Diazonium salts — preparation, Sandmeyer, Gattermann, coupling reactions (azo dyes)

4.4.2 · Chemistry › Nitrogen-Containing Compounds

1. Preparation — Diazotisation

${\text{Ar–NH}}_2+{\text{NaNO}}_2+2\text{HCl}\stackrel{0\text{–}5{}^{\circ }\text{C}}{\to }\text{Ar–}\stackrel{+}{N}\equiv NC{l}^{-}+\text{NaCl}+2{\text{H}}_2\text{O}$

WHY 0–5 °C? Aryldiazonium salts ~5 °C se upar decompose ho jaate hain (woh $N_2$ kho dete hain aur paani ke saath react karke phenol dete hain). Ise zinda rakhne ke liye thanda rakho.

WHY yeh aromatic amines ke liye kaam karta hai par aliphatic ke liye nahi? Ek alkyl chain pe $-{\stackrel{+}{N}}_2$ ke paas koi ring nahi hoti jo uski positive charge share kare, isliye woh turant $N_2$ ugal deta hai → carbocation → mess. Benzene pe, lone pairs/π-system resonance-stabilise karta hai diazonium cation ko, isliye woh low temperature pe survive karta hai.

2. Substitution Reactions — $-{\stackrel{+}{N}}_2$ ko replace karna

Har jagah driving force: $N_2$ ka loss (entropically + thermodynamically bahut bada).

2a. Sandmeyer reaction — Cu(I) salt chahiye

$\text{Ar–}{\stackrel{+}{N}}_{2}C{l}^{-}\stackrel{\text{CuCl}}{\to }\text{Ar–Cl}+N_2$ $\text{Ar–}{\stackrel{+}{N}}_{2}B{r}^{-}\stackrel{\text{CuBr}}{\to }\text{Ar–Br}+N_2$ $\text{Ar–}{\stackrel{+}{N}}_{2}C{l}^{-}\stackrel{\text{CuCN}}{\to }\text{Ar–CN}+N_2$

HOW (mechanism, radical): Cu${}^{+}$ diazonium ko ek electron transfer karta hai → Ar–N=N• → $N_2$ kho deta hai → aryl radical Ar•. Cu(II)–X phir radical ko X deliver karta hai, Cu${}^{+}$ regenerate karta hai. WHY Cu chahiye: yeh aryl radical banane ki barrier ko kam karta hai.

2b. Gattermann reaction — Cu powder + HX (koi special Cu salt nahi)

$\text{Ar–}{\stackrel{+}{N}}_{2}C{l}^{-}\stackrel{\text{Cu/HCl}}{\to }\text{Ar–Cl}+N_2$

2c. Doosre useful substitutions (ek table sab pe raj karne ke liye)

3. Coupling Reactions — Azo Dyes

$\text{Ar–}{\stackrel{+}{N}}_2+\text{H–}\underset{\text{phenol/amine}}{\underbrace{{\text{Ar}}^{\prime }}}\to {\text{Ar–N=N–Ar}}^{\prime }+H^{+}$

WHY $N_2$ loss nahi? Diazonium electrophile ki tarah kaam karta hai dono N atoms intact ke saath; naya C–N bond terminal nitrogen pe banta hai. Dono retained $s{p}^2$ N's conjugated –N=N– chromophore dete hain.

WHY coloured? Extended conjugation Ar–N=N–Ar′ ek lamba π-system create karta hai; iska HOMO→LUMO gap visible range mein aata hai, isliye yeh visible light absorb karta hai → hum complementary colour dekhte hain. Yeh azo dyes hain.

HOW — position & pH control:

• Phenol ke saath: mildly alkaline mein chalao (OH${}^{-}$ phenoxide banata hai, zyada activating) → para pe couple hota hai.
• Aniline/amines ke saath: mildly acidic / neutral mein chalao. Zyada acid amine ko protonate karta hai (ring deactivate hoti hai); zyada basic diazonium ko khatam karta hai → narrow pH window.

Flashcards

Connections

• Aromatic Amines — Aniline (starting material)
• Electrophilic Aromatic Substitution (coupling ka mechanism)
• Activating and Directing Groups (para kyun)
• Phenols — Reactions (coupling partner & hydrolysis product)
• Nucleophilic Substitution vs Radical Substitution (Sandmeyer radical hai)
• Conjugation and Colour — Chromophores (azo dyes)

Concept Map