Q.(a) Write equations involved in the following reactions :

Acylation of Amines – From Intuition to Precision

Imagine you have a primary amine — say, aniline ($\ce{C6H5NH2}$). The nitrogen carries a lone pair and is nucleophilic. Now imagine you bring an acid chloride like acetyl chloride ($\ce{CH3COCl}$) near it. The carbonyl carbon of the acid chloride is electron-deficient (electrophilic). The nitrogen's lone pair attacks that carbon, kicking out the chloride ion. What you get is an amide — a molecule where the nitrogen is now attached to an acyl group ($\ce{-COCH3}$) instead of a hydrogen.

That replacement of an N–H hydrogen by an acyl group ($\ce{RCO-}$) is acylation. It is a nucleophilic acyl substitution reaction.

The precise statement

Primary and secondary amines react with acid chlorides ($\ce{RCOCl}$) or acid anhydrides ($\ce{(RCO)2O}$) to form substituted amides. One N–H hydrogen is replaced by the acyl group. The by-product is HCl (from acid chlorides) or a carboxylic acid (from anhydrides).

For a primary amine ($\ce{RNH2}$):

For a secondary amine ($\ce{R2NH}$):

Tertiary amines have no N–H hydrogen, so they do not undergo acylation.

Why does acylation stop at one step?

After the first acylation, the nitrogen in the amide has its lone pair delocalised into the carbonyl $\pi$-system. This makes the amide nitrogen far less nucleophilic than the original amine. So it does not attack another acyl chloride molecule. This is a huge practical advantage over alkylation, where you often get a messy mixture.

Reagents commonly used

• Acid chlorides (e.g., acetyl chloride, benzoyl chloride) — very reactive, often used in the lab.
• Acid anhydrides (e.g., acetic anhydride) — milder, commonly used in industry (e.g., acetylation of aniline to paracetamol intermediate).
• Esters can also acylate, but much more slowly (requires heating).

The Schotten–Baumann technique

In practice, acylation is often done in the presence of a weak base (like aqueous NaOH or pyridine) to neutralise the HCl produced. This prevents the HCl from protonating the unreacted amine (which would stop the reaction). This method is called the Schotten–Baumann reaction. …

Gabriel Phthalimide Synthesis

You want to make a primary amine — a molecule where an alkyl group is attached to an $NH_2$ group. The obvious route is to react an alkyl halide with ammonia. That gives you a mixture: some primary amine, some secondary amine (two alkyl groups on the nitrogen), some tertiary amine, and even some quaternary ammonium salt. Ammonia is a nucleophile, but once it reacts, the product (the primary amine) is an even better nucleophile than ammonia was. So it keeps attacking more alkyl halide molecules, and you lose control.

The Gabriel synthesis is a clever way to stop that chain reaction. It works by hiding the nitrogen inside a molecule that cannot act as a nucleophile until you want it to.

The intuition

Imagine you have a nitrogen atom that you want to attach exactly one alkyl group to, and then release it as a primary amine. If you put that nitrogen inside a structure that is already "full" — where it has no hydrogen atoms left to be replaced — then it cannot react with more than one alkyl halide molecule. That is the core idea.

Potassium phthalimide is that structure. Phthalimide itself has the formula $C_6H_4(CO)_2NH$. The nitrogen is flanked by two carbonyl groups, which pull electron density away from it. When you treat phthalimide with a base (usually alcoholic KOH), the $N-H$ bond is deprotonated, giving the potassium salt:

This anion is a good nucleophile. It attacks an alkyl halide ($R-X$) in an $S_N2$ reaction, giving $N$-alkylphthalimide:

Now look at that product. The nitrogen already has three bonds: two to the carbonyl carbons and one to the alkyl group. It has no hydrogen left. It cannot react with a second alkyl halide molecule. The secondary and tertiary amine contamination is impossible at this stage.

Releasing the amine

You now have the alkyl group attached to the phthalimide nitrogen. To get the free primary amine, you need to break the two $N-C(=O)$ bonds. This is done by hydrolysis — either acidic or basic.

Alkaline hydrolysis (reflux with aqueous or alcoholic KOH):

The products are potassium phthalate and the primary amine. The amine is liberated as a free base and can be distilled out or extracted.

Acidic hydrolysis (reflux with concentrated HCl):

Here you get phthalic acid and the amine hydrochloride salt. You then treat the salt with a base to free the amine.

The precise statement …