Diazonium salts

Making the ion. Aniline is treated with sodium nitrite and excess hydrochloric acid, kept below $5\,^\circ C$:

where: $NaNO_2 + HCl$ generates nitrous acid ($HNO_2$) in situ, which is the true diazotising agent; the second mole of $HCl$ supplies the chloride counter-ion and keeps the medium acidic. The aryl diazonium ion is unusually stable for a diazonium species because the ring's $\pi$ system delocalises the positive charge, but "stable" here only means stable in cold solution — above about $10\,^\circ C$ it decomposes, often hydrolysing to phenol.

Two families of reaction. Everything the diazonium ion does afterwards falls into exactly two categories: (i) the $-N_2^+$ group leaves as $N_2$ gas and something else takes its place on the ring, or (ii) the whole $-N_2^+$ group survives and bonds intact to another aromatic ring. The first family is a set of substitution reactions; the second, coupling, is discussed below with its own displayed equation.

Substitution — replacement of $-N_2^+$. The choice of reagent fixes the product and gives the reaction its name:

Sandmeyer and Gattermann both give $Ar\text{-}Cl$ or $Ar\text{-}Br$; the difference is only the copper source — a dissolved cuprous salt for Sandmeyer, freshly precipitated copper powder for Gattermann — and only Sandmeyer's cuprous cyanide route extends to $-CN$. The Balz–Schiemann route to $Ar\text{-}F$ is two steps, not one: the diazonium chloride is first converted to the tetrafluoroborate salt, $Ar\text{-}N_2^+BF_4^-$, which is isolated and dried before being heated to release $Ar\text{-}F$, $BF_3$, and $N_2$. No electrophilic method installs $-F$, $-CN$, or removes $-NH_2$ cleanly on a benzene ring; diazonium chemistry is the only route to each.

The chemistry hinges on one fact: $N_2$ is an exceptionally stable, unreactive gas, so any process that can release it is thermodynamically downhill. In the substitution reactions, $-N_2^+$ leaves as $N_2\uparrow$ while a nucleophile or radical fragment from the reagent (a halide ion, cyanide, hydroxide, hydride equivalent) takes its place on the ring carbon. The catalytic copper in Sandmeyer's and Gattermann's reactions works by cycling between $Cu(I)$ and $Cu(II)$, generating an aryl radical that then grabs a halogen atom from the copper halide — this radical pathway is why the reaction needs a copper catalyst at all, unlike the simple ionic substitution with $KI$, where iodide is nucleophilic enough to displace $N_2$ directly. The Balz–Schiemann reaction needs isolation of the solid tetrafluoroborate first because $BF_4^-$ is a poor nucleophile in solution; only on dry heating does the salt decompose cleanly to $Ar\text{-}F$, avoiding the side products a wet ionic mechanism would give. Coupling reactions behave completely differently because nothing leaves. The diazonium ion, though only a weak electrophile, is reactive enough to attack a ring that has been made unusually electron-rich by a strong activating group — phenol's $-OH$ or aniline's $-NH_2$ — in a classic electrophilic aromatic substitution. Because $-OH$ and $-NH_2$ are both powerful ortho/para directors and the para position is normally the less hindered of the two, coupling lands predominantly para (ortho if para is blocked). The product retains every atom of the diazonium ion, joined through the $-N=N-$ (azo) linkage:

where: the mild base converts some phenol to the more nucleophilic phenoxide ion, speeding the reaction without decomposing the diazonium salt (strong base would). Coupling with aniline instead needs near-neutral or mildly acidic conditions, because excess acid would protonate the amine's lone pair and switch off its activating power. The extended conjugation through the $-N=N-$ bridge, linking two aromatic rings' $\pi$ systems, is what gives azo compounds their strong colour and makes them the basis of azo dyes.

Problem. Starting from aniline, propose routes to (a) benzonitrile and (b) 4-hydroxyazobenzene, giving reagents, and state which of the two involves loss of nitrogen gas.

Step 1. Both routes share the same first step: diazotise aniline below $5\,^\circ C$. $C_6H_5NH_2 + NaNO_2 + 2HCl \xrightarrow{0\text{-}5\,^\circ C} C_6H_5N_2^+Cl^- + NaCl + 2H_2O$.

Step 2. For benzonitrile, treat the diazonium salt with cuprous cyanide (Sandmeyer conditions): $C_6H_5N_2^+Cl^- \xrightarrow{CuCN/KCN} C_6H_5CN + N_2\uparrow$. The $-CN$ group replaces $-N_2^+$, and nitrogen gas is lost — this is a substitution reaction, and it is the standard synthesis of benzonitrile since benzene has no direct electrophilic route to a nitrile group.

Step 3. For 4-hydroxyazobenzene, instead couple the same diazonium salt with phenol under mild base: $C_6H_5N_2^+ + C_6H_5OH \xrightarrow{\text{mild base}} C_6H_5\text{-}N=N\text{-}C_6H_4\text{-}OH$ (para product, since $-OH$ directs ortho/para and para is less hindered). Here the diazonium group is not lost — no $N_2$ escapes — the whole ion bonds intact to the phenol ring through the new $-N=N-$ linkage.

Step 4. Contrast the two: benzonitrile formation is a substitution (Sandmeyer, $-N_2^+$ replaced by $-CN$, $N_2$ evolved), while 4-hydroxyazobenzene formation is a coupling (the diazonium ion itself becomes part of the azo product, no gas evolved). Answer: benzonitrile via $CuCN/KCN$ with $N_2$ loss; 4-hydroxyazobenzene via coupling with phenol in mild base, with the diazonium group fully retained in the product.

• Mixing up Sandmeyer's reaction (cuprous salt in solution: $CuCl$, $CuBr$, $CuCN$) with Gattermann's (copper powder plus the halogen acid) — both give the same halide products, but only Sandmeyer's extends to $-CN$.
• Writing the Balz–Schiemann product straight from the chloride salt; the tetrafluoroborate must be isolated and dried first, then heated, to get $Ar\text{-}F$.
• Forgetting the temperature window: diazonium salts are only stable around $0$-$5\,^\circ C$; warming the solution hydrolyses it to phenol even with no coupling partner present.
• Assuming coupling needs strong acid or strong base — phenol coupling wants mild base (to form phenoxide without decomposing the salt), amine coupling wants near-neutral conditions (excess acid deactivates the amine).

Ask the tutor: want to work through a full multi-step synthesis from benzene to a target molecule using the diazonium handle at the key step?