Question

The ratio of \(\sigma\) bonds to \(\pi\) bonds in Q is

• A. 3:1
• B. 1:3
• C. 4:1
• D. 2:1

Hint:

The oxidation of phenols can give different products depending on the reagent. Strong oxidants like chromic acid (prepared from Na\(_2\)Cr\(_2\)O\(_7\)/H\(_2\)SO\(_4\)) oxidize phenol to p-benzoquinone. Milder oxidation or exposure to air can lead to complex colored polymers.

Answer 1

The Correct Option is A

Step 1: Identify compound P.
The starting material is benzenesulfonic acid. It is treated with (i) NaOH, followed by (ii) H\(^+\). This is the process to convert a sulfonic acid into a phenol.
First, \( \text{C}_6\text{H}_5\text{SO}_3\text{H} + \text{NaOH} \rightarrow \text{C}_6\text{H}_5\text{SO}_3\text{Na} + \text{H}_2\text{O} \).
Then, the sodium salt is fused with solid NaOH at high temperature, followed by acidification.
\( \text{C}_6\text{H}_5\text{SO}_3\text{Na} \xrightarrow{\text{NaOH, fuse}} \text{C}_6\text{H}_5\text{ONa} \xrightarrow{\text{H}^+} \text{C}_6\text{H}_5\text{OH} \).
So, compound P is phenol.
Step 2: Identify compound Q.
Phenol (P) is treated with sodium dichromate (Na\(_2\)Cr\(_2\)O\(_7\)) and sulfuric acid (H\(_2\)SO\(_4\)). This is a strong oxidizing agent mixture.
The oxidation of phenol under these conditions yields 1,4-benzoquinone (also known as p-benzoquinone). The ring is oxidized to a conjugated diketone.
The structure of Q is a six-membered ring with two double bonds and two ketone groups at opposite (para) positions. Its formula is C\(_6\)H\(_4\)O\(_2\).
Step 3: Count the \(\sigma\) and \(\pi\) bonds in Q (p-benzoquinone).
Let's analyze the structure:
The ring consists of 4 CH groups and 2 C=O groups.
- \(\pi\) bonds: There are two C=C double bonds and two C=O double bonds. Each double bond contains one \(\pi\) bond. Total \(\pi\) bonds = 2 + 2 = 4.
- \(\sigma\) bonds: - There are 4 C-H single bonds. (4 \(\sigma\)) - In the ring, there are 4 C-C single bonds and 2 C=C double bonds (which also contain 2 \(\sigma\) bonds). Total C-C sigma bonds in the ring = 4 + 2 = 6. - There are 2 C=O double bonds (which contain 2 \(\sigma\) bonds). - Total \(\sigma\) bonds = 4 (C-H) + 4 (C-C) + 2 (C=C) + 2 (C=O) = 12. Let me recount. Let's count edges in the cyclic graph: 4 C-H bonds, 4 C-C single bonds, 2 C=C double bonds (2 \(\sigma\)), 2 C=O double bonds (2 \(\sigma\)). Total \(\sigma\) = 4(C-H) + 4(C-C) + 2(from C=C) + 2(from C=O) = 12.
So we have 12 \(\sigma\) bonds and 4 \(\pi\) bonds.
Step 4: Find the ratio.
The ratio of \(\sigma\) bonds to \(\pi\) bonds is 12 : 4.
Simplifying the ratio by dividing by 4 gives 3 : 1.