Best method for preparing primary amines from alkyl halides without changing the number of carbon atoms in the

NCERT Exemplar · Q21

Q. Best method for preparing primary amines from alkyl halides without changing the number of carbon atoms in the chain is ____.

(i) Hoffmann Bromamide reaction

(ii) Gabriel phthalimide synthesis

(iii) Sandmeyer reaction

(iv) Reaction with

NH_3

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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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The Gabriel phthalimide synthesis is the best method to convert an alkyl halide into a primary amine with the same number of carbon atoms, because it avoids over-alkylation and gives a pure product.

The key problem when you react an alkyl halide directly with ammonia is that the product — a primary amine — is itself a better nucleophile than ammonia. So it immediately attacks another alkyl halide molecule, giving a secondary amine, then a tertiary amine, and finally a quaternary ammonium salt. You end up with a messy mixture, not a pure primary amine.

Gabriel phthalimide synthesis elegantly sidesteps this. It uses a "masked" nitrogen that can only be alkylated once, because the nitrogen is part of a cyclic imide. After alkylation, you hydrolyse the imide to release the pure primary amine.

Let's walk through the reasoning step by step.

What the question asks for

We need a method that starts from an alkyl halide ( $R-X$ ) and produces a primary amine ( $R-NH_2$ ) without changing the carbon chain length. That rules out any reaction that adds or removes carbons from the alkyl group.
Eliminate the wrong options
- Hoffmann Bromamide reaction converts an amide ( $RCONH_2$ ) into an amine with one fewer carbon ( $R-NH_2$ ). The chain shortens by one carbon — not suitable here.
- Sandmeyer reaction replaces a diazonium group with a halogen or other group on an aromatic ring. It doesn't produce amines from alkyl halides at all.
- Reaction with $NH_3$ (direct ammonolysis) gives a mixture of primary, secondary, tertiary amines and quaternary salts. You cannot isolate a pure primary amine easily, and the yield is poor.
Why Gabriel phthalimide synthesis works

The reagent is phthalimide. Its nitrogen is bonded to two carbonyl groups, making the $N-H$ hydrogen quite acidic. In the presence of a base (like alcoholic KOH), it forms the potassium salt of phthalimide — a strong nucleophile.
$\text{Phthalimide} \xrightarrow{\text{KOH (alc.)}} \text{Potassium phthalimide} + H_2O$
This nucleophile attacks the alkyl halide in an $S_N2$ reaction, giving an N-alkylphthalimide. Notice: only one alkyl group can attach because the nitrogen is already bonded to two carbonyls — no further alkylation is possible.

Tip

This is the clever trick: the imide nitrogen can only be alkylated once, so you never get over-alkylation. The product is a single, pure intermediate.
Releasing the amine …

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Here’s a breakdown of the common mistakes students make on this specific question, along with how to avoid each.

1. Mistake: Choosing “Reaction with $NH_3$ ” (Option D) without checking the carbon count

Why students do this:

Students often remember that alkyl halides react with ammonia ( $NH_3$ ) to give primary amines. This is a direct method taught early in the chapter.

Why it’s wrong here:

The reaction of an alkyl halide with $NH_3$ does not stop cleanly at the primary amine. The product ( $RNH_2$ ) is itself a better nucleophile than $NH_3$ , so it attacks another alkyl halide molecule, forming secondary and tertiary amines (and even quaternary ammonium salts). This means you get a mixture of products — not a pure primary amine. Also, the carbon chain length does remain the same, but the lack of purity makes it unsuitable for the “best method” requirement.

How to avoid:

Remember: $NH_3$ + alkyl halide → mixture (primary, secondary, tertiary, quaternary).
If the question asks for the best method to get a pure primary amine, rule out this option immediately.

2. Mistake: Confusing “Hoffmann Bromamide” (Option A) with a method for alkyl halides

Why students do this:

The name “Hoffmann” sounds similar to “Hoffmann elimination” or “Hoffmann rearrangement,” and students sometimes think it’s a general amine preparation.

Why it’s wrong here:

The Hoffmann bromamide reaction converts an amide (RCONH $_2$ ) to a primary amine with one less carbon (due to rearrangement). It does not start from an alkyl halide. So it fails both conditions:

It does not use an alkyl halide as starting material.
It changes the number of carbon atoms (reduces by 1).

How to avoid:

Memorise the starting material for each reaction:
- Hoffmann bromamide: amide → amine (C-chain shortens by 1).
- Gabriel phthalimide: alkyl halide → primary amine (C-chain unchanged).
If the starting material is an alkyl halide, Hoffmann bromamide is irrelevant.

3. Mistake: Picking “Sandmeyer reaction” (Option C) because it involves halides

Why students do this:

The word “Sandmeyer” involves halides (replacement of diazonium group by Cl, Br, CN, etc.), and students think it might be related to alkyl halides.

Why it’s wrong here:

The Sandmeyer reaction converts diazonium salts (from aromatic amines) into aryl halides or other substituted aromatics. It does not produce primary amines from alkyl halides. It’s a method for substituting an amine group, not for introducing one.

How to avoid:

Know that Sandmeyer is for aryl halides from diazonium salts, not for alkyl halides.
If the question says “from alkyl halides,” Sandmeyer is automatically incorrect.

4. Mistake: Forgetting that Gabriel phthalimide synthesis gives pure primary amines

Why students do this: …

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Q.Best method for preparing primary amines from alkyl halides without changing the number of carbon atoms in the chain 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

Concept: Preparation of Primary Amines from Alkyl Halides

Best Method: Gabriel phthalimide synthesis

Why this method?

Steps of Gabriel phthalimide synthesis:

Why other options are incorrect:

1. Mistake: Choosing “Reaction with $NH_3$ ” (Option D) without checking the carbon count

2. Mistake: Confusing “Hoffmann Bromamide” (Option A) with a method for alkyl halides

3. Mistake: Picking “Sandmeyer reaction” (Option C) because it involves halides

4. Mistake: Forgetting that Gabriel phthalimide synthesis gives pure primary amines

Q.Best method for preparing primary amines from alkyl halides without changing the number of carbon atoms in the chain is ____.

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