Question

Which alkyl halide from the given options will undergo S$_N$1 reaction faster?

• A. (CH$_3$)$_3$C–Br
• B. (CH$_3$)$_2$CH–Br
• C. CH$_3$CH$_2$–Br
• D. (CH$_3$)$_3$C–CH$_2$Br

Hint:

In S$_N$1 reactions, the more stable the carbocation, the faster the reaction. Tertiary carbocations are the most stable and hence react the fastest.

Answer 1

The Correct Option is A

The S$_N$1 mechanism involves a two-step process:
1. The leaving group (in this case, Br$^-$) departs, forming a carbocation.
2. The nucleophile (in this case, the solvent or other reactant) attacks the carbocation, resulting in the product. The rate-determining step in an S$_N$1 reaction is the formation of the carbocation. The more stable the carbocation, the faster the reaction. Carbocation stability increases with the number of alkyl groups attached to the positively charged carbon, as these groups provide inductive stabilization and hyperconjugation.
- Tertiary carbocations are the most stable due to the electron-donating effects of the three alkyl groups.
- Secondary carbocations are less stable.
- Primary carbocations are highly unstable and do not readily form.
In this case, the alkyl halide (CH$_3$)$_3$C–Br (option A) forms a tertiary carbocation, making it the most likely to undergo an S$_N$1 reaction fastest. Step 1: The reaction mechanism for S$_N$1 involves the formation of a carbocation. Step 2: Tertiary carbocations are the most stable and form the fastest, which is why (CH$_3$)$_3$C–Br undergoes the reaction faster.