In the following reactions,CH_3COCH_3 xrighta | vedclass

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(A) In the first reaction:$(i)$ $NaBH_4$ acts as a source of hydride ion $(H^-)$,which attacks the carbonyl carbon of acetone $(CH_3COCH_3)$ to form a

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$X = CH_3CH(OD)CH_3, Y = CH_3CD(OH)CH_3$

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(A) In the first reaction:$(i)$ $NaBH_4$ acts as a source of hydride ion $(H^-)$,which attacks the carbonyl carbon of acetone $(CH_3COCH_3)$ to form an alkoxide intermediate.(ii) Upon treatment with $D_3O^+$,the alkoxide oxygen is protonated (deuterated) to form the alcohol $CH_3CH(OD)CH_3$,which is $X$.In the second reaction:$(i)$ $NaBD_4$ acts as a source of deuteride ion $(D^-)$,which attacks the carbonyl carbon of acetone to form an alkoxide intermediate containing a deuterium atom at the alpha position.(ii) Upon treatment with $H_3O^+$,the alkoxide oxygen is protonated to form the alcohol $CH_3CD(OH)CH_3$,which is $Y$.Thus,$X = CH_3CH(OD)CH_3$ and $Y = CH_3CD(OH)CH_3$.

In the following reactions,
$CH_3COCH_3 \xrightarrow[(ii) D_3O^+]{(i) NaBH_4} X$
$CH_3COCH_3 \xrightarrow[(ii) H_3O^+]{(i) NaBD_4} Y$
$X$ and $Y$ are

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In the following reactions,$CH_3COCH_3 \xrightarrow[(ii) D_3O^+]{(i) NaBH_4} X$$CH_3COCH_3 \xrightarrow[(ii) H_3O^+]{(i) NaBD_4} Y$$X$ and $Y$ are