Amines

Structure and classification. Nitrogen in an amine is $sp^3$ hybridised, pyramidal, with the lone pair in the fourth position. This lone pair is the seat of every reaction amines undergo — basicity, nucleophilic attack, salt formation — but for now the important thing is simply how many carbon groups sit on the nitrogen:

• Primary (1°): $R\text{-}NH_2$ — one alkyl/aryl group on nitrogen.
• Secondary (2°): $R_2NH$ — two alkyl/aryl groups, same or different.
• Tertiary (3°): $R_3N$ — three alkyl/aryl groups.

This is a genuinely different rule from alcohols. An alcohol's 1°/2°/3° label depends on the carbon skeleton at the carbon bearing the –OH; an amine's label depends purely on how many carbon groups sit on nitrogen. Carrying the alcohol convention over to amines is the single most common classification error students make.

Nomenclature. IUPAC names a simple amine by adding the suffix "-amine" to the parent alkane name, so $CH_3NH_2$ is methanamine; common usage calls it methylamine. Secondary and tertiary amines are named either as N-substituted derivatives of the longest chain (N-methylethanamine) or, in common nomenclature, by listing the alkyl groups alphabetically before "amine."

Five preparation routes. The syllabus wants aliphatic amines built from five distinct starting points, each worth knowing by its reagents and its characteristic outcome.

where: $R\text{-}X$ is an alkyl halide and excess ammonia is used, yet the reaction still does not stop cleanly at $R\text{-}NH_2$, because that product is itself nucleophilic.

where: the nitrile carbon becomes the new $CH_2$ next to nitrogen, so the product amine carries one more carbon than the alkyl group of the halide originally used to make that nitrile.

where: the amide $R\text{-}CONH_2$ loses its carbonyl carbon as $Na_2CO_3$, so the product has one carbon fewer than the amide — the opposite carbon-count shift to nitrile reduction.

where: a nitro compound is reduced directly to the primary amine with the same carbon skeleton, no chain-length change. The fifth route, the Gabriel phthalimide synthesis, is described fully in the derivation-style section below because it is the syllabus's preferred method for a pure primary amine and deserves a step-by-step walkthrough.

The choice between the five routes really comes down to two questions: does this route stop cleanly at the primary amine, and does it change the carbon count? Ammonolysis fails the first test badly. Once $R\text{-}NH_2$ forms, it has a lone pair just like $NH_3$ does, so it competes with the remaining ammonia for unreacted $R\text{-}X$, giving $R_2NH$, then $R_3N$, and finally a quaternary ammonium salt $R_4N^+X^-$. There is no way to stop the reaction at the primary stage by adjusting conditions alone, which is precisely why chemists needed alternative routes. Nitrile reduction and Hofmann's degradation both stop cleanly at a primary amine, but for different structural reasons. In nitrile reduction, $H_2/Ni$ (or $LiAlH_4$) simply adds hydrogens across the triple bond of $R\text{-}CN$ until it becomes $R\text{-}CH_2NH_2$ — the nitrogen only ever carries one carbon group throughout, so there is no route to a secondary or tertiary product. In Hofmann's degradation, the mechanism itself removes a carbon: bromination of the amide nitrogen followed by base-induced migration of the R group from carbon to nitrogen forms an isocyanate, which hydrolyses and loses $CO_2$ to leave $R\text{-}NH_2$ — again with nitrogen never bonded to more than one carbon group at any stage. The Gabriel synthesis solves the ammonolysis problem by a different trick entirely: it locks nitrogen inside a ring (phthalimide) so it can only be alkylated once. Potassium phthalimide's nitrogen is a good nucleophile but has no N–H left to give up a second time once it reacts with one $R\text{-}X$, so over-alkylation is structurally impossible. Only after this single substitution does hydrolysis release the pure primary amine and regenerate phthalic acid.

Problem: Starting from ethanamide ($CH_3CONH_2$), carry out Hofmann's degradation, and separately convert $CH_3CH_2Br$ to a pure primary amine by the Gabriel synthesis. Name both products and state which route changed the carbon count.

Step 1. Ethanamide reacts with bromine and excess $NaOH$: $CH_3CONH_2 + Br_2 + 4NaOH \rightarrow CH_3NH_2 + Na_2CO_3 + 2NaBr + 2H_2O$. The mechanism brominates the amide nitrogen, base removes the remaining N–H, the alkyl group migrates from carbon to nitrogen (forming an isocyanate $CH_3\text{-}N=C=O$), and alkaline hydrolysis followed by loss of $CO_2$ releases the amine.

Step 2. The Hofmann product is $CH_3NH_2$, methanamine (methylamine) — one carbon fewer than the starting two-carbon amide, the signature carbon-count drop of this route.

Step 3. For the Gabriel route: phthalimide $+ KOH \rightarrow$ potassium phthalimide (its N–H is acidic enough to deprotonate). This reacts with $CH_3CH_2Br$ to give N-ethylphthalimide $+ KBr$ by nucleophilic substitution at the alkyl halide.

Step 4. N-ethylphthalimide is hydrolysed with aqueous $NaOH$ (or treated with hydrazine, the Ing–Manske procedure) to release the amine and regenerate the phthalate derivative. Final answer: Hofmann's degradation gives methanamine (one carbon less than the amide); the Gabriel synthesis gives ethanamine, $CH_3CH_2NH_2$, pure and free of 2°/3° contamination — unlike direct ammonolysis of $CH_3CH_2Br$, which would give a mixture.

Competitive stretch

If asked to convert $CH_3CH_2CN$ (propanenitrile) to a primary amine two carbons longer than the starting alkyl halide it came from, use $H_2/Ni$ reduction: $CH_3CH_2CN \xrightarrow{H_2/Ni} CH_3CH_2CH_2NH_2$ (propan-1-amine) — the nitrile carbon becomes the new terminal $CH_2$, so the amine has one carbon more than the nitrile, and two more than the original alkyl halide used to make that nitrile via cyanide substitution.

• Confusing amines with amides — an amide carries a carbonyl group bonded to nitrogen ($R\text{-}CO\text{-}NH_2$); an amine has no carbonyl at all.
• Assuming ammonolysis of an alkyl halide gives a clean primary amine — it gives a mixture of 1°, 2°, 3° amines and quaternary ammonium salt unless the Gabriel synthesis is used instead.
• Mixing up the direction of the carbon-count shift — nitrile reduction adds one carbon, Hofmann's degradation removes one carbon; do not swap these.
• Applying the alcohol 1°/2°/3° rule to amines — amine classification counts carbon groups on nitrogen, not the skeleton at a particular carbon.

Ask the tutor: Want to work through why the Gabriel synthesis mechanism specifically blocks a second alkylation, or try converting a different starting material through all five routes?