Question

Which compound undergoes the \(S_{N}2\) reaction the fastest?

• A. \(CH_{3}-Br\) (methyl bromide)
• B. \((CH_{3})_{2}CH-Br\) (secondary bromide)
• C. \((CH_{3})_{3}C-Br\) (tertiary bromide)
• D. \(CH_{3}CH_{2}-Br\) (primary bromide)

Hint:

For \(S_N2\) reactions, always remember the order: \[ \boxed{ CH_3X \gt 1^\circ \gt 2^\circ \gt \gt 3^\circ } \] Less steric hindrance means faster backside attack by the nucleophile and therefore a faster \(S_N2\) reaction.

Answer 1

The Correct Option is A

Concept: The \(S_N2\) reaction is a bimolecular nucleophilic substitution reaction that occurs in a single step through backside attack by the nucleophile. The rate law is \[ \boxed{\text{Rate}=k[\text{Alkyl halide}][\text{Nucleophile}]} \] The rate of an \(S_N2\) reaction depends mainly on steric hindrance around the carbon atom attached to the leaving group. The order of reactivity is \[ \boxed{ CH_3X \gt 1^\circ \gt 2^\circ \gt \gt 3^\circ } \] because steric hindrance increases from methyl to tertiary alkyl halides.

Step 1: Identify the nature of each alkyl halide.
\[ CH_3Br \] is a methyl halide. \[ CH_3CH_2Br \] is a primary alkyl halide. \[ (CH_3)_2CHBr \] is a secondary alkyl halide. \[ (CH_3)_3CBr \] is a tertiary alkyl halide.

Step 2: Compare steric hindrance.
Since methyl bromide has no alkyl group attached to the carbon bearing bromine, \[ \boxed{\text{Steric hindrance is minimum}.} \] Therefore, the nucleophile can easily attack the carbon atom. Hence, methyl bromide reacts fastest through the \(S_N2\) mechanism.

Step 3: Determine the correct option.
The order of \(S_N2\) reactivity is \[ CH_3Br \gt CH_3CH_2Br \gt (CH_3)_2CHBr \gt \gt (CH_3)_3CBr. \] Thus, \[ \boxed{CH_3Br} \] undergoes the \(S_N2\) reaction most rapidly. Hence, \[ \boxed{\textbf{Option (A)}} \] is the correct answer.