Question

Which alkyl halide from the following pair would you expect to react more rapidly by an S_N2 mechanism? Explain your answer.

\[\begin{array}{cc}\ce{CH3CHCH2CH2Br}\\|\phantom{.........}\\\ce{CH3}\phantom{......}\end{array}\] or \[\begin{array}{cc}\ce{CH3CH2CHCH2Br}\\\phantom{}|\\\phantom{...}\ce{CH3}\end{array}\]

Answer 1

The S_N2 process involves a transition state with both an incoming nucleophile and a leaving group surrounding the carbon atom. Five atoms are simultaneously bonded together. A transition state requires minimal steric hindrance. Hence, 1° alkyl halides are the most reactive to S_N2, followed by 2° and 3°.

1° RX > 2° RX > 3° RX

Based on the above order, \[\begin{array}{cc}\ce{CH3CHCH2CH2Br}\\|\phantom{.........}\\\ce{CH3}\phantom{......}\end{array}\] is more reactive.

Here, the proximity of the branched chain –CH₃ that determines the reactivity. In \[\begin{array}{cc}\ce{CH3CH2CHCH2Br}\\\phantom{}|\\\phantom{...}\ce{CH3}\end{array}\] the methyl group is closer to the leaving group thereby hindering the transition state.