CH_3-C equiv C-CH_3 xrightarrow[(1) O_3]{(2) | vedclass

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(C) Step $1$: The ozonolysis of $CH_3-C \equiv C-CH_3$ (but$-2-$yne) followed by hydrolysis yields acetic acid $(CH_3COOH)$ as the product $X$.$CH_3-C

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$CH_4$

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(C) Step $1$: The ozonolysis of $CH_3-C \equiv C-CH_3$ (but$-2-$yne) followed by hydrolysis yields acetic acid $(CH_3COOH)$ as the product $X$.$CH_3-C \equiv C-CH_3 \xrightarrow{O_3, H_2O} 2CH_3COOH$.Step $2$: The reaction of $CH_3COOH$ with soda lime $(NaOH + CaO)$ is a decarboxylation reaction.$CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O$.$CH_3COONa + NaOH \xrightarrow{CaO, \Delta} CH_4 + Na_2CO_3$.Thus,the final product $Y$ is methane $(CH_4)$.

$CH_3-C \equiv C-CH_3$ $\xrightarrow[(1) \ O_3]{(2) \ H_2O} X$ $\xrightarrow[\Delta]{NaOH + CaO} Y$; $Y$ is

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