In $\ce{H2C=CH-Br}$ and $\ce{H3C-CH2-Br}$, which will react faster towards a $\mathrm{S_N2}$ reaction?
According to me, as double bond exhibit −I effect, hence the 1st should do a faster reaction. Am I right, or is there any other reason?
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asked Jun 16, 2018 at 3:36
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2 Answers
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There is no sense of comparing the reaction rate when one of the compound doesn't show SN2 mechanism. And, that compound is $\ce{CH2=CHBr}$, due to the fact that it is a vinyl halide.
$\ce{CH2=CHBr}$ will hinder the approach of nucleophile due to the presence of pi-electron cloud around double bond.
And, therefore the obvious answer is $\ce{CH3-CH2Br}$
answered Jun 16, 2018 at 6:58
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1$\begingroup$ I think you mean that the obvious answer is ethyl bromide, rather than vinyl bromide, also considering your answer. Right? $\endgroup$ Commented Jun 16, 2018 at 11:47
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$\begingroup$ @GürayHatipoğlu, +1 for pointing out! $\endgroup$ Commented Jun 16, 2018 at 12:16
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$\ce{CH3-CH-Br}$ will give faster $\mathrm{S_N2}$ reaction because when a nucleophile will approach $\ce{CH2=CH-Br}$ for $\mathrm{S_N2}$ reaction the double bond between $\ce{CH2=CH}$ will hinder its approach (steric effect), but there is no such hindrance in case of $\ce{CH3-CH2-Br}$.
To support the answer we can add one more point that in case of $\ce{CH3-CH2-Br}$ the charge $δ^+$ on the $\ce{C}$ atom of $\ce{CH2}$ will be greater in magnitude than that at the $\ce{C}$ atom of $\ce{CH}$ in case of $\ce{CH2=CH-Br}$ because the double bond has better −I effect than single bond, hence it will be easier for $\ce{-Br}$ to attract the shared electron pair towards it and develop a greater $δ^+$ charge on $\ce{C}$ in case of $\ce{CH3-CH2-Br}$, which will ultimately support the approach of the nucleophile for the $\mathrm{S_N2}$ reaction.
