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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.
CH3CH2CH2CH2Br or \[\begin{array}{cc}
\ce{CH3CH2CHCH3}\\
\phantom{.....}|\\
\phantom{.......}\ce{Br}\
\end{array}\]
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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, CH3CH2CH2CH2Br is more reactive.
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|\phantom{........}\\
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 |
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\[\begin{array}{cc}\ce{CH3CH2CHCH3}\\
\phantom{.....}|\\
\phantom{......}\ce{Br}\\
\end{array}\] or \[\begin{array}{cc}\phantom{.......}\ce{CH3}\\
\phantom{...}|\\
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\phantom{...}|\\
\phantom{.......}\ce{CH3}\\
\end{array}\]
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(i) \[\begin{array}{cc}
\phantom{}\ce{\underset{}{(CH3)2CH - CH2Br} ->[C2H5OH] \underset{}{(CH3)2CH - CH2OC2H5 + HBr}}\\
\end{array}\]
(ii) \[\begin{array}{cc}
\phantom{}\ce{\underset{}{(CH3)2CH - CH2Br} ->[C2H5O-] \underset{}{(CH3)2CH - CH2OC2H5 + Br-}}\\
\end{array}\]
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