Which among the following halides will generate the most stable carbocation in the nucleophilic substitution reaction?

Solution

Chemistry ma’am · 3 steps

Remove Br⁻ from each, then ask which positive charge can spread over the most atoms.

1. Form the cation. In S_N1 substitution, Br⁻ leaves first and a carbocation forms at the carbon that held it. The cation is more stable when its positive charge is shared over more atoms by resonance.

   Halide	Carbocation	Type	Resonance structures
   (a) allyl bromide	CH₂=CH-CH₂⁺	primary, allylic	2
   (b) triphenylmethyl bromide	(C₆H₅)₃C⁺	tertiary, conjugated with three rings	10
   (c) benzyl bromide	C₆H₅CH₂⁺	primary, benzylic	4
   (d) 3-bromocyclohexene	cyclohex-2-enyl cation	secondary, allylic	2 (plus hyperconjugation)

2. Count the spread. In the benzyl cation the charge reaches the ortho and para carbons of one ring, giving 4 structures. In the triphenylmethyl cation each of the three rings does the same, so the charge sits on 1 central carbon plus 9 ring carbons: 10 structures.
3. So. The triphenylmethyl (trityl) cation from (b) is the most stable, because its positive charge is spread over the most atoms.

The idea to keep. For carbocation stability, count where the positive charge can go. Each phenyl ring attached to the cationic carbon adds three more places for it, so more rings means a more stable cation.

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