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(C) $1$. The formation of a white precipitate with ammoniacal $AgNO_3$ indicates the presence of a terminal alkyne $(-C\equiv CH)$.$2$. Oxidation of a

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$(CH_3)_2CH-C\equiv CH$

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(C) $1$. The formation of a white precipitate with ammoniacal $AgNO_3$ indicates the presence of a terminal alkyne $(-C\equiv CH)$.$2$. Oxidation of a terminal alkyne with hot alkaline $KMnO_4$ results in the cleavage of the triple bond,where the terminal carbon is converted to $CO_2$ and the remaining part forms a carboxylic acid.$3$. The product $(CH_3)_2CHCOOH$ (isobutyric acid) contains $4$ carbons. Since the original compound has $5$ carbons,the structure must be $(CH_3)_2CH-C\equiv CH$ ($3$-methyl$-1-$butyne).$4$. Oxidation of $(CH_3)_2CH-C\equiv CH$ gives $(CH_3)_2CHCOOH + CO_2$.

$A$ compound $C_5H_8$ gives a white precipitate with ammoniacal $AgNO_3$. It yields $(CH_3)_2CHCOOH$ upon oxidation with hot alkaline $KMnO_4$. The compound is:

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