Question

Benzene on ozonolysis followed by reaction with Zn + H₂O gives:

• A. 3 moles of glycerol
• B. 3 moles of glyoxal
• C. 3 moles of glyoxalic acid
• D. 3 moles of acetylene

Hint:

Ozonolysis of aromatic compounds like benzene is a powerful method to cleave the entire aromatic ring. The type of workup (reductive or oxidative) determines the final product. A reductive workup (e.g., $\text{Zn}/\text{H}_2\text{O}$ or $\text{Me}_2\text{S}$) yields aldehydes, while an oxidative workup (e.g., $\text{H}_2\text{O}_2$) yields carboxylic acids.

Answer 1

The Correct Option is B

Step 1: Benzene Ozonolysis Overview.
Ozonolysis cleaves unsaturated compounds (alkenes, alkynes, aromatics) using ozone (O₃). For benzene, this breaks all three double bonds, followed by a reductive workup (Zn + H₂O or dimethyl sulfide) to prevent product oxidation.

Step 2: Benzene Ozonolysis Mechanism.
Benzene's cyclic structure contains alternating double bonds. Ozonolysis breaks each carbon-carbon double bond, forming an ozonide intermediate.
Benzene reacts with 3 moles of ozone (O₃) to produce a triozonide.

Step 3: Reductive Workup with Zn + H₂O.
The triozonide undergoes reductive workup, typically with Zn + H₂O. This prevents aldehyde oxidation to carboxylic acids. Each benzene ring fragment containing a C=C bond becomes an aldehyde (glyoxal).
Each of benzene's three carbon-carbon double bonds is cleaved, forming three dialdehyde (glyoxal) molecules.

The overall reaction:
C₆H₆ ⟶ _{1. 3O3} Triozonide ⟶ _{2. Zn/H2O} 3 OHC-CHO
The product, OHC-CHO, is glyoxal. Since benzene has 6 carbon atoms and 3 double bonds, and each glyoxal molecule has 2 carbon atoms, 3 moles of glyoxal are produced from 1 mole of benzene.

Step 4: Analyze the options.

• Option (A): 3 moles of glycerol: Glycerol is a triol (CH₂OH-CHOH-CH₂OH). Ozonolysis produces aldehydes or ketones, not alcohols.
• Option (B): 3 moles of glyoxal: Glyoxal (OHC-CHO) is a dialdehyde formed by cleaving each C₂ unit of the benzene ring. This is correct.
• Option (C): 3 moles of glyoxalic acid: Glyoxalic acid (OHC-COOH) forms with an oxidative workup (e.g., H₂O₂), oxidizing an aldehyde to a carboxylic acid. Zn + H₂O gives aldehydes.
• Option (D): 3 moles of acetylene: Acetylene (CH≡CH) is an alkyne. Ozonolysis cleaves double/triple bonds, forming carbonyl compounds, not unsaturated hydrocarbons.

Step 5: Conclusion.
Benzene ozonolysis followed by reductive workup (Zn + H₂O) yields 3 moles of glyoxal.

\[ \n\boxed{\text{3 moles of glyoxal}} \n\]