Read the passage carefully and answer the questions.
Amines are considered to be derivatives of ammonia obtained by replacement of hydrogen atoms with alkyl or aryl groups. They are usually prepared from nitro compounds, halides, amides, imides etc. The presence of hydrogen bonding influences its physical properties. In alkylamines, a combination of electron releasing steric and hydrogen bonding factors influences the stability of substituted ammonium cations in protic polar solvents and thus affects the basic nature of amines. Alkyl amines are found to be stronger bases than ammonia which is more basic than water. Primary and secondary amines are engaged in intermolecular association due to hydrogen bonding. This intermolecular association is more in primary amines than in secondary amines. Tertiary amines do not have intermoleulcar association due to the absence of hydrogen atom available for hydrogen bonding. Thus, the boiling point of amines depends on hydrogen bonding. In aromatic amines, electron releasing and withdrawing groups increase and decrease their basic character. Aryl diazonium salts usually obtained from arylamine undergo replacement of the diazonium group with a variety of nucleophile. Their coupling reaction of aryl diazonium salts with phenol or arylamine results in formation of dyes.
Complete the following reaction:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 2-bromobenzoic acid
The starting compound is 3-bromo-4-methylaniline: on the ring, –\( \ce{CH3} \) is at C-1, Br at C-2 (next to it) and –\( \ce{NH2} \) at C-4. The three reagents act one after the other: the amino group is first turned into a diazonium group, the diazonium group is then removed and replaced by hydrogen, and finally the methyl group is oxidised to –COOH.
First, sodium nitrite and hydrochloric acid at 273–278 K diazotise the primary aromatic amine:
\( \displaystyle \ce{CH3-C6H3(Br)-NH2 + NaNO2 + 2HCl} \xrightarrow{273-278\ \mathrm{K}} \ce{CH3-C6H3(Br)-N2^+Cl^- + NaCl + 2H2O} \)
Next, hypophosphorous acid reduces the diazonium salt; the –\( \ce{N2^{+}} \) group leaves as nitrogen gas and is replaced by H, giving 2-bromotoluene:
\( \displaystyle \ce{CH3-C6H3(Br)-N2^+Cl^- + H3PO2 + H2O -> CH3-C6H4-Br + N2 + H3PO3 + HCl} \)
Finally, alkaline potassium permanganate oxidises the side-chain methyl group to a carboxylate group, and acidification gives the acid. The bromine stays where it was, next to the side chain:
\( \displaystyle \ce{CH3-C6H4-Br} \xrightarrow[\text{(ii) }\ce{H3O+}]{\text{(i) }\ce{KMnO4},\ \ce{OH-}} \ce{HOOC-C6H4-Br} \)
So the final product is 2-bromobenzoic acid.
NCERT (2026-27 edition):
Chemical Reactions – “… 4. Replacement by H: Certain mild reducing agents like hypophosphorous acid (phosphinic acid) or ethanol reduce diazonium salts to arenes and themselves get oxidised to phosphorous acid and ethanal, respectively.”