Question

Which one of the following reactions does NOT give benzene as the product?

• A. n-hexane \(\xrightarrow{MoO_3, 673K, 10-20 \text{atm}}\)
• B. H\textsubscript{2}C=CH\textsubscript{2} \(\xrightarrow{\text{hot Iron Tube at 873 K}}\)
• C. $C_6H_5N_3$ \(\xrightarrow{H_2O, \text{warm}}\)
• D. $C_6H_5OH$ \(\xrightarrow{\text{soda lime}}\)

Hint:

When dealing with reactions involving aromatic rings, look for reactions that either remove functional groups (like in soda lime reactions) or those that induce cyclization (like ethene cracking).

Answer 1

The Correct Option is C

Let's analyze each reaction:

1. n-Hexane with Mo₂O₃ at 773K and 10-20 atm: This is catalytic reforming or aromatization. n-Hexane can be converted to benzene under these conditions.

2. Acetylene (Ethyne) with Red Hot Iron Tube at 873 K: This is a classic method for synthesizing benzene. Three molecules of acetylene undergo cyclic polymerization to form benzene.

3. Benzenediazonium Ion with Warm Water: Benzenediazonium ion (C₆H₅N₂⁺) reacts with warm water to form phenol (C₆H₅OH), not benzene. The diazonium group is replaced by a hydroxyl group.

4. Benzoate with Sodalime and Heat: This is decarboxylation. The benzoate loses CO₂ to form benzene.

Therefore, the reaction that does not produce benzene as a major product is the reaction of benzenediazonium ion with warm water.