# Which will give faster SN2 reaction

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?

$$\ce{CH3-CH-Br}$$ 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.

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 $\ce{CH3-CH2Br}$

• I think you mean that the obvious answer is ethyl bromide, rather than vinyl bromide, also considering your answer. Right? – Güray Hatipoğlu Jun 16 '18 at 11:47
• @GürayHatipoğlu, +1 for pointing out! – rv7 Jun 16 '18 at 12:16