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.
Why did the hydrolysis of (R) result in racemic mixture
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → It involved a planar carbocation.
A racemic mixture contains equal amounts of the two enantiomers of a chiral product. It forms in the SN1 mechanism, through a planar carbocation.
(R) \( \ce{(CH3)3C-Br} \) is a 3° halide; it ionises readily because the 3° carbocation is the most stable:
\( \displaystyle \ce{(CH3)3C-Br -> (CH3)3C^+ + Br^-}\quad(\text{slow}) \)
\( \displaystyle \ce{(CH3)3C^+ + OH^- -> (CH3)3C-OH}\quad(\text{fast}) \)
So the racemic mixture in the passage is explained by a planar carbocation intermediate.
Why the other options are incorrect
Option (1) It proceeds via SN2 reaction — Incorrect. SN2 is a one-step back-side attack that gives inversion of configuration, not a racemic mixture.
Option (3) It was a free radical substitution — Incorrect. Free radical substitution is the halogenation of alkanes in light; hydrolysis by \( \ce{OH^{-}} \) is an ionic reaction.
Option (4) The product was achiral — Incorrect. An achiral product gives no racemic mixture, since a racemic mixture is made of two chiral enantiomers; so this does not explain racemisation.
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 – “… 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).”