Match List-I with List-II.
| List-I (Conversion) | List-II (Reagent / conditions) |
|---|---|
| (A) Propanamide → ethanamine | (I) Ethanolic potassium hydroxide |
| (B) Propanamide → propan-1-amine | (II) Bromine and aqueous sodium hydroxide |
| (C) Phthalimide → potassium salt of phthalimide | (III) Lithium aluminium hydride, then water |
| (D) Bromoethane → ethanamine | (IV) Ethanolic ammonia, in a sealed tube at 373 K |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → A-II, B-III, C-I, D-IV
Look at the number of carbon atoms. An amide gives an amine with the same number of carbons on reduction, but one carbon less in Hoffmann bromamide degradation.
(A) Propanamide → ethanamine — This matches (II) Bromine and aqueous sodium hydroxide. Ethanamine has one carbon less than propanamide, so this is Hoffmann bromamide degradation; the ethyl group moves from the carbonyl carbon to nitrogen:
\( \displaystyle \ce{CH3CH2CONH2 + Br2 + 4NaOH -> CH3CH2NH2 + Na2CO3 + 2NaBr + 2H2O} \)
(B) Propanamide → propan-1-amine — This matches (III) Lithium aluminium hydride, then water. Reduction turns –CONH2 into –CH2NH2, so the carbon count stays three:
\( \displaystyle \ce{CH3CH2CONH2} \xrightarrow[\text{(ii) }\ce{H2O}]{\text{(i) }\ce{LiAlH4}} \ce{CH3CH2CH2NH2} \)
(C) Phthalimide → potassium salt of phthalimide — This matches (I) Ethanolic potassium hydroxide. This is the first step of Gabriel phthalimide synthesis; the N–H hydrogen of phthalimide is removed by the base:
\( \displaystyle \ce{C6H4(CO)2NH + KOH} \xrightarrow{\text{ethanol}} \ce{C6H4(CO)2NK + H2O} \)
(D) Bromoethane → ethanamine — This matches (IV) Ethanolic ammonia, in a sealed tube at 373 K. This is ammonolysis: ammonia, a nucleophile, replaces the bromine. The ammonium salt formed gives the free amine with a strong base:
\( \displaystyle \ce{C2H5Br + NH3} \xrightarrow[\text{373 K, sealed tube}]{\text{ethanol}} \ce{C2H5NH3^+Br^-} \)
\( \displaystyle \ce{C2H5NH3^+Br^- + NaOH -> C2H5NH2 + H2O + NaBr} \)
NCERT (2026-27 edition):
9.4 Preparation of Amines – “Hoffmann developed a method for preparation of primary amines by treating an amide with bromine in an aqueous or ethanolic solution of sodium hydroxide. … The amine so formed contains one carbon less than that present in the amide.”
“The amides on reduction with lithium aluminium hydride yield amines.”
“Phthalimide on treatment with ethanolic potassium hydroxide forms potassium salt of phthalimide which on heating with alkyl halide followed by alkaline hydrolysis produces the corresponding primary amine.”
“…an alkyl or benzyl halide on reaction with an ethanolic solution of ammonia undergoes nucleophilic substitution reaction in which the halogen atom is replaced by an amino (–NH2) group. This process of cleavage of the C–X bond by ammonia molecule is known as ammonolysis. The reaction is carried out in a sealed tube at 373 K.”