On this page
Making a primary amine from an alkyl halide sounds easy: react the halide with ammonia. In practice the amine you make reacts again, and you get a mixture. The Gabriel phthalimide synthesis gets around this by using a nitrogen that can take exactly one alkyl group. This post explains the steps NCERT gives, why the method stops at primary amines, where it fails, and works a yield calculation.
Where is it in the syllabus?
Gabriel phthalimide synthesis is item 5 under the preparation of amines in Unit 9 of the NCERT Class 12 book, Amines, and intext question 9.12 asks why aromatic primary amines cannot be made this way. The JEE (Advanced) 2026 brochure names "Gabriel phthalimide synthesis" under amines. The Syllabus for JEE (Main) 2026 and the NEET (UG) 2026 syllabus do not name it, but both list general methods of preparation of nitrogen compounds and the classification of amines, which is where NCERT places it. Check the syllabus for your own year.
What are the steps?
NCERT describes three:
Phthalimide is treated with ethanolic potassium hydroxide, forming the potassium salt of phthalimide.
The salt is heated with an alkyl halide, which puts the alkyl group on nitrogen (an N-alkylphthalimide).
Alkaline hydrolysis releases the primary amine.
Siegmund Gabriel published the method in 1887 in a paper whose title translates as a method of preparing primary amines from the corresponding halogen compounds. Ing and Manske (1926) later published a modification of the last step, which a 2004 paper by Ariffin and colleagues describes as hydrazinolysis. For board and JEE answers, NCERT's alkaline hydrolysis is the expected step.
Why does it give only primary amines?
OpenStax Organic Chemistry shows the problem with ammonia. The amine formed first is about as reactive as ammonia, so it is alkylated again. Treating 1-bromooctane with a twofold excess of ammonia gives a mixture with only 45 percent octylamine, nearly as much dioctylamine, and smaller amounts of trioctylamine and tetraoctylammonium bromide.
Phthalimide avoids this. Its N–H sits between two C=O groups, which makes that hydrogen acidic enough for KOH to remove, as OpenStax explains. The resulting anion is a good nucleophile and is alkylated once. After that, the nitrogen has no hydrogen left and its lone pair is shared with the two carbonyl groups, so it does not react with a second alkyl halide. Hydrolysis then frees R–NH₂: one alkyl group on nitrogen, every time.
When does the method fail?
Aryl halides. NCERT: "Aromatic primary amines cannot be prepared by this method because aryl halides do not undergo nucleophilic substitution with the anion formed by phthalimide." So aniline cannot be made from bromobenzene this way.
Tertiary alkyl halides. The alkylation is a nucleophilic substitution, and a tertiary carbon is too crowded for the SN2 attack. A strong base with a tertiary halide gives mainly elimination instead. Our post on SN1 and SN2 reactions of haloalkanes explains why.
Secondary or tertiary amines as targets. The method makes only primary amines, by design.
A naming trap: propan-2-amine, (CH₃)₂CH–NH₂, is a primary amine, because nitrogen carries one alkyl group, even though that group is attached through a secondary carbon. "Primary" describes the nitrogen, not the carbon.
Worked example: benzylamine on paper
A student converts 7.36 g of phthalimide into benzylamine using KOH, then benzyl chloride, then alkaline hydrolysis. Find the theoretical yield, and the percentage yield if 4.02 g is collected (sample numbers).
Phthalimide, C₈H₅NO₂: M = 8(12.011) + 5(1.008) + 14.007 + 2(15.999) ≈ 147.13 g/mol.
Moles: 7.36 ÷ 147.13 ≈ 0.0500 mol.
Each phthalimide gives one benzylamine, so 0.0500 mol of C₇H₉N, with M ≈ 107.16 g/mol.
Theoretical yield: 0.0500 × 107.16 ≈ 5.36 g.
Percentage yield: 4.02 ÷ 5.36 × 100 ≈ 75.0 percent.
Benzyl chloride is a good choice: its carbon is primary and benzylic, so substitution is fast.
Which questions does this answer in the exam?
Can these be made by the Gabriel route?
Propan-1-amine from 1-bromopropane: yes.
Aniline from bromobenzene: no, as NCERT states.
2-Methylpropan-2-amine from 2-bromo-2-methylpropane: no, elimination wins.
N-Methylethanamine: no, it is a secondary amine.
For an amine with one carbon fewer than the starting material, NCERT's next method, Hoffmann bromamide degradation, is the tool. Our post on amines and diazonium salts covers basicity and the tests for primary, secondary and tertiary amines.
Which mistakes cost marks?
Leaving out the KOH step, so no anion forms.
Choosing an aryl or tertiary halide.
Calling the product secondary because the carbon attached to nitrogen is secondary.
Stopping at the N-alkylphthalimide without the hydrolysis step.
If you can follow each step but cannot plan a route from scratch, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning behind each choice of reagent.
A practice routine
Day 1: Write the three Gabriel steps with structures, from memory.
Day 2: For ten target amines, decide whether the Gabriel route works and give the reason in one line.
Day 3: Two yield calculations like the one above, with different alkyl halides.
Today, write the Gabriel route to ethanamine and name the alkyl halide you would use. You should choose bromoethane or chloroethane.
