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

False. Substitutions (Sandmeyer, Gattermann, hydrolysis, deamination) all expel $N_2$, but azo coupling keeps both nitrogens as the –N=N– bridge — that retained $s{p}^2$ nitrogen chain is the chromophore of the dye. (See the fork figure at the top.)

False. Above ~5 °C the aryldiazonium ion hydrolyses to phenol with loss of $N_2$; speed does not save it because the decomposition is the competing reaction, so you must hold 0–5 °C throughout preparation.

True. $HN{O}_2+HCl$ gives $H_{2}O+N{O}^{+}+C{l}^{-}$; the ==$N{O}^{+}$== is what attacks the amine nitrogen (see the mechanism figure below). $HN{O}_2$ is only the reservoir that generates it in situ.

False. An alkyldiazonium has no aromatic ring to resonance-delocalise the positive charge, so it instantly ejects $N_2$ and collapses to a carbocation (giving alcohols/alkenes). Only aryl amines give isolable salts.

False. The roles are reversed: the diazonium ion is a weak electrophile and the electron-rich phenol/amine is the nucleophile in an EAS on the activated ring.

False. It is a radical pathway: Cu(I) donates one electron, $N_2$ leaves to give an aryl radical, and Cu(II)–X delivers the halogen (see the single-electron-transfer cycle figure below, which shows why copper forces the radical route). See Nucleophilic Substitution vs Radical Substitution for the general radical-vs-ionic contrast.

False. Only Sandmeyer with CuCN gives Ar–CN; Gattermann uses Cu metal + HX and installs only halogens (Cl, Br).

False. Colour comes from the extended conjugation Ar–N=N–Ar′ shrinking the HOMO–LUMO gap into the visible range (see Conjugation and Colour — Chromophores); an isolated –N=N– without the aromatic rings would not absorb visible light strongly.

The temperature is wrong. At 40 °C the salt hydrolyses to phenol; the correct condition is 0–5 °C to keep the diazonium alive.

There is no CuF Sandmeyer step. Fluorine requires the Balz–Schiemann route: precipitate the aryldiazonium fluoroborate ($\text{Ar–}{\stackrel{+}{N}}_{2}B{F}_4^{-}$) with $HB{F}_4$, dry it, then heat so it decomposes to Ar–F + $N_2$ + $B{F}_3$ (see the Balz–Schiemann figure below).

Iodide needs no copper at all — just warming the diazonium salt with KI delivers Ar–I. Reaching for CuI mistakenly imposes the Sandmeyer template where it is not needed.

Too basic destroys the diazonium ion — hydroxide adds to give the diazohydroxide/diazotate (defined above), which is not electrophilic. You want only mildly alkaline conditions — enough to form the more nucleophilic phenoxide without killing the electrophile (see the pH map below).

Strong acid protonates the amine to $-\stackrel{+}{N}{H}_3$, a deactivating, meta-directing group that switches off the ring. Amine coupling needs mildly acidic/neutral pH — a narrow window.

Wrong product. Hypophosphorous acid replaces $-{\stackrel{+}{N}}_2$ with –H (deamination); it is water/warm dilute acid that gives –OH.

The –OH is activating and o/p-directing; coupling occurs at the para position (the bulky diazonium avoids the crowded ortho spots). See Activating and Directing Groups.

A secondary aromatic amine reacts with $N{O}^{+}$ to give an N-nitrosamine (Ar–NR–N=O) instead, and it has no N–H left to lose water and form the –N≡N. Only a primary –NH${}_2$ can dehydrate to –N${}_2^{+}$.

When it leaves it becomes ==$N_2$ gas== — a neutral, triple-bonded, extremely stable molecule that escapes the mixture, giving a huge entropic and thermodynamic driving force.

Direct swapping of $-N{H}_2$ for halide/–CN/–OH is not feasible (the amine is a poor leaving group). Diazonium acts as a universal adapter: one easy leaving group that hands the ring position over to almost any incoming group.

Cu(I) is a single-electron donor; it lowers the barrier to form the aryl radical from the diazonium (see the copper-cycle figure). Without it the radical does not form efficiently and the halide is not delivered.

The diazonium ion is only a weak electrophile, so it needs a strongly activated ring (phenol/amine) whose high electron density makes the EAS attack fast enough to occur.

Base converts phenol to the more nucleophilic phenoxide; but that same base would be too harsh for amines, and acid is needed there only up to the point that does not protonate the amine — each partner has its own optimal pH (see the pH map).

You get hydrolysis to phenol (Ar–OH) with $N_2$ loss. This is a deliberate synthesis of phenols, but the same reaction is the failure mode you avoid during preparation.

Sandmeyer uses a pre-made soluble Cu(I) salt (CuCl/CuBr); Gattermann uses Cu metal powder + HX that makes Cu(I) in situ. Products (Ar–Cl/Ar–Br) are the same; Gattermann is cheaper but lower-yielding.

Deamination — the ring position becomes –H. This is used to remove an amino group entirely after it has done its directing job during synthesis.

It can self-couple (homocouple): one diazonium's terminal N acts as electrophile on the electron-rich ring of another diazonium/its diazo-form, building an azobenzene ($\text{Ar–N=N–Ar}$). This is a common decomposition/side pathway, which is another reason coupling reactions are run under carefully controlled pH with a chosen partner.

Too acidic: the amine coupling partner gets protonated and deactivated. Too basic: the diazonium ion is destroyed into the unreactive diazohydroxide/diazotate. The usable window sits between these two failures.

The ring is too electron-poor; it cannot act as the nucleophile, so coupling as the partner fails. (Such a nitro-substituted diazonium, however, is a stronger electrophile and couples more readily onto other activated rings.)

No. Benzene is not activated enough for such a weak electrophile; without a strong donor (–OH, –NH${}_2$, –NR${}_2$) present the coupling does not proceed.

They form, but the electron-poor ring-nitrogen drains density from the diazonium, so a pyridyldiazonium is even more reactive/unstable than the benzene one — it decomposes more readily and, when it survives, is a stronger coupling electrophile. The ring's own basicity (protonation at ring N) also shifts the useful pH window, so conditions differ from aniline chemistry.