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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

Vedclass · Accepted Answer

$CH_3CH_2CH=CHCH_2CH_3$

Vedclass · Answer

(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:

Similar Questions

The decreasing order of reactivity of the following alkenes towards $HBr$ addition is:
$(I) \ CH_3-CH=CH_2$
$(II) \ CF_3-CH=CH_2$
$(III) \ MeOCH=CH_2$
$(IV) \ CH_3-CO-CH=CH_2$

In the following sequence of reactions,the alkene affords the compound $B$.
$CH_3-CH=CH-CH_3$ $\xrightarrow{O_3} A$ $\xrightarrow[Zn]{H_2O} B$
The compound $B$ is:

Propene when heated with chlorine at about $500 \, ^oC$ forms

What is the major product formed by the reaction of $HI$ with $CH_3-C(CH_3)_2-CH=CH_2$?

The stability order of the given alkenes is

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