Which of the following amines can be prepared by Gabriel synthesis? (Two or more options may be correct.) (i)

NCERT Exemplar · Q65

Q. Which of the following amines can be prepared by Gabriel synthesis? (Two or more options may be correct.)

(i) Isobutyl amine

(ii) 2-Phenylethylamine

(iii) N-methylbenzylamine

(iv) Aniline

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Concept understanding — Gabriel Phthalimide Limitation

The Gabriel Phthalimide Limitation – First Principles

Imagine you want to make a primary amine ( $R-NH_2$ ) from an alkyl halide ( $R-X$ ). The obvious idea is to just let ammonia ( $NH_3$ ) attack the halide. But ammonia is a nucleophile and also a base — it will keep reacting. The product amine is even more nucleophilic than ammonia, so it attacks another alkyl halide molecule, giving a secondary amine ( $R_2NH$ ), then tertiary ( $R_3N$ ), and finally a quaternary ammonium salt ( $R_4N^+$ ). You end up with a messy mixture.

The Gabriel synthesis was invented to solve this: it gives only the primary amine, cleanly. But it has a hard limit.

The Intuition: Why the Gabriel Method Works (and Where It Breaks)

The trick is to use phthalimide — a molecule with two carbonyl groups flanking an $N-H$ bond. The $N-H$ is acidic enough to be deprotonated by a mild base (like $KOH$ or $K_2CO_3$ ), giving a phthalimide anion. This anion is a great nucleophile but a terrible base — it won't deprotonate the alkyl halide or cause elimination. It attacks the alkyl halide in an $S_N2$ reaction, forming an $N$ -alkylphthalimide.

Then you hydrolyse (or use hydrazine) to break the two amide bonds, releasing the primary amine and regenerating phthalic acid.

Note

The key limitation is $S_N2$ reactivity. The phthalimide anion is bulky and not very nucleophilic. It can only attack primary alkyl halides (or very reactive secondary ones like allyl/benzyl halides). Tertiary halides? They undergo elimination instead of substitution. Secondary halides? Very slow, often give poor yields.

The Precise Statement of the Limitation

Gabriel phthalimide synthesis fails for alkyl halides that are sterically hindered or prone to elimination. Specifically:

Tertiary alkyl halides ( $R_3C-X$ ) do not react — they undergo $E2$ elimination instead of $S_N2$ substitution.

Secondary alkyl halides ( $R_2CH-X$ ) react very slowly, if at all, and yields are poor.

Aryl halides (like chlorobenzene) do not react because $S_N2$ on an $sp^2$ carbon is impossible.

Alkyl halides with bulky groups near the reaction centre (neopentyl, etc.) also fail.

Watch out

A common mistake: students think the limitation is about the hydrolysis step. No — the limitation is entirely in the alkylation step. The phthalimide anion simply cannot force an $S_N2$ reaction on a hindered carbon.

Why This Matters for Exams

You'll be asked to identify which alkyl halides cannot be used in the Gabriel synthesis. The answer is always: tertiary halides, most secondary halides, and aryl halides. For example:

Alkyl Halide	Works?	Reason
$CH_3CH_2CH_2Br$	Yes	Primary, unhindered
$(CH_3)_2CHBr$	Poor	Secondary, slow $S_N2$
$(CH_3)_3CBr$	No	Tertiary — elimination dominates
$C_6H_5Br$	No	Aryl — $S_N2$ impossible on $sp^2$ carbon
$CH_2=CHCH_2Br$	Yes	Allylic — very reactive $S_N2$

Important

The Gabriel phthalimide synthesis is a reliable method only for preparing primary amines from primary alkyl halides (or very reactive secondary ones). For tertiary amines or hindered substrates, you need alternative methods (like reduction of nitriles or amides).

The Deeper Reason (For the Curious) …

Why this formula?

Gabriel Phthalimide Limitation — Why It Exists

The Gabriel phthalimide synthesis is a classic method to prepare primary amines ( $\text{R-NH}_2$ ) from alkyl halides. However, it has a critical limitation: it fails with secondary and tertiary alkyl halides (and also with aryl halides). Let's understand why this happens — the reasoning is rooted in reaction mechanism and steric hindrance.

1. The Key Reaction Steps (Brief Recap)

The synthesis proceeds in two main steps:

Formation of potassium phthalimide Phthalimide ( $\text{C}_6\text{H}_4(\text{CO})_2\text{NH}$ ) is treated with alcoholic KOH to give the potassium salt:

\text{C}_6\text{H}_4(\text{CO})_2\text{NH} + \text{KOH} \rightarrow \text{C}_6\text{H}_4(\text{CO})_2\text{N}^- \text{K}^+ + \text{H}_2\text{O}

N-alkylation (the critical step) The phthalimide anion acts as a nucleophile and attacks an alkyl halide ( $\text{R-X}$ ) via an $\text{S}_\text{N}2$ mechanism:

\text{C}_6\text{H}_4(\text{CO})_2\text{N}^- + \text{R-X} \rightarrow \text{C}_6\text{H}_4(\text{CO})_2\text{N-R} + \text{X}^-

Hydrolysis to release the primary amine.

2. Why the Limitation Exists — The $\text{S}_\text{N}2$ Bottleneck

The key formula that governs the success of this reaction is the rate law for $\text{S}_\text{N}2$ :

\text{Rate} = k[\text{Nucleophile}][\text{Alkyl halide}]

For the Gabriel synthesis, the nucleophile is the phthalimide anion — a bulky, planar, and resonance-stabilized species. This has two consequences:

A. Steric Hindrance at the Electrophilic Carbon

In an $\text{S}_\text{N}2$ reaction, the nucleophile must approach the backside of the carbon bearing the leaving group.
Primary alkyl halides ( $\text{RCH}_2\text{X}$ ) have a small, unhindered carbon — the nucleophile can easily attack.
Secondary alkyl halides ( $\text{R}_2\text{CHX}$ ) have moderate steric hindrance — the bulky phthalimide anion struggles to approach.
Tertiary alkyl halides ( $\text{R}_3\text{CX}$ ) are severely hindered — the backside is blocked by three alkyl groups. The $\text{S}_\text{N}2$ transition state is impossible to achieve.

B. The $\text{S}_\text{N}2$ Transition State Geometry

The $\text{S}_\text{N}2$ transition state requires a linear arrangement of nucleophile, carbon, and leaving group:

\text{Nu}^- \cdots \text{C} \cdots \text{X}

For the phthalimide anion, this linear approach is sterically impossible when the carbon is tertiary (and difficult for secondary). The bulky phthalimide group cannot fit into the crowded transition state.

3. What Happens Instead? — Elimination Dominates

When a secondary or tertiary alkyl halide is used, the strongly basic phthalimide anion does not perform $\text{S}_\text{N}2$ — it instead acts as a base and promotes $\text{E}2$ elimination:

\text{R}_3\text{C-X} + \text{Phth}^- \rightarrow \text{Alkene} + \text{H-Phth} + \text{X}^-

This is because: …

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Gabriel synthesis uses phthalimide to make primary amines via an $S_N2$ reaction — so only unsubstituted primary amines with no branching at the reaction site work. The correct options are (i) Isobutyl amine and (ii) 2-Phenylethylamine.

Why Gabriel synthesis has a strict limit

Gabriel synthesis is a classic method to prepare primary amines without over-alkylation. The key idea: phthalimide (a strong $N$ -H acid, $pK_a \approx 8.3$ ) is deprotonated by a base like KOH to give the phthalimide anion. This anion acts as a nucleophile and attacks an alkyl halide in an $S_N2$ reaction. After hydrolysis, the primary amine is released.

The limitation is baked into the mechanism: the nucleophile is bulky (the phthalimide anion is planar but sterically hindered), and the reaction is $S_N2$ — so the alkyl halide must be primary (or methyl, or a reasonably unhindered benzylic/allylic). Secondary halides give poor yields; tertiary halides undergo elimination instead. Also, the product is always a primary amine — you cannot make secondary or tertiary amines this way.

Watch out

A common mistake: thinking Gabriel synthesis can make any amine. It cannot make secondary amines (like N-methylbenzylamine) or aromatic amines (like aniline) because aryl halides don't undergo $S_N2$ .

Step-by-step analysis of each option

1. Option (i): Isobutyl amine

Isobutyl amine is $(CH_3)_2CHCH_2NH_2$ . The carbon attached to the $NH_2$ group is a primary carbon (it's bonded to one carbon and two hydrogens). The alkyl halide needed would be isobutyl bromide, $(CH_3)_2CHCH_2Br$ , which is a primary alkyl halide (the bromine is on a primary carbon). This is perfect for $S_N2$ — the steric hindrance from the isopropyl group is far enough away that the reaction proceeds well. So this amine can be prepared.

2. Option (ii): 2-Phenylethylamine

2-Phenylethylamine is $C_6H_5CH_2CH_2NH_2$ . The $NH_2$ is on a primary carbon (the ethyl chain). The corresponding halide is $C_6H_5CH_2CH_2Br$ , a primary alkyl halide with a benzylic group one carbon away. This undergoes $S_N2$ smoothly. In fact, this is a textbook example — 2-phenylethylamine is often prepared via Gabriel synthesis. So this can be prepared.

3. Option (iii): N-methylbenzylamine …

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Here is a breakdown of the common mistakes students make regarding Gabriel synthesis and how to avoid them.

The Core Concept (The "Why")

Gabriel synthesis is a method to prepare primary aliphatic amines ( $1^\circ$ R-NH $_2$ ). It uses phthalimide (which has an acidic N-H) and an alkyl halide (R-X), followed by hydrolysis.

The critical rule: The alkyl halide (R-X) must be primary ( $1^\circ$ ) and aliphatic (not aromatic). The reaction fails with secondary/tertiary halides (due to elimination) and with aryl halides (due to lack of reactivity in S $_N$ 2).

Mistake #1: Forgetting the "Primary Alkyl Halide" Requirement

The Mistake: Students think any alkyl halide works. They try to use a secondary or tertiary halide (like isopropyl bromide or tert-butyl chloride) in the reaction.

Why it fails: Gabriel synthesis is an S $_N$ 2 reaction. Secondary and tertiary halides are sterically hindered and prefer elimination (E2) over substitution. You get an alkene, not the amine.

How to Avoid:

Check the carbon attached to the -NH $_2$ group in the product. In the product amine (R-NH $_2$ ), look at the carbon directly bonded to the nitrogen.
If that carbon is primary (bonded to only one other carbon), the corresponding alkyl halide was primary, and the synthesis works.
If that carbon is secondary or tertiary, the synthesis fails.

Applying to the question:

(A) Isobutyl amine: Structure is (CH $_3$ ) $_2$ CH-CH $_2$ -NH $_2$ . The carbon attached to N is CH $_2$ (primary). Works.
(B) 2-Phenylethylamine: Structure is C $_6$ H $_5$ -CH $_2$ -CH $_2$ -NH $_2$ . The carbon attached to N is CH $_2$ (primary). Works.

Mistake #2: Assuming Aromatic Amines (Anilines) Can Be Prepared

The Mistake: Students see "amine" and assume Gabriel synthesis works for aniline (C $_6$ H $_5$ NH $_2$ ).

Why it fails: To make aniline, you would need chlorobenzene (C $_6$ H $_5$ Cl) as the alkyl halide. Aryl halides do not undergo S $_N$ 2 reactions (the carbon is sp $^2$ hybridized and the $\pi$ bond blocks backside attack).

How to Avoid:

Memorize the exception: Gabriel synthesis is for aliphatic amines only.
Look for a benzene ring directly attached to the -NH $_2$ group. If you see that (like in aniline), it is not possible via Gabriel synthesis.

Applying to the question:

(D) Aniline: The -NH $_2$ is directly on the benzene ring. Does not work.

Mistake #3: Confusing Gabriel Synthesis with Reductive Amination

The Mistake: Students think Gabriel synthesis can make secondary or tertiary amines (like N-methylbenzylamine). …

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

AP EAPCET 2024Set ap-2024-05-17-FN1 markMCQ
$Q. Two statements are given below Statement I: Benzenamine can be prepared from phthalimide Statement II: Benzenamine is less basic than phenyl methanamine Correct answer is (A) Both statements I & II are correct (B) Both statements I & II are not correct (C) Statement I is correct, but statement II is not correct (D) Statement I is not correct, but statement II is correct$
›Reveal solutionSolution
The Gabriel phthalimide synthesis is a classic method for making primary amines, but it fails for aromatic amines like benzenamine because the aryl halide does not undergo nucleophilic substitution. Meanwhile, benzenamine is indeed less basic than phenyl methanamine due to resonance delocalization of the nitrogen lone pair. Thus Statement I is false, Statement II is true — the correct option is (D).

Concept and Intuition

The question tests two separate ideas: a synthetic method (Gabriel phthalimide) and a property (basicity of amines).

Gabriel phthalimide synthesis works beautifully for making aliphatic primary amines: you treat phthalimide with KOH to get the potassium salt, then react it with an alkyl halide (SN2), then hydrolyze to release the amine. But aryl halides (like chlorobenzene) are extremely resistant to SN2 — they need special conditions (e.g., high temperature, copper catalyst) for substitution. So you cannot make benzenamine (aniline) this way.

Basicity of amines depends on how available the nitrogen lone pair is for protonation. In benzenamine, the lone pair is delocalized into the benzene ring (resonance), making it less basic. In phenyl methanamine (benzylamine), the lone pair is on a saturated carbon — no such delocalization — so it is more basic.

Step-by-step reasoning

Evaluate Statement I: “Benzenamine can be prepared from phthalimide”

The Gabriel synthesis uses the phthalimide ion as a nucleophile to attack an alkyl halide.

For benzenamine, you would need to use an aryl halide (e.g., chlorobenzene) as the electrophile.

Aryl halides do not undergo SN2 reactions because the carbon–halogen bond has partial double-bond character and the backside attack is sterically and electronically hindered.

Therefore, the reaction fails — benzenamine cannot be prepared by the Gabriel phthalimide method.

Conclusion: Statement I is incorrect.

Evaluate Statement II: “Benzenamine is less basic than phenyl methanamine”

Benzenamine (aniline): The nitrogen lone pair is conjugated with the π-system of the benzene ring. This resonance delocalization reduces electron density on nitrogen, making it a weaker base. …
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AP EAPCET 2023Set eng-2023-05-19-FN1 markMCQ
$Q. Which of the following amine cannot be prepared by the Gabriel phthalimide synthesis method? (A) Ethylamine (B) Benzylamine (C) Phenylamine (D) Propylamine$
›Reveal solutionSolution
This tests the scope/limitation of the Gabriel phthalimide synthesis; it only works via $\mathrm{S_N2}$ on alkyl halides, so it cannot be used to prepare aromatic amines like aniline (phenylamine), which would require an unreactive aryl halide.

Concept and Intuition

The Gabriel phthalimide synthesis is a classic method for preparing pure primary amines while avoiding the over-alkylation problem seen with direct ammonia alkylation. It proceeds by: (i) deprotonating phthalimide with KOH to form the phthalimide anion (a good nucleophile), (ii) reacting this anion with an alkyl halide via $\mathrm{S_N2}$ substitution to form N-alkylphthalimide, and (iii) hydrolysing (or hydrazinolysing) the N-alkylphthalimide to release the free primary amine.

The crucial mechanistic requirement is step (ii): a genuine backside ( $\mathrm{S_N2}$ ) nucleophilic substitution at a sp³, alkyl carbon bearing the halide. Aryl halides (like chlorobenzene/bromobenzene, which would be needed to make aniline) cannot undergo this kind of substitution under these conditions — the halide is attached to an sp² ring carbon, the C–X bond has partial double-bond character (due to resonance with the ring), and backside attack is sterically blocked by the ring itself. Hence this method simply cannot produce phenylamine (aniline).

Step-by-Step Solution

Recall the Gabriel synthesis mechanism: phthalimide anion + $\mathrm{R-X}$ (alkyl halide, $\mathrm{S_N2}$ ) → N-alkylphthalimide → hydrolysis → $\mathrm{R-NH_2}$ .

Check each target amine's required starting halide: ethylamine needs ethyl halide (alkyl, works); benzylamine needs benzyl halide ( $\mathrm{C_6H_5CH_2X}$ — the halide carbon is an sp³ benzylic carbon, not the aromatic ring itself, so it is a legitimate $\mathrm{S_N2}$ substrate and works); propylamine needs propyl halide (alkyl, works). …
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AP EAPCET 2022Set ap-2022-07-12-FN1 markMCQ
$\xrightleftharpoons{NH_3}$
›Reveal solutionSolution

Ammonia converts the two $\text{--COOH}$ groups to the ammonium salt $\text{--COONH}_4$ (P); heating dehydrates and cyclises it to the cyclic imide (Q) — option (A).

The starting compound is a dimethylbenzene dicarboxylic acid (two $\text{--COOH}$ groups on adjacent ring carbons).

Step 1 — reaction with $\text{NH}_3$ : a carboxylic acid reacts with ammonia to give its ammonium salt, so each $\text{--COOH}$ becomes $\text{--COONH}_4$ .
$\text{P} = \text{the ring carrying two } \text{--COONH}_4 \text{ groups (diammonium carboxylate).}$
Step 2 — heating ( $\Delta$ ): the ammonium salt loses water to the amide and, because the two carbonyls sit on adjacent carbons, cyclises to the cyclic imide (an isoindole-1,3-dione / phthalimide-type ring with $\text{--NH--}$ between the two $\text{C=O}$ groups).

$\text{Q} = \text{the fused cyclic imide.}$ …
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AP EAPCET 2022Set eng-2022-07-05-AN1 markMCQ
$\mathrm{C_6H_5NH_2}$
›Reveal solutionSolution
Gabriel synthesis is restricted to making primary aliphatic amines from alkyl halides; of the four options, only n-propylamine fits that description.

Concept and Intuition

The Gabriel phthalimide synthesis converts an alkyl halide into a primary amine by first alkylating the nitrogen of potassium phthalimide, then hydrolyzing (or hydrazinolysing) the phthaloyl group away. Its key limitation is that it requires an $S_N2$ -reactive alkyl halide — aryl halides don't work (they can't undergo the nucleophilic substitution step), so aromatic amines like aniline cannot be made this way. It's also inherently limited to primary amines, since only one alkylation of the imide nitrogen occurs.

Step-by-Step Solution

(A) Aniline ( $C_6H_5NH_2$ ) is an aromatic primary amine — Gabriel synthesis fails here because aryl halides can't undergo the required nucleophilic substitution with the phthalimide anion.

(B) Pyrrolidine is a cyclic secondary amine (the N is part of the ring, bonded to two carbons) — Gabriel synthesis gives primary amines, not secondary ones. …
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Q.Which of the following amines can be prepared by Gabriel synthesis?
(Two or more options may be correct.)

Concept understanding — Gabriel Phthalimide Limitation

The Gabriel Phthalimide Limitation – First Principles

The Intuition: Why the Gabriel Method Works (and Where It Breaks)

The Precise Statement of the Limitation

Why This Matters for Exams

The Deeper Reason (For the Curious) …

Why this formula?

Gabriel Phthalimide Limitation — Why It Exists

1. The Key Reaction Steps (Brief Recap)

2. Why the Limitation Exists — The $\text{S}_\text{N}2$ Bottleneck

A. Steric Hindrance at the Electrophilic Carbon

B. The $\text{S}_\text{N}2$ Transition State Geometry

3. What Happens Instead? — Elimination Dominates

Why Gabriel synthesis has a strict limit

Step-by-step analysis of each option

1. Option (i): Isobutyl amine

2. Option (ii): 2-Phenylethylamine

3. Option (iii): N-methylbenzylamine …

Concept: Gabriel Synthesis

Key restriction

Method: Identify by amine type

Steps

Applying to the options …

The Core Concept (The "Why")

Mistake #1: Forgetting the "Primary Alkyl Halide" Requirement

Mistake #2: Assuming Aromatic Amines (Anilines) Can Be Prepared

Mistake #3: Confusing Gabriel Synthesis with Reductive Amination

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Q.Which of the following amines can be prepared by Gabriel synthesis? (Two or more options may be correct.)

Why Gabriel synthesis has a strict limit

Step-by-step analysis of each option

1. Option (i): Isobutyl amine

2. Option (ii): 2-Phenylethylamine

3. Option (iii): N-methylbenzylamine …

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Q.Which of the following amines can be prepared by Gabriel synthesis?
(Two or more options may be correct.)

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