Identify the alkyl halide that undergoes SN^2 | vedclass

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(C) The rate of $SN^2$ reaction depends on steric hindrance and the nature of the leaving group. $1$. Steric hindrance: $SN^2$ reactions are fastest f

Vedclass · Accepted Answer

$1-$bromo$-3-$methylpentane

Vedclass · Answer

(C) The rate of $SN^2$ reaction depends on steric hindrance and the nature of the leaving group. $1$. Steric hindrance: $SN^2$ reactions are fastest for primary $(1^{\circ})$ alkyl halides compared to secondary $(2^{\circ})$ or tertiary $(3^{\circ})$ alkyl halides. $2$. Leaving group: For the same alkyl group,the rate of $SN^2$ reaction is faster with a better leaving group. Bromide $(Br^-)$ is a better leaving group than chloride $(Cl^-)$. Comparing the options: - Options $A$ and $B$ are secondary $(2^{\circ})$ alkyl halides. - Options $C$ and $D$ are primary $(1^{\circ})$ alkyl halides. Since primary alkyl halides are less sterically hindered,they react faster than secondary ones. Between $C$ and $D$,$C$ contains a bromide leaving group,which is better than the chloride in $D$. Therefore,$1$-bromo$-3-$methylpentane undergoes $SN^2$ reaction most rapidly.

Identify the alkyl halide that undergoes $SN^2$ reaction most rapidly.

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