Read the passage carefully and answer the Questions.
Haloalkanes undergo nucelophilic substitution reactions owing to the polarity of C-X bond. The nucleophile reacts with the haloalkane on the carbon possessing a partial positive charge holding the halogen atom. The halogen atom X is replaced by a nucleophile. Depending on the kinetics and mode of bond breaking, the mechanism can be either \( S_N1 \) or \( S_N2 \) reaction. The rate of \( S_N1 \) reaction is governed by the stability of carbocation and in \( S_N2 \) reaction, the rate of reaction is governed by steric factor. Chirality is the main factor in both \( S_N1 \) and \( S_N2 \). In \( S_N1 \) reaction, the chirality of alkyl halide is accompanied by racemization of the product, while in \( S_N2 \) reaction, the product is characterized by inversion of configuration. The structure of alkyl halide and the nature of the solvent also governs the mechanism of the substitution.
Hydrolysis of 2-Chlorobutane will result in formation of
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (±) 2-Butanol
The stereochemical result of a substitution at a chiral carbon depends on the mechanism. In an \(S_N2\) reaction the nucleophile attacks from the side opposite to the halogen, so an optically active halide gives a product with the inverted configuration. In an \(S_N1\) reaction the halide first ionises to a carbocation. This carbocation is sp2 hybridised and planar, so the nucleophile can attack it from either face with equal ease and the two enantiomers are formed in equal amounts.
2-Chlorobutane, \( \ce{CH3CH2CHClCH3} \), is a secondary halide, and its C-2 carries four different groups (H, Cl, \( \ce{CH3} \) and \( \ce{C2H5} \)), so it is chiral. In hydrolysis the nucleophile is water, a weak nucleophile that is also a polar protic solvent. Under these conditions the C–Cl bond ionises first and water then attacks the carbocation:
\( \displaystyle \ce{CH3CH2CHClCH3} \xrightarrow{\text{slow}} \ce{CH3CH2CH^+CH3 + Cl^-} \)
\( \displaystyle \ce{CH3CH2CH^+CH3 + H2O} \xrightarrow{\text{fast}} \ce{CH3CH2CH(OH)CH3 + H^+} \)
\( \displaystyle \ce{CH3CH2CHClCH3 + H2O -> CH3CH2CH(OH)CH3 + HCl} \)
So the product is the optically inactive racemic mixture, (±)-2-butanol.
Why the other options are incorrect
Option (1) (+) 2-Butanol — Incorrect. A single enantiomer would need the nucleophile to attack from one side only, as in an \(S_N2\) reaction of a pure enantiomer; the planar carbocation is attacked from both faces.
Option (2) (-) 2-butanol — Incorrect. For the same reason the (−) form is not obtained alone; it makes up only half of the product.
Option (4) 1-Butanol — Incorrect. The –OH group takes the place of the chlorine on C-2, so the product is a butan-2-ol; butan-1-ol, \( \ce{CH3CH2CH2CH2OH} \), would come from 1-chlorobutane.
Note: In the original NTA paper, Option (4) was printed as “2-Butanol”. This is also the product of the hydrolysis, written without its optical form, so it could also be marked correct. So that students are not confused and the question has a single answer, Option (4) has been changed to “1-Butanol”.
NCERT (2026-27 edition):
Stereochemical Aspects of Nucleophilic Substitution Reactions – “… 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.”