Gabriel Phthalimide Synthesis (and Why It Fails for Aryl Amines)
Chemistry · Amines · NEET
Gabriel phthalimide synthesis is a way to make a PURE primary amine (1° amine) from an alkyl halide. Phthalimide is turned into its potassium salt, this salt attacks the alkyl halide, and then hydrolysis sets the amine free. It only gives primary amines (no mixture), but it fails for aromatic amines like aniline because aryl halides do not undergo SN2 substitution. Memory hook: "Gabriel gives ONE clean 1° amine, but ARYL says NO."
Gabriel synthesis: phthalimide → potassium salt → SN2 with R–X → hydrolysis gives a pure primary amine. Aryl halides do not react (no SN2), so aniline cannot be made this way.
Your doubts, answered
Why does Gabriel synthesis give ONLY a primary amine and not a mixture?
The nitrogen in phthalimide is locked inside a ring, connected to two C=O groups on both sides. After the alkyl halide attaches, that nitrogen has no free lone pair or extra N-H to react again. So it cannot pick up a second or third alkyl group. When you finally hydrolyse it, the nitrogen comes off carrying exactly ONE alkyl group. That is why you always get a clean primary amine. This is the big advantage over ammonolysis, which gives a messy mixture of 1°, 2°, 3° amines and quaternary salt.
Why can't aniline (an aromatic amine) be made by Gabriel synthesis?
To make aniline you would need to attach the phthalimide anion to a benzene ring, which means using an aryl halide like chlorobenzene. But the reaction is an SN2 attack, and aryl halides do NOT undergo nucleophilic substitution easily. The C–X bond in aryl halides has partial double-bond character and the ring blocks the back-side attack. So the phthalimide anion cannot replace the halogen on a benzene ring. That is why aromatic primary amines (like aniline) cannot be prepared by Gabriel synthesis. This is a favourite NEET one-liner.
What are the exact steps of Gabriel phthalimide synthesis?
Step 1: Phthalimide + ethanolic KOH → potassium salt of phthalimide (this makes the N nucleophilic). Step 2: The potassium salt is heated with an alkyl halide (R–X) → N-alkyl phthalimide (SN2, the anion replaces the halide). Step 3: Alkaline hydrolysis (or hydrazinolysis) of the N-alkyl phthalimide → the primary amine (R–NH2) is released, along with phthalate/phthalhydrazide. Remember the order: salt first, then alkyl halide, then hydrolysis.
Which alkyl halide works best in Gabriel synthesis?
Primary (1°) alkyl halides work best because the mechanism is SN2, which prefers less crowded carbon. Methyl and primary halides give good yields. Tertiary (3°) alkyl halides work poorly because they favour elimination (E2) over substitution. And aryl halides do not work at all. So for NEET: primary alkyl halide = best, aryl halide = fails.
What is the difference between Gabriel synthesis and ammonolysis of alkyl halides?
Both start from an alkyl halide, but the result is very different. Ammonolysis uses ammonia (NH3), and because the product amine is still nucleophilic, it keeps reacting to give a MIXTURE of primary, secondary, tertiary amines and quaternary salt. Gabriel synthesis uses phthalimide, which stops after ONE alkylation, so it gives only a PURE primary amine. Rule to remember: 'Want a clean primary amine? Use Gabriel, not ammonolysis.'
⚠️ The NEET trap ✗ Thinking Gabriel synthesis can make any primary amine, including aniline, because 'it makes primary amines'. ✓ Gabriel synthesis makes only ALIPHATIC primary amines. It CANNOT make aromatic primary amines like aniline, because aryl halides do not undergo the SN2 substitution needed. 🧠 Gabriel is a specialist: it makes primary amines, but only from the ALKYL (not aryl) side. If the question says aniline + Gabriel, the answer is 'not possible'.
Real NEET questions
NEET 2024
Given below are two statements:
Statement I: Aniline does not undergo Friedel-Crafts alkylation reaction.
Statement II: Aniline cannot be prepared through Gabriel synthesis.
In the light of the above statements, choose the correct answer:
A · Both Statement I and Statement II are false
B · Statement I is correct but Statement II is false
C · Statement I is incorrect but Statement II is true
D · Both Statement I and Statement II are true ✓
Solution: Statement I is TRUE: In Friedel-Crafts, the Lewis acid AlCl3 reacts with aniline's basic nitrogen lone pair to form a salt (C6H5N+H2·AlCl3−). This positive nitrogen strongly deactivates the ring and the catalyst gets used up, so aniline does not undergo Friedel-Crafts alkylation. Statement II is TRUE: Gabriel synthesis works by an SN2 attack of the phthalimide anion on an alkyl halide. To make aniline you would need an aryl halide, but aryl halides do not undergo nucleophilic substitution under these conditions. So aniline cannot be prepared by Gabriel synthesis. Both true → answer (D).
NEET 2017
Which of the following reactions is appropriate for converting acetamide (CH3CONH2) to methanamine (CH3NH2)?
A · Carbylamine reaction
B · Hoffmann bromamide (hypobromamide) reaction ✓
C · Stephen's reaction
D · Gabriel's phthalimide synthesis
Solution: Acetamide has 2 carbons; methanamine has 1 carbon, so we need a reaction that removes one carbon from an amide. That is the Hoffmann bromamide degradation (Br2 + KOH), which turns a primary amide into a primary amine with ONE fewer carbon. Gabriel synthesis (D) makes amines from ALKYL HALIDES, not from amides, so it is wrong here — this is why Gabriel appears as a trap option. Carbylamine (A) is a test, and Stephen's reaction (C) reduces nitriles to aldehydes. Answer: (B).
Solved Amines NEET PYQs
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Does Gabriel synthesis give primary, secondary or tertiary amines?
Only primary (1°) amines, and in pure form. The nitrogen is trapped in the phthalimide ring, so it can only be alkylated once.
Can Gabriel synthesis prepare aniline?
No. Aniline would need an aryl halide, and aryl halides do not undergo the SN2 substitution the reaction needs. So aromatic primary amines cannot be made this way.
What reagent converts phthalimide into its reactive salt?
Ethanolic potassium hydroxide (KOH). It removes the N–H proton to give the potassium salt of phthalimide, which is a good nucleophile.
What is the final step that releases the amine?
Alkaline hydrolysis (or hydrazinolysis with NH2NH2) of the N-alkyl phthalimide releases the free primary amine along with a phthalate/phthalhydrazide by-product.
Why is Gabriel synthesis better than ammonolysis?
Ammonolysis gives a mixture of 1°, 2°, 3° amines and quaternary salt. Gabriel synthesis stops after one alkylation, so it gives a single, pure primary amine.
