Two methods -- reduction of $\text{CH}_3\text{CH}_2\text{CN}$ and the Gabriel synthesis using $1$-bromoethane -- are both proposed for preparing ethylamine. State which one is the more reliable general method for a PRIMARY amine free of secondary/tertiary contamination, and why.

Two methods -- reduction of CH₃CH₂CN and the Gabriel synthesis using 1-bromoethane -- are both proposed for pr

Example · Example 17

\text{CH}_3\text{CH}_2\text{CN}

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Concept understanding — Gabriel Phthalimide Synthesis

Gabriel Phthalimide Synthesis

You want to make a primary amine — a molecule where an alkyl group is attached to an $NH_2$ group. The obvious route is to react an alkyl halide with ammonia. That gives you a mixture: some primary amine, some secondary amine (two alkyl groups on the nitrogen), some tertiary amine, and even some quaternary ammonium salt. Ammonia is a nucleophile, but once it reacts, the product (the primary amine) is an even better nucleophile than ammonia was. So it keeps attacking more alkyl halide molecules, and you lose control.

The Gabriel synthesis is a clever way to stop that chain reaction. It works by hiding the nitrogen inside a molecule that cannot act as a nucleophile until you want it to.

The intuition

Imagine you have a nitrogen atom that you want to attach exactly one alkyl group to, and then release it as a primary amine. If you put that nitrogen inside a structure that is already "full" — where it has no hydrogen atoms left to be replaced — then it cannot react with more than one alkyl halide molecule. That is the core idea.

Potassium phthalimide is that structure. Phthalimide itself has the formula $C_6H_4(CO)_2NH$ . The nitrogen is flanked by two carbonyl groups, which pull electron density away from it. When you treat phthalimide with a base (usually alcoholic KOH), the $N-H$ bond is deprotonated, giving the potassium salt:

C_6H_4(CO)_2NH + KOH \rightarrow C_6H_4(CO)_2N^-K^+ + H_2O

This anion is a good nucleophile. It attacks an alkyl halide ( $R-X$ ) in an $S_N2$ reaction, giving $N$ -alkylphthalimide:

C_6H_4(CO)_2N^-K^+ + R-X \rightarrow C_6H_4(CO)_2N-R + KX

Now look at that product. The nitrogen already has three bonds: two to the carbonyl carbons and one to the alkyl group. It has no hydrogen left. It cannot react with a second alkyl halide molecule. The secondary and tertiary amine contamination is impossible at this stage.

Watch out

This is an $S_N2$ reaction. It works well with primary alkyl halides. Secondary alkyl halides give poor yields due to steric hindrance and elimination side reactions. Tertiary alkyl halides are useless here — they eliminate instead of substituting.

Releasing the amine

You now have the alkyl group attached to the phthalimide nitrogen. To get the free primary amine, you need to break the two $N-C(=O)$ bonds. This is done by hydrolysis — either acidic or basic.

Alkaline hydrolysis (reflux with aqueous or alcoholic KOH):

C_6H_4(CO)_2N-R + 2KOH \rightarrow C_6H_4(COOK)_2 + RNH_2

The products are potassium phthalate and the primary amine. The amine is liberated as a free base and can be distilled out or extracted.

Acidic hydrolysis (reflux with concentrated HCl):

C_6H_4(CO)_2N-R + 2H_2O + HCl \rightarrow C_6H_4(COOH)_2 + RNH_3^+Cl^-

Here you get phthalic acid and the amine hydrochloride salt. You then treat the salt with a base to free the amine.

Tip

A modern alternative to hydrolysis is the Ing-Manske procedure: treat the $N$ -alkylphthalimide with hydrazine ( $NH_2NH_2$ ). This gives phthalhydrazide and the free amine in one step, under milder conditions. The reaction is:

$C_6H_4(CO)_2N-R + NH_2NH_2 \rightarrow C_6H_4(CO)_2(NH)_2 + RNH_2$

The precise statement …

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Reduction of $\text{CH}_3\text{CH}_2\text{CN}$ with $\text{LiAlH}_4$ or $\text{H}_2/\text{Ni}$ does give a clean primary amine PROVIDED the nitrile itself was pure -- but the nitrile-forming step ( $\text{R-X} + \text{KCN} \to \text{R-CN}$ ) is a separate substitution that carries its own risk of side products (elimination competing with substitution on the haloalkane). The Gabriel phthalimide synthesis, by contrast, structurally PREVENTS over-alkylation at its own alkylation step, because the phthalimide nitrogen has no second hydrogen left to react a second time once alkylated -- the purity of the final amine is guaranteed by the mechanism itself, not merely by careful technique. For …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2024Set ANNUAL1 markMCQ
$Q. Amine that can be prepared by Gabriel Phthalimide synthesis -$
(a) Primary
(b) Secondary
(c) Tertiary
(d) None of these
›Reveal solutionSolution

Gabriel phthalimide synthesis gives only primary amines.

In the Gabriel phthalimide synthesis, potassium phthalimide is reacted with an alkyl halide (R-X) to form N-alkylphthalimide, which is then hydrolysed (usually with aqueous NaOH or hydrazine) to give the primary amine (R-NH2) and phthalic acid/hydrazide. Since only one alkylation step is possible (the nitrogen is locked in the imide ring until hydrolysis), this method exclusively yiel …

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CBSE 2019Set ANNUAL1 markMCQ
$Q. Gabriel Pthalimide Synthesis is used for preparation of -$
(a)
(i) 1° amine
(b)
(ii) 2° amine
(c)
(iii) 3° amine
(d)
(iv) All of these
›Reveal solutionSolution

The correct option is (i) 1 $^\circ$ (primary) amine.

Concept. In the Gabriel synthesis, phthalimide is treated with KOH to form potassium phthalimide, which is then alkylated by an alkyl halide. Hydrolysis (or hydrazinolysis) then releases a primary amine and phthalic acid:

$\text{Phthalimide} \xrightarrow{KOH} \text{K-phthalimide} \xrightarrow{R{-}X} N\text{-alkylphthalimide} \xrightarrow{H_2O/H^+} R{-}NH_2 + \text{phthalic acid}$ …

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CBSE 2018Set ANNUAL1 markMCQ
$Q. Gabriel phthalimide reaction is used for the preparation of$
(a) primary aromatic amines
(b) secondary amines
(c) aliphatic primary amines
(d) tertiary amines
›Reveal solutionSolution
Gabriel synthesis converts an alkyl halide into a pure primary aliphatic amine via N-alkylation of potassium phthalimide followed by hydrolysis; it cannot be used for aromatic amines because aryl halides do not undergo the required nucleophilic substitution.

The Gabriel phthalimide synthesis proceeds in two stages:

Phthalimide is treated with ethanolic KOH to form potassium phthalimide (the N–H is acidic enough to be deprotonated). This is then reacted with an alkyl halide, $R-X$ , in an $S_N2$ nucleophilic substitution, giving N-alkylphthalimide.

The N-alkylphthalimide is hydrolysed (acidic or alkaline hydrolysis, or hydrazinolysis) to release the free primary amine, $R-NH_2$ , along with phthalic acid/hydrazide by-product. …
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CBSE 2017Set ANNUAL1 mark
$Q. Explain Gabriel phthalimide synthesis.$
›Reveal solutionSolution

Phthalimide's N-H is converted to a nucleophilic salt, alkylated by an alkyl halide, then hydrolysed to cleanly release a primary amine without any secondary/tertiary amine contamination.

The Gabriel phthalimide synthesis is used to prepare pure primary amines (free of secondary/tertiary amine contamination, unlike direct ammonolysis of alkyl halides):

Step 1: Phthalimide is treated with ethanolic KOH to form its potassium salt (potassium phthalimide), since the N–H hydrogen in phthalimide is acidic (flanked by two carbonyl groups).

Step 2: Potassium phthalimide, now a good nucleophile at nitrogen, is reacted with an alkyl halide (R–X) in an SN2 reaction, giving N-alkylphthalimide.

Step 3: N-alkylphthalimide is hydrolysed (with aqueous acid/alkali, or preferably with hydrazine — the Ing–Manske modification) to release the primary amine R–NH2 and phthalic acid (or phthalhydrazide).

…

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