The compound X (C_5H_8) reacts with ammoniac | vedclass

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(C) $1$. The formation of a white precipitate with ammoniacal $AgNO_3$ indicates that $X$ is a terminal alkyne $(R-C \equiv CH)$.$2$. The reaction wit

Vedclass · Accepted Answer

$(CH_3)_2CH-C \equiv CH$

Vedclass · Answer

(C) $1$. The formation of a white precipitate with ammoniacal $AgNO_3$ indicates that $X$ is a terminal alkyne $(R-C \equiv CH)$.$2$. The reaction with excess $KMnO_4$ (oxidative cleavage) yields $(CH_3)_2CHCOOH$ (isobutyric acid).$3$. The structure of the acid $(CH_3)_2CHCOOH$ contains $4$ carbon atoms. Since the original compound $X$ has $5$ carbon atoms,the terminal alkyne must be $(CH_3)_2CH-C \equiv CH$.$4$. Oxidative cleavage of $(CH_3)_2CH-C \equiv CH$ breaks the triple bond to form $(CH_3)_2CHCOOH$ and $CO_2$.

The compound $X \ (C_5H_8)$ reacts with ammoniacal $AgNO_3$ to give a white precipitate and reacts with excess of $KMnO_4$ to give the acid,$(CH_3)_2CHCOOH$. Therefore,$X$ is:

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