CH equiv CH xrightarrow{O_3/NaOH} X xrightarr | vedclass

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Question

(A) The reaction of ethyne $(CH \equiv CH)$ with $O_3$ in the presence of $NaOH$ forms an ozonide intermediate $(X)$.Upon treatment with $Zn/CH_3COOH$

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$HOCH_2-CH_2OH$

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(A) The reaction of ethyne $(CH \equiv CH)$ with $O_3$ in the presence of $NaOH$ forms an ozonide intermediate $(X)$.Upon treatment with $Zn/CH_3COOH$ (reductive workup),the ozonide is converted into glyoxal $(CHO-CHO)$.Further reduction of glyoxal in the presence of $Zn/CH_3COOH$ leads to the formation of ethylene glycol $(HOCH_2-CH_2OH)$ as the final product '$Y$'.The reaction sequence is:$CH \equiv CH$ $\xrightarrow{O_3} 	ext{Ozonide}$ $\xrightarrow{Zn/CH_3COOH} CHO-CHO$ $\xrightarrow{[H]} HOCH_2-CH_2OH$.

$CH \equiv CH$ $\xrightarrow{O_3/NaOH} X$ $\xrightarrow{Zn/CH_3COOH} Y$. '$Y$' is:

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