Why is ${HCCl}_{3}$ $"HCCl"_3$ a stronger acid than ${HCF}_{3}$ $"HCF"_3$ ?

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Answer 1 · 2017-01-25T01:46:54

Here's my explanation.

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

$HA ⇌ H^{+} + A^{-}$ $"HA" ⇌ "H"^"+" + A^"-"$

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

We have the equilibria

${Cl}_{3} C-H ⇌ H^{+} + {Cl}_{3} {C:}^{-}$ $"Cl"_3"C-H" ⇌ "H"^"+" + "Cl"_3"C:"^"-"$

and

$F_{3} C-H ⇌ H^{+} + F_{3} {C:}^{-}$ $"F"_3"C-H" ⇌ "H"^"+" + "F"_3"C:"^"-"$

$F$ $"F"$ 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 $p K_{a}$ $"p"K_"a"$ of fluoroform is 30.5, while the $p K_{a}$ $"p"K_"a"$ of chloroform is 24.4.

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

The explanation is that $C$ $"C"$ and $F$ $"F"$ atoms are about the same size.

Thus, the lone pairs on $F$ $"F"$ are quite close to the lone pair on the carbanion.

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

In chloroform, the $Cl$ $"Cl"$ atoms can still stabilize by the inductive effect, but the $Cl$ $"Cl"$ atoms are much bigger.

The lone pairs on the $Cl$ $"Cl"$ are so far away that lone pair-lone pair electron repulsion is much less than in fluoroform.

Thus, the ${Cl}_{3} {C:}^{-}$ $"Cl"_3"C:"^"-"$ ion is more stable and ${HCCl}_{3}$ $"HCCl"_3$ is the stronger acid.

Why is "HCCl"_3HCCl3"HCCl"_3 a stronger acid than "HCF"_3HCF3"HCF"_3?