Question

Write short notes on the following:
(i) Gabriel phthalimide reaction
(ii) Hoffmann bromamide reaction
OR
Write the following chemical reactions:
(i) Reaction of ethanolic NH₃ with C₂H₅Cl.
(ii) Reaction of ammonia with benzyl chloride, followed by reaction with two moles of CH₃Cl.

Hint:

Gabriel synthesis makes pure aliphatic primary amines (not aromatic ones) via potassium phthalimide; Hoffmann bromamide degrades an amide with Br₂/NaOH to an amine with one carbon less. In the alternative, ammonolysis gives ethylamine, and benzyl chloride + NH₃ then 2 CH₃Cl gives N,N-dimethylbenzylamine.

Answer 1

Option 1 (short notes)

(i) Gabriel phthalimide reaction
Step 1: Purpose. It is a method to prepare pure aliphatic primary amines (\( R\text{-}NH_2 \)) that are free from secondary and tertiary amines.
Step 2: Form the potassium salt. Phthalimide is treated with alcoholic KOH. The N–H (which is acidic because it is between two carbonyl groups) is removed to give potassium phthalimide.
Step 3: Alkylation. Potassium phthalimide is heated with an alkyl halide (\( R\text{-}X \)). The nitrogen displaces the halide to give N-alkylphthalimide.
Step 4: Release the amine. The N-alkylphthalimide is hydrolysed with aqueous acid or base (or treated with hydrazine, Ing-Manske method) to set free the primary amine along with phthalic acid.
\[ \text{Phthalimide} \xrightarrow{KOH} \text{K-phthalimide} \xrightarrow{R\text{-}X} \text{N-alkylphthalimide} \xrightarrow{\text{hydrolysis}} R\text{-}NH_2 \]
Step 5: Limitation. Aromatic (aryl) primary amines cannot be made this way because aryl halides do not undergo the required nucleophilic substitution with potassium phthalimide.

(ii) Hoffmann bromamide reaction (degradation)
Step 1: Purpose. An amide is converted into a primary amine that has one carbon atom fewer than the starting amide.
Step 2: Reagents. A primary amide (\( R\text{-}CONH_2 \)) is treated with bromine (\( Br_2 \)) and a strong base (NaOH or KOH).
Step 3: Overall reaction.
\[ R\text{-}CONH_2 + Br_2 + 4NaOH \rightarrow R\text{-}NH_2 + Na_2CO_3 + 2NaBr + 2H_2O \]
Step 4: Key feature. The carbonyl carbon is lost (leaves as carbonate), so the amine \( R\text{-}NH_2 \) has one carbon less than the amide. For example, acetamide (\( CH_3CONH_2 \), 2 carbons) gives methylamine (\( CH_3NH_2 \), 1 carbon).

Option 2 (chemical reactions)

(i) Ethanolic NH₃ with C₂H₅Cl.
Step 1: Ethyl chloride undergoes nucleophilic substitution with ammonia (ammonolysis) to give ethylamine.
\[ C_2H_5Cl + NH_3 \rightarrow C_2H_5NH_2 + HCl \]
Step 2: Ethylamine is itself a nucleophile, so with excess \( C_2H_5Cl \) it is further alkylated in steps to diethylamine, triethylamine, and finally the quaternary tetraethylammonium chloride.
\[ C_2H_5NH_2 \xrightarrow{C_2H_5Cl} (C_2H_5)_2NH \xrightarrow{C_2H_5Cl} (C_2H_5)_3N \xrightarrow{C_2H_5Cl} (C_2H_5)_4N^+Cl^- \]
(ii) Ammonia with benzyl chloride, then two moles of CH₃Cl.
Step 1: Benzyl chloride reacts with ammonia to give benzylamine.
\[ C_6H_5CH_2Cl + NH_3 \rightarrow C_6H_5CH_2NH_2 + HCl \]
Step 2: Benzylamine has two N–H bonds; reaction with two moles of methyl chloride methylates both, giving N,N-dimethylbenzylamine.
\[ C_6H_5CH_2NH_2 + 2CH_3Cl \rightarrow C_6H_5CH_2N(CH_3)_2 + 2HCl \]
\[\boxed{\text{Final product: N,N-dimethylbenzylamine}}\]