Compound A undergoes the following sequence o | vedclass

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(B) $1$. The starting material $A$ is $2$-bromo-$3$-methylbutane.$2$. Reaction with $Mg$ in $Et_2O$ forms the Grignard reagent, $3$-methylbutan-$2$-yl

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$2$-Deuterio-$3$-methylbutane, Chiral

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(B) $1$. The starting material $A$ is $2$-bromo-$3$-methylbutane.$2$. Reaction with $Mg$ in $Et_2O$ forms the Grignard reagent, $3$-methylbutan-$2$-ylmagnesium bromide, $(CH_3)_2CH-CH(MgBr)-CH_3$.$3$. Treatment with $D_2O$ replaces the $-MgBr$ group with a deuterium atom $-D$, yielding $2$-deuterio-$3$-methylbutane, $(CH_3)_2CH-CH(D)-CH_3$.$4$. In $2$-deuterio-$3$-methylbutane, the carbon at position $2$ is bonded to four distinct groups: $-H$, $-D$, $-CH_3$, and $-CH(CH_3)_2$. Thus, it is a chiral center, making the molecule chiral.

Compound $A$ undergoes the following sequence of reactions to give compound $B$. The correct structure and chirality of compound $B$ is:
[where $Et$ is $-C_2H_5$]

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Compound $A$ undergoes the following sequence of reactions to give compound $B$. The correct structure and chirality of compound $B$ is:[where $Et$ is $-C_2H_5$]