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

CBSEMCQ· 1mImportance★★★★★

86% · 93/108 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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: …

Start your 14-day free trial to unlock the full solution →

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 …

Start your 14-day free trial to unlock the full solution →

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). …

Start your 14-day free trial to unlock the full solution →

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

CBSE 2026Set 56/3/11 markMCQ
$Q. Assertion (A) : Aromatic primary amines cannot be prepared by Gabriel phthalimide synthesis. Reason (R) : Gabriel phthalimide synthesis is used for the preparation of primary aliphatic amines. Options : (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Assertion (A) is false, but Reason (R) is true.$
›Reveal solutionSolution
Gabriel phthalimide synthesis fails for aromatic primary amines because aryl halides do not undergo nucleophilic substitution under the reaction conditions. The reason correctly states that the method is for aliphatic amines, but it does not explain why aromatic amines fail — so both statements are true, but the reason is not the correct explanation.

The key here is to understand why Gabriel phthalimide synthesis works for some amines and not others — and that comes down to the chemistry of the nucleophilic substitution step.

Gabriel phthalimide synthesis is a two-step method: first, phthalimide is treated with alcoholic KOH to form the potassium salt of phthalimide. This salt is a strong nucleophile. In the second step, it attacks an alkyl halide ( $R-X$ ) in an $S_N2$ reaction, giving an N-alkylphthalimide. Finally, hydrolysis (or hydrazinolysis) liberates the primary amine.

The critical step is the $S_N2$ attack. For this to happen, the carbon bearing the halogen must be able to undergo backside attack — it must be sp³-hybridised and not too sterically hindered. Aryl halides (like chlorobenzene) have the halogen attached directly to an sp² carbon of the benzene ring. Such carbons do not undergo $S_N2$ reactions because the p-orbitals of the double bond block the backside approach, and the C–X bond has partial double-bond character due to resonance.

So the assertion is true: you cannot prepare aromatic primary amines (like aniline) this way. The reason is also true: Gabriel phthalimide synthesis is indeed used for aliphatic primary amines. But the reason does not explain why aromatic amines fail — it merely states what the method is used for. The actual explanation lies in the reactivity of aryl halides, not in the classification of the product.

Let’s walk through the logic step by step.

Understand the assertion.

Assertion (A) says aromatic primary amines cannot be prepared by Gabriel phthalimide synthesis. This is correct. If you try to react potassium phthalimide with chlorobenzene, no reaction occurs under normal conditions. Even if you force the reaction (high temperature, catalyst), the product is not aniline — you get side products.

Understand the reason.

Reason (R) says Gabriel phthalimide synthesis is used for the preparation of primary aliphatic amines. This is also correct. The classic examples are the preparation of ethylamine, benzylamine, etc. from their respective alkyl halides.

Check if (R) explains (A). …
Start your 14-day free trial to see the full solution →
CBSE 2026Set 56/2/11 markMCQ
$Q. Assertion (A) : Aromatic primary amines can be prepared by Gabriel Phthalimide synthesis. Reason (R) : Aryl halides do not undergo nucleophilic substitution with the anion formed by phthalimide.$
›Reveal solutionSolution
Gabriel phthalimide synthesis works only for aliphatic primary amines because aryl halides resist nucleophilic substitution. Assertion is false, Reason is true.

Understanding Gabriel Phthalimide Synthesis

The Gabriel synthesis is a classic method for preparing primary amines. The strategy is elegant: phthalimide (with $\mathrm{p}K_\mathrm{a} \approx 9$ ) is deprotonated by base to form a nucleophilic anion, which then attacks an alkyl halide in an $\mathrm{S_N2}$ displacement. Hydrolysis of the resulting N-alkylphthalimide releases the primary amine.

The entire method hinges on a successful nucleophilic substitution step. That's where the limitation appears.

Why Aryl Halides Don't Cooperate

Aryl halides—compounds where a halogen is directly bonded to a benzene ring—are notoriously unreactive toward nucleophilic substitution under normal conditions. Three factors conspire against the reaction:

Partial double-bond character: The lone pairs on the halogen overlap with the aromatic $\pi$ -system, giving the C–X bond some double-bond character. This strengthens the bond and makes it harder to break.

$\mathrm{S_N2}$ geometry is impossible: An $\mathrm{S_N2}$ mechanism requires backside attack at the carbon bearing the leaving group. In an aryl halide, that carbon is $sp^2$ -hybridized and embedded in a planar ring—there's no accessible backside. The nucleophile would have to attack through the ring itself.

$\mathrm{S_N1}$ is prohibitively unfavorable: Heterolytic cleavage would produce a phenyl cation, an extraordinarily unstable species. The empty $p$ -orbital would be orthogonal to the aromatic $\pi$ -system, offering no stabilization. The activation energy is far too high.

Watch out

A common mistake is assuming that because aryl halides contain a halogen, they behave like alkyl halides. The aromatic ring fundamentally changes the reactivity—standard nucleophilic substitutions simply don't occur.

Evaluating the Statements

Assertion (A): Claims aromatic primary amines can be prepared by Gabriel synthesis.

This is false. The synthesis requires an alkyl halide (or similar electrophile) that undergoes $\mathrm{S_N2}$ substitution with the phthalimide anion. Since aryl halides don't participate in this substitution, you cannot use Gabriel synthesis to make aniline or other aromatic primary amines. …
Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 markMCQ
$Q. Regarding Gabriel synthesis for amines, which of the following statements is correct?$
(a) All types of amines can be synthesized
(b) Only aromatic amines can be synthesized
(c) Only aliphatic amines can be synthesized
(d) Only aliphatic primary amines can be synthesized
›Reveal solutionSolution
Gabriel phthalimide synthesis works only through nucleophilic substitution of an alkyl halide, so it gives only aliphatic primary amines — option (D).

In the Gabriel phthalimide synthesis, phthalimide is treated with $KOH$ to give potassium phthalimide, whose nitrogen then displaces the halogen of an alkyl halide ( $S_N2$ ). Alkaline hydrolysis of the resulting $N$ -alkylphthalimide finally releases a primary amine.

Because the nitrogen ends up bonded to only one alkyl group, the product is always a primary amine (no secondary/tertiary amines). …
Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 mark
$Q. Why cannot aromatic primary amines be prepared by Gabriel phthalimide synthesis?$
›Reveal solutionSolution
Gabriel synthesis's first (and essential) step is an SN2 attack of phthalimide anion on the alkyl/aryl halide; aryl halides are inert to SN2 for structural and electronic reasons, so aromatic primary amines cannot be made this way.

The Gabriel phthalimide synthesis makes a primary amine in two steps: (1) potassium phthalimide is alkylated by an alkyl halide, $R-X$ , through a nucleophilic substitution (SN2) in which the phthalimide nitrogen's lone pair (as its anion) displaces the halide ion; (2) the resulting N-alkylphthalimide is then hydrolysed (or treated with hydrazine) to liberate the primary amine $R-NH_2$ .

Step (1) requires the halide-bearing carbon to be accessible to backside nucleophilic attack, i.e. it must be a genuine SN2-reactive alkyl halide. Aryl halides ( $Ar-X$ ) cannot undergo this step, because:

the halogen-bearing carbon is $sp^2$ -hybridised and lies in the plane of the aromatic ring, so there is no accessible backside for the nucleophile to attack (the ring itself blocks it);

the C–X bond has partial double-bond character (the halogen's lone pair is delocalised into the ring by resonance), making it shorter and stronger than a normal C–X single bond, and a poorer leaving group; …
Start your 14-day free trial to see the full solution →
CBSE 2020Set NC1 mark
$Q. Why aromatic primary amines cannot be prepared by Gabriel phthalimide synthesis?$
›Reveal solutionSolution
Gabriel phthalimide synthesis is an $S_N2$ displacement on an alkyl halide by phthalimide anion; aryl halides are essentially unreactive toward this mechanism, so the method is limited to aliphatic primary amines.

Gabriel synthesis (as normally used, for aliphatic amines): Potassium phthalimide's nitrogen anion is a good nucleophile that displaces a halide from an alkyl halide in an $S_N2$ reaction, giving N-alkylphthalimide, which is then hydrolysed (or hydrazinolysed) to liberate the primary amine and phthalhydrazide:

$\text{Phthalimide}^{-}K^{+} + R\text{–X} \longrightarrow N\text{-alkylphthalimide} \xrightarrow{H_3O^{+}/OH^{-}} R\text{–}NH_2$

Why it fails for aromatic amines. To make an arylamine (e.g. aniline) this way, one would need to react phthalimide anion with an aryl halide (e.g. chlorobenzene) by the same $S_N2$ mechanism. But aryl halides do not undergo nucleophilic substitution under these conditions because:

the carbon–halogen bond in an aryl halide has partial double-bond character (from resonance/conjugation of the halogen lone pair with the ring), making it shorter and stronger than a normal C–X bond; …
Start your 14-day free trial to see the full solution →

←Previous questionQ64 — Which arenium ion(s) (sigma complex) is/are involved in the brominatio…→Next questionQ66 — Which of the following reactions is/are correct? (Two or more options …

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

Understanding Gabriel Phthalimide Synthesis

Why Aryl Halides Don't Cooperate

Evaluating the Statements

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 …

🎓Unlock everything free for 14 days

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

Unlock everything free for 14 days