Question

An organic compound ‘A’ on treatment with ethyl alcohol gives carboxylic acid ‘B’ and compound ‘C’ under acidic condition gives ‘B’ and ‘D’. Oxidation of ‘D’ with \[KMn{{O}_{4}}\] also gives ‘B’. ‘B’ on heating with \[Ca{{(OH)}_{2}}\] gives ‘E’ with the molecular formula \[{{C}_{3}}{{H}_{6}}O\]. ‘E’ does not give Tollens or Fehling test but forms 2,4-dinitrophenylhydrazine. Identify A, B, C, D and E.

Answer 1

Hint: Tollen’s reagent is a chemical reagent used to determine the presence of aldehyde and aromatic aldehyde functional groups this reagent generally gives silver mirror test. Fehling solution used to differentiate between carbohydrate and ketone.

Complete Step by step solution: Discuss the solution with reactions
\[{{(C{{H}_{3}}COO)}_{2}}+C{{H}_{3}}C{{H}_{2}}OH\to C{{H}_{3}}COOH+C{{H}_{3}}COOC{{H}_{2}}C{{H}_{3}}\]
(A)$\hspace{7cm}$(B)$\hspace{3cm}$(C)
Hence according to question ‘A’ which is acetic anhydride reacts with ethanol gives carboxylic acid called acetic acid ‘B’ and compound ‘C’ known by ethyl acetate.
Now C under acidic conditions upon hydrolysis gives
\[C{{H}_{3}}COOC{{H}_{2}}C{{H}_{3}}\to C{{H}_{3}}COOH+{{C}_{2}}{{H}_{5}}OH\]
(C)$\hspace{5cm}$(B)$\hspace{3cm}$(D)
Here D is ethyl alcohol and now oxidation of D with \[KMn{{O}_{4}}\] gives B
\[{{C}_{2}}{{H}_{5}}OH\to C{{H}_{3}}COOH\] (B)
B on heating with \[Ca{{(OH)}_{2}}\] gives
\[C{{H}_{3}}COOH\to C{{H}_{3}}COC{{H}_{3}}\] (E) having molecular formula \[{{C}_{3}}{{H}_{6}}O\] it is ketone.
It does not give Tollens or Fehling test but forms 2,4-dinitrophenylhydrazine it gives Brady's test for the formation of 2,4-DNPH.

Note: 2,4-Dinitrophenylhydrazine can be used to detect the presence of the ketone or aldehyde functional group. A positive test is detected by the formation of a yellow, orange or red precipitate. If the carbonyl compound is aromatic, then the precipitate will be red; if aliphatic, then the precipitate will have a more yellow color.