Which of the following amines can be prepared by Gabriel phthalimide reaction?
(a) n - butylamine
(b

Q: Why can't we prepare aniline using Gabriel synthesis?

Aryl halides (like bromobenzene) do not undergo $S_N2$ nucleophilic substitution because the carbon-halogen bond has partial double bond character and the benzene ring repels the incoming nucleophile.

Q: Why does $t$-butylamine fail to form?

To form $t$-butylamine, we would need to use $t$-butyl halide. However, tertiary halides are highly sterically hindered. When reacted with a strong base/nucleophile like the phthalimide anion, they undergo $E2$ elimination to form an alkene instead of substitution.

Q: Neo-pentyl is a primary group, so why does it fail?

Although the carbon attached to the halogen in a neo-pentyl halide is primary, the adjacent carbon is quaternary (attached to three methyl groups). This creates massive steric bulk right next to the reaction center, completely blocking the backside attack required for an $S_N2$ mechanism.

Q: Why is this method preferred over direct alkylation of ammonia?

Direct alkylation of ammonia with an alkyl halide often leads to a mixture of primary, secondary, and tertiary amines, as well as quaternary ammonium salts. Gabriel synthesis strictly stops at the monoalkylation stage, yielding a pure primary amine.

Q: Can we use hydrazine instead of $NaOH$ in the final step?

Yes! In fact, using hydrazine ($NH_2NH_2$) is often preferred in the laboratory. This is called the Ing-Manske procedure. Hydrazine reacts with the phthalimide derivative to form a stable cyclic byproduct (phthalhydrazide), releasing the primary amine much more efficiently under milder conditions.

Question

Which of the following amines can be prepared by Gabriel phthalimide reaction?

Options:

(A) $n$ - butylamine

(B) triethylamine

(C) $t$-butylamine

(D) neo -pentylamine

Sigma Percentile · Accepted Answer

### The Elegance of Gabriel Synthesis

Imagine you need to synthesize a perfectly pure primary amine, without any pesky secondary or tertiary amines crashing the party. This is where the **Gabriel phthalimide synthesis** shines as an elegant and highly specific chemical tool.

The beauty of this method lies in its clever use of a protecting group. By starting with phthalimide, we effectively "lock" the nitrogen atom so it can only react once, completely preventing the over-alkylation that plagues direct reaction of ammonia with alkyl halides.

### Step 1: Activating the Phthalimide

Our journey begins with **phthalimide**. If you look closely at its structure, the nitrogen atom is sandwiched between two highly electronegative carbonyl groups. These groups pull electron density away from the nitrogen, making the attached hydrogen atom surprisingly acidic.

When we introduce a strong base, such as ethanolic potassium hydroxide ($KOH$), an acid-base reaction occurs instantly. The base snatches the proton, leaving behind the **potassium salt of phthalimide**. This newly formed anion is resonance-stabilized, but more importantly, the nitrogen now bears a full negative charge. It has been transformed into a potent nucleophile, eagerly waiting for an electrophilic target.

### Step 2: The Crucial $S_N2$ Attack

Now we introduce our electrophile: an alkyl halide ($R-X$). The nucleophilic nitrogen attacks the alkyl group from the backside, kicking out the halide leaving group in a classic **$S_N2$ mechanism**.

This step is the absolute heart of the Gabriel synthesis, and it comes with a strict set of rules. Because it is an $S_N2$ reaction, **steric hindrance is the ultimate enemy**.

### The Steric Trap: Why Other Options Fail

Let's look at the options provided in our question to see why only one survives this step:

1. **$n$-butylamine**: To make this, we use $n$-butyl halide. This is a straight-chain, primary alkyl halide. The backside is wide open, and the $S_N2$ attack proceeds flawlessly.
2. **Triethylamine**: This is a tertiary amine. The Gabriel synthesis can **only** produce primary amines because the phthalimide nitrogen only has one site available for alkylation.
3. **$t$-butylamine**: To synthesize this, we would need $t$-butyl halide, a tertiary halide. Tertiary halides are incredibly bulky. When faced with a strong base/nucleophile like our phthalimide anion, they undergo **$E2$ elimination** to form an alkene instead of substitution.
4. **neo-pentylamine**: This is the trickiest one! While neo-pentyl halide is technically a primary halide, the carbon adjacent to the reaction center is a quaternary carbon (a bulky $t$-butyl group). This massive steric wall completely blocks the incoming nucleophile, making the $S_N2$ reaction practically impossible.

### Step 3: Releasing the Amine

Once we have successfully formed our $N$-alkylphthalimide (in our case, $N$-butylphthalimide), the final step is to release the amine from its protective cage.

We do this through hydrolysis, typically using aqueous sodium hydroxide ($NaOH$) or hydrazine ($NH_2NH_2$). The strong base cleaves the two amide bonds, yielding the sodium salt of phthalic acid and our grand prize: **pure $n$-butylamine**.

By understanding the strict $S_N2$ requirements of the alkylation step, you can easily predict which amines can and cannot be synthesized using this masterful technique!

Animated Solution for Chemistry - Organic Chemistry: Which of the following amines can be prepared by Gabriel phthalimide reaction?

Select Answer:

Visualized Solution

$G ab r i e l P h t ha l imi d e S y n t h es i s$

$F or ma t i o n o f P o t a ss i u m P h t ha l imi d e$

$T h e N u c l eo p hi l e$

$A l k y l a t i o n (S_{N} 2 R e a c t i o n)$

$L imi t a t i o n s o f t h e S_{N} 2 S t e p$

$H y d r o l y s i s t o P r ima r y A min e$

$C o n c l u s i o n$

The Sigma Insight: Amines

The Elegance of Gabriel Synthesis

Step 1

Step 2

The Steric Trap

Step 3

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