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Question

A hydrocarbon 'X' on reaction with Br2 in presence of CCl4 decolourises bromine water.
Also 'X' reacts with KMnO4 in the presence of an acid to give two moles of the same carboxylic acid.
The structure of 'X' is:

A
CH3CH=CHCH2CH2CH3
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B
CH3CH2CH=CHCH2CH3
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C
CH3CH2CH2CH=CHCH3
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D
CH2=CHCH2CH2CH3
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Solution

The correct option is B CH3CH2CH=CHCH2CH3
Since it decolorizes bromine water, so X must be an alkene. SO (c) is incorrect.
The reaction of the process:
CH3CH2CH=CHCH2CH3+Br2CH3CH2CHBrCHBrCH2CH3

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.
In presence of acidic KMnO4 alkene is oxidised to form acid.
CH3CH2CH=CHCH2CH3+KMnO4/H+−−−−−−−K2Cr2O72CH3CH2COOH
So the structure of X is CH3CH2CH=CHCH2CH3
Thus option B is correct.

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