Hydrolysis of optically active 2-bromobutane results in formation of
A. (+) Butan-2-ol
B. (–) Butan-2-ol
C. but-2-ene
D. but-1-ene
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → A and B
2-Bromobutane is a secondary halide with a chiral carbon (C-2). Its hydrolysis replaces Br by OH. When this substitution goes by the SN1 mechanism, the bromide ion leaves first and a planar carbocation is formed:
\( \displaystyle \ce{CH3CH2CH(Br)CH3 -> CH3CH2CH^+CH3 + Br^-} \)
The nucleophile (water or \( \ce{OH^{-}} \)) can attack this flat carbocation equally from both faces. Attack from one face keeps the configuration and attack from the other inverts it, so the two enantiomers form in equal amounts. The optically active halide therefore gives a racemic mixture of butan-2-ol. This is called racemisation. The hydrolysis is:
\( \displaystyle \ce{CH3CH2CH(Br)CH3 + H2O -> CH3CH2CH(OH)CH3 + HBr} \)
(A) (+) Butan-2-ol — Correct. One of the two enantiomers of the racemic product.
(B) (–) Butan-2-ol — Correct. The other enantiomer, formed in the same amount, so the product is optically inactive.
(C) but-2-ene — Incorrect. Alkenes are formed by elimination (dehydrohalogenation), which needs alcoholic KOH, not hydrolysis:
\( \displaystyle \ce{CH3CH2CH(Br)CH3 + KOH} \xrightarrow[\text{heat}]{\text{alcohol}} \ce{CH3CH=CHCH3 + KBr + H2O} \)
(D) but-1-ene — Incorrect. This is the minor product of the same elimination, not a hydrolysis product:
\( \displaystyle \ce{CH3CH2CH(Br)CH3 + KOH} \xrightarrow[\text{heat}]{\text{alcohol}} \ce{CH3CH2CH=CH2 + KBr + H2O} \)
NCERT (2026-27 edition):
Reactions of Haloalkanes – “… In case of optically active alkyl halides, SN1 reactions are accompanied by racemisation. … Actually the carbocation formed in the slow step being sp2 hybridised is planar (achiral). The attack of the nucleophile may be accomplished from either side of the plane of carbocation resulting in a mixture of products, one having the same configuration (the –OH attaching on the same position as halide ion) and the other having opposite configuration (the –OH attaching on the side opposite to halide ion). This may be illustrated by hydrolysis of optically active 2-bromobutane, which results in the formation of (±)-butan-2-ol.”