Question 11

$\textbf{Reaction 1:}$

The starting material is $HO_2C-CH(CH_3)-C(=O)-CH(CH_3)-CO_2H$, which is a symmetric di-$\beta$-keto dicarboxylic acid.

In this molecule, both carboxylic acid ($-CO_2H$) groups are located at the $\beta$-position relative to the central ketone group.

Upon heating, $\beta$-keto acids readily undergo decarboxylation (loss of $CO_2$) via a 6-membered cyclic transition state. Because both sides of the ketone possess a $\beta$-carboxyl group, the molecule will undergo double decarboxylation.

Reaction:

$HO_2C-CH(CH_3)-C(=O)-CH(CH_3)-CO_2H \xrightarrow{\Delta} CH_3-CH_2-C(=O)-CH_2-CH_3 + 2CO_2$

The major product $\textbf{X}$ is 3-pentanone.

Number of carbonyl groups ($>C=O$) present in \textbf{X} = 1.

$\textbf{Reaction 2:}$

The starting material is 3,4-dicarboxycyclopentanone. Let's analyze the relative positions of its functional groups:

• The ketone is at C1.

• The carboxylic acid groups are at C3 and C4 of the ring.

For facile thermal decarboxylation, a molecule needs to be a $\beta$-keto acid. However, the $-CO_2H$ group at C3 is attached to the $\beta$-carbon of the ring, meaning the carboxyl carbon itself is at the $\gamma$-position relative to the ketone. Thus, it acts as a $\gamma$-keto acid, which does \textit{not} undergo easy thermal decarboxylation.

Instead, the two $-CO_2H$ groups are located on adjacent carbon atoms (C3 and C4), making it a 1,2-dicarboxylic acid (a vicinal diacid, chemically similar to succinic acid). When heated, 1,2-dicarboxylic acids readily undergo dehydration (loss of a water molecule, $H_2O$) to form a highly stable 5-membered cyclic anhydride.

Reaction:

3,4-dicarboxycyclopentanone $\xrightarrow{\Delta}$ Cyclic Anhydride Derivative + $H_2O$

The major product \textbf{Y} is a bicyclic compound containing the original cyclopentanone ring fused with a newly formed cyclic anhydride ring. The structure of \textbf{Y} contains:

• 1 ketone carbonyl group ($>C=O$) in the original ring.

• 2 carbonyl groups ($>C=O$) as part of the new anhydride linkage ($-C(=O)-O-C(=O)-$ ).

Total number of carbonyl groups ($>C=O$) present in \textbf{Y} = 1 + 2 = 3.

$\textbf{Final Calculation:}$

The sum of the total number of carbonyl groups present in the major products \textbf{X} and \textbf{Y} is:

Sum = (Number of $>C=O$ in X) + (Number of $>C=O$ in Y) = 1 + 3 = 4.