The correct decreasing order of SN^2 reactivi | vedclass

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Question

(B) In $SN^2$ reactions,the reactivity depends primarily on steric hindrance. The general order of reactivity for alkyl halides is $Methyl > Primary (

Vedclass · Accepted Answer

$II > IV > III > I$

Vedclass · Answer

(B) In $SN^2$ reactions,the reactivity depends primarily on steric hindrance. The general order of reactivity for alkyl halides is $Methyl > Primary (1^\circ) > Secondary (2^\circ) > Tertiary (3^\circ)$. $(II)$ $CH_3-Br$ is a methyl halide,which has the least steric hindrance and is most reactive. $(IV)$ $CH_3-CH_2-Br$ is a primary $(1^\circ)$ alkyl halide. $(III)$ $CH_3-CH(Br)-CH_3$ is a secondary $(2^\circ)$ alkyl halide. $(I)$ $CH_2=CH-Br$ is a vinyl halide,which is the least reactive because the $C-Br$ bond has partial double bond character due to resonance,making it very difficult to break. Therefore,the correct decreasing order of reactivity is $II > IV > III > I$.

The correct decreasing order of $SN^2$ reactivity of the following alkyl halides is:
$(I)$ $CH_2=CH-Br$
$(II)$ $CH_3-Br$
$(III)$ $CH_3-CH(Br)-CH_3$
$(IV)$ $CH_3-CH_2-Br$

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The correct decreasing order of $SN^2$ reactivity of the following alkyl halides is: $(I)$ $CH_2=CH-Br$ $(II)$ $CH_3-Br$ $(III)$ $CH_3-CH(Br)-CH_3$ $(IV)$ $CH_3-CH_2-Br$