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

(±) 2-Butanol

The correct answer is Option (3) →(±) 2-Butanol

2-Chlorobutane is a secondary alkyl halide. During hydrolysis in aqueous medium, it generally undergoes an Sₙ1 reaction mechanism.

Step 1: Formation of carbocation

CH₃–CHCl–CH₂–CH₃ → CH₃–CH⁺–CH₂–CH₃ + Cl⁻

A planar secondary carbocation is formed.

Step 2: Nucleophilic attack

Water (or OH⁻) attacks the carbocation. Since the carbocation is planar, the nucleophile can attack from either side of the plane. Thus two products are formed: • (+)-2-butanol • (−)-2-butanol

These two forms are enantiomers.

When equal amounts of both are formed, the product is a racemic mixture, represented as: (±) 2-butanol

Option C: (±) 2-Butanol

Both enantiomers form in equal amounts due to attack on the planar carbocation from either side, producing a racemic mixture. Correct.