Animated Solution for Chemistry - Organic Chemistry: Which of the following compounds can be prepared in good yield by Gabriel phthalimide synthesis?

Imagine you are in a lab, and your mission is to synthesize a pure primary amine. You might think, "Why not just react ammonia with an alkyl halide?" It sounds simple, but there is a major trap: over-alkylation. Ammonia is a nucleophile, but once it reacts to form a primary amine, that primary amine is also a nucleophile—often a better one! This leads to a messy mixture of primary, secondary, tertiary amines, and even quaternary ammonium salts.

To solve this, we need a method that strictly stops at the primary amine stage. Enter the Gabriel Phthalimide Synthesis, a brilliant chemical workaround that uses a bulky protecting group to ensure we get exactly what we want.

Gabriel Synthesis

The genius of this method lies in its starting material: phthalimide.

First, we treat phthalimide with a strong base like potassium hydroxide ($\text{KOH}$). The base removes the acidic proton attached to the nitrogen, creating a potassium phthalimide salt. This leaves the nitrogen atom with a negative charge, turning it into a powerful nucleophile.

Next comes the crucial step. We introduce an alkyl halide ($\text{R-X}$). Our nucleophilic nitrogen attacks the alkyl group, kicking out the halide ion in a classic $S_{N}2$ reaction. Because the phthalimide group is massive, it physically blocks any further alkylation. The nitrogen can only bond to one alkyl group, forming N-alkylphthalimide.

Finally, we need to free our amine from its bulky cage. We treat the intermediate with hydrazine (${\text{NH}}_2{\text{NH}}_2$), which cleaves the bonds to nitrogen, releasing a pure, primary aliphatic amine (${\text{R-NH}}_2$) and a stable byproduct.

Here is where mistakes happen. The success of this synthesis hinges entirely on the $S_{N}2$ attack in the second step. For an $S_{N}2$ reaction to occur, the alkyl halide must be relatively unhindered (like a primary or secondary aliphatic halide) or benzylic.

What if we want to make an aromatic amine, like aniline? We would need to use an aryl halide, such as chlorobenzene. However, the lone pairs on the chlorine atom are in resonance with the benzene ring, giving the carbon-chlorine bond a partial double bond character. This makes the bond incredibly strong. Furthermore, the electron-rich benzene ring repels the incoming nucleophile. As a result, aryl halides do not undergo $S_{N}2$ reactions.

Let's evaluate our options based on this strict constraint:

- (b) ${\text{CH}}_3-{\text{CH}}_2-{\text{NHCH}}_3$: This is a secondary amine. Gabriel synthesis only produces primary amines. - (c) 2-phenylacetamide: This is an amide, not an amine. - (d) Aniline: This is a primary aromatic amine. As we discovered, the required aryl halide will not undergo the necessary $S_{N}2$ reaction. - (a) Benzylamine: The structure shows a primary amine attached to a benzylic carbon ($\text{Ph}-{\text{CH}}_2-{\text{NH}}_2$). The corresponding halide, benzyl chloride, is highly reactive in $S_{N}2$ reactions.

Therefore, benzylamine can be prepared in excellent yield using the Gabriel phthalimide synthesis!