Question

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

Answer 1

Here's my explanation.

Explanation:

We can compare the acidities of acids by examining the stabilities of their conjugate bases.

`"HA" ⇌ "H"^"+" + A^"-"`

The more stable the anion, the greater the acidity will be.

We have the equilibria

`"Cl"_3"C-H" ⇌ "H"^"+" + "Cl"_3"C:"^"-"`

and

`"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.

This is NOT what we observe!

The `"p"K_"a"` of fluoroform is 30.5, while the `"p"K_"a"` of chloroform is 24.4.

Chloroform is a stronger acid them fluoroform by six orders of magnitude!

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 cumulative lone pair-lone pair electron repulsions destabilize the carbanion.

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.