krit.club logo

Organic Compounds Containing Nitrogen - Diazonium salts: Preparation and chemical reactions

Grade 12ICSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

•

Diazonium salts are organic compounds with the general formula ArN2+X−ArN_2^+X^-, where ArAr is an aryl group and X−X^- represents anions like Cl−Cl^-, Br−Br^-, HSO4−HSO_4^-, or BF4−BF_4^-.

•

Preparation (Diazotization): Primary aromatic amines react with nitrous acid (HNO2HNO_2), produced in situ from NaNO2NaNO_2 and HClHCl, at low temperatures (273−278273 - 278 KK). The reaction is C6H5NH2+NaNO2+2HCl→273−278KC6H5N2+Cl−+NaCl+2H2OC_6H_5NH_2 + NaNO_2 + 2HCl \xrightarrow{273-278 K} C_6H_5N_2^+Cl^- + NaCl + 2H_2O.

•

Stability: Benzenediazonium salts are stable for a short time in cold aqueous solution due to the resonance stabilization of the N2+N_2^+ group with the benzene ring.

•

Sandmeyer Reaction: Replacement of the diazo group by ClCl, BrBr, or CNCN using cuprous salts (Cu2Cl2Cu_2Cl_2, Cu2Br2Cu_2Br_2, or CuCNCuCN).

•

Gattermann Reaction: A modification of the Sandmeyer reaction where copper powder and the corresponding halogen acid (HClHCl or HBrHBr) are used instead of cuprous salts.

•

Balz-Schiemann Reaction: Introduction of a fluorine atom by treating the diazonium salt with fluoroboric acid (HBF4HBF_4) to form ArN2+BF4−ArN_2^+BF_4^-, followed by heating to produce ArFArF.

•

Replacement by Hydroxyl group: Warming the diazonium salt solution to 283283 KK or above results in hydrolysis to form phenols: C6H5N2+Cl−+H2O→C6H5OH+N2+HClC_6H_5N_2^+Cl^- + H_2O \rightarrow C_6H_5OH + N_2 + HCl.

•

Coupling Reactions: Diazonium salts react with phenols or aromatic amines to form highly colored azo compounds (Ar−N=N−Ar′Ar-N=N-Ar'). This is an electrophilic aromatic substitution reaction.

📐Formulae

C6H5NH2+NaNO2+2HCl→0−5∘C[C6H5N2]+Cl−+NaCl+2H2OC_6H_5NH_2 + NaNO_2 + 2HCl \xrightarrow{0-5^{\circ}C} [C_6H_5N_2]^+Cl^- + NaCl + 2H_2O

ArN2+Cl−→CuCl/HClArCl+N2ArN_2^+Cl^- \xrightarrow{CuCl/HCl} ArCl + N_2

ArN2+Cl−→CuBr/HBrArBr+N2ArN_2^+Cl^- \xrightarrow{CuBr/HBr} ArBr + N_2

ArN2+Cl−→CuCN/KCNArCN+N2ArN_2^+Cl^- \xrightarrow{CuCN/KCN} ArCN + N_2

ArN2+Cl−+KI→ArI+KCl+N2ArN_2^+Cl^- + KI \rightarrow ArI + KCl + N_2

ArN2+Cl−→HBF4ArN2+BF4−→ΔArF+BF3+N2ArN_2^+Cl^- \xrightarrow{HBF_4} ArN_2^+BF_4^- \xrightarrow{\Delta} ArF + BF_3 + N_2

C6H5N2+Cl−+H3PO2+H2O→C6H6+N2+H3PO3+HClC_6H_5N_2^+Cl^- + H_3PO_2 + H_2O \rightarrow C_6H_6 + N_2 + H_3PO_3 + HCl

C6H5N2+Cl−+C6H5OH→OH−p−HOC6H4N=NC6H5(p−Hydroxyazobenzene)+Cl−+H2OC_6H_5N_2^+Cl^- + C_6H_5OH \xrightarrow{OH^-} p-HOC_6H_4N=NC_6H_5 (p-Hydroxyazobenzene) + Cl^- + H_2O

💡Examples

Problem 1:

How is p−Aminoazobenzenep-Aminoazobenzene prepared from Benzenediazonium chloride?

Solution:

C6H5N2+Cl−+C6H5NH2→pH 4−5p−NH2C6H4−N=N−C6H5+HClC_6H_5N_2^+Cl^- + C_6H_5NH_2 \xrightarrow{pH \, 4-5} p-NH_2C_6H_4-N=N-C_6H_5 + HCl

Explanation:

This is a coupling reaction where Benzenediazonium chloride acts as an electrophile and reacts with the para-position of aniline (a nucleophilic aromatic ring) in a mildly acidic medium to form a yellow-colored azo dye.

Problem 2:

Convert Aniline to Fluorobenzene.

Solution:

C6H5NH2→NaNO2/HCl,273KC6H5N2+Cl−→HBF4C6H5N2+BF4−→ΔC6H5F+BF3+N2C_6H_5NH_2 \xrightarrow{NaNO_2/HCl, 273K} C_6H_5N_2^+Cl^- \xrightarrow{HBF_4} C_6H_5N_2^+BF_4^- \xrightarrow{\Delta} C_6H_5F + BF_3 + N_2

Explanation:

Aniline is first converted to benzenediazonium chloride via diazotization. It is then reacted with fluoroboric acid to precipitate benzenediazonium fluoroborate, which on thermal decomposition yields fluorobenzene (Balz-Schiemann reaction).

Problem 3:

Identify the product formed when Benzenediazonium chloride is treated with H3PO2H_3PO_2 (Hypophosphorous acid).

Solution:

C6H5N2+Cl−+H3PO2+H2O→C6H6+N2+H3PO3+HClC_6H_5N_2^+Cl^- + H_3PO_2 + H_2O \rightarrow C_6H_6 + N_2 + H_3PO_3 + HCl

Explanation:

Hypophosphorous acid acts as a reducing agent, reducing the diazonium salt to benzene while itself getting oxidized to phosphorous acid (H3PO3H_3PO_3).