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(A) The starting material is $3-oxo-3-phenylpropanoic$ acid with a $^{13}C$ label on the methylene carbon. Upon heating,it undergoes decarboxylation t

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(A) The starting material is $3-oxo-3-phenylpropanoic$ acid with a $^{13}C$ label on the methylene carbon. Upon heating,it undergoes decarboxylation to form acetophenone $(E)$ where the methyl group is $^{13}C$ labelled: $Ph-CO-CH_2^*COOH \xrightarrow{\Delta} Ph-CO-CH_3^* + CO_2$. Acetophenone $(E)$ then reacts with $I_2$ in the presence of $NaOH$ (haloform reaction) to produce sodium benzoate $(F)$ and iodoform $(G)$: $Ph-CO-CH_3^* + 3I_2 + 4NaOH \rightarrow Ph-COONa + CHI_3^* + 3NaI + 3H_2O$. Thus,the $^{13}C$ label is present in the methyl group of $E$ and in the iodoform $(G)$.

List-$I$	List-$II$
$A$. $C_6H_6 + CO + HCl \xrightarrow{AlCl_3} C_6H_5CHO$	$a$. Gattermann-Koch reaction
$B$. $C_6H_5CH_3 \xrightarrow{CrO_2Cl_2} C_6H_5CHO$	$b$. Etard reaction
$C$. $C_6H_5CH_2Br + CH_3Br \xrightarrow{Na/ether} C_6H_5CH_2CH_3$	$c$. Wurtz-Fittig reaction
$D$. $C_6H_6 + (CH_3)_2C = CH_2 \xrightarrow{H_2SO_4} C_6H_5C(CH_3)_3$	$d$. Friedel-Crafts alkylation

In the following reaction sequence,the correct structures of $E, F$ and $G$ are

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