The Gabriel phthalimide synthesis is a reaction used to prepare primary amines. In this reaction, phthalimide (an imide compound) is first deprotonated to form a phthalimide anion, which then reacts with an alkyl halide to give an N-alkyl phthalimide. This intermediate can then undergo hydrolysis to yield a primary amine.
The general steps are:
1. Phthalimide is treated with a base (such as potassium hydroxide) to form a phthalimide anion.
2. The phthalimide anion then reacts with an alkyl halide (R-X) to form the N-alkyl phthalimide.
3. Hydrolysis (using acidic or basic conditions) of the N-alkyl phthalimide leads to the formation of a primary aliphatic amine (R-NH\(_2\)).
Thus, the product of the Gabriel phthalimide synthesis is a primary aliphatic amine.
Option (A) Secondary aliphatic amine: Incorrect. The Gabriel synthesis produces primary amines, not secondary amines.
Option (B) Primary aliphatic amine: Correct. As explained, the reaction leads to the formation of a primary aliphatic amine.
Option (C) Aromatic primary amine: Incorrect. This method typically yields aliphatic primary amines, not aromatic primary amines.
Option (D) Tertiary aliphatic amine: Incorrect. The reaction does not result in the formation of tertiary amines.
Thus, the correct answer is (B) Primary aliphatic amine.
Identify A in the following reaction. 
For the reaction, \(N_{2}O_{4} \rightleftharpoons 2NO_{2}\) graph is plotted as shown below. Identify correct statements.
A. Standard free energy change for the reaction is 5.40 kJ \(mol^{-1}\).
B. As \(\Delta G\) in graph is positive, \(N_{2}O_{4}\) will not dissociate into \(NO_{2}\) at all.
C. Reverse reaction will go to completion.
D. When 1 mole of \(N_{2}O_{4}\) changes into equilibrium mixture, value of \(\Delta G = -0.84 \text{ kJ mol}^{-1}\).
E. When 2 mole of \(NO_{2}\) changes into equilibrium mixture, \(\Delta G\) for equilibrium mixture is \(-6.24 \text{ kJ mol}^{-1}\).
Choose the correct answer from the following.
