(i). \(\text{HNO}_3\)/\(\text{H}_2\text{SO}_4\) (ii). \(\text{H}_3\text{PO}_4\) (iii). \(\text{KMnO}_4\) and KOH
(i). \(\text{NaNO}_2\)/HCl (ii). Sn/HCl (iii). \(\text{KMnO}_4\) and KOH
Show Solution
Verified By Collegedunia
The Correct Option isA
Solution and Explanation
Step 1: Understanding the Question:
The reactant is 3-bromo-4-methylaniline. The product is 2-bromobenzoic acid. Comparing the two rings, the -NH2 group has disappeared and the -CH3 group has become -COOH. The bromine stays where it is.
Step 2: Remove the amino group.
First we convert -NH2 into a diazonium salt with NaNO2/HCl at 273-278 K. The low temperature is needed because diazonium salts decompose above about 278 K. Then the diazonium group is replaced by hydrogen using hypophosphorous acid, \(\text{H}_3\text{PO}_2\), and water. The paper prints H3PO4, which is a misprint for H3PO2. This gives 2-bromotoluene, with Br next to the CH3.
Step 3: Oxidise the methyl group.
Hot alkaline KMnO4 oxidises a side-chain CH3 on a benzene ring to COOH. \[ \text{Ar-CH}_3 \xrightarrow{\text{KMnO}_4,\ \text{KOH}} \text{Ar-COOH} \] This gives 2-bromobenzoic acid, the required product.
Step 4: Check option 1.
It has diazotisation at 273-278 K, removal of the diazonium group by the phosphorus reagent in water, and oxidation by KMnO4/KOH. All three steps match the plan. Option 1 is correct.
Step 5: Check option 2.
The last step uses NaOH/H2O. This cannot turn CH3 into COOH. It is wrong.
Step 6: Check option 3.
HNO3/H2SO4 is a nitrating mixture. It would add a nitro group to the ring and would not prepare the diazonium salt. The -NH2 group would not be removed properly. It is wrong.
Step 7: Check option 4.
Sn/HCl is a reducing agent used for nitro groups. It does not replace a diazonium group by hydrogen cleanly (it gives a hydrazine). It is wrong.
Final Answer:
The sequence is diazotisation, removal of the diazonium group, then oxidation of CH3 to COOH. This is option 1.
\[ \boxed{\text{Option 1}} \]
