Read the passage carefully and answer the Questions.
Aldehydes and ketones are important organic compounds prepared mainly by oxidation of alcohols. Primary alcohols give aldehyde and secondary alcohols yield ketones. Aldehydes can also be formed by partial reduction of acid chlorides, reduction of esters or by Gattermann-Koch reaction. Ketones are prepared via Friedel-Craft's acylation, hydration of alkyne and reaction of Grignard's reagent with nitriles. Chemically, aldehydes and ketones are reactive in nature due to the carbonyl group and undergo nucleophilic addition reactions. Aldehydes are more reactive than ketones and can be oxidized to carboxylic acids, while both can be reduced to alcohols or hydrocarbons and exhibit condensation reaction.
Predict the end product (D) of the following reaction sequence;
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → phthalimide
The sequence uses two ideas. First, an acyl side chain on a benzene ring is oxidised to –COOH by hot acidified \( \ce{KMnO4} \); the ring is untouched and only the carbon attached to it stays. Second, a carboxylic acid with ammonia gives an ammonium salt, which on heating loses water to give an amide.
Both acetyl groups of 1,2-diacetylbenzene are oxidised, giving phthalic acid (A), benzene-1,2-dicarboxylic acid:
\( \displaystyle \ce{C6H4(COCH3)2 + 8[O]} \xrightarrow{\ce{KMnO4}/\ce{H+}} \ce{C6H4(COOH)2 + 2CO2 + 2H2O} \)
With ammonia, phthalic acid forms ammonium phthalate (B):
\( \displaystyle \ce{C6H4(COOH)2 + 2NH3 -> C6H4(COONH4)2} \)
On heating, each ammonium carboxylate group loses a molecule of water, giving phthalamide (C):
\( \displaystyle \ce{C6H4(COONH4)2} \xrightarrow{\Delta} \ce{C6H4(CONH2)2 + 2H2O} \)
On stronger heating, the two neighbouring amide groups join into a five-membered ring through one nitrogen atom, with loss of ammonia, giving phthalimide (D):
\( \displaystyle \ce{C6H4(CONH2)2} \xrightarrow{\text{strong heating}} \ce{C6H4(CO)2NH + NH3} \)
So the end product D is phthalimide, the compound used in the Gabriel synthesis of primary amines.
Why the other options are incorrect
Option (1) phthalic acid — Incorrect. This is A, the product of the first (oxidation) step only.
Option (2) phthalamide — Incorrect. This is C; it still has two separate –\( \ce{CONH2} \) groups and changes into the cyclic imide on further heating.
Option (4) benzamide — Incorrect. Benzamide has only one carbonyl group on the ring, while every compound in this sequence keeps both carbon atoms attached to the ring.
Note: It seems that there is a mistake in the original NTA paper. The last step of the reaction sequence (C → D) is printed as −2NH3; it should be −NH3.