Arrange the following molecules in order of increasing pKa:
CH3CH2COOH, CH3CHFCOOH, CH3OCH2COOH, CH3CHBrCOOH
A) CH3CHFCOOH, CH3OCH2COOH, CH3CHBrCOOH, CH3CH2COOH
B) CH3CHBrCOOH, CH3OCH2COOH, CH3CH2COOH, CH3CHFCOOH
C) CH3CHFCOOH, CH3CH2COOH, CH3CHBrCOOH, CH3OCH2COOH
D) CH3CH2COOH, CH3OCH2COOH, CH3CHFCOOH, CH3CHBrCOOH
Increasing pKa means decreasing acid strength. A low pKa, just like a low pH, indicates a stronger acid. Here, we need to arrange the molecules from the strongest acid to the weakest.
An acid is stronger if the conjugate base is stabilized, and one form of stabilization is inductive stabilization. That is, if an electron withdrawing group pulls electron density off of a carbon, that carbon becomes slightly positive. That slight positive charge can then stabilize the negative charge of the conjugate base.
Inductive effects increase with increasing electronegativity. Because fluorine is the most electronegative atom, it has the biggest inductive effect and will most help stabilize the conjugate acid. Thus CH3CHFCOOH is the strongest acid and must be first in our list. That narrows us down to (A) and (C).
Propanoic acid with no inductive stabilization would be the weakest acid here, with the highest pKa. So it must be the last item on the list. That tells us choice (A) is the right answer.