Question

Select the correct set of reagents required for conversion of the following chemical reactions

• A. (i). \(\text{NaNO}_2\)/ HCl, 273-278 K (ii). \(\text{H}_3\text{PO}_4\), \(\text{H}_2\text{O}\) (iii). \(\text{KMnO}_4\) and KOH
• B. (i). \(\text{NaNO}_2\)/HCl (ii). \(\text{H}_3\text{PO}_4\) (iii). NaOH/\(\text{H}_2\text{O}\)
• C. (i). \(\text{HNO}_3\)/\(\text{H}_2\text{SO}_4\) (ii). \(\text{H}_3\text{PO}_4\) (iii). \(\text{KMnO}_4\) and KOH
• D. (i). \(\text{NaNO}_2\)/HCl (ii). Sn/HCl (iii). \(\text{KMnO}_4\) and KOH

Hint:

Remove NH2 through a diazonium salt (then H3PO2). Then oxidise the CH3 group to COOH with KMnO4/KOH.

Answer 1

The Correct Option is A

Step 1: Compare reactant and product.
Reactant: ring with NH2, Br and CH3. Product: ring with Br and COOH only. Two changes are needed: NH2 is lost, and CH3 becomes COOH.

Step 2: Order of steps.
The amine must be handled before oxidation. If we use KMnO4 first, the free -NH2 group would be attacked by the oxidant. So we first deal with the amine and then oxidise the methyl group.

Step 3: Deamination.
Diazotisation, $\text{ArNH}_2 \xrightarrow{\text{NaNO}_2/\text{HCl},\ 273-278\,\text{K}} \text{ArN}_2^+\text{Cl}^-$, followed by hypophosphorous acid ($\text{H}_3\text{PO}_2$) and water gives Ar-H and nitrogen gas. Here the product is 2-bromotoluene. The reagent is written H3PO4 in the question, which is a typing slip for H3PO2.

Step 4: Side chain oxidation.
KMnO4 with KOH, then acidification, turns the benzylic CH3 into COOH and gives 2-bromobenzoic acid.

Step 5: Match.
Only option 1 has this exact order: NaNO2/HCl at low temperature, then the phosphorus reagent in water, then KMnO4/KOH.

Final Answer:
Option 1 gives 2-bromobenzoic acid from 3-bromo-4-methylaniline. \[ \boxed{\text{Option 1}} \]