Substitution in the ring of aniline

An amino group that pushes

Look at the lone pair on the nitrogen of aniline. It is next to the pi system of the ring, so it can move into the ring. This gives three resonance forms with a negative charge on the ring carbon at the two ortho positions and at the para position.

So the ortho and para carbons are rich in electrons. An attacking positive group, an electrophile, goes to those positions. The amino group is an activating, ortho and para directing group. It is one of the strongest.

Bromination

Add bromine water to aniline at room temperature. A white precipitate forms at once. It is 2,4,6-tribromoaniline:

$\text{C}_{6}\text{H}_{5}\text{NH}_{2}\ +\ 3\text{Br}_{2}\ \rightarrow \ \text{C}_{6}\text{H}_{2}\text{Br}_{3}\text{NH}_{2}\ +\ 3\text{HBr}$

All three open ortho and para positions take a bromine. The reaction is so fast that it cannot be stopped at one bromine. This reaction is also used as a test for aniline.

Protecting the amine: acetylation

How do we make p-bromoaniline, with only one bromine? The answer is to turn the NH2 group into an amide first.

1. Acetylate aniline with acetic anhydride, in the presence of pyridine. The product is acetanilide, $\text{C}_{6}\text{H}_{5}\text{NH}\text{-}\text{COCH}_{3}$.
2. Brominate the acetanilide with bromine in ethanoic acid. Bromine goes mostly to the para position.
3. Hydrolyse the amide with aqueous acid or alkali. This gives p-bromoaniline.

Why does acetylation tame the ring? The lone pair on the nitrogen of acetanilide is shared with the C=O group, as in every amide. It is less free to push electrons into the ring. The ring is still activated, but only moderately. The bulky acetamido group also blocks the ortho positions, so para is favoured. The acetyl group is a protecting group: put it on, do the reaction, take it off.

Each step loses some material, and the yields multiply. With 95, 70 and 90 percent, the overall yield is 59.9 percent. A synthesis with a protecting group costs two extra steps, and it is worth it only when the direct reaction does not work.

Practise with new numbers

Nitration

Direct nitration of aniline with concentrated nitric and sulphuric acids gives two problems.

1. Oxidation. Nitric acid oxidises aniline and gives dark tarry products.
2. Protonation. In the strongly acidic mixture, aniline takes a proton and becomes the anilinium ion, $\text{C}_{6}\text{H}_{5}\text{NH}_{3}^{+}$. This group pulls electrons and directs the attack to the meta position.

So direct nitration gives a mixture. NCERT gives about 51 percent para, 47 percent meta and 2 percent ortho. The meta product comes from the anilinium ion.

To get the para product, protect the amine first. Acetylate aniline, nitrate the acetanilide in the cold, and hydrolyse. The amide is only weakly basic, so it does not protonate, and it is hard to oxidise. The main product is p-nitroaniline.

Sulphonation

Add concentrated sulphuric acid to aniline. It gives anilinium hydrogensulphate:

$\text{C}_{6}\text{H}_{5}\text{NH}_{2}\ +\ \text{H}_{2}\text{SO}_{4}\ \rightarrow \ \text{C}_{6}\text{H}_{5}\text{NH}_{3}^{+}\text{HSO}_{4}^{-}$

Heat this salt at 453 to 473 K. The main product is p-aminobenzenesulphonic acid, which is sulphanilic acid. It exists as a zwitterion: the acidic group gives a proton to the basic group in the same molecule, so there is a positive and a negative charge. It melts at a high temperature, dissolves poorly in organic solvents and does not behave like an ordinary amine or acid. It is used to make azo dyes.

Friedel-Crafts reactions fail

Aniline does not undergo Friedel-Crafts alkylation or acylation. The catalyst, aluminium chloride, is a Lewis acid. The amine is a Lewis base. They combine at once. The nitrogen then has a positive charge and strongly pulls electrons from the ring, so the ring is deactivated. No substitution occurs.

Try it in Python

The program follows the yields through the three steps of the nitration route and prints the mass at the end of each step.

Common mistakes

• Expecting one bromine from bromine water. Aniline gives the tribromo compound.
• Nitrating aniline directly and expecting only p-nitroaniline. The anilinium ion gives meta, and oxidation spoils the rest.
• Forgetting to remove the acetyl group. The last step is hydrolysis.
• Trying Friedel-Crafts on aniline. The catalyst binds to the nitrogen.
• Thinking sulphanilic acid is an ordinary amine. It is a zwitterion.

Remember these

• The amino group activates the ring and directs to ortho and para.
• Bromine water gives 2,4,6-tribromoaniline. Acetylate first for the mono-bromo compound.
• Direct nitration gives para, meta and ortho products with oxidation. Protect the amine to get the para product.
• Sulphonation (heat at 453 to 473 K) gives sulphanilic acid, a zwitterion.
• Friedel-Crafts fails, since the amine binds the catalyst.

References

• NCERT Chemistry Class 12, the chapter Amines: electrophilic substitution in aniline.
• OpenStax Organic Chemistry, Chapter 24 Amines and Heterocycles, section on the reactions of arylamines.

Check yourself

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

1. A student adds bromine water to a few drops of aniline. A white precipitate forms. Name it.

2. Why does direct nitration of aniline give a meta product?

3. Give the three steps to make p-bromoaniline from aniline.

4. Why does aniline not give a Friedel-Crafts reaction with chloromethane and aluminium chloride?