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The first step in this synthesis is a Kolbe Schmitt process, but why is two equivalents of base needed? Also, what is the purpose of the two equivalents of methyl iodide? I assumed that there will be the protection of the phenol (to create a methyl ether), which would justify the two equivalents of base while converting the carboxylate into a methyl ester. However, the base is shown in separate steps, which makes me doubt if base would be available in the second step. Also, another problem is that if you follow the sequence (2 Grignard reagents used to convert the ester to a tertiary alcohol, followed by dehydration, followed by hydrogenation of the double bond just formed, followed by a Friedel Crafts acylation), you get to compound R, which only has 13 carbons, when if you had methylated the phenol, you would have 14. Does this mean that the hydrogenation somehow deprotected the methyl ether? Even more confusing, there is methylation occurring from R to S, which tells me that the phenol was protected at some point. How is this possible?

EDIT: Here is the structure of the final product. The problem was about derivatives of natural products from Salvia plants, but I could not find a total synthesis of this compound anywhere. Also, there is no easy name for it so I couldn't search it well. If you can find the source, please let us know. Note that there are other steps in the synthesis that I didn't include in this question. enter image description here

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    $\begingroup$ I wonder why these textbook syntheses tend to be terribly impractical... $\endgroup$
    – Mithoron
    Commented May 14, 2020 at 22:17
  • $\begingroup$ You have a sharp eye. There is no need for 2 eq. of CH3I if you are not forming o-methoxy methyl benzoate. I feel that R is C14H18O4. If not it is the phenolic cmpd C13H16O4. $\endgroup$
    – user55119
    Commented May 14, 2020 at 22:42
  • $\begingroup$ The main thing that I don't like is there is a clear methylation step from R to S and you are definitely not methylation the carboxylic acid, meaning there much be a phenol in R. But how could of the initial ether have been converted back to the phenol? $\endgroup$
    – sweetandtangy
    Commented May 15, 2020 at 0:33

1 Answer 1

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Collecting my comment thoughts together, there is an early demethylation step in this synthesis. Structure 1 is the result of carboxylation while methylation provides ester N. Friedel-Crafts acylations of anisoles at the ortho-position can undergo demethylation via aluminum chelation. Magnesium chelate 2 is a candidate for demethylation by iodide ion. Phenol O undergoes thermal dehydration catalyzed by the acidic phenolic hydroxyl group. Steps P --> Q are straightforward. Step R --> S methylates the phenol hydroxyl and excess KOH insures that any ester that is formed is saponified. Clemmensen reduction of S and remethylation afford T.

Addendum: The OP's comment reveals that T is a carboxylic acid, not a methyl ester. Presumably the Clemmensen reduction effected some demethylation which is the reason for the last methylation step.

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    $\begingroup$ Thank you for your help. The only problem i see is from S to T. The formula says there are 14 carbons but the methyl ester would give 15 carbons. I'm not sure what the methylation after the Clemmensen is supposed to be but you don't add any carbons. Also, the next step in the sequence after T is a SOCl2 step then AlCl3, meaning that a carboxylic acid would be converted to an acid chloride then intramolecular FC acylation, pointing to there being an acid in T. Also, the problem stated that T had acidic character. Maybe methylation is to replace any demethylated methyl ether from Clemmensen? $\endgroup$
    – sweetandtangy
    Commented May 15, 2020 at 18:18
  • $\begingroup$ I think your right about S --> T. Too many carbons in T. Of course if I had known that T was a carboxylic acid as you did, I would have come up with your solution. I'll change the structure for T. BTW: I have done Clemmensen reductions on anisole type cmpds without demethylation. $\endgroup$
    – user55119
    Commented May 15, 2020 at 19:03
  • $\begingroup$ What was the source of this problem? $\endgroup$
    – user55119
    Commented May 15, 2020 at 19:50
  • $\begingroup$ There are two unexpected demethylations in this sequence. I would like to see the original literature it is based on and read the rationale for them. $\endgroup$
    – Waylander
    Commented May 15, 2020 at 21:01
  • $\begingroup$ @Waylander: Agreed. The Grignard demethylation is reasonable given the chelation and the presence of iodide. The Clemmensen less so. I'll do a Chem Abst search. $\endgroup$
    – user55119
    Commented May 15, 2020 at 22:00

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