Question

In the Gabriel's phthalimide synthesis, phthalimide is treated first with

• A. $C_2{H}_{5}I/\text{Alc.}KOH$
• B. Aqueous $KOH$
• C. Ethanol and $H_{2}S{O}_4$
• D. Ether and $LiAl{H}_4$

Answer 1

The correct option is A $C_2{H}_{5}I/\text{Alc.}KOH$

Gabriel phthalimide is firstly treated with alcoholic $KOH$ to from potassium phthalimide which on further reaction with ethyl iodide gives N-Ethyl phthalimide.
The product N-Ethyl phthalimide is hydrolysed with aq. $NaOH$ to form primary amine, i.e. ethylamine. This reaction is known as Gabriel phthalimide synthesis. Thus option (a) is correct.
MECHANISM
Step 1: When potassium hydroxide is introduced to the phthalimide, an acid-base reaction occur. The hydroxide ion deprotonates the imide. The resulting proton is more acidic than any simple amine, generating a strong nucleophile – the potassium phthalimide​ ion.

Step 2:
The nucleophilic potassium imide ion attacks the electrophilic carbon of the alkyl halide. The nitrogen atom subsequently replaces the halogen (Fluorine, Chlorine, Bromine or Iodine) in the alkyl halide and bonds with the carbon itself. This results in the formation of an N-Alkyl Phthalimide.
Step 3:
The product N-Alkyl phthalimide is hydrolysed with aq. $NaOH$ to form primary amine, i.e. Alkylamine.
A short trick to form the product in Gabriel's phthalimide synthesis