Azo coupling and planning a synthesis

Joining two rings

In the last lesson the diazonium ion lost its nitrogen. Here it keeps both nitrogen atoms. The ion is a weak electrophile. It does not attack benzene. It attacks rings that are rich in electrons, such as phenols and aromatic amines.

The ion goes to the para position of the partner. If para is blocked, it goes to the ortho position. The product has the group $\text{-}\text{N}=\text{N}\text{-}$, the azo group, between two rings. This is called a coupling reaction.

• With phenol, in a mildly alkaline solution, the product is p-hydroxyazobenzene, an orange dye.
• With aniline, in a weakly acidic solution, the product is p-aminoazobenzene, a yellow dye.

Why the pH is chosen with care

Two things must be right at the same time. The diazonium ion must be there, and the partner must be rich in electrons.

• Phenol is a weak acid, with $\text{p}K_a$ about 10. In neutral water it is mostly the plain phenol, which is only moderately reactive. As the pH rises, more of it becomes the phenoxide ion, which is far more reactive. So the reaction is run at pH 9 to 10. If the solution were very alkaline, the diazonium ion itself would be turned into an unreactive diazotate, and coupling would stop.
• Aniline couples as the free amine. The anilinium ion is deactivated. In strong acid, almost all of the aniline is the ion. As the pH rises towards 4 to 5, more becomes the free amine. Above that, the diazonium ion again begins to be lost.

Set the pKa to 10.0 and the pH to 9.5. About a quarter of the phenol is the phenoxide ion. At pH 8 it is only 1 percent. Now set the pKa to 4.6 and see how much more free amine you have at pH 4.5.

Practise with new numbers

Why azo compounds are coloured

The azo group joins the two rings into one long system of alternating double bonds. A long system absorbs light of a longer wavelength, in the visible range. The compound therefore shows a colour. Groups such as $\text{-}\text{OH}$, $\text{-}\text{NH}_{2}$ and $\text{-}\text{N}(\text{CH}_{3})_{2}$ on the rings deepen the colour. A group such as the sulphonate, $\text{-}\text{SO}_{3}^{-}$, makes the dye soluble in water, so that it sticks to cloth. Azo compounds make up the largest class of synthetic dyes.

Methyl orange is made by diazotising sulphanilic acid and coupling the salt with N,N-dimethylaniline. It is also a pH indicator: red below about pH 3.1 and yellow above about pH 4.4. Its $\text{p}K_a$ is close to 3.5.

A test for primary aromatic amines. Diazotise the amine in the cold. Add the solution to an alkaline solution of 2-naphthol. A scarlet-red azo dye shows that the sample was a primary aromatic amine. An aliphatic amine gives gas instead.

Planning a synthesis with diazonium salts

The $-\text{NH}_2$ group is a very handy temporary handle. It is put on by nitration and reduction. It directs substitution to the ortho and para positions. And through the diazonium salt it can be swapped for nearly anything, or removed. Here are five routes you should know.

1. Iodobenzene from benzene. Nitrate ($\text{HNO}_{3}/\text{H}_{2}\text{SO}_{4}$), reduce ($\text{Sn}/\text{HCl}$), diazotise ($\text{NaNO}_{2}/\text{HCl}$, 273 to 278 K), then add $\text{KI}$. Direct iodination of benzene does not work well.
2. Benzoic acid from aniline. Diazotise, then $\text{CuCN}/\text{KCN}$ to give benzonitrile, and boil with dilute acid.
3. Phenol from nitrobenzene. Reduce, diazotise, and warm with water.
4. Fluorobenzene from aniline. Diazotise, add $\text{HBF}_{4}$, and heat the solid.
5. 1,3,5-Tribromobenzene from aniline. The three bromine atoms are meta to one another, but direct bromination of benzene never gives that pattern. Brominate aniline with bromine water to get 2,4,6-tribromoaniline, diazotise it, and replace the diazonium group by hydrogen with $\text{H}_{3}\text{PO}_{2}$. The amino group steered the bromine atoms to the right places, and then it left.

How to plan. Work backwards from the product. Ask which group can come from the diazonium ion. Then ask how to make that diazonium ion: from an amine, which comes from a nitro compound, which comes from nitration. Write each step with its reagent.

Try it in Python

The program stores the reactions as a map and searches for the shortest route between two compounds.

Common mistakes

• Coupling in strong acid or strong alkali. The partner or the diazonium ion is lost.
• Forgetting that the dye forms at the para position. Ortho is used only if para is blocked.
• Starting the plan from the wrong end. Work backwards from the product.
• Calling the azo group $\text{-}\text{N}_{2}^{+}$. The azo group is $\text{-}\text{N}=\text{N}\text{-}$ between two rings.
• Expecting an aliphatic amine to give a dye test. It gives nitrogen gas instead.

Remember these

• Phenols couple at pH 9 to 10, and aromatic amines at pH 4 to 5. Both give the para product.
• The azo group $\text{-}\text{N}=\text{N}\text{-}$ links two rings and gives colour.
• A scarlet dye with alkaline 2-naphthol shows a primary aromatic amine.
• The diazonium ion lets you introduce F, I, CN, OH and H, and lets you remove an amino group.

References

• NCERT Chemistry Class 12, the chapter Amines: coupling reactions and the importance of diazonium salts in synthesis.
• 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. Name the product of the coupling of benzenediazonium chloride with phenol.

2. Why is phenol coupled in a mildly alkaline solution?

3. Why does a strongly acidic solution stop the coupling of aniline?

4. Give the steps to make 1,3,5-tribromobenzene from aniline.