Explain the formation of (1,1,1-triphenylmethyl)methyl ether when a solution of 2,2,2-triphenylethanoic acid in concentrated sulfuric acid is poured in methanol.

I cannot figure out where is one carbon atom lost. Has it escaped as $\ce{CO2}$? Or is there something more strange happening.

I wouldn't call this reaction an esterification. There is no ester functional group in the final compound.

I guess there's some special role of the three phenyl groups but I'm unable to come up with any mechanism. I hope dehydration is involved as one of the reaction steps.

  • $\begingroup$ Similar thing as here $\endgroup$ – Mithoron Dec 24 '16 at 22:01
  • $\begingroup$ Note that the name “2,2,2-triphenylethanoic acid” is not wrong but the preferred IUPAC name is triphenylacetic acid. $\endgroup$ – user7951 Jan 22 '17 at 12:33

Disclaimer: The question has pretty much been answered in the comments. I'm just making it substantial.


The triphenylacetic acid is dehydrated in the presence of concentrated sulfuric acid to give an acylium ion, namely triphenylacetylium cation, which is stabilized by resonance.

Then, the triphenylacetylium cation undergoes decarbonylation (removal of carbon monoxide) to give triphenylmethyl cation, which is stabilized by the three phenyl groups.

The oxygen in methanol acts as a nucleophile to attack the carbocation, and is deprotonated to give our desired product triphenylmethyl methyl ether.

It is noted that this reaction should be done in a well-ventilated area because of the evolution of toxic carbon monoxide gas.

  • 2
    $\begingroup$ The answer is great, but it could benefit more from explaining why the regular esterification does not take place. Is it the minor product? Or does it not take place at all? why? $\endgroup$ – Gaurang Tandon Mar 15 '18 at 10:34

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