Amines: Classes, Names and Preparation: Quick Revision

A brief note on Amines: Classes, Names and Preparation. Read it before a test. Try each question at the end before you open its answer.

Classes and structure

• Primary $\text{RNH}_{2}$, secondary $\text{R}_{2}\text{NH}$, tertiary $\text{R}_{3}\text{N}$: the number of carbon groups on the nitrogen.
• Aromatic means the nitrogen is joined straight to a benzene ring. Aniline is aromatic and benzylamine is not.
• A quaternary ammonium ion $\text{R}_{4}\text{N}^{+}$ has no lone pair.
• The nitrogen is $sp^3$ and pyramidal. The lone pair makes an amine a base and a nucleophile.
• A saturated amine with $n$ carbon atoms is $\text{C}_n\text{H}_{2n+3}\text{N}$. $\text{C}_{3}\text{H}_{9}\text{N}$ has 4 isomers and $\text{C}_{4}\text{H}_{11}\text{N}$ has 8.

Names

• IUPAC: the alkane name without the final e, plus amine. Other groups on nitrogen are N-prefixes: N-methylethanamine, N,N-dimethylmethanamine.
• $\text{C}_{6}\text{H}_{5}\text{NH}_{2}$ is aniline, or benzenamine. When another group has priority, NH2 is the prefix amino.

Preparation

• Nitro compound: $\text{H}_{2}/\text{Pd}$, $\text{Sn}/\text{HCl}$ or $\text{Fe}/\text{HCl}$. Same carbon count.
• Nitrile: $\text{LiAlH}_{4}$ or $\text{H}_{2}/\text{Ni}$. Amine has one carbon more than the halide that made the nitrile.
• Amide: $\text{LiAlH}_{4}$. Same carbon count.
• Ammonolysis of a halide: a mixture of all classes and the quaternary salt. Excess ammonia favours the primary amine.
• Gabriel: potassium phthalimide, then a halide, then alkaline hydrolysis. Pure primary aliphatic amines only.
• Hofmann: $\text{RCONH}_{2}\ +\ \text{Br}_{2}\ +\ 4\text{NaOH}\ \rightarrow \ \text{RNH}_{2}\ +\ \text{Na}_{2}\text{CO}_{3}\ +\ 2\text{NaBr}\ +\ 2\text{H}_{2}\text{O}$. One carbon fewer.

Physical properties

• Isomers: primary boils above secondary, secondary above tertiary.
• At similar molar mass: alkane and ether lower than amine, and amine lower than alcohol.
• Low amines dissolve in water. Aniline dissolves only a little, about 3.6 g in 100 mL at $20\ ^\circ\text{C}$.

A worked example

A chemist makes methylamine from ethanamide by the Hofmann degradation. How many grams form from $5.9\ \text{g}$ of ethanamide at a yield of $80\ \%$?

1. Molar masses. Ethanamide $\text{C}_{2}\text{H}_{5}\text{NO}$ is $59\ \text{g/mol}$. Methylamine $\text{CH}_{5}\text{N}$ is $31\ \text{g/mol}$.
2. Amount. $\dfrac{5.9}{59} = 0.100\ \text{mol}$.
3. Mass. $0.100 \times 31 \times 0.80 = 2.48\ \text{g}$.

Practice with new numbers

A program to try

The program counts the carbon atoms that the starting compound needs for each route to an amine with five carbon atoms.

Mistakes to avoid

• Naming the class from the carbon, not from the nitrogen.
• Gabriel for an aromatic amine, or ammonolysis for a pure one.
• Counting the carbons wrongly: the nitrile route adds one, Hofmann removes one.
• A tertiary amine does not dissolve: it does, since it accepts a hydrogen bond from water.

Check yourself

1. Name $\text{CH}_{3}\text{-}\text{NH}\text{-}\text{CH}_{2}\text{CH}_{2}\text{CH}_{3}$ and give its class.

2. Which amide gives propan-1-amine by the Hofmann degradation?

3. Why is the boiling point of a tertiary amine lower than that of its primary isomer?

4. How many grams of aniline form from $0.50\ \text{mol}$ of nitrobenzene at a yield of $100\ \%$?