The source of nitrogen in Gabriel synthesis of amines is __. (i) Sodium azide, NaN₃ (ii) Sodium nitrite, NaNO₂

NCERT Exemplar · Q8

Q. The source of nitrogen in Gabriel synthesis of amines is ____.

(i) Sodium azide,

NaN_3

(ii) Sodium nitrite,

NaNO_2

(iii) Potassium cyanide, KCN

(iv) Potassium phthalimide,

C_6H_4(CO)_2N^-K^+

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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 phthalimide synthesis introduces the amino group ( $-NH_2$ ) using potassium phthalimide as the nitrogen source. The correct answer is (D).

Why Gabriel synthesis works — the key idea

Gabriel synthesis is a classic method to make primary amines ( $R-NH_2$ ) without over-alkylation (which plagues direct $NH_3$ + alkyl halide reactions). The trick is to "protect" the nitrogen in a stable, non-basic form, then release it cleanly at the end.

The nitrogen atom in the final amine comes entirely from the phthalimide ion — a nucleophile that attacks the alkyl halide. So the source of nitrogen is the reagent that supplies this ion.

Step-by-step reasoning

What is Gabriel synthesis?

It’s a two-step reaction:
- Step 1: Potassium phthalimide ( $\ce{C6H4(CO)2N^-K+}$ ) reacts with an alkyl halide ( $R-X$ ) via $S_N2$ to form $N$ -alkylphthalimide.
- Step 2: Hydrolysis (usually with aqueous acid or base) cleaves the phthalimide ring, releasing the primary amine $R-NH_2$ and phthalic acid.
Where does the nitrogen come from?

The nitrogen in the final amine is the same nitrogen that was originally part of the phthalimide ring. Potassium phthalimide has a nitrogen anion ( $\ce{N^-}$ ) that acts as the nucleophile. That single nitrogen atom is the only source of nitrogen in the entire sequence.
Check each option:
- (A) Sodium azide — used in the Schmidt reaction or azide reduction to make amines, not in Gabriel synthesis.
- (B) Sodium nitrite — used in diazotization (to make diazonium salts), not as a nitrogen source here.
- (C) Potassium cyanide — used in the Strecker synthesis of amino acids, not in Gabriel synthesis.
- (D) Potassium phthalimide — this is the actual reagent that supplies the nitrogen. The phthalimide ion ( $\ce{C6H4(CO)2N^-}$ ) is the nucleophile that gets alkylated. …

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Common Mistakes in Gabriel Synthesis of Amines

The question asks for the source of nitrogen in the Gabriel synthesis. The correct answer is (D) Potassium phthalimide, $C_6H_4(CO)_2N^-K^+$ .

Here are the most frequent errors students make and how to avoid them:

Mistake 1: Confusing Gabriel synthesis with other amine preparation methods

What students do wrong:

They pick Sodium azide ( $NaN_3$ ) or Sodium nitrite ( $NaNO_2$ ) because these are also nitrogen sources in other reactions.

Why it’s wrong:

$NaN_3$ is used in the Hoffmann rearrangement or Curtius rearrangement, not Gabriel synthesis.
$NaNO_2$ is used in diazotization (to make diazonium salts), not for making primary amines via Gabriel.

How to avoid:

Memorise the reagent–reaction mapping clearly:

Reaction	Nitrogen Source
Gabriel synthesis	Potassium phthalimide
Hoffmann rearrangement	Amide ( $RCONH_2$ ) + $Br_2$ + $NaOH$
Reduction of nitriles	$KCN$ or $NaCN$ (but nitrogen comes from cyanide)
Diazotization	$NaNO_2$ + $HCl$

Mistake 2: Thinking the nitrogen comes from the alkyl halide or the base

What students do wrong:

They assume the nitrogen is supplied by the alkyl halide ( $R-X$ ) or the base ( $KOH$ ).

Why it’s wrong:

The alkyl halide contains no nitrogen.
$KOH$ is only used to hydrolyse the intermediate N-alkylphthalimide to release the amine. It does not supply nitrogen.

How to avoid:

Trace the source of the $NH_2$ group in the final product. In Gabriel synthesis:

Potassium phthalimide ( $C_6H_4(CO)_2N^-K^+$ ) already has the nitrogen.
It attacks the alkyl halide → forms N-alkylphthalimide.
Hydrolysis (with $KOH$ or $H_3O^+$ ) breaks the phthalimide ring, releasing $RNH_2$ .

So the nitrogen enters via the phthalimide ion.

Mistake 3: Picking Potassium cyanide (KCN) because it contains nitrogen

What students do wrong:

They see KCN has nitrogen and think it’s the source.

Why it’s wrong:

KCN is used in the reduction of nitriles to amines ( $RCN \xrightarrow{H_2/Ni} RCH_2NH_2$ ).
In Gabriel synthesis, KCN is not used at all. The reaction uses an alkyl halide + potassium phthalimide.

How to avoid:

Remember the key difference:

Gabriel synthesis: $RX$ + potassium phthalimide → primary amine (no cyanide).
Nitrile reduction: $RX$ + $KCN$ → $RCN$ → $RCH_2NH_2$ (different mechanism, different nitrogen source).

Mistake 4: Forgetting that Gabriel synthesis gives only primary amines …

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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.The source of nitrogen in Gabriel synthesis of amines is ____.

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 works — the key idea

Step-by-step reasoning

Concept: Gabriel Phthalimide Synthesis

Method Name

Steps of the Method

Common Mistakes in Gabriel Synthesis of Amines

Mistake 1: Confusing Gabriel synthesis with other amine preparation methods

Mistake 2: Thinking the nitrogen comes from the alkyl halide or the base

Mistake 3: Picking Potassium cyanide (KCN) because it contains nitrogen

Mistake 4: Forgetting that Gabriel synthesis gives only primary amines …

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Q.The source of nitrogen in Gabriel synthesis of amines is ____.

Why Gabriel synthesis works — the key idea

Step-by-step reasoning

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