An alcohol X (C_4H_{10}O) on dehydration give | vedclass

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(B) $1$. Dehydration of $X$ $(C_4H_{10}O)$ gives $C_4H_8$ (but$-2-$ene as major product). This implies $X$ is butan$-2-$ol.$2$. Bromination of but$-2-

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butan$-2-$ol; $(i) alc. KOH, (ii) NaNH_2$

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(B) $1$. Dehydration of $X$ $(C_4H_{10}O)$ gives $C_4H_8$ (but$-2-$ene as major product). This implies $X$ is butan$-2-$ol.$2$. Bromination of but$-2-$ene gives $2,3-$dibromobutane.$3$. Treatment of $2,3-$dibromobutane with $Y$ $(NaNH_2)$ causes double dehydrohalogenation to form but$-2-$yne $(CH_3-C \equiv C-CH_3)$.$4$. But$-2-$yne is an internal alkyne and does not have an acidic hydrogen,so it does not react with sodium metal.$5$. Thus,$X$ is butan$-2-$ol and $Y$ is $(i) alc. KOH, (ii) NaNH_2$.

An alcohol $X$ $(C_4H_{10}O)$ on dehydration gives an alkene $(C_4H_8)$ as the major product,which on bromination followed by treatment with $Y$ gives an alkyne $C_4H_6$. The alkyne $C_4H_6$ does not react with sodium metal. What are $X$ and $Y$?

Similar Questions

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