Which among the following halide/s will not show S_N 1 reaction:
(A) H₂ C=CH-CH₂ Cl
(B) CH₃-CH=CH-Cl
(C)
(D)
Choose the most appropriate answer from the options given below :

Solution

Chemistry ma’am · 3 steps

SN1 lives or dies by the carbocation: check which cation each chloride would leave behind.

1. What SN1 needs. In S_N1 the slow first step is the C–Cl bond breaking on its own to give a carbocation. If that carbocation is too unstable to form, the halide cannot react by S_N1.

   Halide	Cation formed	Stability	S_N1?
   (A) allyl chloride	CH₂=CH-CH₂⁺ (allylic)	spread over two carbons by resonance	yes
   (B) 1-chloroprop-1-ene	CH₃-CH=CH⁺ (vinylic)	very unstable, no resonance help	no
   (C) benzyl chloride	C₆H₅CH₂⁺ (benzylic)	spread into the ring by resonance	yes
   (D) 2-chloropropane	(CH₃)₂CH⁺ (secondary)	held up by two methyl groups (hyperconjugation)	yes

2. Why vinylic fails. In (B) the Cl sits on a double-bonded carbon. Its lone pair conjugates with the π bond, giving the C–Cl bond partial double-bond character, so it is hard to break. The vinylic cation it would give is also very unstable.
3. So. Only (B) does not show S_N1, which is option (c).

The idea to keep. Allylic and benzylic halides love S_N1 because their cations are resonance stabilised. Vinylic and aryl halides avoid it because their C–X bond has partial double-bond character.

Answer from the NTA final answer key. Solution by PrepWiser.