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Addition Reactions

A hydrocarbon...

Question

A hydrocarbon 'X' on reaction with $B r_{2}$ in presence of $C C l_{4}$ decolourises bromine water.
Also 'X' reacts with $K M n O_{4}$ in the presence of an 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 B $C H_{3} C H_{2} C H = C H C H_{2} C H_{3}$
Since it decolorizes bromine water, so X must be an alkene. SO (c) is incorrect.
The reaction of the process:
$C H_{3} C H_{2} C H = C H C H_{2} C H_{3} + B r_{2} \to C H_{3} C H_{2} C H B r - C H B r C H_{2} C H_{3}$

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.
In presence of acidic $K M n O_{4}$ alkene is oxidised to form acid.
$C H_{3} C H_{2} C H = C H C H_{2} C H_{3} + K M n O_{4} / H^{+} - ------- \to K_{2} C r_{2} O_{7} 2 C H_{3} C H_{2} C O O H$
So the structure of X is $C H_{3} C H_{2} C H = C H C H_{2} C H_{3}$
Thus option B is correct.

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