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.
The reversal of reactivity order in high concentration of NaOH in aqueous medium is because the mechanism shifts to
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → E2
Note 1: The given answer is as per the official NTA Answer Key. However, the correct answer should be Option (1) SN1, as explained below.
Alkyl halides react with \( \ce{OH^{-}} \) by substitution (SN1, SN2), which gives an alcohol (hydrolysis), or by elimination (E1, E2), which gives an alkene:
SN2: one step, back-side attack; reactivity CH3X > 1° > 2° > 3°; inversion of configuration.
SN1: slow ionisation to a planar carbocation, then attack from either face; reactivity 3° > 2° > 1° > CH3X (carbocation stability); racemisation of a chiral substrate.
E2: one step; the base removes a β-hydrogen while X leaves, forming a C=C bond.
In the second set of conditions the order is R > Q > P, that is 3° > 2° > \( \ce{CH3Br} \). The passage calls the reaction hydrolysis and says (R) gave a racemic mixture, which needs a planar carbocation. Q43 of this paper asks “Why did the hydrolysis of (R) result in racemic mixture”, and its answer is “It involved a planar carbocation.” Both point to SN1.
Option (1) SN1 — Correct as per NCERT. Its reactivity order, 3° > 2° > 1° > CH3X, is the reversed order of the second set of conditions, and its planar carbocation explains the racemic mixture.
\( \displaystyle \ce{(CH3)3C-Br -> (CH3)3C^+ + Br^-}\quad(\text{slow}) \)
Option (2) SN2 — Incorrect. This is the mechanism of the first set of conditions, where the order was P > Q > R (see Q41).
Option (3) E1 — Incorrect. It passes through a carbocation like SN1, but the carbocation loses \( \ce{H^{+}} \) to give an alkene, not a hydrolysis product; \( \ce{CH3Br} \) cannot eliminate either.
Option (4) E2 — The answer given by NTA, but incorrect. A strong base at high concentration can push a 3° halide towards E2, but E2 gives an alkene, not the alcohol of hydrolysis; it has no carbocation, so it cannot give a racemic mixture; and \( \ce{CH3Br} \) has no β-hydrogen, so a series R > Q > P for one mechanism cannot be E2.
Note 2: 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.