MP Board · Class 12 · Chemistry · AminesDiscuss the preparation, physical properties, and chemical properties of benzenediazonium chloride, emphasizing its importance in synthetic organic chemistry (Sandmeyer reaction, Gattermann reaction, and Coupling reactions).
1. Preparation (Diazotization)\nBenzenediazonium chloride is prepared by the reaction of aniline with nitrous acid (generated in situ from sodium nitrite and hydrochloric acid) at a low temperature of 0-5°C (273-278 K).
$$C_6H_5NH_2 + NaNO_2 + 2HCl \xrightarrow{273-278 K} C_6H_5N_2^+Cl^- + NaCl + 2H_2O$$\nThis conversion of primary aromatic amines into diazonium salts is known as diazotization.
2. Physical Properties
- It is a colourless crystalline solid.
- It is readily soluble in water and is stable in cold conditions.
- It decomposes easily in the dry state and can explode when heated.
3. Chemical Properties and Synthetic Applications\nBenzenediazonium chloride is an extremely useful intermediate in organic synthesis due to the easy displacement of the diazo group ($-N_2^+$) by various nucleophiles.
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Sandmeyer Reaction: Replacement of the diazonium group by $-Cl$, $-Br$, or $-CN$ using cuprous halides or cuprous cyanide dissolved in the corresponding acid. $$C_6H_5N_2^+Cl^- \xrightarrow{Cu_2Cl_2/HCl} C_6H_5Cl + N_2$$
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Gattermann Reaction: Similar to Sandmeyer, but using copper powder in the presence of the corresponding halogen acid or acid. $$C_6H_5N_2^+Cl^- \xrightarrow{Cu/HCl} C_6H_5Cl + N_2 + HCl$$
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Replacement by Iodine and Fluorine: Treatment with potassium iodide gives iodobenzene ($C_6H_5I$). Treatment with fluoroboric acid followed by heating gives fluorobenzene (Balz-Schiemann reaction).
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Coupling Reactions: Benzenediazonium chloride reacts with electron-rich aromatic compounds like phenol and aniline to form brightly coloured azo dyes. In these reactions, the diazo group is retained. For example, coupling with phenol in alkaline medium gives p-hydroxyazobenzene (orange dye). $$C_6H_5N_2^+Cl^- + C_6H_5OH \xrightarrow{OH^-} C_6H_5-N=N-C_6H_5-OH + HCl$$