Question

Rank these acids according to their expected pKa values.

ClCH₂COOH
BrCH₂COOH
FCH₂COOH
F₃CCOOH

Answer 1

We know that electronegativity of F > Cl > Br

F3CCOOH has 3 Fluorine atoms. Its conjugate base F3CCOO- is the most stable due to electron withdrawing by Fluorine (dispersion of -ve charge on carboxylate ion). This makes F3CCOOH most stable amongst all.

Now,

Among ClCH₂COOH
BrCH₂COOH
FCH₂COOH

F stabilizes the conjugate base of FCH2COOH (i.e FCH2COOH-) most amongst the three (since it is most electronegative) by dispersing the -ve charge on carboxylate ion. The highest stability of the conjufate base makes FCH2COOH most acidic among three.

Second most acidic among the three would be ClCH2COOH and the least acidic would be BrCH2COOH

Hence, the final acidity order would be

F3CCOOH > FCH2COOH > ClCH2COOH > BrCH2COOH

pKa is inversely proportional to the acidity (Higher the acidity, lower the pKa)

=> Order of pKa (High pKa to Low pKa)

BrCH2COOH > ClCH2COOH > FCH2COOH > F3CCOOH