Question

An organic compound (A) with molecular formula $C_8{H}_{8}O$ forms an orange-red precipitate with $2,4-\text{DNP}$ reagent and gives yellow precipitate on heating with iodine in the presence of sodium hydroxide. It neither reduces Tollens' or Fehling's reagent, nor does it decolourise bromine water or Baeyer's reagent. On drastic oxidation with chromic acid, it gives a carboxylic acid (B) having molecular formula $C_7{H}_6{O}_2$. Identify the compounds (A) and (B) and explain the reactions involved.

Answer 1

Tollens reagent, Fehling reagent, Baeyer's reagent

• Tollens Reagent $[AgN{O}_3+N{H}_{4}OH]$ is a chemical reagent which is used to detect an aldehyde functional group, an aromatic aldehyde functional group, or an alpha hydroxy ketone functional group in an organic compound.
• Fehling's solution is prepared by combining two separate solutions: Fehling's A, which is a deep blue aqueous solution of copper (II) sulfate, and Fehling's B, which is a colorless solution of aqueous potassium sodium tartrate (also known as Rochelle salt) made strongly alkali with sodium hydroxide.

1. Fehling reagent is used for the detection of reducing sugars and non-reducing sugars. All the aldehydes which having alpha hydrogen gives positive Fehling's test.

• Baeyer’s reagent $\left(KMn{O}_4\right)$ is an alkaline solution of cold potassium permanganate $\left(KMn{O}_4\right)$ of violet colour solution. It is a strong oxidizing agent.

1. As this solution interacts with a double bond compound the colour disappears and becomes colourless

Identification of compound (A):
Compound (A) forms $2,4-\text{DNP}$ derivative which means that it may be an aldehyde or a ketone.

Compound (A) does not reduce Tollens' or Fehling reagent which indicates that compound (A) must be a ketone.

Compound (A) giving positive iodoform test. As a result, it should be a methyl ketone.
The molecular formula $C_8{H}_{8}O$ of (A) indicates high degree of unsaturation even though it does not decolorize the reagent with bromine water or Baeyer's reagent which means that the molecule contains unsaturation due to an aromatic ring.

Compound (A) on reaction with $NaOH+I_2$ gives iodoform test. Which indicates it must be methyl ketone with aromatic ring.

According to molecular formula $C_8{H}_{8}O$ of (A) indicates that it should be phenyl methyl ketone (acetophenone).

So that the compound (A) is phenyl methyl ketone (acetophenone).

Identification of compound (B):
Compound (B) with molecular formula $C_7{H}_6{O}_2$ forms carboxylic acid due to oxidation of ketone.

So, the compound (B) should be benzoic acid. So that the compound (B) is benzoic acid.
This can be represented as given below-