Making and replacing diazonium salts

A salt that is meant to lose nitrogen

Pour hydrochloric acid on aniline and cool the mixture in ice. Add a solution of sodium nitrite slowly. The oily aniline dissolves, and you get a clear, colourless solution. Nothing seems to have happened. But the solution holds a very useful salt, benzenediazonium chloride. Warm it a little, and gas bubbles out. The gas is nitrogen, and a new compound is left behind. Most of this lesson is about what that new compound can be.

Diazotisation

A diazonium salt has the group $\text{-}\text{N}_{2}^{+}$ on a carbon. A primary aromatic amine reacts with nitrous acid at 273 to 278 K to give one. This is diazotisation. We prepare the nitrous acid in the flask, from sodium nitrite and hydrochloric acid, because nitrous acid is unstable.

$\text{C}_{6}\text{H}_{5}\text{NH}_{2}\ +\ \text{NaNO}_{2}\ +\ 2\text{HCl}\ \rightarrow \ \text{C}_{6}\text{H}_{5}\text{N}_{2}^{+}\text{Cl}^{-}\ +\ \text{NaCl}\ +\ 2\text{H}_{2}\text{O}$

• The mixture is kept at 273 to 278 K (0 to 5 °C) in an ice bath. Once it warms above 278 K the diazonium salt reacts with water and gives phenol and nitrogen. Nitrous acid also breaks down when warm.
• Excess hydrochloric acid is used. The equation needs 2 mol of acid for each mole of amine, but in practice about 2.5 to 3 mol are used. The extra acid keeps unreacted aniline as its salt, so it cannot couple with the diazonium ion that has already formed.
• A drop of the mixture on starch-iodide paper turns blue-black when a slight excess of nitrous acid is present. This shows that all the amine has reacted.
• The solution is used at once. A dry diazonium salt can explode.

Use the default 9.3 g, which is 0.100 mol of aniline. You need 6.90 g of sodium nitrite. The acid comes to 0.250 mol, or 83 mL of 3 mol/L acid. Slide the mass up to 18.6 g and watch every number double.

Practise with new numbers

Why the salt is stable in the cold

The group $\text{-}\text{N}\equiv \text{N}^{+}$ lies in a straight line with the ring carbon. The positive charge is not all on the nitrogen. It spreads into the ring by resonance. This is why an arene diazonium salt can be kept in cold solution.

A diazonium salt from an aliphatic amine, $\text{R}\text{-}\text{N}_{2}^{+}$, has no ring to share the charge. It decomposes at once, even in the cold, and nitrogen leaves. That is why aliphatic amines give gas with nitrous acid, and why aromatic amines do not at first.

Benzenediazonium chloride is a colourless crystalline solid, soluble in water, stable in the cold and explosive when dry. Benzenediazonium fluoroborate, $\text{C}_{6}\text{H}_{5}\text{N}_{2}^{+}\text{BF}_{4}^{-}$, is different. It does not dissolve in water and is stable at room temperature.

Replacing the diazonium group

Nitrogen is an excellent leaving group. It leaves as a gas, and something else takes its place on the ring. These reactions let you introduce groups that cannot be added to a ring directly.

Halides and cyanide.

• Sandmeyer reaction. A copper(I) salt in the matching acid replaces the diazonium group. $\text{CuCl}/\text{HCl}$ gives chlorobenzene, $\text{CuBr}/\text{HBr}$ gives bromobenzene, and $\text{CuCN}/\text{KCN}$ gives benzonitrile. Copper(I) passes one electron to the diazonium ion. Nitrogen leaves, and the aryl radical takes the halide from copper.
• Gattermann reaction. Copper powder with $\text{HCl}$ or $\text{HBr}$ does the same job. The yield is usually lower than with a copper(I) salt.
• Iodide. Potassium iodide alone gives iodobenzene. No copper is needed: $\text{C}_{6}\text{H}_{5}\text{N}_{2}^{+}\text{Cl}^{-}\ +\ \text{KI}\ \rightarrow \ \text{C}_{6}\text{H}_{5}\text{I}\ +\ \text{N}_{2}\ +\ \text{KCl}$.
• Fluoride. Add fluoroboric acid, $\text{HBF}_{4}$. Insoluble $\text{C}_{6}\text{H}_{5}\text{N}_{2}^{+}\text{BF}_{4}^{-}$ separates. Dry it and heat it: $\text{C}_{6}\text{H}_{5}\text{N}_{2}^{+}\text{BF}_{4}^{-}\ \rightarrow \ \text{C}_{6}\text{H}_{5}\text{F}\ +\ \text{BF}_{3}\ +\ \text{N}_{2}$. This is the Balz-Schiemann reaction.
• Nitro. Heat the fluoroborate with sodium nitrite and copper. Nitrobenzene forms.

Hydroxyl. Warm the solution of the salt. Above 278 K it reacts with water: $\text{C}_{6}\text{H}_{5}\text{N}_{2}^{+}\text{Cl}^{-}\ +\ \text{H}_{2}\text{O}\ \rightarrow \ \text{C}_{6}\text{H}_{5}\text{OH}\ +\ \text{N}_{2}\ +\ \text{HCl}$.

Hydrogen. Hypophosphorous acid, $\text{H}_{3}\text{PO}_{2}$, or ethanol replaces the diazonium group by hydrogen:

$\text{C}_{6}\text{H}_{5}\text{N}_{2}^{+}\text{Cl}^{-}\ +\ \text{H}_{3}\text{PO}_{2}\ +\ \text{H}_{2}\text{O}\ \rightarrow \ \text{C}_{6}\text{H}_{6}\ +\ \text{N}_{2}\ +\ \text{H}_{3}\text{PO}_{3}\ +\ \text{HCl}$

$\text{C}_{6}\text{H}_{5}\text{N}_{2}^{+}\text{Cl}^{-}\ +\ \text{CH}_{3}\text{CH}_{2}\text{OH}\ \rightarrow \ \text{C}_{6}\text{H}_{6}\ +\ \text{N}_{2}\ +\ \text{CH}_{3}\text{CHO}\ +\ \text{HCl}$

Reactions that keep both nitrogen atoms. Mild reducing agents, such as tin(II) chloride with hydrochloric acid, reduce the salt to phenylhydrazine. Coupling with phenols and amines gives azo dyes. We take that up in the next lesson.

Try it in Python

The program lists each reagent with its product, and works out the mass of each product from 9.3 g of aniline at an overall yield of 70 percent.

Common mistakes

• Keeping the mixture warm. Above 278 K you start to get phenol.
• Using too little acid. The diazonium ion couples with the remaining aniline.
• Calling the reaction of potassium iodide a Sandmeyer reaction. It needs no copper.
• Thinking the fluoride comes from copper(I) fluoride. It comes from the fluoroborate.
• Expecting an aliphatic amine to give a stable diazonium salt. It decomposes at once.

Remember these

• Diazotisation: $\text{ArNH}_{2}\ +\ \text{NaNO}_{2}\ +\ 2\text{HCl}\ \rightarrow \ \text{ArN}_{2}^{+}\text{Cl}^{-}\ +\ \text{NaCl}\ +\ 2\text{H}_{2}\text{O}$ at 273 to 278 K.
• Sandmeyer: $\text{CuCl}/\text{HCl}$, $\text{CuBr}/\text{HBr}$, $\text{CuCN}/\text{KCN}$. Gattermann: copper powder. Iodide: $\text{KI}$. Fluoride: $\text{HBF}_{4}$ then heat.
• Warm water gives phenol. $\text{H}_{3}\text{PO}_{2}$ or ethanol gives the arene.
• Arene diazonium salts are stable in the cold, and aliphatic ones are not.

References

• NCERT Chemistry Class 12, the chapter Amines: diazonium salts, their preparation and chemical reactions.
• OpenStax Organic Chemistry, Chapter 24 Amines and Heterocycles, section on the reactions of arylamines (diazonium salts).

Check yourself

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

1. Why is the diazotisation of aniline carried out in an ice bath?

2. Give the reagent that converts benzenediazonium chloride into iodobenzene.

3. How is fluorobenzene made from benzenediazonium chloride?

4. Which reagent replaces the diazonium group by hydrogen?