Question

An aromatic compound (X) (${C_8}{H_8}O$) gives $2,4$ DNP Test. It gives a yellow precipitate of compound (Y) on reaction with iodine and sodium hydroxide solution. (X) does not give Tollens Test on oxidation under drastic conditions. It gives a carboxylic acid (Z) (${C_7}{H_6}{O_2}$). (Z) is also formed with (Y) during the reaction . (X), (Y), (Z) respectively are
A. ${C_6}{H_5}OC{H_3}\,,\,CH{I_3}\,,\,{C_6}{H_5}COOH$
B. $C{H_3}COC{H_3}\,,\,CH{I_3}\,,\,C{H_3}COOH$
C. ${C_6}{H_5}COC{H_3}\,,\,CH{I_3}\,,\,C{H_3}COOH$
D. $C{H_3}CHO\,,\,CH{I_3}\,,\,{C_6}{H_5}COOH$

Answer 1

As compound X does not give Tollens Test, so it is not an Aldehyde. Also, it gives a positive $2,4$ DNP Test, so it must be ketone or aldehyde. In reaction with compound X with iodine and sodium hydroxide, yellow precipitate is formed.

Complete step by step answer: $2,4$ DNP Test: $2,4$ DNP Test can be used to detect ketone or aldehyde functional groups. When formation of a yellow, orange or red precipitate occurs, the test is considered as positive.

Tollen’s Test: Tollen’s test is also known as the silver-mirror test. It is used to differentiate an aldehyde and a ketone. If an aldehyde is present $A{g^ + }$ is reduced to $A{g^0}$ which forms precipitates, as a silver mirror. On the other hand, in the case of Ketones, no precipitation takes place.
As ${C_8}{H_8}O$ gives $2,4$ DNP Test so it should be a ketone. If formation of yellow,orange or red precipitates takes place, we can say that X is ketone.
When ${C_8}{H_8}O$ reacts with iodine and sodium hydroxide, benzoic acid and iodoform are formed.
When ${C_8}{H_8}O$ reacts with potassium permanganate, benzoic acid is formed.

Hence, after seeing the above given reactions, we find that, X compound is ${C_6}{H_5}OC{H_3}$ , Y compound is $CH{I_3}$ and Z is ${C_6}{H_5}COOH$.

So, the correct answer is “Option A”.

Note: Except Tollen’s Test and $2,4$ DNP test, the following test can also distinguish between aldehydes and ketones:
- Fehling’s Test
- Schiff’s Test
- Test with Chromic Acid