Consider the following statements about the reactivity of some bromides in SN1 and SN2 reactions.
(A) \( \ce{(CH3)2CHCH2Br} \) is more reactive than \( \ce{CH3CH2CH2CH2Br} \) in SN1 reactions because of the greater electron withdrawing inductive effect of the \( \ce{(CH3)2CH-} \) group.
(B) Among the four isomeric bromobutanes, \( \ce{(CH3)3CBr} \) is the least reactive in SN2 reactions because the steric hindrance around its electrophilic carbon is the greatest.
(C) \( \ce{C6H5CH(C6H5)Br} \) is more reactive than \( \ce{C6H5CH(CH3)Br} \) in SN1 reactions because its carbocation is stabilised by two phenyl groups through the inductive effect.
(D) \( \ce{C6H5CH(C6H5)Br} \) is less reactive than \( \ce{C6H5CH(CH3)Br} \) in SN2 reactions because a phenyl group is bulkier than a methyl group.
Which of the statements given above are incorrect? Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A) and (C) only
SN1 reactivity depends on the stability of the carbocation formed in the slow step. SN2 reactivity depends on how easily the nucleophile can reach the carbon, so steric hindrance decides it.
(A) \( \ce{(CH3)2CHCH2Br} \) is more reactive than \( \ce{CH3CH2CH2CH2Br} \) in SN1 reactions because of the greater electron withdrawing inductive effect of the \( \ce{(CH3)2CH-} \) group. — Incorrect statement. The comparison is right, but the reason is not. The (CH3)2CH– group has a greater electron donating inductive effect; an electron donating group stabilises the carbocation, while an electron withdrawing group would destabilise it.
(B) Among the four isomeric bromobutanes, \( \ce{(CH3)3CBr} \) is the least reactive in SN2 reactions because the steric hindrance around its electrophilic carbon is the greatest. — Correct statement. Three methyl groups crowd the carbon bearing bromine, so the tertiary bromide comes last in the SN2 order.
(C) \( \ce{C6H5CH(C6H5)Br} \) is more reactive than \( \ce{C6H5CH(CH3)Br} \) in SN1 reactions because its carbocation is stabilised by two phenyl groups through the inductive effect. — Incorrect statement. The two phenyl groups stabilise the carbocation by resonance: the positive charge spreads over both rings. The comparison is right, but the effect named is wrong.
(D) \( \ce{C6H5CH(C6H5)Br} \) is less reactive than \( \ce{C6H5CH(CH3)Br} \) in SN2 reactions because a phenyl group is bulkier than a methyl group. — Correct statement. The bulkier phenyl group hinders the approach of the nucleophile, so this bromide reacts more slowly by SN2 though it is faster by SN1.
NCERT (2026-27 edition):
6.7.1 Reactions of Haloalkanes – “(i) CH3CH2CH2CH2Br < (CH3)2CHCH2Br < CH3CH2CH(Br)CH3 < (CH3)3CBr (SN1)
CH3CH2CH2CH2Br > (CH3)2CHCH2Br > CH3CH2CH(Br)CH3> (CH3)3CBr (SN2)”
“Of the two primary bromides, the carbocation intermediate derived from (CH3)2CHCH2Br is more stable than derived from CH3CH2CH2CH2Br because of greater electron donating inductive effect of (CH3)2CH- group. … The reactivity in SN2 reactions follows the reverse order as the steric hinderance around the electrophilic carbon increases in that order.”
“…Of the two secondary bromides, the carbocation intermediate obtained from C6H5CH(C6H5)Br is more stable than obtained from C6H5CH(CH3)Br because it is stabilised by two phenyl groups due to resonance. Therefore, the former bromide is more reactive than the latter in SN1 reactions. A phenyl group is bulkier than a methyl group. Therefore, C6H5CH(C6H5)Br is less reactive than C6H5CH(CH3)Br in SN2 reactions.”