# Selective alkylation of carboxylic acid versus phenol

Starting from the ketone shown, I plan to use a haloform reaction to transform the acetyl group into a carboxylic acid.

Following that, there are two alkylations to be carried out: one esterification on the carboxylic acid (using propargyl bromide) and one etherification on the alcohol (using 6-bromohexan-1-ol). In what order should these be performed?

Using the haloform reaction to convert the methyl ketone 1 to the corresponding carboxylic acid 2 is a neat idea. According to an old review by R. C. Fuson and B. A. Bull, published in Chem. Rev., 1934, 15, 275-309 (DOI), this seems to work for 4-hydroxyacetophenone. Your starting material 1 is a phenyl homolog of that.

In 2, you have two acidic protons ($\ce{-COOH}$ and $\ce{-OH}$). Based on our knowledge and by comparison with 4-hydroxybiphenyl ($\mathrm pK_\mathrm a = 9.55$), benzoic acid ($\mathrm pK_\mathrm a = 4.2$) and 2-phenylbenzoic acid ($\mathrm pK_\mathrm a = 3.46$) (all data are taken from the Rubber Book), you know that

• the carboxylic acid is more acidic than the phenol
• the difference is acidity is quite large

Potassium carbonate ($\ce{K2CO3}$) in a polar aprotic solvent, such as N,N-dimethylformamide (DMF) should be sufficient to deprotonate both centres.

Back to your question on the order of events:

I would add 1 equivalent $\ce{K2CO3}$ of to a solution of 2, stir at room temperature for a while and then add the propargyl bromide, assuming that the large differences in acidities will make sure that the carboxylic acid is deprotonated, while the phenol is not.

Monitoring the reaction by TLC will show whether the esterification is quantitative.

If this is the case, I would dare to run a one-pot reaction, add another equivalent of $\ce{K2CO3}$, stir again and then add the bromohexanol to form the ether.

It might be a good idea to try this first with very small amounts, just enough to check via TLC.