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

1. Preparation — Diazotisation

$\text{Ar–NH}_2 + \text{NaNO}_2 + 2\,\text{HCl} \xrightarrow{0\text{–}5\,^\circ\text{C}} \text{Ar–}\overset{+}{N}\!\equiv\! N\ Cl^- + \text{NaCl} + 2\,\text{H}_2\text{O}$

WHY 0–5 °C? Aryldiazonium salts decompose above ~5 °C (they lose $N_2$ and react with water to give phenol). Keep it cold to keep it alive.

WHY does it work for aromatic amines but not aliphatic? The $-\overset{+}{N}_2$ on an alkyl chain has no ring to share its positive charge with, so it instantly spits out $N_2$ → carbocation → mess. On benzene, the lone pairs/π-system resonance-stabilise the diazonium cation, so it survives at low temperature.

Formula How HNO

$_2$ becomes the active species (derive the attacking agent) $\text{NaNO}_2 + \text{HCl} \rightarrow \text{HNO}_2 + \text{NaCl}$ $\text{HNO}_2 + \text{HCl} \rightarrow \text{H}_2\text{O} + \underbrace{\text{NO}^+}_{\text{nitrosonium}} + Cl^-$ The real electrophile is the ==nitrosonium ion $NO^+$ ==. It attacks the amine N, then a sequence of proton transfers and a dehydration yields the diazonium. $\text{Ar–NH}_2 \xrightarrow{NO^+} \text{Ar–NH–N=O} \rightleftharpoons \text{Ar–N=N–OH} \xrightarrow{-H_2O,\,+H^+} \text{Ar–}\overset{+}{N}\!\equiv\!N$

2. Substitution Reactions — replacing $-\overset{+}{N}_2$

The driving force everywhere: loss of $N_2$ (entropically + thermodynamically huge).

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

$\text{Ar–}\overset{+}{N}_2 Cl^- \xrightarrow{\;\text{CuCl}\;} \text{Ar–Cl} + N_2$ $\text{Ar–}\overset{+}{N}_2 Br^- \xrightarrow{\;\text{CuBr}\;} \text{Ar–Br} + N_2$ $\text{Ar–}\overset{+}{N}_2 Cl^- \xrightarrow{\;\text{CuCN}\;} \text{Ar–CN} + N_2$

HOW (mechanism, radical): Cu $^+$ transfers an electron to the diazonium → Ar–N=N• → loses $N_2$ → aryl radical Ar•. The Cu(II)–X then delivers X to the radical, regenerating Cu $^+$ . WHY Cu is needed: it lowers the barrier for forming the aryl radical.

2b. Gattermann reaction — Cu powder + HX (no special Cu salt)

$\text{Ar–}\overset{+}{N}_2 Cl^- \xrightarrow{\;\text{Cu/HCl}\;} \text{Ar–Cl} + N_2$

2c. Other useful substitutions (one table to rule them all)

Reagent	Product	Note
H $_2$ O, warm (>5 °C, dil. acid)	Ar–OH (phenol)	the unwanted side-reaction we avoid during prep
KI (just warm, no Cu)	Ar–I	iodide works without Cu!
HBF $_4$ , then heat (Balz–Schiemann)	Ar–F	fluoride needs the fluoroborate
H $_3$ PO $_2$ (hypophosphorous) / EtOH	Ar–H	replaces $-N_2$ with H (deamination)

3. Coupling Reactions — Azo Dyes

$\text{Ar–}\overset{+}{N}_2 + \text{H–}\!\underbrace{\text{Ar}'}_{\text{phenol/amine}} \rightarrow \text{Ar–N=N–Ar}' + H^+$

WHY no $N_2$ loss? The diazonium acts as the electrophile with both N atoms intact; the new C–N bond forms at the terminal nitrogen. The two retained $sp^2$ N's give the conjugated –N=N– chromophore.

WHY coloured? The extended conjugation Ar–N=N–Ar′ creates a long π-system; its HOMO→LUMO gap falls in the visible range, so it absorbs visible light → we see the complementary colour. These are azo dyes.

HOW — position & pH control:

With phenol: run in mildly alkaline (OH $^-$ makes phenoxide, more activating) → couples at para.
With aniline/amines: run in mildly acidic / neutral. Too much acid protonates the amine (deactivates ring); too basic kills the diazonium → narrow pH window.

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

Common mistake Steel-man the common errors

(1) "Sandmeyer also gives Ar–F and Ar–I." Feels right because they're halogens too. Fix: I $^-$ needs only KI (no Cu); F needs Balz–Schiemann (HBF $_4$ , heat). Sandmeyer = Cl, Br, CN specifically. (2) "Coupling releases N $_2$ like the other reactions." Feels right — every other diazonium reaction loses $N_2$ . Fix: Coupling is EAS; both N's are kept as the –N=N– bridge. That's the whole point of a dye. (3) "Diazotise aliphatic amines the same way." Feels right — it's still a primary amine. Fix: alkyldiazonium is unstable → instantly loses $N_2$ giving alcohols/alkenes; no stable salt. (4) "Warm it to speed up the prep." Feels right — heat speeds reactions. Fix: above 5 °C it hydrolyses to phenol. Keep 0–5 °C during preparation.

Recall Feynman: explain to a 12-year-old

Imagine the benzene ring is a backpack. The $-NH_2$ is a sticker that's hard to peel off. So we change it into a " $N_2$ rocket booster" ( $-N_2^+$ ). This booster wants to fly off as a puff of nitrogen gas. If we hand the backpack a piece of chlorine (with a copper helper), as the rocket flies off, chlorine sticks in its place. But if we instead bring another friendly ring nearby, the rocket doesn't fire — instead the two rings hold hands through the nitrogen, and that hand-holding chain is so special it shows up as a bright colour. That's how we make dyes!

Flashcards

What temperature range is required for diazotisation and why?

0–5 °C; above 5 °C the diazonium salt hydrolyses to phenol with loss of N₂.

What is the actual electrophile formed from HNO₂ + HCl in diazotisation?

The nitrosonium ion, NO⁺.

Why can't stable diazonium salts be made from primary aliphatic amines?

No ring to resonance-stabilise the −N₂⁺; it instantly loses N₂ to give carbocation products.

Sandmeyer reagents and products?

CuCl→Ar–Cl, CuBr→Ar–Br, CuCN→Ar–CN (cuprous salts as radical catalyst).

Gattermann reaction reagent?

Copper metal powder + HX (HCl/HBr) to give Ar–Cl/Ar–Br.

How do you put I, F, OH, and H onto the ring from a diazonium salt?

I: KI (no Cu); F: HBF₄ then heat (Balz–Schiemann); OH: warm with water; H: H₃PO₂ (hypophosphorous acid).

In azo coupling, is N₂ lost?

No — both nitrogens are retained as the –N=N– azo bridge.

Why are azo compounds coloured?

Extended conjugation (Ar–N=N–Ar′) gives a small HOMO–LUMO gap absorbing visible light.

pH conditions for coupling with phenol vs amine?

Phenol: mildly alkaline (forms phenoxide). Amine: mildly acidic/neutral. Coupling occurs at para.

What is the mechanistic type of azo coupling?

Electrophilic aromatic substitution, diazonium as the weak electrophile.

Connections

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

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, diazonium salt ka funda simple hai: aromatic amine (Ar–NH $_2$ ) ko thanda (0–5°C) rakh ke NaNO $_2$ + HCl daalo, to ban jaata hai Ar–N $_2^+$ Cl $^-$ . Yeh $-N_2^+$ group ek "rocket" jaisa hai — usko $N_2$ gas ban ke udd jaana hai (kyunki N $_2$ bahut stable hai). Isi udne ki tendency ki wajah se hum benzene ring pe almost koi bhi group laga sakte hain — jo seedha $-NH_2$ se possible nahi tha. Temperature 0–5°C zaroori hai, warna warm hone pe paani attack karke phenol bana dega.

Substitution reactions me yaad rakho: Sandmeyer me cuprous salt (CuCl, CuBr, CuCN) lagta hai aur Cl/Br/CN aata hai. Gattermann me bas copper powder + HCl/HBr — sasta version, halogen ke liye. Iodine ke liye sirf KI kaafi (Cu ki zaroorat nahi), fluorine ke liye HBF $_4$ phir heat (Balz–Schiemann), aur agar $-N_2^+$ ko H se replace karna ho to H $_3$ PO $_2$ . In sab reactions me $N_2$ gas nikal jaati hai — yahi driving force hai.

Coupling reaction thoda alag aur sabse important hai dyes ke liye. Yahan diazonium ek weak electrophile ki tarah phenol ya amine (electron-rich ring) pe attack karta hai, aur ban jaata hai Ar–N=N–Ar′ — yaani azo dye. Sabse bada point: coupling me $N_2$ nahi nikalta, dono nitrogen –N=N– bridge ki form me rehte hain. Yeh long conjugation visible light absorb karti hai isliye colour dikhta hai. Phenol ke saath mild base (NaOH) me, amine ke saath mild acid/neutral me, aur coupling hamesha para position pe hota hai. Bas yeh "N $_2$ stays vs N $_2$ goes" wala difference exam me confuse mat karna!

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

1. Preparation — Diazotisation

2. Substitution Reactions — replacing $-\overset{+}{N}_2$

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

2b. Gattermann reaction — Cu powder + HX (no special Cu salt)

2c. Other useful substitutions (one table to rule them all)

3. Coupling Reactions — Azo Dyes

Flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Nitrogen-Containing Compounds

Connections