Azo coupling needs pH 9-10 for phenol but pH 4-5 for aniline, why? Rest of my diazonium table is done!!
Tanvi Deshmukh12th Sample
26 Sept 2026 Β· 1 min read
Hii everyone!! I finally finished my diazonium chart (colour coded, obviously) and I have everything except the reasoning behind one thing, which is eating my brain.
My chart so far, benzene diazonium chloride made from aniline + + HCl at 273β278 K:
- with CuCl/HCl gives chlorobenzene (Sandmeyer), same with CuBr/HBr, CuCN/KCN
- with Cu powder + HCl gives chlorobenzene too (Gattermann), usually lower yield
- with KI gives iodobenzene
- with then heat gives fluorobenzene
- with and water gives benzene
- with phenol gives p-hydroxyazobenzene (orange), with aniline gives p-aminoazobenzene (yellow)
My teacher says coupling with phenol is done in mildly basic medium and with aniline in mildly acidic medium. I can copy it but I don't understand it. If the diazonium ion is the electrophile, why does pH matter? And why does the product always seem to be para? Also, why do we need 273β278 K, is it just because things decompose?
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[+][β]Nisha Verma12thSampleΒ· 24d ago
Phenol part: in mildly basic medium phenol is converted partly into phenoxide ion, , which is a far stronger activator than neutral βOH. The diazonium ion is a pretty weak electrophile, so the ring needs to be very electron-rich to attack it. Phenoxide does that job.
Why not very strongly basic though? Because in strong base the diazonium ion itself gets converted into a diazotate/diazohydroxide which is no longer an electrophile, and coupling stops. So you want just enough base to make some phenoxide and not enough to kill the diazonium ion.
[+][β]Siddharth RaoDropperSampleΒ· 24d ago
Aniline part is the mirror image. You need the free β (lone pair available) to activate the ring. In strongly acidic medium aniline sits as anilinium, , which is deactivating and no longer reactive for this, so coupling is poor. In basic medium again the diazonium is destroyed. Mildly acidic (pH 4β5) keeps enough free amine and still keeps diazonium alive.
So the pattern is: activating group must be "on" (phenoxide / free amine) and the diazonium ion must be "on", and the pH window is where both are true.
[+][β]Aman YadavDropperSampleΒ· 24d ago
Your chart has Sandmeyer and Gattermann swapped. Sandmeyer is Cu powder. Gattermann is Cu(I) chloride. Fix it.
[+][β]Rahul DubeyDropperSampleΒ· 23d ago
Aman, other way round actually, tanvi's chart is right. Sandmeyer uses copper(I) salts (CuCl/HCl, CuBr/HBr, CuCN/KCN). Gattermann uses copper powder with HCl or HBr, and gives lower yield. A trick for remembering: S for Sandmeyer, "Salt" of copper; G for Gattermann, "Grains" of copper powder.
Also for iodine you don't need copper at all, just KI. And fluorine goes through the fluoroborate salt (heat it), because direct substitution doesn't work well. These are the lines NEET picks as "which reagent is NOT used" type options.
[+][β]Aman YadavDropperSampleΒ· 23d ago
ok my bad, thanks. sandmeyer = salt, noted.
[+][β]Mehak AroraDropperSampleΒ· 23d ago
The temperature bit: aromatic diazonium salts are stable only in that cold window (273β278 K) in solution, above that they start losing and you get phenol. They are more stable than aliphatic ones because the positive charge gets delocalised into the benzene ring by resonance. Aliphatic diazonium salts decompose instantly.
And the colour is because of the azo group linking two aromatic rings, a long conjugated system which absorbs visible light. That's why these are dyes.
[+][β]Tanvi Deshmukh12thSampleΒ· 23d agoAuthor
omg thank you all, this is going straight into the chart in green ink!! And is para just because ortho is crowded?
[+][β]Rahul DubeyDropperSampleΒ· 23d ago
Yes mostly steric, the diazonium ion is big. Para is the preferred position, ortho only if para is already blocked.
