
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
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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., \[C = O\].
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., \[C = O\].
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