Making amines

Six ways to make an amine

A chemist who wants an amine asks one question first. Which compound do I have, and how many carbon atoms must the amine have? The answer picks the route.

The picture shows all six routes arriving at the same primary amine $\text{R}\text{-}\text{NH}_{2}$. Let us take them one at a time.

1. Reduce a nitro compound

A nitro compound, $\text{R}\text{-}\text{NO}_{2}$ or $\text{Ar}\text{-}\text{NO}_{2}$, is reduced to the amine. Hydrogen gas with a palladium, platinum or nickel catalyst does it:

$\text{C}_{6}\text{H}_{5}\text{NO}_{2}\ +\ 3\text{H}_{2}\ \rightarrow \ \text{C}_{6}\text{H}_{5}\text{NH}_{2}\ +\ 2\text{H}_{2}\text{O}$

A metal in acid also works. Tin with hydrochloric acid is common in the laboratory, and iron with hydrochloric acid is common in industry. Iron is preferred for a reason. The iron(II) chloride that forms is hydrolysed and gives back some hydrochloric acid, so only a small amount of acid is needed to start the reaction.

This is the usual way to make aniline, because nitrobenzene is easy to make from benzene. The amine has the same number of carbon atoms as the nitro compound.

2. Reduce a nitrile

A nitrile, $\text{R}\text{-}\text{C}\equiv \text{N}$, is reduced to a primary amine by lithium aluminium hydride, $\text{LiAlH}_{4}$, or by hydrogen with a nickel catalyst. Sodium in ethanol also works.

$\text{R}\text{-}\text{C}\equiv \text{N}\ +\ 2\text{H}_{2}\ \rightarrow \ \text{R}\text{-}\text{CH}_{2}\text{-}\text{NH}_{2}$

The carbon of the nitrile group stays in the amine. So the amine has one carbon atom more than the alkyl halide that made the nitrile. Chemists call this ascending the series. A halide $\text{CH}_{3}\text{CH}_{2}\text{Br}$ gives $\text{CH}_{3}\text{CH}_{2}\text{CN}$ with potassium cyanide, then $\text{CH}_{3}\text{CH}_{2}\text{CH}_{2}\text{NH}_{2}$.

3. Reduce an amide

An amide $\text{R}\text{-}\text{CO}\text{-}\text{NH}_{2}$ is reduced by $\text{LiAlH}_{4}$ followed by water. The C=O group becomes $\text{CH}_{2}$:

$\text{R}\text{-}\text{CO}\text{-}\text{NH}_{2}\ \rightarrow \ \text{R}\text{-}\text{CH}_{2}\text{-}\text{NH}_{2}$

The carbon count does not change. Compare this with route 6, where the carbon count falls by one.

4. Ammonolysis of an alkyl halide

Ammonia is a nucleophile. It attacks an alkyl halide and pushes out the halide ion. The first product is the salt of a primary amine, and sodium hydroxide frees the amine:

$\text{R}\text{-}\text{X}\ +\ \text{NH}_{3}\ \rightarrow \ \text{R}\text{-}\text{NH}_{3}^{+}\text{X}^{-}$, then $\text{R}\text{-}\text{NH}_{3}^{+}\text{X}^{-}\ +\ \text{NaOH}\ \rightarrow \ \text{R}\text{-}\text{NH}_{2}\ +\ \text{NaX}\ +\ \text{H}_{2}\text{O}$

The trouble is that the amine formed is itself a nucleophile, and a better one than ammonia. It attacks another halide molecule. So the product is a mixture of primary, secondary and tertiary amines and a quaternary ammonium salt. A large excess of ammonia makes the primary amine the main product, because most halide molecules meet ammonia. Reaction in a sealed tube at 373 K is the textbook condition. An iodide reacts faster than a bromide, and a bromide faster than a chloride.

Aryl halides, such as chlorobenzene, do not react in this way under ordinary conditions. So route 4 does not make aniline.

5. Gabriel phthalimide synthesis

This route gives a pure primary aliphatic amine. It avoids the mixture of route 4. Phthalimide has a hydrogen on its nitrogen that is acidic, because two C=O groups pull on the nitrogen.

1. Ethanolic potassium hydroxide removes that hydrogen and gives potassium phthalimide.
2. The phthalimide ion attacks an alkyl halide, $\text{R}\text{-}\text{X}$, and gives N-alkylphthalimide. Only one alkyl group can be added, because the nitrogen has no more hydrogen to lose.
3. Heating with aqueous sodium hydroxide splits the imide. It gives disodium phthalate and the primary amine $\text{R}\text{-}\text{NH}_{2}$.

Gabriel synthesis cannot make an aromatic primary amine. The phthalimide ion would have to attack an aryl halide, and aryl halides are inert to this kind of attack. It also cannot make a secondary or a tertiary amine.

6. Hofmann bromamide degradation

An amide is treated with bromine and sodium hydroxide, or with sodium hypobromite, $\text{NaOBr}$. The product is a primary amine with one carbon atom less:

$\text{R}\text{-}\text{CO}\text{-}\text{NH}_{2}\ +\ \text{Br}_{2}\ +\ 4\text{NaOH}\ \rightarrow \ \text{R}\text{-}\text{NH}_{2}\ +\ \text{Na}_{2}\text{CO}_{3}\ +\ 2\text{NaBr}\ +\ 2\text{H}_{2}\text{O}$

The carbonyl carbon leaves as carbonate. Along the way the group R moves from the carbon to the nitrogen, and an isocyanate, $\text{R}\text{-}\text{N}=\text{C}=\text{O}$, forms for a moment. Water and base then break it down. R can be an alkyl or an aryl group. So benzamide gives aniline, and ethanamide gives methylamine.

The amine you want determines the amide: it needs one carbon more than the amine.

Set the amide to 4 carbon atoms and 20 g. The amine has 3 carbon atoms. Now raise the carbon number to 8. The molar masses grow, but the fraction of mass that is kept grows too. Why does a lower yield only scale the line?

Practise with new numbers

Try it in Python

The program picks a target amine with 4 carbon atoms and shows what each route needs to start with. Look at the change in the carbon count that each route carries.

Which route when?

• An aromatic primary amine, such as aniline: reduce the nitro compound, or use Hofmann on a benzamide. Not ammonolysis, not Gabriel.
• A pure primary aliphatic amine: Gabriel, or reduce a nitrile or an amide.
• A primary amine with one more carbon than the halide: the nitrile route.
• A primary amine with one carbon less than the acid derivative: Hofmann.
• A mixture is acceptable: ammonolysis, using excess ammonia for the primary amine.

Common mistakes

• Using ammonolysis for a clean primary amine. It gives a mixture.
• Trying Gabriel on chlorobenzene. Aryl halides do not take part.
• Getting the carbon count wrong. The nitrile route adds one, the Hofmann route removes one.
• Forgetting the carbonate. In the Hofmann equation the carbon that leaves forms $\text{Na}_{2}\text{CO}_{3}$, with four moles of sodium hydroxide.
• Reducing an amide and expecting an alcohol. $\text{LiAlH}_{4}$ gives the amine.

Remember these

• Reduce a nitro compound (same carbon count), a nitrile (one more) or an amide (same).
• Ammonolysis gives a mixture. Excess ammonia favours the primary amine.
• Gabriel gives 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.

References

• NCERT Chemistry Class 12, the chapter Amines: methods of preparation.
• OpenStax Organic Chemistry, Chapter 24 Amines and Heterocycles, section on the synthesis of amines.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Which route turns $\text{CH}_{3}\text{CH}_{2}\text{CH}_{2}\text{Br}$ into butan-1-amine?

2. Which amine does the Hofmann degradation of propanamide give?

3. Why does the Gabriel synthesis not give a secondary amine?

4. Why is iron with hydrochloric acid preferred to tin with hydrochloric acid for reducing nitrobenzene in industry?