Question

A hydrocarbon X adds on one mole of hydrogen to give another hydrocarbon and decolorized bromine water. X reacts with\[KMn{O_4}\] in presence of acid to give two moles of the same carboxylic acid. The structure of X is
A. \[{\rm{C}}{{\rm{H}}_{\rm{2}}}{\rm{ = CHC}}{{\rm{H}}_{\rm{2}}}{\rm{C}}{{\rm{H}}_{\rm{2}}}{\rm{C}}{{\rm{H}}_{\rm{3}}}\]
B. \[{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{C}}{{\rm{H}}_{\rm{2}}}{\rm{C}}{{\rm{H}}_{\rm{2}}}{\rm{CH = CHC}}{{\rm{H}}_{\rm{3}}}\]
C. \[{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{C}}{{\rm{H}}_{\rm{2}}}{\rm{CH = CHC}}{{\rm{H}}_{\rm{2}}}{\rm{C}}{{\rm{H}}_{\rm{3}}}\]
D. \[{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{CH = CHC}}{{\rm{H}}_{\rm{2}}}{\rm{C}}{{\rm{H}}_{\rm{2}}}{\rm{C}}{{\rm{H}}_{\rm{3}}}\]

Answer 1

Hint: The reaction of an alkene with bromine is an addition reaction. Alkene when reacted with bromine forms vicinal dihalides or 1,2-dihalides. Alkene on reaction with acidic potassium permanganate undergoes oxidation.

Complete Step by Step Answer:
Here in this question, we are given a hydrocarbon X which adds on one mole of hydrogen to give another hydrocarbon and decolorized bromine water.
It then reacts with KMnO₄ in presence of acid to give two moles of the same carboxylic acid.
We have to find out the compound.
A. \[{\rm{C}}{{\rm{H}}_{\rm{2}}}{\rm{ = CHC}}{{\rm{H}}_{\rm{2}}}{\rm{C}}{{\rm{H}}_{\rm{2}}}{\rm{C}}{{\rm{H}}_{\rm{3}}}\]
This compound when added with bromine will give \[BrC{H_2}CHBr - C{H_2}C{H_2}C{H_3}\].
\[C{H_2} = CH - C{H_2}C{H_2}C{H_3} + {\rm{ }}B{r_2} \to C{H_2}BrCHBr - C{H_2}C{H_2}C{H_3}\]
This compound on oxidation with acidified KMnO4 will give methanoic acid and butanoic acid.
The reaction will happen as follows:-

Image: Reaction with acidified potassium permanganate.
So, A is incorrect.

B. \[{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{C}}{{\rm{H}}_{\rm{2}}}{\rm{C}}{{\rm{H}}_{\rm{2}}}{\rm{CH = CHC}}{{\rm{H}}_{\rm{3}}}\]
This compound when added with bromine will give\[C{H_3}C{H_2}C{H_2} - CHBrCHBrC{H_3}\].
\[C{H_3}C{H_2}C{H_2} - CH = CHC{H_3} + {\rm{ }}B{r_2} \to C{H_3}C{H_2}C{H_2} - CHBrCHBrC{H_3}\]
This compound on oxidation with acidified KMnO4 will give butanoic acid and ethanoic acid.
The reaction will happen as follows:-

Image: Reaction with acidified potassium permanganate.
So, B is incorrect.
C. \[{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{C}}{{\rm{H}}_{\rm{2}}}{\rm{CH = CHC}}{{\rm{H}}_{\rm{2}}}{\rm{C}}{{\rm{H}}_{\rm{3}}}\]
This compound when added with bromine will give\[C{H_3}C{H_2}CHBrCHBrC{H_2}C{H_3}\].
\[C{H_3}C{H_2}CH = CHC{H_2}C{H_{3\;}} + {\rm{ }}B{r_2} \to C{H_3}C{H_2}CHBrCHBrC{H_2}C{H_3}\]
This compound on oxidation with acidified KMnO4 will give two moles of propanoic acid.
The reaction will happen as follows:-

Image: Reaction with acidified potassium permanganate.
Since this compound reacts with KMnO4 in presence of acid to give two moles of the same carboxylic acid.
So, C is correct.
D. \[{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{CH = CHC}}{{\rm{H}}_{\rm{2}}}{\rm{C}}{{\rm{H}}_{\rm{2}}}{\rm{C}}{{\rm{H}}_{\rm{3}}}\]
This compound when added with bromine will give\[C{H_3}CHBrCHBrC{H_2}C{H_2}C{H_3}\].
\[C{H_3}CH = CHC{H_2}C{H_2}C{H_3} + B{r_2} \to C{H_3}CHBrCHBrC{H_2}C{H_2}C{H_3}\]
This compound on oxidation with acidified KMnO4 will give ethanoic acid and butanoic acid.
The reaction will happen as follows:-

Image: Reaction with acidified potassium permanganate.
So, D is incorrect.

So, option C is correct.

Note: Out of the given options, the only option C is a symmetrical alkene. So, it will form two same carboxylic acids on oxidation. Acidified potassium permanganate is a strong oxidising agent. It oxidises alkene by breaking down the carbon-carbon double bond and converting it into two carbon-oxygen double bonds i.e., \[C = O\].