Reactions at the nitrogen atom

One lone pair, many jobs

In the last lesson the lone pair took a proton. It can also attack a carbon atom that is short of electrons. An attacker of that kind is a nucleophile. This lesson follows the nucleophilic reactions of amines. Each reaction has a twist that makes it a test for one class.

1. Alkylation

An amine attacks an alkyl halide and replaces the halogen. A primary amine becomes a secondary amine:

$\text{CH}_{3}\text{CH}_{2}\text{NH}_{2}\ +\ \text{CH}_{3}\text{CH}_{2}\text{Br}\ \rightarrow \ (\text{CH}_{3}\text{CH}_{2})_{2}\text{NH}_{2}^{+}\text{Br}^{-}$

Sodium hydroxide frees the secondary amine, $(\text{CH}_{3}\text{CH}_{2})_{2}\text{NH}$. The new amine has a lone pair too, so it attacks another halide. The reaction goes on to the tertiary amine and then to a quaternary ammonium salt, $\text{R}_{4}\text{N}^{+}\text{X}^{-}$. With enough halide, the quaternary salt is the end of the line. It cannot go further, as it has no lone pair.

So alkylation does not stop at one step. It is the same behaviour as ammonolysis in the first course.

2. Acylation

An acyl group is $\text{R}\text{-}\text{CO}\text{-}$. Acid chlorides, acid anhydrides and esters carry one. The amine attacks the carbonyl carbon and replaces the chlorine (or the other group). The N-H hydrogen is lost, and an amide forms. This is acylation of the amine.

$\text{CH}_{3}\text{NH}_{2}\ +\ \text{CH}_{3}\text{COCl}\ \rightarrow \ \text{CH}_{3}\text{NH}\text{-}\text{COCH}_{3}\ +\ \text{HCl}$

$\text{C}_{6}\text{H}_{5}\text{NH}_{2}\ +\ (\text{CH}_{3}\text{CO})_{2}\text{O}\ \rightarrow \ \text{C}_{6}\text{H}_{5}\text{NH}\text{-}\text{COCH}_{3}\ +\ \text{CH}_{3}\text{COOH}$

The second equation makes acetanilide. A base such as pyridine is added to take up the hydrogen chloride. This pushes the equilibrium to the right, and it keeps the amine from being turned into its unreactive salt.

With benzoyl chloride, $\text{C}_{6}\text{H}_{5}\text{COCl}$, in the presence of aqueous sodium hydroxide, the product is a benzamide. This is called benzoylation, or the Schotten-Baumann reaction.

Only primary and secondary amines can be acylated. They have an N-H to lose. A tertiary amine has none. Notice also that an amide is far less basic than the amine it came from, and its N-H is hardly able to react again. So acylation stops after one step. That is a big difference from alkylation.

3. The carbylamine reaction

Warm a primary amine with chloroform and alcoholic potassium hydroxide:

$\text{RNH}_{2}\ +\ \text{CHCl}_{3}\ +\ 3\text{KOH}\ \rightarrow \ \text{RNC}\ +\ 3\text{KCl}\ +\ 3\text{H}_{2}\text{O}$

The product is an isocyanide, $\text{R}\text{-}\text{N}\equiv \text{C}$, also called a carbylamine. It has an extremely unpleasant smell. The reaction works for primary amines only, aliphatic or aromatic. Secondary and tertiary amines give nothing. It is the standard test for a primary amine. Take care: isocyanides are toxic.

4. Nitrous acid

Nitrous acid, $\text{HNO}_{2}$, is unstable. So we make it in the flask from sodium nitrite and hydrochloric acid. The result depends on the class of the amine and on whether the nitrogen is on a ring.

• Primary aliphatic. Nitrogen gas bubbles off and an alcohol forms:

$\text{CH}_{3}\text{CH}_{2}\text{NH}_{2}\ +\ \text{HNO}_{2}\ \rightarrow \ \text{CH}_{3}\text{CH}_{2}\text{OH}\ +\ \text{N}_{2}\ +\ \text{H}_{2}\text{O}$

• Primary aromatic. In the cold, at 273 to 278 K, a diazonium salt forms. It does not decompose at once. Course a3 follows it.
• Secondary, aliphatic or aromatic. A yellow oily liquid forms, an N-nitrosoamine, $\text{R}_{2}\text{N}\text{-}\text{N}=\text{O}$. No gas comes off. Treated with phenol and concentrated sulphuric acid, it gives a green colour that turns red and then blue or green with alkali. This is Liebermann's nitroso reaction.
• Tertiary aliphatic. Only a salt forms, a trialkylammonium nitrite, $\text{R}_{3}\text{NH}^{+}\text{NO}_{2}^{-}$. It dissolves, and there is no gas.
• Tertiary aromatic. The nitroso group enters the ring, at the para position. N,N-Dimethylaniline gives green crystals of p-nitroso-N,N-dimethylaniline.

Collecting the nitrogen of a primary aliphatic amine measures the amount of the NH2 group. This is a standard method for analysing amino groups, named after Van Slyke.

The sim treats the gas as ideal. At 273 K and 1 atm, one mole of any gas has a volume close to 22 400 mL. Try 0.59 g of propan-1-amine: that is 0.0100 mol, so you get about 224 mL.

Practise with new numbers

5. Hinsberg's test

Benzenesulphonyl chloride, $\text{C}_{6}\text{H}_{5}\text{SO}_{2}\text{Cl}$, is called Hinsberg's reagent. It attacks an amine in the same way as an acid chloride, and the product is a sulphonamide. The test uses aqueous alkali to tell the three classes apart.

• Primary: $\text{C}_{6}\text{H}_{5}\text{SO}_{2}\text{Cl}\ +\ \text{RNH}_{2}\ \rightarrow \ \text{C}_{6}\text{H}_{5}\text{SO}_{2}\text{NHR}\ +\ \text{HCl}$. The sulphonamide has an N-H next to the strongly pulling SO2 group, so this hydrogen is acidic. Alkali removes it, and the sodium salt dissolves. The solution is clear. Acid then brings the sulphonamide back as a precipitate.
• Secondary: $\text{C}_{6}\text{H}_{5}\text{SO}_{2}\text{Cl}\ +\ \text{R}_{2}\text{NH}\ \rightarrow \ \text{C}_{6}\text{H}_{5}\text{SO}_{2}\text{NR}_{2}\ +\ \text{HCl}$. The sulphonamide has no N-H. It does not dissolve in alkali and stays as a precipitate.
• Tertiary: it forms no stable sulphonamide. The amine stays as an insoluble oil in the alkali, and dissolves when dilute acid is added.

Telling the classes apart

• Carbylamine reaction: a primary amine gives the smell. Secondary and tertiary amines give nothing.
• Hinsberg's test in alkali: primary gives a clear solution, secondary gives a precipitate, tertiary gives no product.
• Acetyl chloride: primary and secondary amines give an amide. A tertiary amine gives none.
• Nitrous acid (aliphatic amines): primary gives a gas, secondary gives a yellow oil, tertiary gives a soluble salt.

Try it in Python

The program holds each test as a small function. It prints the result of every test for six amines.

Common mistakes

• Expecting the carbylamine smell from a secondary amine. Only primary amines give it.
• Expecting gas from aniline and nitrous acid in the cold. It gives a diazonium salt.
• Calling the Hinsberg product of a primary amine insoluble. It dissolves in alkali.
• Acylating a tertiary amine. There is no N-H to lose.
• Stopping at the tertiary amine in alkylation. It goes on to the quaternary salt.

Remember these

• Alkylation does not stop. Acylation stops after one step.
• Carbylamine reaction: primary amines only. $\text{RNH}_{2}\ +\ \text{CHCl}_{3}\ +\ 3\text{KOH}\ \rightarrow \ \text{RNC}\ +\ 3\text{KCl}\ +\ 3\text{H}_{2}\text{O}$.
• Nitrous acid: primary aliphatic gives gas, secondary gives a yellow oil, tertiary aliphatic gives a salt.
• Hinsberg: primary gives a clear solution in alkali, secondary gives a precipitate, tertiary gives no product.

References

• NCERT Chemistry Class 12, the chapter Amines: chemical reactions.
• OpenStax Organic Chemistry, Chapter 24 Amines and Heterocycles, section on the reactions 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. An amine $\text{C}_{4}\text{H}_{11}\text{N}$ gives a precipitate with benzenesulphonyl chloride in alkali, and the precipitate does not dissolve. What class is it? Name one isomer it could be.

2. Why does a tertiary amine fail the acylation reaction?

3. How many millilitres of nitrogen at 273 K and 1 atm come from $0.0200\ \text{mol}$ of ethylamine and excess nitrous acid?

4. Why is pyridine added to an acylation with an acid chloride?