Identify the incorrect statement for nucleophilic substitution reactions of alkyl halides.
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → Allylic and benzylic halides are inert towards SN1 reaction.
Alkyl halides undergo nucleophilic substitution by two mechanisms:
SN2 (one step): the nucleophile attacks from the side opposite to the leaving group, passing through a transition state in which both are partly bonded. Rate = k[RX][Nu−]; the product has an inverted configuration.
SN1 (two steps): the C–X bond first breaks slowly to give a carbocation, which the nucleophile then attacks quickly. Rate = k[RX]; the more stable the carbocation, the faster the reaction.
Allylic and benzylic halides give carbocations that are stabilised by resonance: the positive charge spreads over the double bond or the benzene ring.
\( \ce{CH2=CH-CH2^{+}} \) ↔ \( \ce{^{+}CH2-CH=CH2} \)
Option (1) SN2 reaction proceeds with inversion of configuration. — A correct statement. Back-side attack turns the molecule inside out, like an umbrella in the wind.
Option (2) SN2 reaction involves formation of a transition state. — A correct statement. In the single step the carbon is briefly bonded to both the incoming and the leaving group.
Option (3) Allylic and benzylic halides are inert towards SN1 reaction. — The incorrect statement. Their carbocations are stabilised by resonance, so they are highly reactive in SN1 reactions.
Option (4) In SN1 reaction, the rate of reaction depends on the concentration of only one reactant. — A correct statement. Only the alkyl halide takes part in the slow step, so rate = k[RX].
NCERT (2026-27 edition):
Reactions of Haloalkanes – “… Since the rate of reaction depends upon the slowest step, the rate of reaction depends only on the concentration of alkyl halide and not on the concentration of hydroxide ion. …”
“For the same reasons, allylic and benzylic halides show high reactivity towards the SN1 reaction. The carbocation thus formed gets stabilised through resonance as shown below:
H2C=CH–CH2+ ↔ +H2C–CH=CH2”