I encountered a question in my textbook which asked the following:
At first, i thought that i can approach it in the same way i sort the effect of EDG and EWG on the acidity of phenols, that is, methoxide makes phenols a weaker acid, while halides make them stronger. But I was surprised to learn that its not the case for carboxylic acids. After conducting few research, I learnt that for methoxides, there is some sort of competition between the inductive and resonance effect, whereby resonance is what reduces the acidity by stabilising the conjugate base, and induction does the opposite. I also learnt that when both effects act at the same time, resonance always wins, that is, it's effect is stronger. I thought that in carboxylic acids, the methoxide cannot contribute to the resonance of the carboxylate ion, and its lone pair remains on the methoxide oxygen, and accordingly, the 2-methoxypropanoic acid is more acidic than propanoic acid due to the inductive effect of oxygen (high E.N.).
Sounds fine so far, right? Not until when i decided to test this hypothesis by comparing the pKa values of the following:
A website i found offered pKa values for them, which appeared to be:
Compound 1 (2-chloropropanoic acid) = 2.96
Compound 2 (2-floropropanoic acid) = 2.68
Compound 3 (2-methoxypropanoic acid) = 3.59
Compound 4 (2-aminopropanoic acid) = 2.34
And those numbers do not confirm my hypothesis at all. I expected the order of acidity to be: 2-floropropanoic acid>2-methoxypropanoic acid>2-chloropropanoic acid>2-aminopropanoic acid. Due to their respective E.N. values.
How can i establish a trend of acidity here? Please help me understand this.
