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(A) $1$. Reductive ozonolysis of an alkene $R_1R_2C=CR_3R_4$ yields carbonyl compounds $R_1R_2C=O$ and $R_3R_4C=O$.$2$. $Y$ is obtained by Etard's rea

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Styrene $(C_6H_5CH=CH_2)$

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(A) $1$. Reductive ozonolysis of an alkene $R_1R_2C=CR_3R_4$ yields carbonyl compounds $R_1R_2C=O$ and $R_3R_4C=O$.$2$. $Y$ is obtained by Etard's reaction,which is the oxidation of toluene to benzaldehyde $(C_6H_5CHO)$. Thus,$Y$ is benzaldehyde.$3$. $Z$ undergoes a disproportionation reaction (Cannizzaro reaction) with concentrated alkali. Formaldehyde $(HCHO)$ is an aldehyde without $\alpha$-hydrogens,so it undergoes the Cannizzaro reaction.$4$. Combining $Y$ $(C_6H_5CHO)$ and $Z$ $(HCHO)$,the original alkene $X$ must be styrene $(C_6H_5CH=CH_2)$.$5$. The reaction is: $C_6H_5CH=CH_2 \xrightarrow{	ext{O}_3, 	ext{Zn/H}_2	ext{O}} C_6H_5CHO + HCHO$.

$X \underset{\text{Reductive}}{\stackrel{\text{Ozonolysis}}{\longrightarrow}} Y + Z$. $Y$ can be obtained by Etard's reaction,$Z$ undergoes disproportionation reaction with concentrated alkali. $X$ could be:

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