By which mechanism does isopropyl chloride (2-chloropropane) undergo…

The correct answer is Option (1) → S_N1 and S_N2 mechanisms

Isopropyl chloride, \( \ce{(CH3)2CHCl} \), is a secondary (2°) halide: the carbon carrying Cl is bonded to two other carbons.

• S_N2: the nucleophile attacks the carbon from the back, so the rate falls as the carbon gets more crowded: 1° > 2° > 3°. Two methyl groups hinder the attack only moderately, so a 2° halide can still react by S_N2.

• S_N1: the C–Cl bond first breaks to give a carbocation, so the rate rises with carbocation stability: 3° > 2° > 1°. A secondary carbocation is stable enough to form, so a 2° halide can also react by S_N1.

Lying in the middle of both reactivity orders, 2-chloropropane is hydrolysed by both mechanisms:

    \( \displaystyle \ce{(CH3)2CHCl + OH^- -> (CH3)2CHOH + Cl^-} \)

Why the other options are incorrect

Option (2) Neither S_N1 nor S_N2 mechanisms — Incorrect. The halogen is on an sp³ carbon, so the halide does undergo nucleophilic substitution and gives propan-2-ol.

Option (3) S_N1 mechanism only — Incorrect. Substitution almost only by S_N1 is shown by tertiary halides, whose bulky groups block S_N2 attack.

Option (4) S_N2 mechanism only — Incorrect. Substitution almost only by S_N2 is shown by methyl and primary halides, whose carbocations are too unstable for S_N1.

NCERT (2026-27 edition):
6.7.1 Reactions of Haloalkanes – “… Tertiary halides are the least reactive because bulky groups hinder the approaching nucleophiles. Thus the order of reactivity followed is: Primary halide > Secondary halide > Tertiary halide.”

“… Further, greater the stability of carbocation, greater will be its ease of formation from alkyl halide and faster will be the rate of reaction. In case of alkyl halides, 3° alkyl halides undergo S_N1 reaction very fast because of the high stability of 3° carbocations. We can sum up the order of reactivity of alkyl halides towards S_N1 and S_N2 reactions as follows:”