are shared all around the ring, not just between two atoms.

🎯 Exam Tip: The C-C bond length in benzene is intermediate between a typical C-C single bond (1.54 Å) and a C=C double bond (1.34 Å), indicating resonance stabilization.

 

Question 86. The resonance energy Benzene is
(a) 36 kcal / mol
(b) 85.8 kJ / mole
(c) 150.48 kJ / mole
(d) Both (a) and (c)
Answer: (d) Both (a) and (c)
In simple words: Benzene is extra stable because of a special sharing of electrons called resonance, and this extra stability is measured by its resonance energy, which is 36 kilocalories per mole or 150.48 kilojoules per mole. These are two different units for the same amount of energy.

🎯 Exam Tip: Be mindful of units when stating resonance energy values, as both kcal/mol and kJ/mol are commonly used, and ensure you convert correctly if needed.

 

Question 87. In Huckel’s (4n + 2) π rule for aromaticity, ‘n’ represents
(a) Number of carbon atoms
(b) Number of rings
(c) Whole number
(d) Fractional number (or) integer (or) zero
Answer: (c) Whole number
In simple words: In Huckel's rule, 'n' is a simple counting number (0, 1, 2, 3, and so on) that helps decide if a chemical ring is special and stable because of its electrons. It ensures the molecule has the correct number of pi electrons to be aromatic.

🎯 Exam Tip: For aromaticity, 'n' must be a non-negative whole number (0, 1, 2, 3...) when applying the \( (4n+2)\pi \) electron rule.

 

Question 88. Coal tar is obtained as a by product during
(a) Destructive distillation of wood
(b) Destructive distillation of coal
(c) Destructive distillation of bones
(d) steam distillation of light oil
Answer: (b) Destructive distillation of coal
In simple words: Coal tar is a thick, black liquid made when coal is heated very strongly without any air. This process breaks down the coal into many useful things, and coal tar is one of them.

🎯 Exam Tip: Destructive distillation of coal is a key industrial process for producing various useful products, including coal gas, coke, and coal tar.

 

Question 89. Gammaxene is ________ isomer of benzene hexa chloride.
(a) \( \alpha \)
(b) \( \beta \)
(c) \( \gamma \)
(d) \( \delta \)
Answer: (c) \( \gamma \)
In simple words: Gammaxene is a specific type of chemical that is one of the many different shapes or versions (isomers) of benzene hexa chloride. It is specifically known as the gamma isomer.

🎯 Exam Tip: Remember that Gammaxene is also known as Lindane, and it is the most potent insecticidal isomer of benzene hexachloride.

 

Question 90. The empirical formula of benzene and acetylene is/are
(a) \( CH_2, CH \)
(b) \( CH_2, CH_2 \)
(c) CH, CH
(d) \( CH_3, CH_3 \)
Answer: (c) CH, CH
In simple words: The simplest ratio of atoms in both benzene and acetylene is one carbon atom to one hydrogen atom. This means their empirical formula is CH.

🎯 Exam Tip: The empirical formula shows the simplest whole-number ratio of atoms in a compound, while the molecular formula shows the actual number of atoms.

 

Question 91. Chemical name of the insecticide gammaxene is
(a) DDT
(b) Benzene hexa chloride
(c) Chloral
(d) Hexa chloro ethane
Answer: (b) Benzene hexa chloride
In simple words: Gammaxene is another name for a chemical called benzene hexa chloride, which is used to kill insects. It is a powerful pesticide.

🎯 Exam Tip: Gammaxene is also known as Lindane, specifically the gamma-isomer of benzene hexachloride, and is a potent insecticide.

 

Question 92. Which is non benzenoidal aromatic compound?
(a) Benzene
(b) Pyridine
(c) Toluene
(d) Phenol
Answer: (b) Pyridine
In simple words: Pyridine is a special type of stable ring chemical that acts like benzene but has a nitrogen atom inside its ring instead of only carbon atoms. This makes it "non-benzenoidal".

🎯 Exam Tip: Non-benzenoid aromatic compounds possess aromatic character but do not contain a benzene ring, like pyridine or azulene, still obeying Huckel's rule.

 

Question 93. Benzene is purified by
(a) distillation
(b) fractional distillation
(c) Evaporation
(d) sublimation
Answer: (b) fractional distillation
In simple words: Benzene is cleaned and separated from other mixed liquids by a special heating process called fractional distillation. This process uses different boiling points to get a pure form of benzene.

🎯 Exam Tip: Fractional distillation is effective for purifying liquids with close boiling points, making it ideal for separating benzene from coal tar fractions.

 

Question 94. Preparation of benzene from phenol is
(a) Reduction
(b) Oxidation
(c) Addition
(d) Dehydrogenation
Answer: (a) Reduction
In simple words: You can make benzene from phenol by removing oxygen from it, which is a process called reduction. This usually involves reacting phenol with zinc dust to remove the oxygen.

🎯 Exam Tip: Remember the specific reagent used for this reduction: heating phenol with zinc dust efficiently removes the hydroxyl group, yielding benzene.

 

Question 95. The true statement about benzene is
(a) Because of unsaturation benzene easily undergoes addition reactions
(b) There are two types C – C bonds in benzene molecule
(c) There is a cyclic delocalization of pi – electrons in benzene
(d) Mono substitution of benzene gives three isomeric products
Answer: (c) There is a cyclic delocalization of pi – electrons in benzene
In simple words: The electrons in benzene are not fixed in one place but spread out in a circle above and below its flat ring. This makes benzene very stable and gives it special chemical behaviors.

🎯 Exam Tip: Cyclic delocalization of pi electrons is the defining characteristic of aromaticity in benzene, explaining its exceptional stability and preference for substitution over addition reactions.

 

Question 96. – COOH group in electrophilic substitution directs the incoming group to
(a) o – position
(b) p – position
(c) m – position
(d) o – and p – position
Answer: (c) m – position
In simple words: When a \( -COOH \) group is attached to a benzene ring, it guides new chemical groups that want to join the ring to the "meta" position. This group pulls electrons away, making the meta position the most likely place for new additions.

🎯 Exam Tip: Carboxylic acid groups (\( -COOH \)) are meta-directing and deactivating groups because they withdraw electron density from the ortho and para positions, making the meta position relatively more reactive for electrophilic attack.

 

Question 97. Which of the following is not meta directing group?
(a) \( -SO_3H \)
(b) \( -NO_2 \)
(c) \( -CN \)
(d) \( -NH_2 \)
Answer: (d) \( -NH_2 \)
In simple words: The \( -NH_2 \) group does not make new chemicals attach at the "meta" spot on a benzene ring. Instead, it guides them to the "ortho" and "para" spots because it gives electrons to the ring.

🎯 Exam Tip: The \( -NH_2 \) group is a strong ortho-para directing and activating group due to the donation of its lone pair of electrons to the benzene ring via resonance.

 

Question 98. Which among the following is very strong o – , p – directing group?
(a) \( -Cl \)
(b) \( -OR \)
(c) \( -NH_2 \)
(d) \( -NHR \)
Answer: (c) \( -NH_2 \)
In simple words: Among the options, the \( -NH_2 \) group is very good at directing new chemical groups to the "ortho" and "para" spots on a benzene ring. It pushes electrons into the ring strongly, making these positions very attractive.

🎯 Exam Tip: Amino groups (\( -NH_2 \)) are among the strongest activating and ortho-para directing groups in electrophilic aromatic substitution reactions due to their powerful electron-donating resonance effect.

 

Question 99. Cyclo butadiene is said to be
(a) Aromatic
(b) Aliphatic
(c) anti aromatic
(d) heterocyclic
Answer: (c) anti aromatic
In simple words: Cyclobutadiene is a ring-shaped molecule that has a specific number of electrons (\( 4\pi \) electrons) that actually make it less stable than if it were not a ring at all. This special kind of instability is called anti-aromatic.

🎯 Exam Tip: Cyclobutadiene is anti-aromatic because it is cyclic, planar, fully conjugated, but possesses \( 4\pi \) electrons, violating Huckel's rule.

 

Question 100. In the reaction Answer:
(i) propene → propane
\( \text{CH}_3 – \text{CH} = \text{CH}_2 + \text{H}_2 \xrightarrow{\text{Pt}} \text{CH}_3 – \text{CH}_2 – \text{CH}_3 \)
propene \( \quad \quad \quad \quad \quad \quad \) propane
(ii) ethene → ethane
\( \text{CH}_2 = \text{CH}_2 + \text{H}_2 \xrightarrow{\text{Pt}} \text{CH}_3 – \text{CH}_3 \)
ethene \( \quad \quad \quad \quad \) ethane
(iii) prop – 1 – yne → propane
\( \text{CH}_3 – \text{C} \equiv \text{CH} + 2\text{H}_2 \xrightarrow{\text{Pt}} \text{CH}_3 – \text{CH}_2 – \text{CH}_3 \)
prop – 1 – yne \( \quad \quad \quad \quad \quad \quad \quad \) propane
In simple words: To convert alkenes or alkynes into alkanes, hydrogen gas is added in the presence of a platinum catalyst. This process is called hydrogenation, making the molecule saturated.

🎯 Exam Tip: Remember the specific catalysts (like Pt, Pd, or Ni) and reaction conditions (temperature, pressure) for hydrogenation reactions, as they are crucial for full marks.

 

Question 2. write the IUPAC name of the following compounds.
Answer:

In simple words: To name these compounds, first find the longest carbon chain. Then, number the carbons to give the lowest possible numbers to any branches (methyl, ethyl groups). Finally, list the branches in alphabetical order before the main chain name.

🎯 Exam Tip: Always follow the IUPAC rules strictly: find the longest continuous carbon chain, number it correctly, identify and name substituents, and list them alphabetically with their position numbers.

 

Question 3. Write the structural formula for the following compounds.
(i) 3 – Ethyl – 4, 5 – dipropyl octane
(ii) 2, 3 – Dimethyl pentane
(iii) 4 – Ethyl – 2,7 – Dimethyl octane
Answer:
(i) 3 – Ethyl – 4, 5 – dipropyl octane
\( \text{CH}_3 – \text{CH}_2 – \text{CH} – \text{CH} – \text{CH} – \text{CH}_2 – \text{CH}_2 – \text{CH}_3 \)
\( \quad \quad \quad \quad \quad | \quad \quad | \quad \quad | \)
\( \quad \quad \quad \quad \quad \text{CH}_2 \quad \text{CH}_2 \quad \text{CH}_2 \)
\( \quad \quad \quad \quad \quad | \quad \quad | \quad \quad | \)
\( \quad \quad \quad \quad \quad \text{CH}_3 \quad \text{CH}_3 \quad \text{CH}_3 \)
(ii) 2, 3 – Dimethyl pentane
\( \text{CH}_3 – \text{CH} – \text{CH} – \text{CH}_2 – \text{CH}_3 \)
\( \quad \quad \quad | \quad \quad | \)
\( \quad \quad \quad \text{CH}_3 \quad \text{CH}_3 \)
(iii) 4 – Ethyl – 2,7 – Dimethyl octane
\( \text{CH}_3 – \text{CH} – \text{CH}_2 – \text{CH} – \text{CH}_2 – \text{CH}_2 – \text{CH} – \text{CH}_3 \)
\( \quad \quad \quad | \quad \quad \quad \quad | \quad \quad \quad \quad \quad | \)
\( \quad \quad \quad \text{CH}_3 \quad \quad \quad \text{CH}_2 \quad \quad \quad \quad \quad \text{CH}_3 \)
\( \quad \quad \quad \quad \quad \quad \quad | \)
\( \quad \quad \quad \quad \quad \quad \quad \text{CH}_3 \)
In simple words: To draw the structural formula, first draw the longest carbon chain given in the name. Then, add the branches (ethyl, methyl, propyl) at their correct numbered positions. Finally, add enough hydrogen atoms to each carbon so that every carbon has four bonds in total.

🎯 Exam Tip: Always double-check that each carbon atom in your drawn structure has exactly four bonds, and that all branches are at their correct positions as indicated by the IUPAC name.

 

Question 4. Write a short note on
(i) Wurtz reaction
(ii) Corey – House Mechanism
Answer:
(i) Wurtz reaction:
The Wurtz reaction is a chemical reaction where haloalkanes (alkyl halides) react with sodium metal in dry ether. This reaction produces higher alkanes, which means the carbon chain gets longer. It is especially useful for making symmetrical alkanes, where the two parts joined are identical.
Example:
\( \text{CH}_3 – \text{Br} + 2\text{Na} + \text{Br} – \text{CH}_3 \xrightarrow{\text{dry ether}} \text{CH}_3 – \text{CH}_3 + 2\text{NaBr} \)
methyl bromide \( \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \) ethane
(ii) Corey-House Mechanism:
The Corey-House synthesis is a method to create alkanes by reacting an alkyl halide with a lithium dialkylcuprate reagent. This method is very versatile and can produce both symmetrical and unsymmetrical alkanes, making it more flexible than the Wurtz reaction. This reaction helps in forming new carbon-carbon bonds.
Example:
\( \text{CH}_3\text{CH}_2\text{Br} + (\text{CH}_3)_2\text{LiCu} \rightarrow \text{CH}_3\text{CH}_2\text{CH}_3 + \text{CH}_3\text{Cu} + \text{LiBr} \)
Ethyl bromide
In simple words: Wurtz reaction joins two similar alkyl halides using sodium to make a longer alkane. Corey-House is a more flexible way to make alkanes, joining an alkyl halide with a special copper-lithium compound, which can make both same and different sized chains.

🎯 Exam Tip: Remember that Wurtz reaction mainly makes symmetrical alkanes, while Corey-House synthesis is useful for both symmetrical and unsymmetrical alkanes due to its broader applicability with different alkyl halides.

 

Question 5. How is methane prepared from Grignard reagent?
Answer: Methyl chloride reacts with magnesium in the presence of dry ether to form methyl magnesium chloride, which is a Grignard reagent. Methyl magnesium chloride then reacts with water to produce methane. This is a common way to prepare specific alkanes from Grignard reagents.
\( \text{CH}_3 – \text{Cl} + \text{Mg} \xrightarrow{\text{Dry ether}} \text{CH}_3\text{MgCl} \)
chloromethane \( \quad \quad \quad \quad \quad \quad \quad \) methyl magnesium chloride
\( \text{CH}_3\text{MgCl} + \text{H}_2\text{O} \rightarrow \text{CH}_4 + \text{Mg(OH)Cl} \)
methane
In simple words: Methane is made by first mixing methyl chloride with magnesium in dry ether. This creates a special chemical called methyl magnesium chloride. Then, when you add water to this special chemical, it turns into methane gas.

🎯 Exam Tip: Grignard reagents are highly reactive and sensitive to moisture, so ensure the reaction is carried out in dry conditions to prevent premature reaction with water.

 

Question 6. What is meant by pyrolysis? Explain the pyrolysis reaction of Ethane and propane.
Answer: Pyrolysis is the thermal decomposition of organic compounds into smaller fragments. This happens in the absence of air and requires heat. The word 'pyro' means fire, and 'lysis' means separating, so it's like breaking things apart with heat. This process is also known as cracking, and it is widely used in the petroleum industry to convert large hydrocarbons into smaller, more valuable ones.
In the absence of air, when alkane vapors are passed through a red-hot metal tube, they break down into simpler hydrocarbons.
1) Propane pyrolysis:
\( \text{CH}_3 – \text{CH}_2 – \text{CH}_3 \xrightarrow{773 \text{ k}} \text{CH}_3 – \text{CH} = \text{CH}_2 + \text{H}_2 \)
\( \quad \quad \quad \quad \quad \quad \quad \quad \quad \) Propylene
\( \quad \quad \quad \quad \quad \quad \quad \quad \quad \text{CH}_2 = \text{CH}_2 + \text{CH}_4 \)
\( \quad \quad \quad \quad \quad \quad \quad \quad \quad \) ethylene
2) Ethane pyrolysis:
\( 2\text{CH}_3 – \text{CH}_3 \xrightarrow{773 \text{ K}} \text{CH}_2 = \text{CH}_2 + 2\text{CH}_4 \)
Ethane \( \quad \quad \quad \quad \quad \quad \) Ethylene
In simple words: Pyrolysis is when big molecules are broken into smaller ones just by using heat, without air. For example, propane can break into propylene and hydrogen, or into ethylene and methane. Ethane also breaks down into ethylene and methane when heated.

🎯 Exam Tip: Remember that pyrolysis reactions typically occur at high temperatures and yield a mixture of smaller alkanes and alkenes, which is why the products can vary.

 

Question 7. Mention the uses of alkanes.
Answer: The combustion of alkanes produces a lot of heat, which is why they are widely used as fuels. Methane, found in natural gas, is used for home heating and cooking. A mixture of propane and butane is known as LPG (Liquefied Petroleum Gas), which is also used for domestic cooking. Gasoline, a complex mixture of many hydrocarbons, powers internal combustion engines. Alkanes are also crucial as raw materials in the chemical industry.
Carbon black, made from alkanes, is used to produce ink, printer ink, and black pigments. It is also used as a filler in various materials, such as tires. Various alkanes are also used as solvents in laboratories and industries. Small alkanes like methane, ethane, propane, and butane are gases. Pentane through octadecane are liquids, and alkanes with more than eighteen carbon atoms are solids.

No of Carbon Atoms	State at room temperature	Major uses
1-4	Gas	Heating fuel,Cooking fuel
5-7	Low boiling liquid	Solvents, Gasoline
6-12	Liquid	Gasoline
12-24	Liquid	Jet fuel- portable stove fuel
18-50	High boiling liquid	Diesel fuel, lubricant, heating oil
50+	Solid	Petroleum jelly and paraffin wax

In simple words: Alkanes are used as fuels because they burn well and give off heat. Things like natural gas, LPG, and gasoline are all alkanes. They are also used to make carbon black (for inks) and as solvents.

🎯 Exam Tip: When listing uses, categorize them (e.g., fuels, raw materials, solvents) to ensure comprehensive coverage and clarity in your answer.

 

Question 8. Write all possible structural isomers with the molecular formula C₄H₁₀ and name them.
Answer: The molecular formula C₄H₁₀ represents two structural isomers. Structural isomers are compounds that have the same molecular formula but different arrangements of atoms. These compounds can be named according to the IUPAC (International Union of Pure and Applied Chemistry) rules, which provide a systematic way to name organic compounds based on their structure.
(i) Butane (n-butane): This is a straight-chain alkane with four carbon atoms.
\( \text{CH}_3 – \text{CH}_2 – \text{CH}_2 – \text{CH}_3 \)
(ii) 2-Methylpropane (isobutane): This is a branched-chain alkane with a methyl group attached to the second carbon atom of a three-carbon chain.
\( \text{CH}_3 – \text{CH} – \text{CH}_3 \)
\( \quad \quad \quad | \)
\( \quad \quad \quad \text{CH}_3 \)
These structures (i) and (ii) are chain isomers because they differ in the arrangement of the carbon skeleton.
In simple words: For the formula C₄H₁₀, there are two ways to arrange the atoms: one is a straight line of four carbons called butane, and the other is a three-carbon chain with a branch, called 2-methylpropane. These are called chain isomers because their carbon skeletons are different.

🎯 Exam Tip: When identifying isomers, always draw the full structural formula to ensure no atom is missed and that all valencies (bonds) for each atom are correct. For alkanes, focus on the different carbon chain arrangements.

 

Question 9. How is ethene prepared by Kolbe’s electrolytic method?
Answer: Ethene can be prepared using Kolbe's electrolytic method. In this method, an aqueous solution of potassium succinate is electrolyzed. This means an electric current is passed through the solution. The succinate ions move towards the anode (positive electrode), where they lose electrons and undergo decarboxylation (lose carbon dioxide), forming ethene gas. This method is effective for preparing alkenes from dicarboxylic acid salts.
\( \text{H}_2\text{C} – \text{COOK} \)
\( | \quad \quad \quad \quad \quad \quad \) Electrolysis \( \quad \quad \quad \text{H}_2\text{C} – \text{COO}^- \)
\( \text{H}_2\text{C} – \text{COOK} \quad \quad \quad \quad \quad \quad \quad \quad | \quad \quad \quad \quad + 2\text{K}^+ \)
Potassium Succinate \( \quad \quad \quad \text{H}_2\text{C} – \text{COO}^- \)
At anode:
\( \text{H}_2\text{C} – \text{COO}^- \)
\( | \quad \quad \quad \quad \quad \quad \rightarrow \text{CH}_2 \quad \quad \quad + \text{CO}_2(\text{g}) + 2\text{e}^- \)
\( \text{H}_2\text{C} – \text{COO}^- \quad \quad \quad \quad \quad \text{CH}_2 \)
\( \quad \quad \quad \quad \quad \quad \quad \quad \quad \) Ethene
In simple words: To make ethene, a solution of potassium succinate is put into an electric current. At the positive side, the succinate loses electrons and carbon dioxide, leaving behind ethene gas.

🎯 Exam Tip: Remember that Kolbe's electrolysis is a decarboxylation reaction, meaning carbon dioxide is removed, and it produces a symmetrical product at the anode. The process is key for synthesizing hydrocarbons like alkanes and alkenes.

 

Question 10. How are the following compounds prepared by ozonolysis method?
(i) Formaldehyde
(ii) Acetaldehyde
Answer: Ozonolysis is a chemical reaction that involves the cleavage of an alkene or alkyne with ozone, forming an ozonide, which is then typically reduced to produce carbonyl compounds like aldehydes or ketones. This reaction is very useful for determining the position of double or triple bonds in unsaturated compounds.
(i) Formaldehyde preparation:
Formaldehyde is prepared from ethene by ozonolysis. Ethene reacts with ozone to form an ozonide intermediate, which is then cleaved by zinc dust and water to yield formaldehyde.
\( \text{CH}_2 = \text{CH}_2 + \text{O}_3 \rightarrow \) (ethene ozonide) \( \xrightarrow{\text{Zn} / \text{H}_2\text{O}} 2\text{HCHO} \)
ethene \( \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \) Formaldehyde
(ii) Acetaldehyde preparation:
Acetaldehyde is prepared from 2-butene by ozonolysis. 2-Butene reacts with ozone to form an ozonide, which is then broken down by zinc and water to produce acetaldehyde.
\( \text{CH}_3 – \text{CH} = \text{CH} – \text{CH}_3 + \text{O}_3 \rightarrow \) (2-butene ozonide) \( \xrightarrow{\text{Zn} / \text{H}_2\text{O}} 2\text{CH}_3 – \text{CHO} \)
\( \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \) Acetaldehyde
In simple words: Formaldehyde is made by reacting ethene with ozone, then breaking it with zinc and water. Acetaldehyde is made the same way, but starting with 2-butene instead of ethene. Ozonolysis helps break double bonds to make smaller aldehyde molecules.

🎯 Exam Tip: Remember that ozonolysis always breaks carbon-carbon double bonds, and the products depend on the substituents on the original alkene. Zinc is used to prevent the further oxidation of aldehydes to carboxylic acids.

 

Question 11. How ozone reacts with 2 – methyl propene?
Answer: When ozone reacts with 2-methyl propene, it undergoes an ozonolysis reaction. First, ozone adds across the double bond of 2-methyl propene to form an unstable intermediate called an ozonide. This ozonide is then treated with a reducing agent, usually zinc dust and water. The ozonide breaks down, cleaving the carbon-carbon double bond and forming carbonyl compounds. In the case of 2-methyl propene, the products are acetone and formaldehyde.
\( \text{CH}_3 – \text{C} = \text{CH}_2 + \text{O}_3 \rightarrow \) (ozonide intermediate) \( \xrightarrow{\text{Zn} / \text{H}_2\text{O}} \text{CH}_3 – \text{C} = \text{O} + \text{HCHO} \)
\( \quad \quad \quad | \)
\( \quad \quad \quad \text{CH}_3 \)
2 – methyl propene \( \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \) acetone \( \quad \) formaldehyde
In simple words: When 2-methyl propene reacts with ozone, its double bond breaks apart. After that, by adding zinc and water, it turns into two new chemicals: acetone and formaldehyde.

🎯 Exam Tip: For ozonolysis, clearly identify the original alkene and predict the products by mentally "cutting" the double bond and adding oxygen atoms. Remember the reductive work-up with zinc to get aldehydes/ketones.

 

Question 12. How is acetylene prepared from ethylene?
Answer: Acetylene can be prepared from ethylene through a two-step process: halogenation followed by dehydrohalogenation. This conversion demonstrates the interconversion between alkenes and alkynes, which are fundamental in organic synthesis. It's a key method for lengthening carbon chains.
(i) Halogenation of alkenes to form vicinal dihalides:
First, ethylene reacts with bromine (\( \text{Br}_2 \)) to form 1,2-dibromoethane. This is an addition reaction where bromine adds across the double bond.
\( \text{CH}_2 = \text{CH}_2 + \text{Br}_2 \rightarrow \text{CH}_2 – \text{CH}_2 \)
ethylene \( \quad \quad \quad \quad \quad \quad \quad \quad | \quad | \)
\( \quad \quad \quad \quad \quad \quad \quad \quad \text{Br} \quad \text{Br} \)
1, 2 dibromoethane
(ii) Dehalogenation of vicinal dihalides to form alkynes:
Next, 1,2-dibromoethane is treated with a strong base like sodamide (\( \text{NaNH}_2 \)) or alcoholic KOH. This causes the removal of two hydrogen atoms and two bromine atoms, leading to the formation of a triple bond and thus producing acetylene.
\( \text{CH}_2 – \text{CH}_2 \xrightarrow{\text{NaNH}_2} \text{CH} \equiv \text{CH} \)
\( | \quad | \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \) Acetylene
\( \text{Br} \quad \text{Br} \)
bromoethene \( \quad \quad \quad \quad \quad – \text{NH}_3 \)
\( \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad – \text{NaBr} \)
In simple words: To make acetylene from ethylene, first add bromine to ethylene to make 1,2-dibromoethane. Then, use a strong chemical like sodamide to take away two hydrogen and two bromine atoms from this compound, which creates a triple bond and gives you acetylene.

🎯 Exam Tip: Remember that vicinal dihalides (halogens on adjacent carbons) are crucial intermediates for alkyne synthesis via dehydrohalogenation with strong bases like NaNH₂.

 

Question 13. How is acetylene prepared by Kolbe’s electrolytic method?
Answer: Acetylene can be prepared by Kolbe's electrolytic method, specifically through the electrolysis of the sodium or potassium salt of maleic acid or fumaric acid. This electrochemical process is valuable for synthesizing alkynes. During electrolysis, the dicarboxylate ions migrate to the anode where they lose electrons and undergo decarboxylation, leading to the formation of the triple bond.
Example using potassium maleate:
\( \text{CHCOOK} \)
\( || \quad \quad \quad \quad \quad \) Electrolysis \( \quad \quad \quad \text{CHCOO}^- \)
\( \text{CHCOOK} \quad \quad \quad \quad \quad \quad \quad \quad || \quad \quad \quad \quad + 2\text{K}^+ \)
Potassium maleate \( \quad \quad \quad \text{CHCOO}^- \)
At anode:
\( \text{CHCOO}^- \)
\( || \quad \quad \quad \quad \quad \rightarrow \text{CH} \quad \quad \quad + 2\text{CO}_2 + 2\text{e}^- \)
\( \text{CHCOO}^- \quad \quad \quad \quad \quad \text{CH} \)
\( \quad \quad \quad \quad \quad \quad \quad \quad \quad \) acetylene
In simple words: Acetylene is made using Kolbe's method by passing electricity through a solution of a salt like potassium maleate. At the positive end, the salt loses carbon dioxide and forms acetylene.

🎯 Exam Tip: Kolbe's electrolysis is a decarboxylation and dimerization reaction that is particularly useful for synthesizing symmetrical hydrocarbons (alkanes or alkynes) at the anode.

 

Question 14. Write Ozonolysis reaction of Propyne?
Answer: The ozonolysis of propyne involves the reaction of propyne with ozone, followed by hydrolysis, to cleave the carbon-carbon triple bond. This reaction is important for identifying the position of triple bonds and for synthesizing carboxylic acids. The products formed after the breakdown of the ozonide are ethanoic acid and methanoic acid.
\( \text{CH}_3 – \text{C} \equiv \text{CH} \xrightarrow{\text{O}_3} \)
Propyne
\( \quad \quad \quad \quad \text{CH}_3 – \text{C} – \text{C} – \text{H} \)
\( \quad \quad \quad \quad \quad | \quad | \)
\( \quad \quad \quad \quad \text{O} – \text{O} \)
\( \quad \quad \quad \quad \quad | \)
\( \quad \quad \quad \quad \text{O} \)
(Ozonide intermediate)
\( \xrightarrow{\text{H}_2\text{O}_2} \text{CH}_3\text{COOH} + \text{HCOOH} \)
Ethanoic Acid \( \quad \quad \) Methanoic Acid
In simple words: When propyne reacts with ozone, its triple bond breaks. Then, by adding water, it turns into two acids: ethanoic acid and methanoic acid.

🎯 Exam Tip: Ozonolysis of alkynes with an oxidative work-up (like with H₂O₂) generally produces carboxylic acids, while a reductive work-up typically yields aldehydes or ketones, or a mixture of both if the alkyne is unsymmetrical.

 

Question 15. How is BHC prepared? Give its uses.
Answer: BHC stands for Benzene Hexa Chloride. It is prepared by the chlorination of benzene. In this reaction, benzene reacts with three molecules of chlorine (\( \text{Cl}_2 \)) in the presence of sunlight or ultraviolet (UV) light. This leads to an addition reaction where chlorine atoms add to the benzene ring, breaking the aromaticity and forming a saturated cyclic compound. BHC is also known as gammexane or Lindane, and it is a powerful insecticide.
Chlorination of Benzene:
Benzene reacts with three molecules of \( \text{Cl}_2 \) in the presence of sun light or UV light to yield Benzene Hexa Chloride (BHC) \( \text{C}_6\text{H}_6\text{Cl}_6 \). This is known as gammexane or Lindane which is a powerful insecticide.
\[ \text{Benzene} + 3\text{Cl}_2 \xrightarrow{\text{UV light}} \text{BHC} \]
Uses: BHC (gammexane or Lindane) is primarily used as an insecticide to control various pests in agriculture, public health, and veterinary applications. It is effective against a wide range of insects, including lice, ticks, and mites. However, due to its persistence in the environment and potential toxicity, its use has been restricted or banned in many countries.
In simple words: BHC is made by mixing benzene with chlorine gas under sunlight. It's a strong insect killer also known as gammexane, used to fight pests.

🎯 Exam Tip: Note that BHC formation is an addition reaction, not a substitution, because it occurs under UV light and breaks the aromaticity of benzene. The conditions (UV light, not Lewis acid) are key.

 

Question 16. How propane is prepared form 1, 2 – dichloro propane?
Answer: Propane can be prepared from 1,2-dichloropropane by a reduction reaction. This conversion involves removing the chlorine atoms and replacing them with hydrogen atoms, typically using a metal and an acid. This is a common method for reducing alkyl halides to their corresponding alkanes.
The preparation involves treating 1,2-dichloropropane with zinc and hydrochloric acid (HCl), or zinc and ethanol. The zinc metal reduces the dichloride by removing the chlorine atoms and forming zinc chloride, while hydrogen atoms replace the chlorines, yielding propane. This type of reduction is effective for converting vicinal dihalides into alkanes.
\[ \text{CH}_3 – \text{CH} – \text{CH}_2 \xrightarrow{\text{Zn} / \text{CH}_3\text{OH}} \text{CH}_3 – \text{CH}_2 – \text{CH}_3 + \text{ZnCl}_2 \]
\( \quad \quad \quad | \quad | \)
\( \quad \quad \quad \text{Cl} \quad \text{Cl} \)
1, 2 – dichloropropane \( \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \) propane
In simple words: To make propane from 1,2-dichloropropane, you use zinc metal and hydrochloric acid. The zinc removes the chlorine atoms, and hydrogen atoms take their place, turning the compound into propane.

🎯 Exam Tip: Remember that zinc metal in the presence of an acid (like HCl) is a common reducing agent for alkyl halides, effectively converting C-X bonds to C-H bonds.

 

Question 17. Explain the polymerization reaction of alkenes.
Answer: Polymerization is a process where many small molecules, called monomers, combine to form a very large molecule, known as a polymer. In the case of alkenes, this process is called addition polymerization. Alkenes, which contain a carbon-carbon double bond, can react with each other at high temperatures and pressures, and in the presence of a catalyst. The double bonds break, and the alkene molecules link together end-to-end to form long chains. This reaction is fundamental to the plastics industry.
Example:
(i) Ethylene polymerization:
When ethylene is heated in a red-hot iron tube at 573 K, it polymerizes to form polyethylene, a common plastic. In this reaction, many ethylene units join together to form a long chain.
\[ \text{n}(\text{CH}_2 = \text{CH}_2) \xrightarrow{\text{red hot iron tube, } 573 \text{ K}} (\text{CH}_2 – \text{CH}_2)\text{n} \]
ethene \( \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \) polyethylene or polythene
(ii) Propene polymerization:
Propene can also polymerize to form polypropylene, another widely used plastic. This often occurs under specific catalytic conditions.
\[ \text{n}(\text{CH}_3 – \text{CH} = \text{CH}_2) \xrightarrow{473 \text{ K}} (\text{CH}_3 – \text{CH} – \text{CH}_2)\text{n} \]
prop-1-ene \( \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \) polypropylene
(iii) Styrene polymerization:
Styrene polymerizes to form polystyrene, a plastic used in many products, from disposable cups to insulation. This reaction typically occurs via a free radical mechanism.
\[ \text{n}(\text{C}_6\text{H}_5 – \text{CH} = \text{CH}_2) \xrightarrow{\text{free radical polymerization}} (\text{C}_6\text{H}_5 – \text{CH} – \text{CH}_2)\text{n} \]
styrene \( \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \) polystyrene
In simple words: Polymerization is like linking many small alkene molecules together into one very long chain to make a big molecule called a polymer. This happens when alkenes are heated with a catalyst, causing their double bonds to break and join up. For example, ethylene becomes polyethylene.

🎯 Exam Tip: Key aspects to remember for alkene polymerization are the breaking of the double bond, the formation of a long chain, and the general conditions of high temperature, pressure, and catalysts.

 

Question 18. Write a note on acidic nature of Alkynes.
Answer: Alkynes, especially terminal alkynes (those with a triple bond at the end of the carbon chain), can show acidic properties. This acidic nature is due to the presence of a hydrogen atom attached to a carbon atom that is part of a triple bond (\( \text{C} \equiv \text{C} – \text{H} \)). The explanation lies in the hybridization of the carbon atom.
The carbon atom involved in a triple bond is sp hybridized. An sp orbital has a higher s-character (50%) compared to \( \text{sp}^2 \) hybridized orbitals in alkenes (33%) and \( \text{sp}^3 \) hybridized orbitals in alkanes (25%). This higher s-character means that the sp hybrid orbital is closer to the nucleus and is more electronegative. As a result, the sp hybridized carbon atom pulls the electron density of the C-H bond more strongly towards itself. This weakens the C-H bond and makes the hydrogen atom more likely to be removed as a proton (\( \text{H}^+ \)), hence exhibiting acidic behavior. This property allows terminal alkynes to react with strong bases or active metals to form metal acetylides.
\[ \text{CH} \equiv \text{CH} + \text{Na}^+ \rightarrow \text{Na} – \text{C} \equiv \text{C} – \text{Na} + \text{H}_2 \quad \text{(Disodium Acetylide)} \]
\[ \text{CH} \equiv \text{CH} + \text{Ag}^+ \rightarrow \text{Ag} – \text{C} \equiv \text{C} – \text{Ag} + \text{H}_2 \quad \text{(Disilver Acetylide)} \]
\[ \text{CH} \equiv \text{CH} + \text{Cu}^+ \rightarrow \text{Cu} – \text{C} \equiv \text{C} – \text{Cu} + \text{H}_2 \quad \text{(Dicopper Acetylide)} \]
In simple words: Terminal alkynes (those with a hydrogen on the triple bond) are slightly acidic. This is because the carbon in the triple bond holds electrons very tightly, making it easier for the hydrogen to leave as an \( \text{H}^+ \) ion. This lets them react with metals.

🎯 Exam Tip: Focus on the sp hybridization of the carbon in terminal alkynes, which results in increased s-character and electronegativity, leading to the acidic nature of the terminal hydrogen.

 

Question 19. Write a short note on
(i) Wurtz – Fittig reactions
(ii) Friedel Craft’s reaction
Answer:
(i) Wurtz – Fittig reactions:
The Wurtz-Fittig reaction is a variation of the Wurtz reaction. It is used to synthesize alkylarenes (compounds containing both an alkyl group and an aromatic ring). In this reaction, an aryl halide (like bromobenzene) and an alkyl halide (like iodomethane) react with metallic sodium in dry ether. The sodium metal couples the aryl group and the alkyl group, forming a new carbon-carbon bond between them. This is a useful method for attaching an alkyl chain to an aromatic ring.
When a solution of bromo benzene and iodo methane in dry ether is treated with metallic sodium, toluene is formed.
\( \text{C}_6\text{H}_5\text{Br} + 2\text{Na} + \text{ICH}_3 \xrightarrow{\text{ether}} \text{C}_6\text{H}_5 – \text{CH}_3 + \text{NaBr} + \text{NaI} \)
Bromo benzene \( \quad \) Iodo methane \( \quad \quad \quad \) Toluene
(ii) Friedel Craft’s reaction:
Friedel-Crafts reactions are a group of electrophilic aromatic substitution reactions that involve the alkylation or acylation of aromatic compounds. These reactions are typically catalyzed by a Lewis acid, such as anhydrous aluminum chloride (\( \text{AlCl}_3 \)). They are very important for introducing alkyl or acyl groups onto aromatic rings.
When benzene is treated with methyl chloride in the presence of anhydrous aluminium chloride, toluene is formed.
\[ \text{Benzene} + \text{CH}_3\text{Cl} \xrightarrow{\text{Anhydrous AlCl}_3} \text{Toluene} + \text{HCl} \]
In simple words: Wurtz-Fittig reaction joins an aromatic ring to an alkyl chain using sodium. Friedel-Crafts reaction adds an alkyl or acyl group to a benzene ring, usually with a chemical like anhydrous aluminum chloride.

🎯 Exam Tip: Remember the key difference: Wurtz-Fittig couples an aryl halide with an alkyl halide, while Friedel-Crafts alkylation uses an alkyl halide directly on an aromatic compound, both typically with a metal or Lewis acid catalyst.

 

IV. Long Question and answers (5 Marks):

 

Question 1. How to draw structural formula for given IUPAC name with suitable example.
Answer: Drawing the structural formula from an IUPAC name is a systematic process that reverses the steps of naming. It ensures that the compound's structure accurately represents its given name. This is a fundamental skill in organic chemistry as it translates a systematic name into a visual representation of the molecule.
Let's draw the structural formula for **3 – ethyl – 2, 3 – dimethyl pentane**
Step 1: Identify and draw the parent hydrocarbon chain.
The parent hydrocarbon is "pentane," which means it's an alkane with five carbon atoms. Draw a chain of five carbon atoms and number them. Numbering provides a reference for attaching substituents.
\( \text{C} – \text{C} – \text{C} – \text{C} – \text{C} \)
\( 1 \quad 2 \quad 3 \quad 4 \quad 5 \)
Step 2: Attach the alkyl groups (substituents) at their specified positions.
The name indicates "3-ethyl" and "2,3-dimethyl." This means an ethyl group (\( \text{CH}_2\text{CH}_3 \)) is attached to carbon 3, and two methyl groups (\( \text{CH}_3 \)) are attached to carbon 2 and carbon 3, respectively.
\( \text{CH}_3 – \text{C} – \text{C} – \text{CH}_2 – \text{CH}_3 \)
\( \quad \quad | \quad | \)
\( \quad \quad \text{CH}_3 \quad \text{C} \)
\( \quad \quad \quad \quad | \)
\( \quad \quad \quad \quad \text{CH}_3 \)
\( \quad \quad \quad \quad | \)
\( \quad \quad \quad \quad \text{CH}_2 \)
\( \quad \quad \quad \quad | \)
\( \quad \quad \quad \quad \text{CH}_3 \)
Step 3: Add hydrogen atoms to complete the valency of each carbon atom.
Each carbon atom must form a total of four bonds. Add enough hydrogen atoms to each carbon in the skeleton to satisfy this requirement. This makes the molecule stable and complete. For example, carbons at the ends of the main chain and carbons with fewer substituents will have more hydrogens.
\[ \text{CH}_3 – \text{CH} – \text{C} – \text{CH}_2 – \text{CH}_3 \]
\( \quad \quad | \quad | \)
\( \quad \quad \text{CH}_3 \quad \text{CH}_2 \)
\( \quad \quad \quad \quad | \)
\( \quad \quad \quad \quad \text{CH}_3 \)
In simple words: To draw a molecule from its IUPAC name, first draw the main carbon chain. Then, put all the side branches at the correct carbon numbers. Finally, add enough hydrogen atoms to each carbon so that every carbon has four bonds.

🎯 Exam Tip: Always start with the parent chain, then add substituents, and finally, complete hydrogen atoms to satisfy carbon's tetravalency (four bonds). This systematic approach helps avoid errors.

 

Question 2. Explain the conformations of ethane.
Answer: Conformations refer to the different spatial arrangements of a molecule that can be interconverted by rotation about single bonds. For ethane (\( \text{CH}_3 – \text{CH}_3 \)), rotation about the carbon-carbon single bond leads to various conformations. The two extreme and most important conformations are the staggered and eclipsed conformations, along with intermediate forms known as skew conformations. These different arrangements have varying energy levels due to torsional strain and steric hindrance. The flexibility of single bonds allows these molecules to continuously change their shape, though some shapes are more stable than others.
Eclipsed conformation:
In this conformation, the hydrogen atoms on the front carbon are directly aligned with (eclipsed by) the hydrogen atoms on the back carbon. This arrangement results in the minimum distance between electron clouds of the hydrogen atoms, leading to maximum repulsion. Therefore, the eclipsed conformation is the least stable and has the highest energy.

In this conformation, the hydrogen atoms of one carbon are directly behind those of the other. The repulsion between the atoms is maximum and it is the least stable conformer.
In simple words: The different shapes ethane can take by spinning around its middle bond are called conformations. In the "eclipsed" shape, the hydrogen atoms on the front carbon line up perfectly with the ones behind them. This causes the most pushing away (repulsion) between them, making it the least stable shape.

🎯 Exam Tip: When explaining conformations, always highlight the relative positions of substituents and correlate them with energy levels (e.g., eclipsed = high energy, staggered = low energy).

 

Question 3. Explain the steps involved in the mechanism of halogenations of alkane. Halogenation:
Answer: Halogenation is a chemical reaction where one or more hydrogen atoms in an alkane are replaced by halogen atoms. This reaction follows a free radical chain mechanism, which includes three main steps: initiation, propagation, and termination. The process involves a sequence of steps where reactive free radicals are formed and then react with other molecules to form new products, until the radicals are eventually consumed.
In simple words: Halogenation means swapping hydrogen atoms in an alkane for halogen atoms. It happens in three steps: starting, spreading, and stopping, using very reactive particles called free radicals.

🎯 Exam Tip: Remember that halogenation is a substitution reaction, not an addition reaction, and requires light or heat to initiate the free radical mechanism.

 

(i) Chain Initiation:
Answer: The chain starts when UV light causes a chlorine molecule to break apart into two chlorine free radicals. This homolytic fission (equal breaking) creates highly reactive chlorine atoms. We choose the \( \text{Cl} - \text{Cl} \) bond for this breaking because it is weaker than the \( \text{C} - \text{C} \) and \( \text{C} - \text{H} \) bonds found in alkanes.
\( \text{Cl}_2 \xrightarrow{\text{Light or Heat}} 2 \text{Cl}^\bullet \)
In simple words: UV light breaks a chlorine molecule into two chlorine atoms, each with a free electron, starting the reaction.

🎯 Exam Tip: The initiation step always involves the generation of free radicals from a stable molecule, often by light energy.

 

(ii) Propagation:
Answer: This step involves the reaction spreading through a series of steps where free radicals react to form new radicals, continuing the chain.

(a) A chlorine free radical attacks a methane molecule, breaking a \( \text{C} - \text{H} \) bond and creating a methyl free radical and \( \text{HCl} \). This initial attack creates a new radical that can continue the reaction.
\( \text{CH}_4 + \text{Cl}^\bullet \xrightarrow{\text{hv}} \text{CH}_3^\bullet + \text{HCl} \)

(b) The methyl free radical then attacks another chlorine molecule, forming chloromethane (\( \text{CH}_3\text{Cl} \)) and generating a new chlorine free radical. This new chlorine radical can then repeat step (a), ensuring the reaction continues.
\( \text{CH}_3^\bullet + \text{Cl}_2 \rightarrow \text{CH}_3\text{Cl} + \text{Cl}^\bullet \)

(c) The chlorine free radical cycles back to step (a), and steps (a) and (b) are repeated many times. This continuous cycle means the chain reaction keeps going, producing many product molecules from only a few initial radicals.
In simple words: Free radicals react with molecules to make new products and new free radicals, keeping the reaction going in a cycle.

🎯 Exam Tip: Propagation steps are crucial because they regenerate the active species (free radicals), allowing the reaction to proceed multiple times with minimal initiation.

 

(iii) Chain termination:
Answer: Over time, the reaction slows down and stops because the reactants are used up. This happens when two free radicals combine with each other, leading to the end of the chain. This combination removes the reactive species from the system, stopping the propagation steps.
\( \text{Cl}^\bullet + \text{Cl}^\bullet \rightarrow \text{Cl}_2 \)
\( \text{CH}_3^\bullet + \text{CH}_3^\bullet \rightarrow \text{CH}_3 - \text{CH}_3 \)
\( \text{CH}_3^\bullet + \text{Cl}^\bullet \rightarrow \text{CH}_3\text{Cl} \)
In simple words: The reaction stops when two free radicals bump into each other and join, stopping the chain.

🎯 Exam Tip: Termination steps are important for controlling the overall reaction rate and preventing unwanted side reactions.

 

Question 4. Write the IUPAC name of the following compounds.
1) \( \text{CH}_3 - \text{CH} = \text{CH}_2 \)
2) \( \text{CH}_3 - \text{CH} = \text{CH} - \text{CH}_3 \)
3) Answer: Huckel's rule helps to determine if a compound is aromatic. For a compound to be aromatic, it must follow these rules:
(i) The molecule needs to be flat (co-planar).
(ii) It must have electrons that are completely delocalized across the ring structure.
(iii) It must have \((4n+2)\) pi electrons in the ring, where 'n' can be any whole number like 0, 1, 2, and so on. This is called Huckel's rule. For example, a compound with 6 pi electrons for \(n=1\) would be aromatic.

Here are some examples that follow Huckel's rule:
1. **Benzene:** The benzene ring is flat and has 6 delocalized pi electrons. It follows Huckel’s rule because for \(n=1\), \( (4 \times 1) + 2 = 6 \) pi electrons. So, benzene is an aromatic compound.
2. **Naphthalene:** This compound has a flat, two-ring structure with 10 delocalized pi electrons. It follows Huckel’s rule because for \(n=2\), \( (4 \times 2) + 2 = 10 \) pi electrons. Therefore, naphthalene is an aromatic compound.
3. **Anthracene:** This molecule is a flat, three-ring structure with 14 delocalized pi electrons. It follows Huckel’s rule because for \(n=3\), \( (4 \times 3) + 2 = 14 \) pi electrons. Thus, anthracene is an aromatic compound.

Here are some examples that do NOT follow Huckel's rule:
4. **Cyclopentadiene:** This compound has a five-membered ring structure. However, it only has 4 pi electrons, not a \((4n+2)\) number. For \(n=1\), it should have 6 pi electrons to be aromatic. So, it is not an aromatic compound. This means it doesn't have the special stability of aromatic compounds.
5. **Cyclooctatetraene:** This compound has an eight-membered ring structure which is "tub-shaped," meaning it is not flat. Even though it has 8 pi electrons (which follows the \(4n\) rule), it is not aromatic because it is not planar. It does not obey Huckel's rule for aromaticity.
In simple words: Huckel's rule helps us check if a chemical ring is "aromatic" (very stable). The ring must be flat, have electrons that move freely around the whole ring, and have a special number of electrons: 2, 6, 10, 14, and so on. If any of these conditions are not met, the compound is not aromatic.

🎯 Exam Tip: Remember the three key criteria for aromaticity: cyclic, planar, and \((4n+2)\) pi electrons. All three must be met for a compound to be aromatic.

 

Question 15. Explain the Kekule’s structure of benzene.
Answer: In 1865, August Kekulé proposed a structure for benzene. He suggested that benzene has a ring of six carbon atoms, with single and double bonds alternating between them. This is a cyclic and planar structure.

However, there were two main problems with Kekulé's initial structure:
(i) **Number of Isomers:** Kekulé's structure predicted that benzene should form two different "ortho" disubstituted products (where two groups are on adjacent carbons). This is because the position of the double bond would create two slightly different forms. But in reality, benzene only forms one type of ortho disubstituted product. The difference would be whether the adjacent substituents are connected by a single bond or a double bond.
(ii) **Reactivity:** Kekulé's structure suggested that benzene, with its three double bonds, should behave like other alkenes and easily undergo addition reactions. However, benzene is known to be very stable and primarily undergoes substitution reactions, not addition reactions, under normal conditions. This difference in reactivity could not be explained by the alternating single and double bonds alone.

To fix these issues, Kekulé later suggested that benzene is actually a mixture of two forms (like two ways of drawing the alternating bonds) that are quickly changing between each other. This idea of fast equilibrium helped explain some of the observed properties, like the single ortho isomer. This early model laid the groundwork for understanding resonance in benzene. The benzene molecule has a unique stability due to its electron arrangement.
In simple words: Kekulé thought benzene was a ring of six carbons with single and double bonds taking turns. But this idea had two problems: it predicted too many kinds of compounds and didn't explain why benzene was so stable and didn't act like other molecules with double bonds. He then updated his idea, suggesting benzene was a mix of two such alternating structures that quickly changed back and forth.

🎯 Exam Tip: Focus on Kekulé's initial proposal (alternating single/double bonds) and the two key objections (isomer count and reactivity) that led to the idea of resonance or oscillating structures.

 

Question 16. Explain the resonance in benzene.
Answer: Resonance is a concept where a single chemical structure cannot fully explain a molecule's properties. Instead, a molecule's true structure is a "hybrid" of two or more possible structures, called resonance structures. In benzene's case, Kekulé's initial structures (with alternating single and double bonds, shown as forms I and II) are theoretical resonance structures. The actual structure of benzene is a blend, or hybrid, of these forms, often represented with a circle inside the hexagon to show the delocalized electrons (form III).

This means the double bonds are not fixed in one place but are spread out over all six carbon atoms. Spectroscopic measurements confirm this, showing that all carbon-carbon bonds in benzene are exactly the same length (1.40 Å). This length is between a typical carbon-carbon single bond (1.54 Å) and a carbon-carbon double bond (1.34 Å). This equal bond length is strong evidence for resonance, as it shows the electrons are delocalized, making the molecule very stable. Benzene's stability from resonance is a key reason for its unique chemical behavior.
In simple words: Benzene's true structure is not like one drawing with fixed double bonds, but a mix of a few possible drawings. This means its electrons are spread out evenly over all carbon atoms, making all its bonds the same length and making the molecule very stable.

🎯 Exam Tip: Highlight that resonance explains the uniform bond lengths in benzene and its exceptional stability, which is greater than what simple alternating double bonds would suggest.

 

Question 17. Write the industrial preparation of benzene from coal tar.
Answer: Coal tar is a thick, dark liquid produced when coal is heated without air (a process called pyrolysis). Industrially, benzene is prepared from coal tar through fractional distillation.

During this process, coal tar is heated, and different compounds boil and turn into vapor at different temperatures. These vapors are then collected at various levels of a fractionating column. Benzene, along with toluene and xylene, are among the volatile compounds that distill over in a specific temperature range, typically from 350 to 443 Kelvin. This temperature fraction is called "light oil" and is a rich source of benzene and its derivatives. Fractional distillation separates these compounds based on their boiling points, allowing for the collection of relatively pure benzene. This is a very important industrial process, as benzene is a fundamental building block for many other chemicals.

Here is a table summarizing the fractions obtained from coal tar distillation:

NAME OF THE FRACTION	TEMPERATURE RANGE	NAME OF THE COMPONENTS
1. Crude light oil	350 - 443 K	Benzene, Toluene, Xylenes
2. Middle oil	443 - 503 K	Phenol, Naphthalene
3. Heavy oil	503 - 543 K	Naphthalene, Cresol
4. Green oil	543 - 633 K	Anthracene
5. Pitch	Above 633 K	Residue

In simple words: Benzene is made from coal tar, which is a thick liquid from heating coal. We heat the coal tar, and benzene boils off as a vapor at a certain temperature. This vapor is collected and cooled to get benzene.

🎯 Exam Tip: Remember that fractional distillation separates components based on their boiling points and that "light oil" is the fraction containing benzene.

 

Question 18. How is benzene prepared from
(i) Acetylene
(ii) Decarboxylation
(iii) Phenol
Answer: Benzene can be prepared using different methods:

(i) **From Acetylene (Cyclic Polymerization)**:
When acetylene is passed through a red-hot iron tube at 873 K, three molecules of acetylene combine to form one molecule of benzene. This process is called cyclic polymerization or trimerization. This is an efficient way to convert simple alkyne units into an aromatic ring. The red hot iron tube acts as a catalyst for this reaction.
\[ 3 \text{CH} \equiv \text{CH} \xrightarrow{\text{Red Hot Iron tube}} \text{Benzene} \]
(ii) **From Decarboxylation of Aromatic Acid (e.g., Sodium Benzoate)**:
Benzene can be formed by heating the sodium salt of an aromatic carboxylic acid (like sodium benzoate) with soda lime (\(\text{NaOH} + \text{CaO}\)). In this reaction, a carboxyl group (\(-\text{COOH}\)) is removed as carbon dioxide, which then reacts to form sodium carbonate. This method reduces the carbon chain and produces a simpler hydrocarbon.
\[ \text{C}_6\text{H}_5\text{COONa} + \text{NaOH} \xrightarrow{\text{CaO}} \text{C}_6\text{H}_6 + \text{Na}_2\text{CO}_3 \]
(iii) **From Phenol (Reduction)**:
Phenol can be converted to benzene by heating it with zinc dust. The zinc removes the hydroxyl group (\(-\text{OH}\)) from the phenol, acting as a reducing agent. This is a direct method to get benzene from a common aromatic compound.
\[ \text{C}_6\text{H}_5\text{OH} + \text{Zn} \xrightarrow{\Delta} \text{C}_6\text{H}_6 + \text{ZnO} \]
In simple words: Benzene can be made in a few ways: by joining three acetylene molecules with heat, by removing a carbon dioxide part from a sodium salt of an acid, or by taking the oxygen part off phenol using zinc dust. Each method helps create the stable benzene ring.

🎯 Exam Tip: For each preparation method, clearly state the reactants, reagents (like red hot tube, soda lime, zinc dust), and key conditions (like temperature) to score full marks.

 

Question 19. Explain Electrophilic substitution reaction of benzene.
Answer: Benzene is known for undergoing electrophilic substitution reactions. In these reactions, an electrophile (an electron-deficient species) replaces a hydrogen atom on the benzene ring. The delocalized pi electrons in the benzene ring make it electron-rich, attracting electrophiles. Here are some examples:

(i) **Nitration**:
Nitration involves replacing a hydrogen atom with a nitro group (\(-\text{NO}_2\)). Benzene reacts with a nitrating mixture (concentrated \(\text{HNO}_3\) and concentrated \(\text{H}_2\text{SO}_4\)) at temperatures below 333 K. The concentrated \(\text{H}_2\text{SO}_4\) generates the nitronium ion (\(\text{NO}_2^+\)), which is the electrophile. This reaction produces nitrobenzene.
\[ \text{Benzene} + \text{HNO}_3 \xrightarrow[\text{330K}]{\text{Conc. H}_2\text{SO}_4} \text{Nitrobenzene} + \text{H}_2\text{O} \]
(ii) **Halogenation**:
In halogenation, a hydrogen atom is replaced by a halogen atom (like chlorine or bromine). Benzene reacts with halogens (\(\text{Cl}_2\) or \(\text{Br}_2\)) in the presence of a Lewis acid catalyst (like \(\text{FeCl}_3\), \(\text{FeBr}_3\), or \(\text{AlCl}_3\)). For example, with chlorine and \(\text{FeCl}_3\), chlorobenzene is formed.
\[ \text{Benzene} + \text{Cl}_2 \xrightarrow{\text{FeCl}_3} \text{Chlorobenzene} + \text{HCl} \]
(iii) **Sulphonation**:
Sulphonation introduces a sulfonic acid group (\(-\text{SO}_3\text{H}\)) onto the benzene ring. Benzene reacts with fuming sulfuric acid (concentrated \(\text{H}_2\text{SO}_4\) containing \(\text{SO}_3\)) to produce benzenesulfonic acid. The electrophile in this case is sulfur trioxide (\(\text{SO}_3\)).
\[ \text{Benzene} + \text{H}_2\text{SO}_4 \text{ (fuming)} \rightarrow \text{Benzenesulfonic acid} + \text{H}_2\text{O} \]
(iv) **Friedel-Crafts Alkylation (Methylation)**:
This reaction introduces an alkyl group onto the benzene ring. Benzene reacts with an alkyl halide (like methyl chloride) in the presence of an anhydrous Lewis acid catalyst (\(\text{AlCl}_3\)). For example, with methyl chloride, toluene (methylbenzene) is formed.
\[ \text{Benzene} + \text{CH}_3\text{Cl} \xrightarrow{\text{Anhydrous AlCl}_3} \text{Toluene} + \text{HCl} \]
(v) **Friedel-Crafts Acylation (Acetylation)**:
Acylation introduces an acyl group (\(\text{RCO}-\)) onto the benzene ring. Benzene reacts with an acyl chloride (like acetyl chloride, \(\text{CH}_3\text{COCl}\)) in the presence of an anhydrous Lewis acid catalyst (\(\text{AlCl}_3\)). For example, with acetyl chloride, acetophenone is formed.
\[ \text{Benzene} + \text{CH}_3\text{COCl} \xrightarrow{\text{Anhydrous AlCl}_3} \text{Acetophenone} + \text{HCl} \]
In simple words: In these reactions, a special positive-charged chemical (called an electrophile) comes and takes the place of one hydrogen atom on the benzene ring. Benzene's ring of electrons makes it attractive to these positive chemicals. This lets us attach different groups like nitro, halogen, sulfonic acid, or carbon chains to benzene.

🎯 Exam Tip: Remember that Lewis acid catalysts are typically used in Friedel-Crafts reactions and halogenation, and concentrated sulfuric acid is crucial for generating the electrophile in nitration.

 

Question 20. How is benzene converted into
a) Cyclo hexane
b) maleic anhydride
c) 1, 4 cyclo hexadiene
Answer: Benzene can be converted into various other compounds through different reactions:

a) **Cyclohexane (Hydrogenation)**:
Benzene can be converted to cyclohexane by adding hydrogen atoms across its double bonds. This reaction is called hydrogenation and occurs in the presence of a catalyst like platinum (\(\text{Pt}\)) or palladium (\(\text{Pd}\)) at high temperatures. Three molecules of hydrogen are added to one molecule of benzene.
\[ \text{C}_6\text{H}_6 + 3\text{H}_2 \xrightarrow{\text{Pt/Pd}} \text{C}_6\text{H}_{12} \]
b) **Maleic Anhydride (Oxidation)**:
Although benzene is very stable, it can undergo oxidation under strong conditions to form maleic anhydride. This happens when benzene vapor is mixed with oxygen and passed over a vanadium pentoxide (\(\text{V}_2\text{O}_5\)) catalyst at high temperatures (773 K). This reaction breaks the benzene ring and forms a cyclic anhydride.
\[ \text{C}_6\text{H}_6 \xrightarrow[\text{773 K}]{\text{O}_2, \text{V}_2\text{O}_5} \text{Maleic anhydride} \]
c) **1,4-Cyclohexadiene (Birch Reduction)**:
Benzene can be reduced to non-conjugated dienes like 1,4-cyclohexadiene. This is achieved by treating benzene with sodium (\(\text{Na}\)) or lithium (\(\text{Li}\)) in a mixture of liquid ammonia and an alcohol. This specific reduction method creates double bonds that are separated by more than one single bond, resulting in a non-conjugated system.
\[ \text{Benzene} \xrightarrow[\text{Liq. NH}_3]{\text{Na/Li} + \text{R-OH}} \text{Cyclohexa-1,4-diene} \]
In simple words: We can change benzene into cyclohexane by adding hydrogen with a catalyst. We can turn it into maleic anhydride by burning it with oxygen and a special catalyst. And we can make 1,4-cyclohexadiene by using special metals like sodium or lithium in liquid ammonia and alcohol. Each reaction changes the benzene ring in a different way.

🎯 Exam Tip: Pay attention to the specific reagents and conditions for each conversion, as these are crucial for determining the product and mechanism.