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Question
Which would undergo SN2 reaction faster in the following pair and why ?
CH3 – CH2 – Br and CH3 – CH2 – I
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Solution
CH3 – CH2 – I would undergo an SN2 reaction faster than CH3 – CH2 – Br. Since iodine is a better leaving group because of its large size, it will be released at a faster rate in the presence of an incoming nucleophile as compared to bromine.
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RELATED QUESTIONS
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(i) (a) and (e) both are nucleophiles.
(ii) In (c) carbon atom is sp3 hybridised.
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(iv) (a) and (e) both are electrophiles.
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\[\begin{array}{cc}
\ce{CH3CHCH2CH2Br}\\
|\phantom{.............}\\
\ce{CH3}\phantom{..........}\\
\end{array}\] or \[\begin{array}{cc}
\ce{CH3CH2CHCH2Br}\\
|\\
\phantom{...}\ce{CH3}\\
\end{array}\]
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(i) \[\begin{array}{cc}
\phantom{}\ce{\underset{}{(CH3)2CH - CH2Br} ->[C2H5OH] \underset{}{(CH3)2CH - CH2OC2H5 + HBr}}\\
\end{array}\]
(ii) \[\begin{array}{cc}
\phantom{}\ce{\underset{}{(CH3)2CH - CH2Br} ->[C2H5O-] \underset{}{(CH3)2CH - CH2OC2H5 + Br-}}\\
\end{array}\]
The mechanisms of reactions (i) and (ii) are respectively:
