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Question
An organic compound 'A' with the molecular formula C4H8O2 undergoes acid hydrolysis to form two compounds 'B' and 'C'. Oxidation of 'C' with acidified potassium permanganate also produces 'B'. Sodium salt of 'B' on heating with soda lime gives methane.
- Identify 'A', 'B' and 'C'.
- Out of 'B' and 'C', which will have higher boiling point? Give reason.
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Solution
- A = C4H8O2 [CH3COOC2H5] ester Reactions involved are
\[\ce{\underset{A}{CH3-COOC2H5} + H2O ->[dil H2SO4] \underset{B}{CH3COOH} + \underset{C}{CH3CH2OH}}\]
Then
\[\ce{\underset{C}{CH3CH2OH} ->[KMnO4][{[O]}] \underset{B}{CH3COOH}}\]
So A = CH3COOC2H5, B = CH3COOH, C = CH3CH2OH - B has a higher boiling point than C. Because of their propensity to generate intermolecular H-bonds, carboxylic acids have higher boiling temperatures than alcohols.
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