A hydrocarbon $^{'} X^{'}$ adds on one mole of hydrogen to give another hydrocarbon and decolourised bromine water. $^{'} X^{'}$ reacts with ${K M n O}_{4}$ in the presence of acid to give two moles of the same carboxylic acid. The structure of $^{'} X^{'}$ is:

C H_{3} C H = C H C H_{2} C H_{2} C H_{3}

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C H_{3} C H_{2} C H = C H C H_{2} C H_{3}

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C H_{3} C H_{2} C H_{2} - C H = C H C H_{3}

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C H_{2} = C H C H_{2} C H_{2} C H_{3}

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Solution

The correct option is A $C H_{3} C H_{2} C H = C H C H_{2} C H_{3}$
The reaction of a process:
${C H}_{3} {C H}_{2} C H = C H C H_{2} {C H}_{3} + B r_{2} ⟶ {C H}_{3} {C H}_{2} C H B r - C H B r C H_{2} {C H}_{3}$
And in presence of acidic $K M n O_{4}$ alkene is oxidised to form acid.
${C H}_{3} {C H}_{2} C H = C H C H_{2} {C H}_{3} + (H_{2} O + O) ⟶ 2 {C H}_{3} C H_{2} C O O H$
So the structure of X is ${C H}_{3} {C H}_{2} C H = C H C H_{2} {C H}_{3}$
Since it gives two moles of same carboxylic acid on reaction with oxidizing agent potassium permanganate hence it must be symmetrical alkene. Here the only alkene in option B is symmetrical alkene. Thus option B is correct.

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