Gabriel Phthalimide Synthesis, Mechanism - Reaction with FAQs

Gabriel Phthalimide Synthesis

The Gabriel Phthalimide Synthesis is a method for synthesizing primary amines from alkyl halides using phthalimide as a reagent. This reaction is widely used due to its ability to selectively produce primary amines, avoiding the formation of secondary or tertiary amines. The Gabriel phthalimide synthesis has applications in the alkylation of sulfonamides and imides, followed by their deprotection, to obtain amines. The alkylation of ammonia is frequently an extensive and inefficient route to amines. In the Gabriel phthalimide synthesis, phthalimide anion is recruited as a proxy of H₂N⁻.

Gabriel Phthalimide Reaction

The Gabriel Phthalimide Reaction is an important organic reaction used for the preparation of primary aliphatic amines from alkyl halides. In this reaction, potassium phthalimide reacts with an alkyl halide (R–X) to form N-alkyl phthalimide, which on hydrolysis gives a primary amine (R–NH₂).

$\begin{gathered}\mathrm{C}_6 \mathrm{H}_4(\mathrm{CO})_2 \mathrm{NK}+\mathrm{R}-\mathrm{X} \rightarrow \mathrm{C}_6 \mathrm{H}_4(\mathrm{CO})_2 \mathrm{NR}+\mathrm{KX} \\ \mathrm{C}_6 \mathrm{H}_4(\mathrm{CO})_2 \mathrm{NR} \xrightarrow{\text { Hydrolysis }} \mathrm{R}-\mathrm{NH}_2+\text { Phthalic Acid }\end{gathered}$

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Gabriel Synthesis Mechanism

Step 1: Formation of Potassium Phthalimide

Phthalimide is deprotonated by KOH/NaOH to form potassium phthalimide.

Phthalimide + KOH → Potassium Phthalimide

• The N–H proton of phthalimide is acidic.
• OH⁻ removes this proton.
• Nitrogen acquires a negative charge and becomes nucleophilic.

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

Step 2: Alkylation of Potassium Phthalimide (S_N² Step)

Potassium Phthalimide + R–X → N-Alkyl Phthalimide

• The negatively charged nitrogen attacks the electrophilic carbon of the alkyl halide.
• Halide ion leaves simultaneously.
• The reaction proceeds through an SN2 mechanism.

$\mathrm{C}_6 \mathrm{H}_4(\mathrm{CO})_2 \mathrm{~N}^{-} \mathrm{K}^{+}+\mathrm{R}-\mathrm{X} \rightarrow \mathrm{C}_6 \mathrm{H}_4(\mathrm{CO})_2 \mathrm{NR}+\mathrm{KX}$

Step 3: Nucleophilic Attack by Hydrazine

N-Alkyl Phthalimide + NH₂NH₂

• One nitrogen atom of hydrazine attacks the carbonyl carbon.
• A tetrahedral intermediate is formed.
• The carbonyl π-bond shifts to oxygen.

Step 4: Proton Transfer (Deprotonation)

• The phthalimide nitrogen removes a proton from the protonated hydrazine nitrogen.
• An uncharged intermediate is generated.
• This stabilizes the reaction intermediate.

Step 5: Second Nucleophilic Acyl Substitution

• The remaining NH₂ group of hydrazine attacks the second carbonyl group.
• Ring rearrangement occurs.
• The N–R bond is weakened and cleavage begins.
• The leaving group is RNH⁻.

Step 6: Formation of Primary Amine

• Proton transfer converts RNH⁻ into RNH₂.
• Phthalhydrazide is formed as a stable by-product.
• Primary amine is liberated.

Also Read

Some Solved Examples

Question 1: Gabriel phthalimide reaction is used in the synthesis of:

1) Primary aromatic amines

2) Secondary amines

3) (correct) Primary aliphatic amines

4) Tertiary amines

Solution:

As we learn

Gabriel phthalimide row is used in the synthesis of primary aliphatic amines.

Hence, the answer is option (3).

Question 2: Which of the following compounds can be prepared in good yield by Gabriel phthalimide synthesis?

1)

2) $\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{NHCH}_3$

3)

4)

Solution:

Gabriel Phthalimide Synthesis is used for the preparation of 1° Aliphatic amine from an Alkyl halide.

Thus, only Benzylamine can be prepared by Gabriel Phthalimide Synthesis

Hence, the answer is the option (1).

Question 3: Number of isomeric aromatic amines with molecular formula $\mathrm{C}_8 \mathrm{H}_{11} \mathrm{~N}_{, \text {, }}$ which can be synthesized by Gabriel Phthalimide synthesis is $\qquad$

Solution:

By Gabriel phthalimide synthesis $\rightarrow$ i-amine is prepared

$\mathrm{C}_8 \mathrm{H}_{11} \mathrm{~N} \rightarrow$ Should be aromatic & i-amine

$\begin{aligned} \text { Degree of unsaturation } & =\mathrm{C}+1-\frac{\mathrm{H}-\mathrm{N}}{2} \\ & =8+1-\frac{11-1}{2} \\ & =9-\frac{10}{2}=9-5=4\end{aligned}$

It means benzene ring

Hence, the correct answer is (6).

Question 4: Which of the following compounds cannot be prepared by Gabriel Phthalimide Synthesis?
A. $\mathrm{CH}_3 \mathrm{NH}_2$
B. $\mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_2$
C. $\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2$
D. $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{NH}_2$

Solution:

Gabriel synthesis proceeds through an SN2 mechanism.
For successful SN2 attack, the substrate must be an alkyl halide.
To prepare aniline $\left(\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2\right)$, one would require an aryl halide such as chlorobenzene. $\mathrm{C}_6 \mathrm{H}_5 \mathrm{Cl}$
Aryl halides do not undergo S_N² reactions because:

• The carbon attached to halogen is $\mathbf{s p}^{\mathbf{2}}$ hybridized.
• Backside attack is not possible.
• Partial double-bond character exists in the $\mathrm{C}-\mathrm{X}$ bond.

Therefore, aniline cannot be prepared by Gabriel synthesis.

Hence, the correct answer is option (C)

Question 5:

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