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

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, the more reactive alkyl halide is:

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

Here, the proximity of the branched chain –CH₃ that determines the reactivity.

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

Here, the methyl group is closer to the leaving group, thereby hindering the transition state.