JEE Chemistry Hydrocarbons MCQs Set 29

Answer: (d) All of these

Explanation:
1. Alkenes are hydrogenated to alkanes over finely divided Ni, Pd or Pt.
2. Hence all three catalysts work.

Teacher's Note:
1. Ni, Pd and Pt are the standard hydrogenation catalysts.

Answer: (a) any of 2-methylbut-1-ene, 2-methylbut-2-ene or 3-methylbut-1-ene

Explanation:
1. Each of the three alkenes has the 2-methylbutane carbon skeleton.
2. Hydrogenation does not change the skeleton, so all three give 2-methylbutane.

Teacher's Note:
1. Hydrogenation only saturates the double bond.

Answer: (a) Zn-Hg/conc. HCl

Explanation:
1. H₂/Pd does not reduce a ketone C=O to CH₂ under normal conditions, and LiAlH₄ stops at the alcohol.
2. Clemmensen reduction (Zn-Hg/conc. HCl) converts C=O into CH₂.

Teacher's Note:
1. Clemmensen: acidic medium. Wolff-Kishner: basic medium.

Answer: (c) an alcohol, an alkene or an ether group (any of these)

Explanation:
1. Conc. HCl dehydrates alcohols, adds to alkenes and cleaves ethers.
2. So acid-sensitive groups such as these should not be present.

Teacher's Note:
1. Use Wolff-Kishner for acid-sensitive substrates.

Answer: (d) ethylcyclohexane

Explanation:
1. Wolff-Kishner reduction converts C=O into CH₂.
2. The ring and the ethyl group are untouched, so ethylcyclohexane forms.

Teacher's Note:
1. Wolff-Kishner: hydrazone forms first, then loses N₂ in strong base.

Answer: (b) NaOH + CaO (soda lime)

Explanation:
1. Heating a sodium carboxylate with soda lime removes CO₂ as Na₂CO₃.
2. CH₃COONa + NaOH → CH₄ + Na₂CO₃.

Teacher's Note:
1. The product has one carbon fewer than the acid salt.

Answer: (c) 2-methylbutane

Explanation:
1. Decarboxylation removes the COONa group and puts H in its place.
2. (CH₃)₂CH–CH₂–CH₂–COONa gives (CH₃)₂CH–CH₂–CH₃, which is 2-methylbutane.

Teacher's Note:
1. Count carbons: the product has one carbon less than the acid.

Answer: (a) butane

Explanation:
1. This is the Wurtz reaction: 2R–X + 2Na → R–R + 2NaX.
2. Two ethyl groups join to give butane.

Teacher's Note:
1. Wurtz gives symmetrical alkanes with an even number of carbons.

Answer: (b) 1-chloropropane

Explanation:
1. Two propyl groups must couple to give a six-carbon straight chain.
2. 2 CH₃CH₂CH₂Cl + 2Na → CH₃(CH₂)₄CH₃ + 2NaCl.

Teacher's Note:
1. 2-Chloropropane would give 2,3-dimethylbutane.

Answer: (a) alkyl halides

Explanation:
1. Wurtz coupling needs an alkyl halide and sodium in dry ether.
2. Acid halides are not used.

Teacher's Note:
1. Methane cannot be made by this method.

Answer: (a) propane

Explanation:
1. CH₃CH₂Cl + 2Li gives CH₃CH₂Li, then CuI gives (CH₃CH₂)₂CuLi.
2. Corey-House coupling with CH₃Cl transfers one ethyl group to CH₃, giving CH₃CH₂CH₃.

Teacher's Note:
1. Corey-House works well with 1° alkyl halides as the second halide.

Answer: (c) F₂ > Cl₂ > Br₂ > I₂

Explanation:
1. Reactivity decreases down the group as the halogen radical becomes less reactive.
2. So F₂ is the most reactive and I₂ the least.

Teacher's Note:
1. Bromine is more selective than chlorine.

Answer: (a) 1-chloropentane, 2-chloropentane and 3-chloropentane (all)

Explanation:
1. Free radical chlorination replaces any of the three types of hydrogen in n-pentane.
2. Hence all three monochloro isomers form.

Teacher's Note:
1. Chlorine is not very selective.

Answer: (b) benzyl chloride (C₆H₅CH₂Cl)

Explanation:
1. Light and no Lewis acid favour free radical substitution in the side chain.
2. The benzylic hydrogen is replaced to give C₆H₅CH₂Cl.

Teacher's Note:
1. Cl₂/FeCl₃ in dark gives ring substitution; Cl₂/hν gives side-chain substitution.

Answer: (a) isomerisation of alkane

Explanation:
1. A straight chain alkane is converted into a branched isomer under these conditions.
2. This is isomerisation.

Teacher's Note:
1. Quaternary carbon cannot be generated by this method.

Answer: (a) cis-hex-2-ene

Explanation:
1. A poisoned (Lindlar-type) catalyst stops at the alkene stage.
2. Syn addition of H₂ gives the cis alkene.

Teacher's Note:
1. Lindlar gives cis; Na/liquid NH₃ gives trans.

Answer: (b) trans-hex-3-ene

Explanation:
1. Birch-type (dissolving metal) reduction adds H₂ in an anti manner.
2. This gives the trans alkene.

Teacher's Note:
1. This reduction is not shown by terminal alkynes.

Answer: (d) 2,5-dimethylhex-2-ene

Explanation:
1. Alcoholic KOH causes dehydrohalogenation.
2. By Saytzeff's rule the more substituted alkene forms: (CH₃)₂C=CH–CH₂–CH(CH₃)₂.

Teacher's Note:
1. Aqueous KOH gives substitution; alcoholic KOH gives elimination.

Answer: (a) 2,5-dimethylhex-3-ene

Explanation:
1. Zinc removes two vicinal bromine atoms (dehalogenation).
2. A C=C forms between C-3 and C-4.

Teacher's Note:
1. Vicinal dihalide + Zn gives an alkene.

Answer: (c) Ph–CHBr–CH₃

Explanation:
1. Markovnikov addition: H goes to the CH₂ carbon.
2. The benzylic carbocation is the more stable one, so Br attaches to the benzylic carbon.

Teacher's Note:
1. Benzylic carbocations are stabilised by delocalisation into the ring.

Answer: (d) butan-2-ol

Explanation:
1. Acid-catalysed hydration adds H and OH across the double bond.
2. Both carbons of the double bond are equivalent, giving CH₃CH(OH)CH₂CH₃.

Teacher's Note:
1. Hydration follows Markovnikov's rule.

Answer: (b) 1,2-dibromopropane

Explanation:
1. Halogens add across C=C in an inert solvent.
2. Br₂ adds to both carbons to form a vicinal dibromide.

Teacher's Note:
1. The reaction is used as a test for unsaturation (bromine water or Br₂/CCl₄ decolourises).

Answer: (c) 1-methylcyclopentanol

Explanation:
1. Oxymercuration-demercuration gives Markovnikov hydration.
2. OH attaches to the more substituted carbon (the one carrying the methyl group).

Teacher's Note:
1. No rearrangement occurs in oxymercuration.

Answer: (a) Hg(OAc)₂, H₂O followed by NaBH₄

Explanation:
1. Butan-2-ol is the Markovnikov hydration product.
2. Oxymercuration-demercuration gives Markovnikov alcohols; hydroboration-oxidation gives the anti-Markovnikov alcohol.

Teacher's Note:
1. Markovnikov: Hg(OAc)₂. Anti-Markovnikov: B₂H₆.

Answer: (a) B₂H₆ followed by H₂O₂/OH⁻

Explanation:
1. Butan-1-ol requires anti-Markovnikov addition of water.
2. Hydroboration-oxidation gives this.

Teacher's Note:
1. The OH ends up on the less substituted carbon.