Read the passage carefully and answer.
Organic compounds derived from \( NH_3 \) by replacing one or more H-atoms with alkyl or aryl groups will form primary, secondary and tertiary amines. The common methods which include their preparations are reduction of nitro group, amide group and nitrile group, Gabriel phthalimide synthesis and ammonolysis of alkyl halides. Amines are basic in nature due to the presence of a lone pair of electrons on the nitrogen atom. Important chemical properties of amines include acylation, alkylation and their interaction with nitrous acid. There are various tests which can distinguish different types of amines from each other, for e.g. carbylamine test, Hinsberg's test, coupling reaction etc. Another important class of compounds which can be derived from primary amines is azo compounds. These compounds are intensely colored and widely used as dyes in textile, leather and food industries.
In the following sequence of the conversion, identify the final product (X)
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 2-bromobenzoic acid
In a multi-step conversion, follow the groups one step at a time. The amino group made here is only a temporary handle: it is turned into a diazonium group and then removed, so its only role is to decide where the bromine sits.
First, 4-nitrotoluene is brominated (with the usual \( \ce{FeBr3} \) catalyst). The –\( \ce{CH3} \) group directs ortho/para and the –\( \ce{NO2} \) group directs meta; both point to the carbon ortho to CH3 and meta to NO2:
\( \displaystyle \text{4-nitrotoluene} + \ce{Br2} \xrightarrow{\ce{FeBr3}} \text{2-bromo-4-nitrotoluene} + \ce{HBr} \)
Next, Sn/HCl reduces the nitro group to an amino group:
\( \displaystyle \text{2-bromo-4-nitrotoluene} \xrightarrow{\ce{Sn/HCl}} \text{3-bromo-4-methylaniline} \)
The amine is diazotised in the cold (273–278 K):
\( \displaystyle \ce{ArNH2 + NaNO2 + 2HCl} \xrightarrow{\text{273--278 K}} \ce{ArN2^+Cl^- + NaCl + 2H2O} \)
Hypophosphorous acid replaces the diazonium group by hydrogen, which leaves 2-bromotoluene:
\( \displaystyle \ce{ArN2^+Cl^- + H3PO2 + H2O -> ArH + N2 + H3PO3 + HCl} \)
Finally, acidified \( \ce{KMnO4} \) oxidises the side-chain –\( \ce{CH3} \) to –COOH:
\( \displaystyle \text{2-bromotoluene} \xrightarrow{\ce{KMnO4/H^+}} \text{2-bromobenzoic acid} \)
So X is 2-bromobenzoic acid, with Br ortho to the COOH group.
Why the other options are incorrect
Option (1) 4-bromobenzoic acid — Incorrect. The para position of \( \ce{CH3} \) is held by the nitro group during bromination, so bromine cannot enter there.
Option (2) 3-bromobenzoic acid — Incorrect. Bromine meta to \( \ce{CH3} \) would be ortho to \( \ce{NO2} \); neither group directs the electrophile to that carbon.
Option (4) 3-bromoaniline — Incorrect. The amino group does not survive: it is converted into the diazonium group and removed by \( \ce{H3PO2} \), and the last step makes a carboxylic acid.
Note: It seems that there is a mistake in the original NTA paper, which prints the diazotisation temperature as 273-278 °C when it should read 273-278 K.