In the sequence of the following reactions,X | vedclass

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(A) $1$. The reaction sequence starts with methanol $(CH_3OH)$,which reacts with $HI$ to form methyl iodide $(CH_3I)$.$2$. $CH_3I$ reacts with $KCN$ t

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$CH_3CH_2NH_2$ and $CH_3CH_2OH$

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(A) $1$. The reaction sequence starts with methanol $(CH_3OH)$,which reacts with $HI$ to form methyl iodide $(CH_3I)$.$2$. $CH_3I$ reacts with $KCN$ to form methyl cyanide $(CH_3CN)$.$3$. Reduction of methyl cyanide $(CH_3CN)$ using a reducing agent (like $LiAlH_4$ or $H_2/Ni$) yields ethylamine $(CH_3CH_2NH_2)$,which is $X$.$4$. Ethylamine $(CH_3CH_2NH_2)$ reacts with nitrous acid $(HNO_2)$ to form ethyl alcohol $(CH_3CH_2OH)$,which is $Y$.

In the sequence of the following reactions,$X$ and $Y$ are respectively: $CH_3OH$ $\xrightarrow{HI} CH_3I$ $\xrightarrow{KCN} CH_3CN$ $\xrightarrow{\text{reduction}} X$ $\xrightarrow{HNO_2} Y$

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