Question

Which one of the following will undergo Nucleophilic substitution, by $S_N^1$ mechanism, fastest?

• A. Br$^{-}$ – CH$_2$ = CH – CH$_2$
• B. C$_6$H$_5$Br
• C. CH$_3$ – CH = CHBr
• D. C$_6$H$_{11}$Br

Hint:

For $S_N^1$ reactions, more stable carbocations lead to faster substitution reactions. Allyl and benzyl carbocations are more stable than primary carbocations, thus the rate of reaction increases with increasing stability of the carbocation.

Answer 1

The Correct Option is A

The $S_N^1$ mechanism involves two steps: the formation of a carbocation intermediate and then the attack by the nucleophile. This mechanism generally occurs more rapidly when the carbocation formed is more stable. Thus, the stability of the carbocation plays a key role in determining the rate of the reaction. - Option A: Br$^{-}$ – CH$_2$ = CH – CH$_2$ will form a relatively stable carbocation due to the allyl structure (the positive charge is delocalized over the two carbons of the double bond). This stabilization through resonance makes this substrate undergo $S_N^1$ substitution fastest. - Option B: C$_6$H$_5$Br: A benzyl halide would form a highly stable carbocation as well, but the allylic carbocation in Option A is slightly more stabilized due to resonance. - Option C: CH$_3$ – CH = CHBr: This is a simple alkyl halide with a secondary carbocation, which is less stable than an allylic or benzyl carbocation. - Option D: C$_6$H$_{11}$Br: This is a primary alkyl halide, leading to a less stable carbocation, thus slower in undergoing $S_N^1$ substitution.