A hydrocarbon X adds on one mole of hydrogen to give another hydrocarbon and decolourised bromine water. X react with ${KMnO}_{4}$ in presence of acid to give two moles of the same carboxylic acid. The structure of X is:

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${CH}_{3} CH = {CHCH}_{2} {CH}_{2} {CH}_{3}$

${CH}_{3} {CH}_{2} {CH}_{2} — CH = {CHCH}_{3}$

${CH}_{2} = CH — {CH}_{2} {CH}_{2} {CH}_{3}$

${CH}_{3} {CH}_{2} CH = {CHCH}_{2} {CH}_{3}$

answer is B.

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Detailed Solution

The reaction of a process:
CH₃CH₂CH=CHCH₂CH₃+Br₂⟶CH₃CH₂CHBr−CHBrCH₂CH₃

And in presence of acidic KMnO4 alkene is oxidised to form acid.

CH₃CH₂CH=CHCH₂CH₃+(H₂O+O)⟶2CH₃CH₂COOH

So the structure of X is CH₃CH₂CH=CHCH₂CH₃

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