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(B) The reaction involves the ozonolysis of an unsaturated lactone. $1$. Ozonolysis of the given cyclic compound breaks the double bond to form formal

Vedclass · Accepted Answer

$Ph-NH-CO-CH_2-CO_2H$

Vedclass · Answer

(B) The reaction involves the ozonolysis of an unsaturated lactone. $1$. Ozonolysis of the given cyclic compound breaks the double bond to form formaldehyde $(HCHO)$ and a cyclic anhydride/lactone intermediate $(A)$,which is malonic anhydride (or a related cyclic structure). $2$. The nucleophilic attack of aniline $(Ph-NH_2)$ on the carbonyl group of the intermediate $(A)$ leads to ring opening. $3$. The reaction follows the mechanism of nucleophilic acyl substitution,resulting in the formation of $N$-phenylmalonamic acid,which is $Ph-NH-CO-CH_2-CO_2H$.

The product $(B)$ is:

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