Benzonitrile is a benzene ring that carries a CN group, and no electrophilic aromatic substitution can make it. The reactions in Electrophilic aromatic substitution can add a halogen, a nitro group, a sulfonic acid, an alkyl group or an acyl group, but no practical electrophile delivers a CN to benzene. The same is true for an OH and for an F. Iodine goes on only when an oxidant is added along with the I2. And no substitution can take a group off a ring once it is there.
Diazonium chemistry gets around every one of these limits through a single intermediate, the aryl diazonium salt, Ar–N2+ X−. Ar stands for an aryl group, a benzene ring bonded through one of its carbons. X− is the negative ion that balances the charge, usually chloride. The N2+ group can be swapped for CN, OH, F, I, Cl or Br, or for a plain hydrogen.
Three steps to the salt
Start from benzene. Nitration with HNO3 and H2SO4 gives nitrobenzene. Reduction with Fe and HCl, followed by NaOH, gives aniline. Diazotization with NaNO2 and HCl at 0–5 °C then turns the NH2 group into an N2+ group, and the product is benzenediazonium chloride.
You have met all three steps before. Nitration is in Electrophilic aromatic substitution. The reduction is in Making amines, where Sn with HCl, or H2 over Pd, does the same job as Fe with HCl. Diazotization is in Amine reactions, which describes its mechanism in words. Here it is drawn.
The N2+ group sits where nitration put the nitro group. So the directing effects of that first nitration decide where every product below carries its new group.
The salt also has to be made cold and used straight away, because it decomposes above about 5 °C. Some of the reactions that use it are then warmed on purpose. Keep the two temperatures apart: the cold belongs to making the salt, and any heat belongs to using it.
Why N2+ is such a good leaving group
Warm a solution of benzenediazonium chloride in water and it fizzes. The gas is nitrogen, N2, released from the diazonium group. N2 is one of the most stable molecules there is, and as a gas it bubbles out of the solution and never comes back. That makes N2+ the best leaving group you will meet (Leaving groups).
When N2 leaves on its own, it takes both electrons of the C–N bond with it. The ring carbon left behind has a positive charge and an empty orbital: it is an aryl cation. The empty orbital is an sp2 orbital that points straight out from the ring, in the ring plane. The ring’s π electrons sit in p orbitals above and below that plane, at 90° to the empty orbital, so they cannot share the charge. That makes an aryl cation very unstable (Carbocations).
This is why an aryl halide does not ionize. A halide ion does not leave well enough to form an aryl cation. N2 does. Of the reactions in this course, only the loss of N2 from a diazonium salt forms an aryl cation. Water or fluoride can then bond to it. The copper reactions reach the same carbon by a different path, shown below.
The substitution menu
Add a cold solution of benzenediazonium chloride to copper(I) cyanide, CuCN, and warm it gently. Nitrogen bubbles off, and benzonitrile is left: the CN has taken the place of the N2+. Other reagents put other groups on the same carbon.
| Reagent | Replaces N2+ with | Name |
|---|---|---|
| CuCl | –Cl | Sandmeyer |
| CuBr | –Br | Sandmeyer |
| CuCN | –CN | Sandmeyer |
| KI | –I | — |
| HBF4, then heat | –F | Schiemann |
| H2O/H2SO4, warm | –OH | — |
| H3PO2 | –H | — |
Three of these groups are ones electrophilic substitution cannot put on a ring at all: CN, OH and F. Electrophilic substitution can add iodine only with an oxidant; KI here needs none. The last row is different. It removes the group entirely, and a later section shows why that is so useful.
What the copper does
The three reactions that use a copper(I) salt are the Sandmeyer reactions, and the copper takes part in each one. Take CuBr. First, the Cu(I) hands the diazonium ion one electron. The ion loses N2 and becomes a phenyl radical, a ring carbon with one unpaired electron (radicals are in Radical halogenation). The copper, now Cu(II), picks up a bromide ion from the solution.
Second, the phenyl radical takes a bromine atom from the Cu(II). Bromobenzene forms, and the copper is Cu(I) again. The same two steps deliver Cl from CuCl and CN from CuCN.
Because the copper ends where it started, the Sandmeyer reaction is described as a copper redox cycle: Cu(I) is oxidized to Cu(II) when it gives up the electron, and reduced back to Cu(I) when it hands over the bromine atom. No aryl cation forms along the way.
The hydrolysis and the Schiemann reaction do go through the aryl cation, which is why both need heat. In the hydrolysis, water bonds to the cation’s empty orbital and then loses H+, giving a phenol. In the Schiemann reaction, HBF4 first turns the salt into the diazonium tetrafluoroborate, Ar–N2+ BF4−, a salt stable enough to collect as a solid. Heating it releases N2, and fluoride from the BF4− bonds to the cation.
KI needs no copper. Iodide gives up an electron more easily than chloride or bromide does, so it can start the radical steps by itself, and the reaction runs at room temperature.
Aryl nitriles, and where they lead
CuCN is the usual way to put a nitrile on a benzene ring. Nitriles showed that an SN2 cannot do it, because a ring carbon cannot be attacked from behind. The Sandmeyer route starts from an aryl amine instead. The amine sits where nitration put the nitro group.
Step 1. NaNO2, HCl, 0–5 °C. The diazonium salt, with its N2+ para to the methyl group.
Step 2. CuCN. 4-Methylbenzonitrile. The nitrile carbon is a new carbon, brought in by the cyanide.
Step 3. H3O+, heat. The nitrile hydrolyzes through the amide to 4-methylbenzoic acid, exactly as in Nitriles. LiAlH4 and then water would give 4-methylbenzylamine instead, with the nitrile carbon turned into a CH2.
The move to remember: once a nitrile is on the ring, everything the Nitriles chapter does with a nitrile is available. An aryl amine is three steps from an aryl carboxylic acid: diazotize, add CuCN, then hydrolyze.
Other routes to an aryl nitrile exist. Heating an aryl bromide with CuCN at high temperature swaps the Br for CN, and a palladium-catalyzed coupling (the family in Organolithiums, Cuprates & Cross-Coupling) can join an aryl halide to a cyanide salt. Both start from an aryl halide.
Removing a group is a synthetic tool
Suppose you need 3,5-dibromotoluene, a toluene with both bromines meta to the methyl group. Brominating toluene cannot give it, because a methyl group directs ortho and para. Start from 4-methylaniline instead. Its NH2 group is a far stronger director than the methyl, and the carbon para to it already carries the methyl, so bromine water puts both bromines ortho to the NH2. Those two carbons are meta to the methyl. Diazotize, add H3PO2, and the NH2 is replaced by a hydrogen.
That is what the H3PO2 row is for. Replacing N2+ with H deletes the whole substituent, which sounds like undoing your own work. But while the NH2 group was on the ring, it was a powerful ortho/para director. So you can put an amino group on a ring just to steer the next substitution, and then take it off. A group used this way is a removable directing group.
Why not just brominate three times? Bromine is an ortho/para director. The second bromine goes ortho or para to the first, and the third does the same. The route ends at 1,2,4-tribromobenzene, and the 1,3,5 pattern is out of reach.
Step 1. HNO3/H2SO4. Nitrobenzene.
Step 2. Fe/HCl, then NaOH. Aniline. The ring now carries the strongest ortho/para director available.
Step 3. Br2 in water, no catalyst. 2,4,6-Tribromoaniline. Aniline is so activated that all three positions react at once, without a Lewis acid, and the product precipitates. Phenols showed the same runaway bromination as a nuisance. Here it is exactly what you want.
Step 4. NaNO2/HCl, 0–5 °C. The diazonium salt. Keep it cold.
Step 5. H3PO2. The nitrogen is replaced by H. The bromines were on carbons 2, 4 and 6, each meta to the other two. With the NH2 gone, the ring is numbered again from a bromine, and the product is 1,3,5-tribromobenzene.
The move to generalize: the NH2 group was never part of the target. When a substitution pattern looks forbidden, ask which temporary director would have allowed it.
An amine is a director, not a blocker. A blocker only holds a position so the next group cannot go there. The usual one is the reversible SO3H group from Ortho/meta/para directing effects. An NH2 group cannot just hold a position: it is too strong a director, and it sends the next group ortho or para to itself. In the figure, SO3H fills the para position, so the nitro group goes ortho to the methyl.
Azo coupling, and why dyes are colored
Add a cold solution of benzenediazonium chloride to phenol in dilute NaOH, and an orange solid forms. It is 4-(phenylazo)phenol: two benzene rings joined through an N=N unit. No N2 is lost this time. The diazonium ion keeps both of its nitrogens and acts as an electrophile.
It is a weak electrophile. It cannot attack benzene, but it does react with a strongly activated ring, such as a phenoxide ion or an aromatic amine. The reaction is called azo coupling, and its product is an azo compound, Ar–N=N–Ar′, where Ar′ is the second ring. Its mechanism is the electrophilic aromatic substitution you already know.
Coupling goes para to the OH or amino group. If the para carbon is already taken, it goes ortho. If the para carbon and both ortho carbons are taken, it does not go at all.
A phenol is coupled in mildly basic solution, pH 8–10, because the phenoxide ion is nucleophilic enough to react and the neutral phenol is not (Phenols). Above about pH 10 the coupling fails for another reason. Hydroxide adds to the diazonium ion and turns it into a diazotate, Ar–N=N–O−, which is no longer an electrophile.
An aromatic amine such as N,N-dimethylaniline is coupled in mildly acidic solution, pH 4–7. Some acid keeps the diazonium salt stable in solution. Too much acid protonates the amine nitrogen. The protonated amine, an anilinium ion (for N,N-dimethylaniline, ArNH(CH3)2+), has no lone pair to give the ring and deactivates it instead.
The same pH logic explains the acid in the hydrolysis row of the menu. Warm the salt in basic water and the phenol you are making becomes phenoxide, which couples with the diazonium salt still in the flask, so you get an azo dye instead of a phenol. In acid the phenol stays neutral and leaves the salt alone.
The two rings and the N=N bridge form one long conjugated π system. A long π system has a small HOMO–LUMO gap, so it absorbs visible light (UV-Vis spectroscopy). That is why azo compounds are strongly colored, and they were the foundation of the synthetic dye industry. Methyl orange, the acid–base indicator, is one: it is made by coupling diazotized sulfanilic acid with N,N-dimethylaniline.
What carries forward
Nitrate, reduce, diazotize cold, then substitute. The diazonium group leaves as N2, the best leaving group on a ring. That makes it the route to OH, CN and F, and an easy route to I. An amine put on just to direct, then deleted, is a removable directing group, and coupling a diazonium ion onto an activated ring makes a dye. When a substitution pattern looks impossible by directing effects alone, check whether a diazonium salt reaches it.