Which of the following reacts with HBr at the fastest rate?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 1-Methylcyclohexan-1-ol (a tertiary alcohol)
An alcohol reacts with HBr to give an alkyl bromide; the –OH group is replaced by Br. The rate depends on the class of the alcohol, in the order 3° > 2° > 1°, because a tertiary alcohol forms the most stable (tertiary) carbocation most easily.
In Option (2) the –OH group sits on a ring carbon that also carries a methyl group, so that carbon is bonded to three other carbons: a tertiary alcohol.
\( \displaystyle \ce{CH3C6H10OH + HBr -> CH3C6H10Br + H2O} \)
The product is 1-bromo-1-methylcyclohexane.
Why the other options are incorrect
Option (1) Cyclohexanol — Incorrect. The –OH carbon carries one H and two ring carbons: a secondary alcohol, which reacts more slowly.
Option (3) 3-Methylcyclohexan-1-ol — Incorrect. The methyl group is not on the –OH carbon, which still carries one H: a secondary alcohol.
Option (4) 4-Methylcyclohexan-1-ol — Incorrect. Again the –OH carbon carries one H: a secondary alcohol.
NCERT (2026-27 edition):
6.4.1 From Alcohols – “The hydroxyl group of an alcohol is replaced by halogen on reaction with concentrated halogen acids, phosphorus halides or thionyl chloride. … The order of reactivity of alcohols with a given haloacid is 3°>2°>1°.”