Question

An organic compound with the molecular formula \( \mathrm{C_9H_{10}O} \) forms 2,4-DNP derivative, reduces Tollens' reagent and undergoes Cannizzaro reaction. On vigorous Oxidation, it gives phthalic acid. Identify the compound. Write the reaction of this compound with Tollens' reagent and Cannizzaro reaction.

Hint:

The compound is an aromatic aldehyde with no alpha hydrogen, and oxidation to phthalic acid means the groups are ortho. It is 2-ethylbenzaldehyde.

Answer 1

Step 1: Read the clues:
A 2,4-DNP derivative shows that the compound has a carbonyl group.
Reducing Tollens' reagent shows that it is an aldehyde.
The Cannizzaro reaction shows that the aldehyde has no alpha hydrogen, so the -CHO is on a benzene ring.

Step 2: Use the formula and the oxidation product:
\( \mathrm{C_9H_{10}O} \) is a benzene ring (\( \mathrm{C_6H_4} \)) with -CHO and one more group, \( \mathrm{C_2H_5} \).
Vigorous oxidation turns both side groups into -COOH. Phthalic acid is benzene-1,2-dicarboxylic acid.
So the two groups must be ortho to each other.
The compound is 2-ethylbenzaldehyde (o-ethylbenzaldehyde).

Step 3: Reaction with Tollens' reagent:
The aldehyde is oxidised to a carboxylate ion and silver is deposited as a mirror.
\[ 2\text{-}\mathrm{C_2H_5C_6H_4CHO} + 2[\mathrm{Ag(NH_3)_2}]^{+} + 3\,\mathrm{OH^-} \rightarrow 2\text{-}\mathrm{C_2H_5C_6H_4COO^-} + 2\,\mathrm{Ag}\downarrow + 4\,\mathrm{NH_3} + 2\,\mathrm{H_2O} \]

Step 4: Cannizzaro reaction:
With concentrated NaOH, one molecule is oxidised and the other is reduced.
\[ 2\,(2\text{-}\mathrm{C_2H_5C_6H_4CHO}) + \mathrm{NaOH} \rightarrow 2\text{-}\mathrm{C_2H_5C_6H_4CH_2OH} + 2\text{-}\mathrm{C_2H_5C_6H_4COONa} \]
The products are 2-ethylbenzyl alcohol and sodium 2-ethylbenzoate.

Final Answer:
The compound is 2-ethylbenzaldehyde (o-ethylbenzaldehyde).
With Tollens' reagent it gives 2-ethylbenzoate and a silver mirror. In the Cannizzaro reaction it gives 2-ethylbenzyl alcohol and sodium 2-ethylbenzoate.