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ChemistryInteger+4Question 21 of 60

JEE Advanced 2015 · Paper 2, Chemistry question 21

The number of hydroxyl group(s) in Q is

A bicyclic alcohol (octahydronaphthalene skeleton): the left ring carries a secondary OH (H wedged) next to a gem-dimethyl carbon at the ring fusion; the right ring has one C=C. Treated with H⁺, heat to give P, then aqueous dilute KMnO₄ (excess) at 0 °C to give Q.
Answer and worked solution

Official answer: 4

H⁺, heat. The secondary alcohol ionises. A 1,2-methyl shift from the neighbouring gem-dimethyl carbon gives a tertiary cation, which loses H⁺ to form a tetrasubstituted C=C. P is a diene: this new double bond plus the original one in the other ring.

Cold dilute KMnO₄ (excess). Syn-dihydroxylation of both C=C bonds gives Q, a tetraol.

Answer: 4

Topics:AlkenesAlcoholsReactive intermediates
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