3.3.11 Amines

Two required routes give primary aliphatic amines. A halogenoalkane reacts with excess NH₃ by nucleophilic substitution; alternatively, reducing a nitrile converts –C≡N into –CH₂NH₂.

When you choose between the routes, watch the carbon count. CN⁻ substitution creates the nitrile with an extra carbon before reduction; the ammonia/halogenoalkane route keeps the original carbon skeleton.

If you want an aromatic amine from benzene chemistry, nitration followed by reduction is the useful route to remember. Reducing an aromatic –NO₂ group gives –NH₂, producing compounds such as phenylamine.

Reducing an aromatic nitro group converts –NO₂ into –NH₂, providing aromatic amines such as phenylamine that are useful intermediates in dye chemistry.

Amines are weak Brønsted–Lowry bases because the nitrogen lone pair can accept H⁺, forming an ammonium ion.

When you compare amine basicity, follow that lone pair: the more available it is to form a bond to H⁺, the stronger the base.

For the usual basicity comparison, I’d learn the order primary aliphatic amine > NH₃ > primary aromatic amine. Then explain it by asking how available the nitrogen lone pair is to accept H⁺.

The aromatic end of the comparison has a clear structural reason: the nitrogen lone pair is delocalised into the benzene ring, so it is less available to accept H⁺.

When you explain amine basicity, keep your attention on the nitrogen lone pair and how available it is to bond to H⁺. It is more available in a simple primary aliphatic amine than in NH₃, and less available in a primary aromatic amine.

For the aromatic amine, make the electronic link explicit: delocalisation of the lone pair into the ring reduces its availability at nitrogen. Keep the explanation centred on lone-pair availability.

The same nitrogen lone pair also makes amines nucleophiles. Keep the two roles distinct: base behaviour targets H⁺; nucleophilic behaviour attacks an electron-deficient carbon.

If you draw the substitution mechanism, start the curly arrow at the nitrogen lone pair and send it to the δ⁺ carbon bonded to the halogen. Then remove the extra proton from the initially formed alkylammonium ion.

Alkylating ammonia with a halogenoalkane first gives a primary amine, but that product still has a nucleophilic lone pair. It can react again, so successive alkylation can take you through secondary and tertiary amines to a quaternary ammonium salt.

Using excess ammonia increases the chance that NH₃ is the more likely collision partner for the halogenoalkane, so excess ammonia favours the primary amine when that is the desired product.

For a quaternary ammonium surfactant, picture two very different regions in the same ion: a permanently charged ammonium head that interacts well with water and long hydrocarbon groups that interact with non-polar material.

A quaternary ammonium ion has a permanent positive charge; when attached to long hydrophobic alkyl chains it can interact with water through the charged head while the hydrocarbon region interacts with non-polar material.

Ammonia or a primary amine attacks the acyl carbon of an acyl chloride/anhydride by nucleophilic addition–elimination, forming an amide. Track any HCl or carboxylic-acid by-product required by the reagent.

For example, ethanoyl chloride with NH₃ gives ethanamide, while ethanoyl chloride with CH₃NH₂ gives an N-substituted amide. Keep the nitrogen substituent when you name or draw the product.

When you draw the amine mechanisms, keep the reaction centre clear. With a halogenoalkane the nitrogen lone pair attacks the δ⁺ carbon and C–X breaks; with an acyl chloride it attacks the carbonyl carbon, gives a tetrahedral intermediate and then C=O reforms as Cl⁻ leaves.

The two mechanisms both use nitrogen as a nucleophile but have different reaction centres and bond changes. Do not copy a halogenoalkane substitution mechanism onto an acyl chloride.