A benzene derivative did not produce a white | vedclass

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(C) The compound is $C_8H_8$ (Styrene).$1$. Ammoniacal silver nitrate (Tollens' reagent) reacts with terminal alkynes to form a white precipitate. Sin

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$C_8H_8$

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(C) The compound is $C_8H_8$ (Styrene).$1$. Ammoniacal silver nitrate (Tollens' reagent) reacts with terminal alkynes to form a white precipitate. Since the compound does not produce this precipitate,it does not contain a terminal $-C \equiv CH$ group.$2$. Cold dilute alkaline $KMnO_4$ (Baeyer's reagent) is a test for unsaturation (alkenes or alkynes). Decolorization indicates the presence of a double or triple bond.$3$. Styrene $(C_6H_5-CH=CH_2)$ has the molecular formula $C_8H_8$. It contains a double bond,which reacts with $KMnO_4$ to decolorize it,but it lacks a terminal alkyne group,so it does not react with Tollens' reagent.

$A$ benzene derivative did not produce a white precipitate with ammoniacal silver nitrate solution (Tollens' reagent) but decolorized cold dilute alkaline $KMnO_4$ solution (Baeyer's reagent). The compound is:

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