JEE Advanced 2015 · Paper 1 · Chemistry Q30 · Dronacharya Exams

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ChemistryMultiple correct+4 / -2Question 30 of 60

The major product of the following reaction is

2-(3-Oxopentyl)cyclohexanone treated with i. KOH, H₂O and then ii. H⁺, heat.

A

1-Methyl-4,4a,5,6,7,8-hexahydronaphthalen-2(3H)-one: a bicyclic α,β-unsaturated ketone with the methyl on the double-bond carbon next to C=O.

B

A bicyclic ketone with the ring-fusion double bond (β,γ-unsaturated) and a methyl on the carbon between C=O and the double bond.

C

A spiro compound: cyclohexanone spiro-fused at C2 to a cyclobutane bearing an exocyclic ethylidene.

D

A bicyclo[4.2.0]octene (cyclohexane fused to a cyclobutene) bearing a propanoyl group on the cyclobutene double bond.

Answer and worked solution

Official answer: A

KOH forms the enolate at the CH₂ of the ethyl ketone.
It attacks the ring C=O intramolecularly, closing a new six-membered ring (an aldol, as in Robinson annulation).
Acid and heat dehydrate the aldol to the conjugated enone.

The methyl from the ethyl group ends up on the α-carbon of the C=C-C=O: structure (A).

(A)

Topics:Aldol and related condensations

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