The isomeric deuterated bromide with molecula | vedclass

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(C) chiral carbon atom is one that is bonded to four different groups. Let us analyze the structures: $A$. $2-$Bromo$-1-$deuterobutane: $CH_3-CH_2-CH(

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$2-$Bromo$-3-$deuterobutane

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(C) chiral carbon atom is one that is bonded to four different groups. Let us analyze the structures: $A$. $2-$Bromo$-1-$deuterobutane: $CH_3-CH_2-CH(Br)-CH_2D$. Here,$C-2$ is chiral (bonded to $H, Br, CH_3, CH_2D$). Only one chiral center. $B$. $2-$Bromo$-2-$deuterobutane: $CH_3-CH_2-C(Br)(D)-CH_3$. Here,$C-2$ is chiral (bonded to $Br, D, CH_3, CH_2CH_3$). Only one chiral center. $C$. $2-$Bromo$-3-$deuterobutane: $CH_3-CH(Br)-CH(D)-CH_3$. Here,$C-2$ is bonded to $H, Br, CH_3, CH(D)CH_3$ (chiral). $C-3$ is bonded to $H, D, CH_3, CH(Br)CH_3$ (chiral). This molecule has two chiral carbon atoms. $D$. $2-$Bromo$-1-$deutero$-2-$methylpropane: $(CH_3)_2C(Br)-CH_2D$. No chiral carbon atom. Therefore,the correct option is $C$.

The isomeric deuterated bromide with molecular formula $C_4H_8DBr$ having two chiral carbon atoms is

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