# Why is CF3COOH exceptionally acidic?

The pKa values of almost all carboxylic acids lie much above 0. But this is violated by trifluoroacetic acid(-0.25). How can this be justified?

• There's nothing strange in this, normal consequence of inductive effect. – Mithoron Feb 19 '17 at 15:51
• – user7951 Jan 10 '19 at 18:25

Try drawing the structure of the conjugate base $\ce{CF_3COO^-}$. In general, the more stabilized the negative charge of the conjugate base is, the more the equilibrium favors that form, thus the more the acid dissociates, thus the "stronger" the acid is.

So for a stronger acid (lower $\ce{pK_a}$), the negative charge must be more stabilized. When comparing $\ce{CF_3COOH}$ (TFA) to $\ce{CH_3COOH}$ (we'll use acetic acid to represent a "typical" carboxylic acid), TFA is stronger than acetic acid because its conjugate base can better stabilize the negative charge. If we draw the structures of the conjugate bases (I encourage you to do so), we will see that they both can stabilize the negative charge via resonance in the carboxylate functional group. However, TFA also has three highly electronegative fluorine atoms which withdraw electron density "through the single bonds" via induction.

As stated above, the $\ce{-CF_3}$ moiety is an electron withdrawing group, by the inductive effect. So our negative charge (rather, the electron density it represents) will be slightly drawn through the bonds toward the fluorines, thus it is further delocalized and thus more stable.

Note that we do not have convenient structural diagrams to show induction like we do resonance, but you should be able to imagine the electrons "sloshing" more towards one side of a bond/functional group. It's the same concept as a good old polar bond like $\ce{H-Cl}$; we say Cl "hogs" the electron density and pulls it away from H. In the same way $\ce{-CF_3}$ will draw electron density toward itself.

So to summarise, in general:

Stronger acid ~ more stable conjugate base. Whenever you want to qualitatively rationalize the relative strengths of two acids, draw their conjugate bases and determine which is more stable. (If your acid was neutral, its conjugate will be negative, and it will be easier to use this method than for a positive acid/neutral conjugate.) Stabilizing factors are size and electronegativity of the atom bearing the charge, resonance delocalization of the charge, hybridization of the orbital holding the charge, induction delocalization of the charge, and hyperconjugation which donates electron density and may destabilized a negative charge. SERHIH, roughly in that order of significance.

In this example, TFA and acetic acid are tied at size & electronegativity because both place the charge on an oxygen atom, tied at resonance because both may use the carboxylate functionality, tied at hybridization because the atoms in a carboxylate are necessarily $\mathrm{sp^2}$, but now induction is present in TFA and not acetic acid, so we expect TFA will be a stronger acid.

If $\ce{CF3COOH}$ loses a hydrogen atom, the $\ce{CF3COO-}$ ion is highly stabilized by the electron withdrawing power (-I) of the fluorine atoms.

The acidic strength of a acid is directly proportional to the stability of its conjugate base.

• Your second sentence sounds a lot like a mathematical statement, is there an equation I'm not aware of? – electronpusher Feb 28 '17 at 4:43
• No..It's just a statement..More stable is the conjugate base more strong the acid will be...That's what directly proportional means.. – Mitchell Feb 28 '17 at 4:45
• Directly proportional is a mathematical statement with a specific meaning. Just because Y increases when X increases does not mean they are proportional, the relationship may not even be linear at all. For example, when temperature increases the equilibrium constant Ka increases, but the the relationship is quite nonlinear: K = Ce^(-a/T). Your statement literally implies there is some equation pKa = (constant)*∆H[base]. This would be very nice, and I hope it's true, but personally I'm not aware of such a model. – electronpusher Mar 1 '17 at 0:15
• @electronpusher for that matter, recall the definition of $K_a$ for the reaction $$\ce{HA<=>H+ + A-}$$ would be $$K_a=\frac{\ce{[H+][A-]}}{\ce{[HA]}}$$ You can say that $K_a$ is directly proportional to concentration of conjugate base, which is $\ce{[A-]}$ – Pritt says Reinstate Monica Apr 24 '17 at 7:54