Which of the following will not lose asymmetr | vedclass

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(B) Asymmetry is lost if the reduction of a functional group results in a product where the central carbon atom is bonded to two identical groups,maki

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(B) Asymmetry is lost if the reduction of a functional group results in a product where the central carbon atom is bonded to two identical groups,making it achiral.$LiAlH_4$ reduces aldehydes $(-CHO)$ to primary alcohols $(-CH_2OH)$,carboxylic acids $(-COOH)$ to primary alcohols $(-CH_2OH)$,and nitriles $(-CN)$ to primary amines $(-CH_2NH_2)$.Let us analyze the options:$(A)$ The central carbon is bonded to $-CHO$,$-CH_2OH$,$-CH_2CH_3$,and $-CH=CH_2$. Upon reduction,$-CHO$ becomes $-CH_2OH$. The central carbon will then be bonded to two identical $-CH_2OH$ groups,thus losing asymmetry.$(B)$ The central carbon is bonded to $-CHO$,$-CH_3$,$-CH_2CH_3$,and $-OCH=CH_2$. Upon reduction,$-CHO$ becomes $-CH_2OH$. The groups attached are $-CH_2OH$,$-CH_3$,$-CH_2CH_3$,and $-OCH=CH_2$. All four groups are different,so it remains chiral (does not lose asymmetry).$(C)$ The central carbon is bonded to $-COOH$,$-CH_2OH$,$-CH_3$,and $-C\equiv CH$. Upon reduction,$-COOH$ becomes $-CH_2OH$. The central carbon will then be bonded to two identical $-CH_2OH$ groups,thus losing asymmetry.$(D)$ The central carbon is bonded to $-CHO$,$-CH_3$,$-C\equiv N$,and $-CH_2NH_2$. Upon reduction,$-CHO$ becomes $-CH_2OH$ and $-C\equiv N$ becomes $-CH_2NH_2$. The central carbon is bonded to $-CH_2OH$,$-CH_3$,$-CH_2NH_2$,and $-CH_2NH_2$. It has two identical $-CH_2NH_2$ groups,thus losing asymmetry.Therefore,the correct option is $(B)$.

Which of the following will not lose asymmetry on reduction with $LiAlH_4$?

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