The final product [D] of the reaction isCHCl_ | vedclass

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(C) Step $1$: $CHCl_3$ reacts with $O_2$ in the presence of light $(hv)$ to form phosgene,$[A] = COCl_2$.Step $2$: $COCl_2$ reacts with ethanol $(EtOH

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$(CH_3COO)_2Ca$

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(C) Step $1$: $CHCl_3$ reacts with $O_2$ in the presence of light $(hv)$ to form phosgene,$[A] = COCl_2$.Step $2$: $COCl_2$ reacts with ethanol $(EtOH)$ to form diethyl carbonate,$[B] = (EtO)_2CO$.Step $3$: Diethyl carbonate reacts with $2 \ mole$ of $MeMgBr$ to form acetone,$[C] = CH_3COCH_3$.Step $4$: Acetone reacts with $CaOCl_2$ (bleaching powder) to form calcium acetate,$[D] = (CH_3COO)_2Ca$.Step $5$: Calcium acetate on heating $(\Delta)$ undergoes decarboxylation to yield acetone,$CH_3-C(=O)-CH_3$.

The final product $[D]$ of the reaction is
$CHCl_3$ $\xrightarrow[hv]{O_2} [A]$ $\xrightarrow{EtOH} [B]$ $\xrightarrow{2 \ mole \ MeMgBr} [C]$ $\xrightarrow{CaOCl_2} [D]$ $\xrightarrow{\Delta} CH_3-C(=O)-CH_3$

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The final product $[D]$ of the reaction is$CHCl_3$ $\xrightarrow[hv]{O_2} [A]$ $\xrightarrow{EtOH} [B]$ $\xrightarrow{2 \ mole \ MeMgBr} [C]$ $\xrightarrow{CaOCl_2} [D]$ $\xrightarrow{\Delta} CH_3-C(=O)-CH_3$