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

SN2 reactions proceed by backside attack of a nucleophile on the electrophilic carbon; rate decreases sharply with increasing steric crowding at that carbon. $CH_3Br$ has no alpha substituents and the least steric hindrance among the four options, making it the fastest in SN2 reactions.