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(D) The reaction shown is the ozonolysis of an alkene.$1$. The ozonide intermediate formed is $CH_3-CH-O-CH_2$ (with an $O-O$ bridge).$2$. Reductive o

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propene and ethanal

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(D) The reaction shown is the ozonolysis of an alkene.$1$. The ozonide intermediate formed is $CH_3-CH-O-CH_2$ (with an $O-O$ bridge).$2$. Reductive ozonolysis with $Zn + H_2O$ cleaves the $C=C$ bond.$3$. The products are $B$ and $HCHO$ (methanal).$4$. The structure of the ozonide indicates the original alkene was propene $(CH_3-CH=CH_2)$.$5$. Cleavage of propene $(CH_3-CH=CH_2)$ yields ethanal $(CH_3CHO)$ and methanal $(HCHO)$.$6$. Therefore,$A$ is propene and $B$ is ethanal.

The given reaction represents the ozonolysis of an alkene $A$. Identify $A$ and the product $B$ in the following reaction sequence:
$A + O_3$ $\rightarrow \text{ozonide intermediate}$ $\xrightarrow{Zn + H_2O} B + HCHO$

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The given reaction represents the ozonolysis of an alkene $A$. Identify $A$ and the product $B$ in the following reaction sequence:$A + O_3$ $\rightarrow \text{ozonide intermediate}$ $\xrightarrow{Zn + H_2O} B + HCHO$