Diazonium Salts and Other Nitrogen Compounds: Quick Revision

A brief note on Diazonium Salts and Other Nitrogen Compounds. Read it before a test. Try each question at the end before you open its answer.

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. Excess acid stops coupling with unreacted amine.
• Arene diazonium salts are stable in cold solution, since the ring shares the charge. Aliphatic ones lose nitrogen at once.

Replacing the diazonium group

• $\text{CuCl}/\text{HCl}$, $\text{CuBr}/\text{HBr}$, $\text{CuCN}/\text{KCN}$ (Sandmeyer). Copper powder with $\text{HCl}$ or $\text{HBr}$ (Gattermann).
• $\text{KI}$ gives the iodide, with no copper. $\text{HBF}_{4}$ then heat gives the fluoride.
• Warm water gives phenol. $\text{H}_{3}\text{PO}_{2}$ or ethanol gives the arene.
• Each mole of salt gives one mole of nitrogen gas.

Coupling

• A weak electrophile attacks the para position of a phenol (pH 9 to 10) or an aromatic amine (pH 4 to 5) and gives an azo dye, $\text{Ar}\text{-}\text{N}=\text{N}\text{-}\text{Ar}$.
• Methyl orange: diazotised sulphanilic acid with N,N-dimethylaniline. Red below about pH 3.1 and yellow above 4.4.
• Scarlet dye with alkaline 2-naphthol shows a primary aromatic amine.
• Planning: nitrate, reduce, diazotise, replace. Work backwards.

Cyanides, isocyanides and nitro compounds

• $\text{KCN}$ gives $\text{R}\text{-}\text{CN}$. $\text{AgCN}$ gives $\text{R}\text{-}\text{NC}$. $\text{AgNO}_{2}$ gives $\text{R}\text{-}\text{NO}_{2}$. $\text{KNO}_{2}$ gives $\text{R}\text{-}\text{ONO}$.
• Nitrile: hydrolysis to the acid, reduction to $\text{RCH}_{2}\text{NH}_{2}$. Isocyanide: acid hydrolysis to $\text{RNH}_{2}$ and $\text{HCOOH}$, no alkali hydrolysis, reduction to $\text{RNHCH}_{3}$.
• Nitrobenzene: aniline in acid (6 electrons), N-phenylhydroxylamine when neutral (4), hydrazobenzene in alkali (5).

A worked example

Find the mass of sodium nitrite for $4.65\ \text{g}$ of aniline.

1. Amount. $\dfrac{4.65}{93} = 0.0500\ \text{mol}$.
2. Nitrite. One mole for each mole of amine: $0.0500\ \text{mol}$.
3. Mass. $0.0500 \times 69 = 3.45\ \text{g}$.

Practice with new numbers

A program to try

The program lists each reagent with its product and the mass of the product from aniline.

Mistakes to avoid

• Warming the diazotisation. It gives phenol.
• Copper for the iodide. Potassium iodide alone is enough.
• Mixing up silver and potassium salts.
• Coupling at the wrong pH.
• Alkali hydrolysing an isocyanide. It does not.

Check yourself

1. Why is diazotisation done at 273 to 278 K?

2. What reagent turns the diazonium salt into benzene?

3. Which salt gives the isocyanide from an alkyl halide?

4. What does nitrobenzene give with zinc and ammonium chloride?