Question

What is the product 'Z' in the given sequence of reactions?

• A. A phenyl ether with an ortho -OH group
• B. A phenyl ether with an ortho -COOH group
• C. A phenyl ether with a meta -OH group
• D. A phenyl ether with a para -OH group

Hint:

Remember the key reactions of phenol: - \textbf{Kolbe-Schmitt Reaction:} Phenol + NaOH, then CO\(_2\), then H\(^+\) \(\rightarrow\) Salicylic acid (o-hydroxybenzoic acid). - \textbf{Reimer-Tiemann Reaction:} Phenol + CHCl\(_3\)/NaOH \(\rightarrow\) Salicylaldehyde (o-hydroxybenzaldehyde). - \textbf{Acetylation:} Phenol + Acetic Anhydride \(\rightarrow\) Phenyl acetate. The reaction of Salicylic acid with acetic anhydride gives Aspirin.

Answer 1

The Correct Option is B

Let's follow the reaction sequence starting from Aniline.
Step 1: Formation of X.
Aniline (C\(_6\)H\(_5\)NH\(_2\)) is treated with NaNO\(_2\)/HCl at 273-278 K (0-5 °C). This is the diazotization reaction, which converts the primary amine group into a diazonium salt.
The product is benzenediazonium chloride, [C\(_6\)H\(_5\)N\(_2\)]\(^+\)Cl\(^-\).
The diazonium salt is then treated with warm water (H\(_2\)O/warm). This is a substitution reaction where the diazonium group is replaced by a hydroxyl group (-OH).
The product X is Phenol (C\(_6\)H\(_5\)OH).
Step 2: Formation of Y.
Phenol (X) is treated with (i) NaOH and (ii) CO\(_2\), followed by (iii) H\(^+\) (acidification). This is the Kolbe-Schmitt reaction.
Phenol is first converted to its more reactive phenoxide ion by NaOH. The phenoxide ion then undergoes electrophilic substitution with the weak electrophile CO\(_2\). The reaction predominantly occurs at the ortho position. Acidification then protonates the resulting carboxylate.
The product Y is Salicylic acid (o-hydroxybenzoic acid).
Step 3: Formation of Z.
Salicylic acid (Y) is treated with acetic anhydride, (\(CH_3CO)_2O\). Acetic anhydride is an acetylating agent. It will react with the hydroxyl group of salicylic acid. The carboxylic acid group is less reactive towards acetylation.
The phenolic -OH group is acetylated to form an ester group (-OCOCH\(_3\)).
The product Z is Acetylsalicylic acid, commonly known as Aspirin.
The structure shown in option B is O-acetylsalicylic acid, where the carboxylic acid group is ortho to the acetylated hydroxyl group. This matches our derived structure for Z. The structure has a -COOH group ortho to an OCOCH3 group. This matches option B's structure.