Alkyl halide
The reactivity of alkyl halides is influenced by their chemical structure and the reaction conditions, leading to different mechanistic pathways. A chemist was studying the hydrolysis of the different bromo-alkanes (P) CH3Br (Q) (CH3)2CHBr and (R) (CH3)3C-Br. When reacted at a low concentration of NaOH in aqueous acetone medium, the rate of reaction was found to be P > Q > R. However, when the same reaction was carried out with a high conc. NaOH in purely aqueous solution, the order of reactivity reversed, becoming R > Q > P. Furthermore, the hydrolysis of (R) resulted in the formation of a racemic mixture while (P) led to the inversion of configuration.
Methyl chloride, methyl bromide, ethyl chloride and some chlorofluoromethanes are gases at room temperature. Higher members are liquids or solids. Molecules of organic halogen compounds are generally polar. Due to greater polarity as well as higher molecular mass as compared to the parent hydrocarbon, the intermolecular forces of attraction (dipole-dipole and van der Waals) are stronger in the halogen derivatives. This is the reason, the boiling points of chlorides, bromides and iodides are considerably higher than those of the hydrocarbons of comparable molecular mass.
The attractions get stronger as the molecules get bigger in size and have more electrons. For the same alkyl group, the boiling points of alkyl halides decrease in the order: RI > RBr > RCl > RF. This is because with the increase in size and mass of halogen atom, the magnitude of van der Waals forces increases.
In the first set of conditions (low concentration NaOH in aqueous acetone) the predominant mechanism operating is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → SN2
Alkyl halides undergo nucleophilic substitution by two mechanisms:
SN2 (one step): the nucleophile attacks from the side opposite to the halogen while the C–X bond breaks. Rate = k\( \ce{[RX][OH^{-}]} \). The attack is blocked by bulky groups, so reactivity is CH3X > 1° > 2° > 3°, and the product shows inversion of configuration.
SN1 (two steps): the C–X bond first ionises slowly to a planar carbocation, which the nucleophile then attacks from either side. Rate = k[RX]. Reactivity follows carbocation stability, 3° > 2° > 1° > CH3X, and a chiral substrate gives a racemic mixture.
In the first set of conditions the rate was P > Q > R, that is \( \ce{CH3Br} \) > \( \ce{(CH3)2CHBr} \) (2°) > \( \ce{(CH3)3CBr} \) (3°). The least crowded halide reacts fastest, which is the SN2 order. The passage also links (P) with inversion of configuration, the hallmark of SN2.
For the fastest halide, (P), the substitution is:
\( \displaystyle \ce{CH3-Br + OH^- -> CH3-OH + Br^-} \)
So the predominant mechanism is SN2.
Why the other options are incorrect
Option (1) SN1 — Incorrect. Its order is the reverse, 3° > 2° > 1°, because it depends on carbocation stability.
Option (3) E1 — Incorrect. An elimination through a carbocation; it gives alkenes, not hydrolysis products, and also favours 3° halides.
Option (4) E2 — Incorrect. An elimination giving alkenes; \( \ce{CH3Br} \), the fastest here, has no β-hydrogen and cannot eliminate at all.
Note: It seems that there is a mistake in the original NTA paper, which says that the hydrolysis of (R) (CH3)3C-Br gave a racemic mixture and that (P) CH3Br showed inversion of configuration; neither compound is chiral, so neither observation can be made on them.
NCERT (2026-27 edition):
Reactions of Haloalkanes – …
