JEE Chemistry Haloalkanes and Haloarenes MCQs Set 18

Answer: (c)

Explanation:
1. 3-Hydroxycyclohexanone is a beta-hydroxy ketone: the hydrogen on the carbon alpha to C=O is acidic, so loss of water through an enol/enolate (E1cB) gives the conjugated enone (cyclohex-2-enone) very easily.
2. 2-Hydroxycyclohexanone has OH on the alpha carbon (no suitable beta-hydrogen route), and the diol and methylcyclohexanol are simple secondary alcohols.

Teacher's Note:
1. Beta-hydroxy carbonyl compounds dehydrate easily to give conjugated alpha,beta-unsaturated carbonyl compounds.

Answer: (d)

Explanation:
1. Ease of dehydration follows carbocation stability: tertiary > secondary > primary.
2. 3-Methylpentan-3-ol is tertiary; the others are secondary or primary.

Teacher's Note:
1. Reactivity of alcohols in E1 dehydration: 3 degree > 2 degree > 1 degree.

Answer: (d) ethene

Explanation:
1. Hofmann elimination removes the most accessible and most acidic beta-hydrogen, giving the least substituted alkene.
2. The beta-hydrogens of the ethyl group (primary, unhindered) are removed preferentially over those of the butyl group or the ring, so ethene is formed.

Teacher's Note:
1. Hofmann rule: the least substituted alkene is major; the ethyl group is the best leaving alkyl group in such salts.

Answer: (a)

Explanation:
1. Dehydration is E1 and gives the most stable alkene; elimination towards the benzylic carbon gives an alkene conjugated with the ring.
2. Among the stereoisomers the trans (E) isomer, with the large phenyl and isopropyl groups on opposite sides, is more stable than the cis.

Teacher's Note:
1. Conjugation with the benzene ring and trans geometry give the most stable alkene.

Answer: (d) ethylene

Explanation:
1. \( \ce{LiAlH4} \) reduces acetic acid to ethanol (A); \( \ce{PCl5} \) converts it to ethyl chloride (B); alcoholic KOH eliminates HCl to give ethene (C).

Teacher's Note:
1. Alcoholic KOH causes dehydrohalogenation.

Answer: (c)

Explanation:
1. Decolourisation of bromine water needs an alkene to be formed by elimination (E2) with the base.
2. In cyclohexyl bromomethyl ether, the carbon bearing Br (OCH2Br) has no beta-hydrogen (its only neighbour is oxygen), so no alkene can form.
3. The other three each have a beta-hydrogen and give an alkene, which decolourises bromine water.

Teacher's Note:
1. E2 needs a hydrogen on the carbon next to the carbon bearing the leaving group.

Answer: (c) 3,3-dimethylcyclohexene

Explanation:
1. The carbon next to the bromine-bearing carbon on one side is the gem-dimethyl carbon (no hydrogen), so E2 can only remove a hydrogen from the other neighbouring ring carbon.
2. That gives the ring double bond next to the quaternary carbon: 3,3-dimethylcyclohexene.

Teacher's Note:
1. E2: look for the beta-hydrogens first; a quaternary carbon has none.

Answer: (d) hexa-2,4-diene (\( \ce{CH3CH=CH-CH=CH-CH3} \))

Explanation:
1. In 2,4-dibromohexane, \( \ce{CH3-CHBr-CH2-CHBr-CH2-CH3} \), loss of HBr using the hydrogen on C-3 gives the C2=C3 double bond, and loss of the second HBr using a hydrogen on C-5 gives C4=C5.
2. The result is the conjugated hexa-2,4-diene, which is the most stable diene (Saytzeff product); a cumulated or isolated diene would be less stable.

Teacher's Note:
1. With a vicinal pair of halogen sites, the conjugated diene is the more stable product.

Answer: (b) 4-hydroxypentan-2-one

Explanation:
1. 4-Hydroxypentan-2-one is a beta-hydroxy ketone: the hydrogen on C-3 is acidic and loss of water gives the conjugated enone pent-3-en-2-one easily.
2. The alpha-hydroxy ketones cannot do this, and a primary alcohol dehydrates only under forcing conditions.

Teacher's Note:
1. Beta-hydroxy carbonyl compounds dehydrate readily.

Answer: (d) but-1-yne

Explanation:
1. Alcoholic KOH removes HBr to give a vinyl bromide, and sodamide removes a second HBr to give the alkyne.
2. A terminal alkyne is formed by elimination towards C-1 and is trapped as the acetylide by the very strong base, so but-1-yne results after work-up.

Teacher's Note:
1. Vicinal dihalide + strong base (2 eq) gives an alkyne; a very strong base gives the terminal alkyne.

Answer: (c) concentrated \( \ce{H3PO4} \)

Explanation:
1. HCl, HBr and HCl/\( \ce{ZnCl2} \) convert the alcohol into the alkyl halide because their anions are good nucleophiles.
2. Phosphoric acid has a poorly nucleophilic anion, so dehydration (elimination) is the result.

Teacher's Note:
1. Use an acid with a non-nucleophilic conjugate base (H3PO4, H2SO4) for dehydration.

Answer: (b) E2

Explanation:
1. A primary isotope effect (C-H breaks faster than C-D) shows that the C-H bond is broken in the rate-determining step.
2. A primary bromide cannot ionise to a carbocation (so not E1), and a concerted base-induced loss of H and Br is the E2 pathway.

Teacher's Note:
1. Kinetic isotope effect means C-H cleavage is rate determining: E2 (or slow-step E1cB).

Answer: (a) E1cB (a reversible carbanion forms first)

Explanation:
1. Exchange of hydrogen for deuterium in the recovered starting material means the carbon is deprotonated reversibly to a carbanion before bromide leaves.
2. This two-step sequence (carbanion first, then loss of the leaving group) is E1cB.

Teacher's Note:
1. H/D exchange in recovered substrate is evidence for E1cB.

Answer: (c) 3,3-dimethylbut-1-yne, \( \ce{(CH3)3C-C#CH} \)

Explanation:
1. Acetone as the only ozonolysis product means the alkene is \( \ce{(CH3)2C=C(CH3)2} \).
2. That alkene can come from 3,3-dimethylbutan-2-ol: the secondary cation \( \ce{(CH3)3C-CH+-CH3} \) undergoes a 1,2-methyl shift to the tertiary cation \( \ce{(CH3)2CH-C+(CH3)2} \), which loses a proton.
3. Q is that alcohol, from reduction of the ketone obtained by hydration of the terminal alkyne \( \ce{(CH3)3C-C#CH} \) (Markovnikov hydration gives \( \ce{(CH3)3C-CO-CH3} \)).

Teacher's Note:
1. Work backwards from the ozonolysis product.
2. Hg2+/H+ hydration of a terminal alkyne gives a methyl ketone.

Answer: (d) \( \ce{CH3CH=C(CH3)COOH} \)

Explanation:
1. Cyanide adds to the carbonyl group to give the cyanohydrin \( \ce{CH3CH2C(OH)(CH3)CN} \).
2. Hot concentrated sulphuric acid dehydrates the tertiary alcohol (Saytzeff, towards the more substituted carbon) and hydrolyses the nitrile to the acid, giving 2-methylbut-2-enoic acid (this is the textbook route to alpha,beta-unsaturated acids).

Teacher's Note:
1. Cyanohydrin + hot acid: dehydration plus nitrile hydrolysis to an unsaturated acid.