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(C) The reaction is an oxymercuration-demercuration process. Step $1$: Oxymercuration involves the electrophilic attack of $Hg(OAc)^+$ on the double b

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$CH_3-CH(OH)-CH_2D$

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(C) The reaction is an oxymercuration-demercuration process. Step $1$: Oxymercuration involves the electrophilic attack of $Hg(OAc)^+$ on the double bond of propene $(CH_3-CH=CH_2)$,followed by the attack of water $(H_2O)$ to form a mercurial alcohol. The $OH$ group attaches to the more substituted carbon (Markovnikov addition),resulting in $CH_3-CH(OH)-CH_2-HgOAc$. Step $2$: Demercuration with $NaBD_4$ replaces the $-HgOAc$ group with a deuterium atom $(D)$. Therefore,the final product is $CH_3-CH(OH)-CH_2D$.

Predict the product of the following reaction: $CH_3-CH=CH_2 \xrightarrow[(2) NaBD_4]{(1) H_2O, Hg(OCOCH_3)_2} \text{Product}$

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