Question

[Figure]

• A. OHC-CHO
• B. CH3-CHO
• C. CH3-COOH
• D. H-COOH

Answer 1

Answer: (D)

H-COOH

Answer 2

Step-by-step derivation:

1. Ozonolysis of Benzene:
   The starting material is benzene. Reductive ozonolysis of benzene ($O_3/Zn/\Delta, H_2O$) cleaves all three double bonds. This reaction converts each carbon-carbon double bond into two carbonyl groups. For benzene, this leads to the formation of three molecules of glyoxal.

   Therefore, product (A) is Glyoxal (OHC-CHO).

2. Reaction of (A) with Fehling's solution:
   Product (A) is Glyoxal (OHC-CHO). Glyoxal contains two aldehyde groups. Aldehydes are oxidized by Fehling's solution (a mild oxidizing agent containing Cu(II) ions) to carboxylic acids, while the Cu(II) ions are reduced to Cu(I) oxide ($Cu_2O$), which is a red precipitate.
   The observation "Red colour" (C) confirms that (A) is an aldehyde and that the reaction with Fehling's solution is positive, forming $Cu_2O$.

3. Reaction of (A) with Periodic Acid ($HIO_4$):
   Product (A) is Glyoxal (OHC-CHO). Periodic acid ($HIO_4$) is a reagent used for the oxidative cleavage of 1,2-diols, 1,2-diketones, 1,2-hydroxyaldehydes, and 1,2-dialdehydes. Glyoxal is a 1,2-dialdehyde.
   The cleavage of a 1,2-dialdehyde by periodic acid results in the formation of carboxylic acids.
   In the case of glyoxal (OHC-CHO), both aldehyde groups are on adjacent carbons. The cleavage occurs between these two carbons, and each carbon is oxidized to a carboxylic acid group.
   $OHC-CHO \xrightarrow{HIO_4} HCOOH + HCOOH$
   Therefore, product (B) is Formic acid (HCOOH).

Conclusion:
Based on the reaction sequence, (B) is Formic acid (HCOOH).

Benzene undergoes reductive ozonolysis to form glyoxal (OHC-CHO), which is product (A). Glyoxal, being an aldehyde, gives a positive Fehling's test (red precipitate (C)). Glyoxal (a 1,2-dialdehyde) is cleaved by periodic acid ($HIO_4$) to produce two molecules of formic acid (HCOOH), which is product (B).