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Question
Which alkyl halide from the following pair would you expect to react more rapidly by an SN2 mechanism? Explain your answer.
\[\begin{array}{cc}\ce{CH3CHCH2CH2Br}\\|\phantom{.........}\\\ce{CH3}\phantom{......}\end{array}\] or \[\begin{array}{cc}\ce{CH3CH2CHCH2Br}\\\phantom{}|\\\phantom{...}\ce{CH3}\end{array}\]
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Solution
The SN2 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 SN2, 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.
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