Determine Structures of Products E F and G in Labelled Decarboxylation and Haloform Reaction

To determine the structures of products $\mathbf{E}$, $\mathbf{F}$, and $\mathbf{G}$, we analyze the reaction sequence step-by-step:

Step 1: Decarboxylation to form Product [E]

The starting material is a $\beta$-keto acid labelled with $^{13}\text{C}$ at the $\alpha$-position:
$\text{Ph}-\text{C}(=\text{O})-\overset{*}{\text{C}}\text{H}_2-\text{COOH}$

Upon heating, $\beta$-keto acids undergo thermal decarboxylation via a six-membered cyclic transition state to form an enol intermediate:
$\text{Ph}-\text{C}(=\text{O})-\overset{*}{\text{C}}\text{H}_2-\text{COOH} \xrightarrow{\Delta} \text{Ph}-\text{C}(\text{OH})=\overset{*}{\text{C}}\text{H}_2 + \text{CO}_2$

Tautomerization of the enol intermediate yields labelled acetophenone:
$\mathbf{E} = \text{Ph}-\text{C}(=\text{O})-\overset{*}{\text{C}}\text{H}_3$

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Step 2: Haloform Reaction to form Products [F] and [G]

When compound $\mathbf{E}$ ($\text{Ph}-\text{C}(=\text{O})-\overset{*}{\text{C}}\text{H}_3$) reacts with iodine in the presence of sodium hydroxide ($\text{I}_2 / \text{NaOH}$), it undergoes the haloform reaction:

1. The acidic hydrogens on the labelled methyl group ($\overset{*}{\text{C}}\text{H}_3$) are replaced by iodine atoms to form a triiodomethyl ketone intermediate:
   $\text{Ph}-\text{C}(=\text{O})-\overset{*}{\text{C}}\text{I}_3$
2. Nucleophilic attack by hydroxide ion ($\text{OH}^-$) on the carbonyl carbon leads to the cleavage of the $\text{C}-\overset{*}{\text{C}}$ bond, producing sodium benzoate ($\mathbf{F}$) and labelled iodoform ($\mathbf{G}$):
   $\text{Ph}-\text{C}(=\text{O})-\overset{*}{\text{C}}\text{I}_3 + \text{NaOH} \rightarrow \text{Ph}-\text{C}(=\text{O})\text{O}^-\text{Na}^+ + \overset{*}{\text{C}}\text{HI}_3$

Thus, the reaction products are:

• $\mathbf{F} = \text{Ph}-\text{COO}^-\text{Na}^+$
• $\mathbf{G} = \overset{*}{\text{C}}\text{HI}_3$

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Conclusion:

• $\mathbf{E} = \text{Ph}-\text{C}(=\text{O})-\overset{*}{\text{C}}\text{H}_3$
• $\mathbf{F} = \text{Ph}-\text{C}(=\text{O})\text{O}^-\text{Na}^+$
• $\mathbf{G} = \overset{*}{\text{C}}\text{HI}_3$

This matches the structures provided in Option C.