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(C) $1$. The hydrocarbon $X$ decolourises bromine water,indicating it is an unsaturated hydrocarbon (alkene).$2$. $X$ reacts with acidic $KMnO_4$ (a s

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$CH_3CH_2CH=CHCH_2CH_3$

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(C) $1$. The hydrocarbon $X$ decolourises bromine water,indicating it is an unsaturated hydrocarbon (alkene).$2$. $X$ reacts with acidic $KMnO_4$ (a strong oxidizing agent) to produce two moles of the same carboxylic acid. This indicates that the double bond is located in the middle of a symmetrical alkene.$3$. The reaction is: $CH_3CH_2CH=CHCH_2CH_3 + [O] \xrightarrow{H^+} 2 CH_3CH_2COOH$ (Propanoic acid).$4$. Among the given options,$CH_3CH_2CH=CHCH_2CH_3$ (Hex$-3-$ene) is the only symmetrical alkene that cleaves to form two identical carboxylic acid molecules.

$A$ hydrocarbon $X$ adds one mole of hydrogen to give another hydrocarbon and decolourises bromine water. $X$ reacts with $KMnO_4$ in the presence of acid to give two moles of the same carboxylic acid. The structure of $X$ is:

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