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

WHAT: CuCN is the Sandmeyer reagent that installs $-CN$. WHY: Cu(I) hands one electron to the diazonium, it sheds $N_2$ as an aryl radical $C_6{H}_5\bullet$, and the copper delivers the $-CN$ group to that radical. Product: benzonitrile, $C_6{H}_{5}CN$. Yes, one molecule of $N_2$ leaves per molecule of product. $C_6{H}_5{\stackrel{+}{N}}_{2}C{l}^{-}\stackrel{\text{CuCN}}{\to }{C}_6{H}_{5}CN+N_2$

Temperature: $0\text{–}5{}^{\circ }\text{C}$. Why it fails above $5{}^{\circ }\text{C}$: the diazonium ion hydrolyses — water attacks, $N_2$ leaves, and you get phenol $C_6{H}_{5}OH$ instead of a usable salt. Cold keeps the fragile salt alive.

No $N_2$ is evolved. Coupling is an electrophilic aromatic substitution in which the diazonium acts as the electrophile with both nitrogen atoms intact; the terminal N forms a new C–N bond, giving the $-N=N-$ azo bridge. Losing $N_2$ would destroy the very chromophore we want.

Reagent: KI (just warm, no copper). Why not CuI-Sandmeyer? Iodide is special: the $I^{-}$ ion is polarisable and reactive enough to displace $N_2$ without a copper catalyst. So you simply warm the diazonium salt with KI. Why not HBF${}_4$? That is the Balz–Schiemann route for fluorine, not iodine. $C_6{H}_5{\stackrel{+}{N}}_{2}C{l}^{-}+KI\stackrel{\text{warm}}{\to }{C}_6{H}_{5}I+N_2+KCl$

Balz–Schiemann reaction.

1. Add fluoroboric acid $HB{F}_4$ → precipitate the diazonium fluoroborate $C_6{H}_5{\stackrel{+}{N}}_{2}B{F}_4^{-}$.
2. Heat the dry salt → it decomposes to fluorobenzene, $N_2$, and $B{F}_3$. (Recall $\Delta$ = heat.) $C_6{H}_5{\stackrel{+}{N}}_{2}B{F}_4^{-}\stackrel{\Delta }{\to }{C}_6{H}_{5}F+N_2+B{F}_3$ Why the detour? Direct $F^{-}$ substitution doesn't work well; the insoluble fluoroborate is stable enough to isolate and then thermolyse cleanly.

1. Diazotise: $C_6{H}_{5}N{H}_2\underset{0\text{–}{5}^{\circ }C}{\overset{NaN{O}_2/HCl}{\to }}{C}_6{H}_5{\stackrel{+}{N}}_{2}C{l}^{-}$.
2. Treat with hypophosphorous acid $H_{3}P{O}_2$ (or warm ethanol). The $-{\stackrel{+}{N}}_2$ is replaced by $-H$; $N_2$ leaves. $C_6{H}_5{\stackrel{+}{N}}_{2}C{l}^{-}+H_{3}P{O}_2+H_{2}O\to C_6{H}_6+N_2+H_{3}P{O}_3+HCl$ This is deamination — useful when you needed $-N{H}_2$ only temporarily as a directing group and now want it gone.

It will NOT work. Gattermann (Cu metal powder + HX) only delivers the halogen from HX — chlorine or bromine — never a nitrile. What's needed: the $-CN$ group comes from CuCN, which is the Sandmeyer reagent. $C_6{H}_5{\stackrel{+}{N}}_{2}C{l}^{-}\stackrel{\text{CuCN (Sandmeyer)}}{\to }{C}_6{H}_{5}CN+N_2$ Analysis of the difference: Sandmeyer uses a pre-formed soluble Cu(I) salt that carries the group to be installed (Cl, Br, or CN). Gattermann generates Cu(I) in situ from Cu⁰ + HX, so the only group available is the halide of that acid.

Aniline: the $-{\stackrel{+}{N}}_2$ nitrogen sits next to the benzene $\pi$-system. The ring's $\pi$-electrons delocalise onto the diazonium, spreading the positive charge over the whole ring. This resonance stabilisation lets the salt survive at $0\text{–}5{}^{\circ }\text{C}$. Ethylamine: an $s{p}^3$ alkyl carbon has no $\pi$-system to share the charge. The $C{H}_{3}C{H}_2{\stackrel{+}{N}}_2$ ion is a naked, high-energy cation → it instantly expels $N_2$, giving a carbocation that grabs water/eliminates. No stable salt exists. Conclusion: diazonium chemistry is essentially an aromatic phenomenon. (See Aromatic Amines — Aniline.)

Why strong acid kills it: aniline's ring is activated by its lone pair. In strong acid the $-N{H}_2$ is protonated to $-\stackrel{+}{N}{H}_3$, which is a deactivating, meta-directing group — the ring is now electron-poor and won't accept the weak diazonium electrophile. But you can't go strongly basic either: high $O{H}^{-}$ converts the diazonium into unreactive diazohydroxide/diazotate, destroying the electrophile. Correct condition: mildly acidic to neutral (a narrow window) so that free (unprotonated) aniline is present and the diazonium survives. (For phenol coupling the choice is mildly alkaline — see the parent note — because we want the more nucleophilic phenoxide $C_6{H}_5{O}^{-}$.)

The trick: use $-NHCOC{H}_3$/$-N{H}_2$'s para-directing power to place Br, then convert the nitrogen to $-OH$.

1. Protect & brominate: acetylate aniline to acetanilide (moderates reactivity), then $B{r}_2$ → $p$-bromoacetanilide (para, o/p-director, para less hindered). Hydrolyse back → $p$-bromoaniline.
2. Diazotise: $NaN{O}_2/HCl,0\text{–}5{}^{\circ }C$ → $p$-Br–$C_6{H}_4$–${\stackrel{+}{N}}_{2}C{l}^{-}$.
3. Hydrolyse (warm in dilute acid/water): the $-{\stackrel{+}{N}}_2$ is replaced by $-OH$, $N_2$ leaves → $p$-bromophenol. $p\text{-Br-}{C}_6{H}_4{\stackrel{+}{N}}_{2}C{l}^{-}\underset{\text{dil. acid}}{\overset{H_{2}O,\Delta }{\to }}p\text{-Br-}{C}_6{H}_4\text{OH}+N_2+HCl$ Why not brominate phenol directly? $-OH$ is so strongly activating that $B{r}_2$ gives 2,4,6-tribromophenol, not the mono product. Going through the milder amine intermediate gives clean mono-para substitution. (See Phenols — Reactions and Activating and Directing Groups.)

1. Tribrominate: aniline + excess $B{r}_2$ (water) → 2,4,6-tribromoaniline immediately (the $-N{H}_2$ is a powerful o/p-activator, so it lands Br at 2, 4, 6 in one step).
2. Diazotise: $NaN{O}_2/HCl,0\text{–}5{}^{\circ }C$ → 2,4,6-tribromobenzenediazonium salt.
3. Deaminate: treat with $H_{3}P{O}_2$ → replace $-{\stackrel{+}{N}}_2$ by $-H$. Result: the three Br's remain at positions 2, 4, 6 relative to where N was, which are exactly 1, 3, 5 relative to each other → 1,3,5-tribromobenzene. Why diazonium is essential: you can't put three meta-Br's on benzene by direct EAS (Br is o/p-directing). The $-N{H}_2$ acts as a removable o/p template, and deamination erases it afterward.

(a) Product & position: the diazonium is a weak electrophile and attacks the most electron-rich, least hindered site of $N,N$-dimethylaniline — the para position (the $-N(C{H}_3{)}_2$ is a strong o/p-activator, para is unhindered): $C_6{H}_5{\stackrel{+}{N}}_{2}C{l}^{-}+C_6{H}_{5}N(C{H}_3{)}_2\underset{\text{mild}}{\overset{\text{neutral/mild acid}}{\to }}{C}_6{H}_5\text{–}N=N\text{–}{C}_6{H}_4\text{–}N(C{H}_3{)}_2+HCl$ (b) Why coloured: the $-N=N-$ bridge joins the two rings into one long conjugated $\pi$-system. This narrows the HOMO→LUMO gap so it falls in the visible range; the molecule absorbs visible light and we see the complementary (orange) colour. (See Conjugation and Colour — Chromophores.) (c) Moles of $N_2$:

• Diazotisation of $0.10$ mol aniline → $0.10$ mol diazonium (no $N_2$ released here; the two N's stay in the salt).
• Coupling → keeps both N's as the azo bridge → releases $0$ mol $N_2$. Total $N_2$ released $=0.00$ mol. The dye traps all the nitrogen — that is the signature of coupling.

Diazotisation gives $0.20$ mol diazonium (no gas yet).

• Half (0.10 mol) → hydrolysis to $p$-cresol: each molecule loses one $N_2$ → $0.10$ mol $N_2$.
• Half (0.10 mol) → Sandmeyer with CuBr → $p$-bromotoluene: each loses one $N_2$ → $0.10$ mol $N_2$. Total $N_2=0.10+0.10=0.20$ mol. Contrast with L5·Q1: here both branches are substitutions that dump $N_2$; coupling would have released none. This is the quantitative face of "substitution loses $N_2$, coupling keeps it."

1. Nitrate toluene: $HN{O}_3/H_{2}S{O}_4$ → mainly $p$-nitrotoluene ($-C{H}_3$ is o/p-directing; para is less hindered). Fork: why nitrate, not iodinate directly? Direct ring iodination is sluggish/reversible; we route through nitrogen instead.
2. Reduce: $Sn/HCl$ (or $H_2$/cat.) → $p$-toluidine ($p$-CH${}_3$C${}_6$H${}_4$NH${}_2$).
3. Diazotise: $NaN{O}_2/HCl,0\text{–}5{}^{\circ }C$ → $p{\text{-CH}}_3{\text{C}}_6{\text{H}}_4{\stackrel{+}{N}}_{2}C{l}^{-}$. Fork: why cold? Above $5{}^{\circ }\text{C}$ it hydrolyses to $p$-cresol — the wrong product.
4. Add KI, warm: → 4-iodotoluene + $N_2$ + KCl. Fork: why KI, not CuI or HBF${}_4$? Iodide substitutes without copper; CuI is not the standard reagent and $HB{F}_4$ would give the fluoride. $p{\text{-CH}}_3{\text{C}}_6{\text{H}}_4{\stackrel{+}{N}}_{2}C{l}^{-}+KI\stackrel{\text{warm}}{\to }p{\text{-CH}}_3{\text{C}}_6{\text{H}}_{4}I+N_2+KCl$

Which single reaction of a diazonium salt does not release $N_2$? ::: Azo coupling — both N atoms stay as the –N=N– bridge.