JEE Main 2022 · 28 Jul · Shift 2 · Chemistry Q42

← JEE Main 2022 · 28 Jul · Shift 2

ChemistrySingle correct+4 / -1Question 42 of 90

The major product in the given reaction is

2,6-Dimethylhept-5-en-2-ol, (CH3)2C=CH-CH2CH2-C(CH3)2OH, treated with (1) H⁺, heat; (2) HBr.

Open-chain tertiary bromide keeping the C=C

Option A: open chain (CH3)2C=CH-CH2CH2-C(CH3)2Br.

1-Bromo-1,2,2-trimethylcyclopentane

1-(2-Bromopropan-2-yl)-1-methylcyclopentane

Option C: cyclopentane bearing a methyl and a 2-bromopropan-2-yl group on one ring carbon.

Open-chain secondary bromide

Option D: open chain (CH3)2CH-CHBr-CH2CH2-CH(CH3)2.

Answer and worked solution

Official answer: C

Acid, heat. The tertiary alcohol ionises to a tertiary cation, which the trisubstituted C=C attacks to close a five-membered ring, leaving a new tertiary cation on the side chain.

HBr then gives the tertiary bromide: 1-(2-bromopropan-2-yl)-1-methylcyclopentane in NTA's drawing.

(C)

Topics:Alcohols Reactive intermediates

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JEE Main 2022 · 28 Jul · Shift 2, Chemistry question 42