S_{N}2 reaction involving inversion of config | vedclass

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(C) For a compound to be optically active,it must possess a chiral carbon atom (a carbon atom bonded to four different groups).In $S_{N}2$ reactions,t

Vedclass · Accepted Answer

$CH_3CH_2CH(CH_3)X$

Vedclass · Answer

(C) For a compound to be optically active,it must possess a chiral carbon atom (a carbon atom bonded to four different groups).In $S_{N}2$ reactions,the nucleophile attacks the electrophilic carbon from the side opposite to the leaving group,leading to an inversion of configuration (Walden inversion).Let's analyze the options:$(A)$ $CH_3CH_2X$: The carbon attached to $X$ is bonded to two identical $H$ atoms. It is achiral.$(B)$ $(CH_3)_2CHX$: The carbon attached to $X$ is bonded to two identical $CH_3$ groups. It is achiral.$(C)$ $CH_3CH_2CH(CH_3)X$: The carbon attached to $X$ is bonded to $H$,$CH_3$,$CH_2CH_3$,and $X$. Since all four groups are different,this carbon is chiral,making the compound optically active.$(D)$ $(CH_3)_3CX$: The carbon attached to $X$ is bonded to three identical $CH_3$ groups. It is achiral.Therefore,the optically active compound $Z$ is $CH_3CH_2CH(CH_3)X$.

$S_{N}2$ reaction involving inversion of configuration takes place with an optically active compound $Z$. The compound $Z$ is

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