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प्रश्न
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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उत्तर
- 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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You are given four organic compounds “A”, “B” , “C” and “D”. The compounds “A”, “B” and “C” form an orange-red precipitate with 2, 4 DNP reagent. Compounds “A” and “B” reduce Tollen’s reagent while compounds “C” and “D” do not. Both “B” and “C” give a yellow precipitate when heated with iodine in the presence of NaOH. Compound “D” gives brisk effervescence with sodium bicarbonate solution. Identify “A”, “B”, “C” and “D” given the number of carbon atoms in three of these carbon compounds is three while one has two carbon atoms. Give an explanation for our answer.
Match List-I with List-II:
| List-I (Reaction) |
List-II (Reagents/Condition) |
||
| A. |  |
I. |  |
| B. |  |
II. | CrO3 |
| C. |  |
III. | KMnO4/KOH, Δ |
| D. |  |
IV. | (i) O3 (ii) Zn-H2O |
Choose the correct answer from the options given below:
