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4-Methoxyphenol (mequinol) is treated with CHCl3 and NaOH (Reimer-Tiemann reaction), followed by acidification with H+. The product A is:

1. 2-hydroxy-4-methoxybenzaldehyde
2. 3-methoxy-2-hydroxybenzaldehyde
3. 2-hydroxy-4-methoxybenzoic acid
4. 3-methoxy-2-hydroxybenzoic acid

Correct answer: 2-hydroxy-4-methoxybenzaldehyde

Solution

The Reimer-Tiemann reaction: CHCl3 + NaOH generates:CCl2 (dichlorocarbene) which attacks the ortho position of the phenol ring. Starting material: 4-methoxyphenol (OH at C1, OCH3 at C4). Electrophilic attack at C2 (ortho to OH) gives a dichloromethyl intermediate at C2, which on alkaline hydrolysis and acidification yields the aldehyde CHO at C2. Product A = 2-hydroxy-4-methoxybenzaldehyde (also known as isovanillin). This matches option A.

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