Ph-C(=O)-CH_3 xrightarrow{NaOCl} A + text{sal | vedclass

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(C) Acetophenone $(Ph-C(=O)-CH_3)$ undergoes a haloform reaction with $NaOCl$ to produce chloroform $(CHCl_3)$ and sodium benzoate $(Ph-COONa)$. Thus,

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$CH_3-C(OH)(CCl_3)-CH_3$

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(C) Acetophenone $(Ph-C(=O)-CH_3)$ undergoes a haloform reaction with $NaOCl$ to produce chloroform $(CHCl_3)$ and sodium benzoate $(Ph-COONa)$. Thus,$A$ is $CHCl_3$. Chloroform $(CHCl_3)$ reacts with acetone $(CH_3-C(=O)-CH_3)$ in the presence of a base (nucleophilic addition) to form chloretone $(CH_3-C(OH)(CCl_3)-CH_3)$. Therefore,$B$ is $CH_3-C(OH)(CCl_3)-CH_3$.

$Ph-C(=O)-CH_3 \xrightarrow{NaOCl} A + \text{salt}$
$A + CH_3-C(=O)-CH_3 \rightarrow B$
$B$ is

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$Ph-C(=O)-CH_3 \xrightarrow{NaOCl} A + \text{salt}$ $A + CH_3-C(=O)-CH_3 \rightarrow B$ $B$ is