The Gattermann-Koch Reaction involves the formation of an aryl aldehyde using carbon monoxide and hydrochloric acid as reactants. The reaction is catalyzed by aluminium trichloride in the presence of trace amounts of cuprous chloride.

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What I don't understand about this reaction is that according to Wikipedia along with other online texts I've used, the Gattermann-Koch reaction is inapplicable to phenol substrates. The drawback has been left unexplained by all my resources (online and textbooks) and even my professor was oblivious to it. I am unable to understand why the reaction doesn't apply to phenol and what exact side reactions might occur.

Additionally, why is the explanation so obscure?

[EDIT] I found the explanation. The Gattermann-Koch Reaction is not applicable to phenol and phenol ethers because they "could not be successfully formylated at atmospheric pressure in benzene as a solvent." "This failure to react was attributed to the insolubility of the cuprous chloride in the reaction mixture."

Can someone please explain this in simple terms?

  • $\begingroup$ This question might be of interest to you as it is closely related. $\endgroup$
    – bon
    May 24, 2015 at 18:09
  • $\begingroup$ You should try to consult it from books $\endgroup$
    – user143337
    Feb 7 at 5:15

3 Answers 3


You have outlined that the formyl chloride is generated as the reactive, electrophilic intermediate under these conditions.

You are planning to perform an electrophilic aromatic substitution, but in a phenol, there's high electron density on the $\ce{O}$ atom. Isn't it conceivable that your main reaction product is phenylformate here, resulting from $\ce{O}$-acylation?

  • 1
    $\begingroup$ I didn't consider that, but then why would that only apply for phenol substrates? Wouldn't amines and ethers also be reactive in the same manner? $\endgroup$
    – tkhanna42
    May 24, 2015 at 7:55
  • $\begingroup$ @tennispro1213 Primary and secondary amines will undergo N-acylation as well, Ethers, such as anisole, on the other hand, can be formylated in a Gattermann-Koch reaction. $\endgroup$ May 24, 2015 at 8:07
  • $\begingroup$ can we simply say that the Gattermann-Koch Reaction is generally not used in compounds containing strongly basic groups? $\endgroup$
    – tkhanna42
    May 24, 2015 at 8:23
  • $\begingroup$ @tennispro1213 We could say that the Gattermann-Koch or the Vilsmeier-Haack reaction can lead to heteroatom acylation. In the case of phenols, we'd use the Reimer-Tiemann or the Duff reaction instead. $\endgroup$ May 24, 2015 at 9:46
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    $\begingroup$ How sure can we be that we actually have formyl chloride as a reactive intermediate? I could imagine to have only protonated carbonmonoxide, according to $$\ce{CO + HCl + AlCl3 <=> H+CO + [AlCl4]^-}$$. Also cuprous chloride seems to play an important role in the mechanism. $\endgroup$ May 26, 2015 at 3:24

Phenol can act as Lewis base and form adduct with AlCl3, using a lone pair on the O atom, hence deactivating the ring for further reaction.

  • 1
    $\begingroup$ Welcome to Chemistry.SE! Take the tour to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. If you have a NEW question, please ask it by clicking the Ask Question button. If you have sufficient reputation, you may upvote the question. Alternatively, "star" it as a favourite and you will be notified of any new answers. $\endgroup$ Nov 13, 2015 at 8:53

I suspect that, in the presence of the Lewis Acid catalyst, even if a formyl ester is formed, it will rearrange the the acylbenzene, which is irreversible. This may have consequences for the o,p-product ratio, And as far as the cuprous chloride, use aluminum trichloride. And you could make the Scylla bromide first, with Formica acid an bromine. Much more reactive will likely form the sculpture ester first, in catalyze. Then catalyst to rearrange it

  • $\begingroup$ Are you simply trying to put obscure words in last sentences? I know what Scylla is and it's not some wacky mythology chemistry here... $\endgroup$
    – Mithoron
    May 3, 2017 at 21:22
  • 2
    $\begingroup$ I have to agree, this looks like an autocorrect nightmare. Can you polish it up a bit? $\endgroup$
    – jonsca
    May 3, 2017 at 23:29

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