Why is #"HCCl"_3# a stronger acid than #"HCF"_3#?

Question

Brooklyn Barber · Accepted Answer

Here's my explanation.

By comparing the stabilities of the conjugate bases of acids, we can compare how acidic they are. #"HA" ⇌ "H"^"+" + A^"-"# The acidity will increase with the stability of the anion. The equilibria exist. #"Cl"_3"C-H" ⇌ "H"^"+" + "Cl"_3"C:"^"-"# additionally #"F"_3"C-H" ⇌ "H"^"+" + "F"_3"C:"^"-"# #"F"# is highly electronegative, so we would expect it to stabilize the negative charge on the carbanion and make fluoroform the stronger acid. It's not what we see at all! The #"p"K_"a"# of fluoroform is 30.5, while the #"p"K_"a"# of chloroform is 24.4. Six orders of magnitude stronger than fluoroform is chloroform! The explanation is that #"C"# and #"F"# atoms are about the same size. Thus, the lone pairs on #"F"# are quite close to the lone pair on the carbanion. The carbanion becomes unstable due to the cumulative lone pair-lone pair electron repulsions. In chloroform, the #"Cl"# atoms can still stabilize by the inductive effect, but the #"Cl"# atoms are much bigger. The lone pairs on the #"Cl"# are so far away that lone pair-lone pair electron repulsion is much less than in fluoroform. Thus, the #"Cl"_3"C:"^"-"# ion is more stable and #"HCCl"_3# is the stronger acid.

Ryan Andrews · Answer

undefined HCCl₃ is a stronger acid than HCF₃ because chlorine is more electronegative than fluorine. The greater electronegativity of chlorine results in a more stable conjugate base (CCl₃⁻) compared to the conjugate base of HCF₃ (CF₃⁻). Additionally, the larger size of chlorine compared to fluorine allows for better dispersal of negative charge in the conjugate base, further stabilizing it.