Question

A compound \(A\) having molecular formula \(C_7H_8O\) gives a violet colour with neutral \(FeCl_3\). On treatment with excess \(CH_3I/K_2CO_3\), it forms compound \(B\). Ozonolysis of \(B\) followed by reductive workup \((Zn/H_2O)\) gives one mole of methanal and one mole of anisaldehyde. The number of phenolic compounds among \(A\), \(B\), and the ozonolysis products is:

• A. 0
• B. 1
• C. 2
• D. 3

Hint:

Methylation of phenols converts \(-OH\) into \(-OCH_3\), destroying the phenolic nature and eliminating the \(FeCl_3\) test.

Answer 1

The Correct Option is B

Concept: This question combines several high-weightage NCERT topics:
• Phenols and \(FeCl_3\) test
• Williamson ether synthesis concept
• Ozonolysis
• Functional group identification A student must identify the structure first and then count phenolic compounds carefully.

Step 1: Interpret the \(FeCl_3\) test. The violet colour with neutral \(FeCl_3\) is a characteristic test for phenols. Therefore compound \(A\) must contain: \[ -ArOH \] group. Hence: \[ A=\text{phenolic compound} \]

Step 2: Analyze methylation reaction. Treatment with: \[ CH_3I/K_2CO_3 \] converts phenol into methyl ether. Thus: \[ ArOH \longrightarrow ArOCH_3 \] Therefore compound \(B\) is no longer a phenol. It is an ether.

Step 3: Identify compound B from ozonolysis data. Formation of: \[ \text{Methanal} \] and \[ \text{Anisaldehyde} \] indicates that \(B\) is anisole containing a terminal double bond. The exact structure need not be drawn because the key observation is: \[ B \] contains an ether and not a phenolic OH group.

Step 4: Examine ozonolysis products. Products: \[ HCHO \] and \[ \text{anisaldehyde} \] Neither possesses a phenolic OH group. Hence neither is phenolic.

Step 5: Count phenolic compounds. Compound \(A\): Phenolic \[ \checkmark \] Compound \(B\): Not phenolic \[ \times \] Ozonolysis products: Not phenolic \[ \times \] Total phenolic compounds: \[ 1 \]

Step 6: Final conclusion. \[ \boxed{1} \] Therefore: \[ \boxed{\text{Option (B)}} \]