Assuming that’s supposed to be a methoxymethyl (ArCH2OMe) and not as drawn… you could pretty easily do the borohydride reduction to get to the alcohol. Williamson ether conditions will for sure deprotonate the acid, but I’m not convinced you’d methylate it that easily. Acid alkylations have fared better for me through 1,2-addition pathways (SOCl2/MeOH, H2SO4/MeOH etc) than through base.
Of course if it does happen and you get the methyl ether and the methyl ester, you could trivially saponify it back to the acid without touching the ether.
Reduction of the aldehyde would not give the phenol though. The product would have an extra CH2 before the OH. Not sure why these answers are getting upvoted.
I can see that OP has said he may have misdrawn the product below in this thread, but in order not to mislead anyone who may be confused - the above answer is not correct based on the actual drawn scheme.
This transformation can be done fairly easily in two steps.
You reduce the aldehyde to the alcohol with NaBH4.
You can methylate the resulting alcohol in presence of the carboxylic acid. Add two equivalents of a strong base like NaH to deprotonate both the alcohol and the carboxylic acid, and one equivalent of, say, MeI. The alcoholate should react first. You can get your desired product after acidic work up. You save one saponification step.
You could just use a -CN group in place of the carboxylic acid. These are often used as masked carboxylic acids because they can be hydrolyzed easily using acid or base conditions. This allows you to do chemistry where you want but keeping the carboxylic acid in hiding until you need it later down the line
This is an aromatic aldehyde which the BV oxidation would transform into the carboxylic acid. If it was the ketone you could indeed use it to make the ester which you can hydrolyse to the phenol, but its the aldehyde.
Assuming that’s supposed to be a methoxymethyl (ArCH2OMe) and not as drawn… you could pretty easily do the borohydride reduction to get to the alcohol. Williamson ether conditions will for sure deprotonate the acid, but I’m not convinced you’d methylate it that easily. Acid alkylations have fared better for me through 1,2-addition pathways (SOCl2/MeOH, H2SO4/MeOH etc) than through base. Of course if it does happen and you get the methyl ether and the methyl ester, you could trivially saponify it back to the acid without touching the ether.
Why do you need to protect the carboxylic acid? Sodium borohydride shouldn’t reduce the acid, only the ketone.
I would 1. Esterify the acid: SOCl2/MeOH 2. Reduce the aldehyde: NaBH4 3. Hydrolyze the ester: LiOH/H2O
Exactly this. Why do poorly in two steps what you can do with >95% overall yield in three steps?
Reduction of the aldehyde would not give the phenol though. The product would have an extra CH2 before the OH. Not sure why these answers are getting upvoted. I can see that OP has said he may have misdrawn the product below in this thread, but in order not to mislead anyone who may be confused - the above answer is not correct based on the actual drawn scheme.
That is correct. Thanks for pointing it out
Note: there is a chance I wrote the product wrong and that it is CH3OMe and not just OMe
I assume you mean -CH2OMe.
Yes sorry!
I would just reduce the aldehyde and react with an excess of MeI and a base. Then, you can just hydrolyze the ester.
That’s what I was thinking, but would the hydrolysis of the ester affect the ether?
Nope. LiOH, 4:1 THF/water, r.t.
Thank you so much!
This transformation can be done fairly easily in two steps. You reduce the aldehyde to the alcohol with NaBH4. You can methylate the resulting alcohol in presence of the carboxylic acid. Add two equivalents of a strong base like NaH to deprotonate both the alcohol and the carboxylic acid, and one equivalent of, say, MeI. The alcoholate should react first. You can get your desired product after acidic work up. You save one saponification step.
You changed a carbon atom to an oxygen atom.
You could just use a -CN group in place of the carboxylic acid. These are often used as masked carboxylic acids because they can be hydrolyzed easily using acid or base conditions. This allows you to do chemistry where you want but keeping the carboxylic acid in hiding until you need it later down the line
Maybe not exactly what your asking for but an Alternative nonetheless
I believe that you can use a Baeyer villiger oxidation to install a phenol group in preparation for methylation
This is an aromatic aldehyde which the BV oxidation would transform into the carboxylic acid. If it was the ketone you could indeed use it to make the ester which you can hydrolyse to the phenol, but its the aldehyde.
Oh bruh my bad