In the given reaction sequence,Z is a mixture | vedclass

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(C) $1$. Dehydration of ethanol: $CH_3CH_2OH \xrightarrow[443 \ K]{	ext{Conc. } H_2SO_4} CH_2=CH_2$ ($X$ is ethene).$2$. Ozonolysis of ethene: $CH_2=

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Cannizzaro reaction

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(C) $1$. Dehydration of ethanol: $CH_3CH_2OH \xrightarrow[443 \ K]{	ext{Conc. } H_2SO_4} CH_2=CH_2$ ($X$ is ethene).$2$. Ozonolysis of ethene: $CH_2=CH_2 \xrightarrow[(2) \ Zn/H_2O]{(1) \ O_3} 2HCHO$ ($Y$ is formaldehyde).$3$. Cannizzaro reaction: Formaldehyde $(HCHO)$ has no $\alpha$-hydrogen,so it undergoes the Cannizzaro reaction in the presence of concentrated $NaOH$ to form methanol $(CH_3OH)$ and sodium formate $(HCOONa)$.$4$. Thus,the conversion of $Y$ to $Z$ is the Cannizzaro reaction.

In the given reaction sequence,$Z$ is a mixture of alcohol and acid. The reaction for the conversion of $Y$ to $Z$ is:
$CH_3CH_2OH$ $\xrightarrow[443 \ K]{\text{Conc. } H_2SO_4} X$ $\xrightarrow[(2) \ Zn/H_2O]{(1) \ O_3} Y$ $\xrightarrow[(ii) \ H^+]{(i) \ \text{Conc. } NaOH} Z$

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In the given reaction sequence,$Z$ is a mixture of alcohol and acid. The reaction for the conversion of $Y$ to $Z$ is:$CH_3CH_2OH$ $\xrightarrow[443 \ K]{\text{Conc. } H_2SO_4} X$ $\xrightarrow[(2) \ Zn/H_2O]{(1) \ O_3} Y$ $\xrightarrow[(ii) \ H^+]{(i) \ \text{Conc. } NaOH} Z$