Does steric hindrance affect Brønsted basicity?

I understand that basicity is a thermodynamic quality. Not a kinetic factor. So it doesn't matter if it takes a million years for the proton to reach the site of basicity; time is nothing in thermodynamics. We only care about the initial and final state. So this suggests to me that steric hindrance does NOT affect Brønsted basicity

In addition, the hydrogen proton is quite small. So I wouldn't expect it to be particularly sterically hindered by anything (unless we're in solution and the proton is heavily solvated). Still, once the proton reaches the site of basicity, it's reached it, and if I understand thermo correctly, thermo doesn't deal with anything in the middle of the energy graph. Only the points on the two ends.

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Also it would be nice if one could comment on the difficulty of this problem in general. Just curious. Personally I think it's a fascinating problem. (On the other hand I'm not the one actually doing the problem).

  • $\begingroup$ You've got it right. $\endgroup$ Jul 31, 2014 at 0:38
  • $\begingroup$ That's very interesting because my professor asked us to consider steric effects on basicity. Picture of problem posted. Something's fishy here. $\endgroup$
    – Dissenter
    Jul 31, 2014 at 0:45
  • $\begingroup$ Waiting for the intellectual smackdown. $\endgroup$
    – Dissenter
    Jul 31, 2014 at 0:53
  • $\begingroup$ @SilvioLevy I disagree (a little). Steric effects change the geometry of the molecule, and therefore also the bonding of each atom to another. This will also affect the lone pairs, that bond to the proton. Based on steric reasons that bond can be weaker or stronger than ordinarily expected. Since the bond dissociation energy is also a thermodynamic property, it should not be neglected. Consider dimethyl and diethyl amine, the latter is slightly (tiny!!) more basic mostly because of steric considerations. $\endgroup$ Jul 31, 2014 at 2:41
  • $\begingroup$ Why is the latter more basic, specifically? From what I understand alkyl groups are electron releasing (slightly). $\endgroup$
    – Dissenter
    Jul 31, 2014 at 2:44

2 Answers 2


Anything that stabilize one of the species in an acid-base equilibrium influences the free energy difference between the reactants and products, which alters the equilibrium constant:

$$\ce{HA <=> H+ + A-}$$

$$\Delta_r G^\circ = -RT\ln K_{eq}$$

You forget the solvent. Acidity and basicity depend on the interaction between the acid/base and the solvent. While steric hindrance will decrease the rate solvent access, it more importantly also decreases solvation of the acidic or basic site. More steric hindrance prevents additional solvent molecules from reaching the site to stabilize an ionic conjugate acid or base.

Consider the difference in $\mathrm{p}K_a$ between methanol, ethanol, 1-butanol, and tert-butanol. These are equilibrium constants, so they most certainly derived from thermodynamic phenomena!

Methanol $\ce{CH3OH}$ has a $\mathrm{p}K_a = 15.5$ (in water)

Ethanol $\ce{CH3CH2OH}$ has $\mathrm{p}K_a = 15.9$ (in water)

1-Butanol $\ce{CH3CH2CH2CH2OH}$ has $\mathrm{p}K_a = 16.1$ (in water)

t-Butanol $\ce{(CH3)3COH}$ has $\mathrm{p}K_a = 17$ (in water) (I've also seen 18 as its $\mathrm{p}K_a$)

The difference between methanol, ethanol, and 1-butanol (all primary alcohols) is small, while t-butanol has a larger pKa. If sterics had no effect, then both butanol isomers should have similar acidities since both isomers have similar inductive destabilization of the conjugate base by electron donating alkyl groups.

However in methanol, ethanol, and 1-butanol, more water molecules can solvate the alkoxide anion conjugate base, stabilizing that conjugate base more than the conjugate base of tert-butanol, which due to sterics cannot have as much solvation of the anion.

Now, gas phase acidities, in which there is no solvent, should have no solvent and solvation effects, and thus no (or at least reduced) steric effects.

  • 1
    $\begingroup$ And although my answer focuses on acidity, acidity and basicity follow the same rules. $\endgroup$
    – Ben Norris
    Jul 31, 2014 at 1:02
  • $\begingroup$ Damn, that is a good angle of looking at this problem. You know, I doubt my professor thought of this himself. I'll have to ask him. $\endgroup$
    – Dissenter
    Jul 31, 2014 at 1:04
  • $\begingroup$ "If sterics had no effect, then both butanol isomers should have similar acidities since both isomers have similar inductive destabilization of the conjugate base by electron donating alkyl groups." Similar doesn't mean the same. Consider that $pK_a$ of something in $\ce{H2O}$ is different from $pK_a$ of the same thing in $\ce{D2O}$. Hard to argue sterics there. (link to an example, variation up to 0.6 pH units but dependent on the exact solute: pubs.acs.org/doi/pdf/10.1021/j100631a015) $\endgroup$ Jul 31, 2014 at 1:09
  • $\begingroup$ I'm guessing the extra proton in D has to do with the differing acidities? $\endgroup$
    – Dissenter
    Jul 31, 2014 at 1:12
  • 1
    $\begingroup$ I think words like "steric effects" are not precise enough to warrant too much ink spilled in arguments. That said, @Ben Norris has a point that there could be an influence of sterics on $pK_a$ due to solvation. I don't agree that his examples prove that effect, because with 3 attached methyl groups, each slightly electron-donating, the tert-butoxide ion is more destabilized than the 1-butoxide ion anyway. If there were an example running the other way, the case would be altered -- for instance if $\ce{CF3OH}$ dissociated less than $\ce{CH3OH}$, the sterics argument would be persuasive. $\endgroup$ Jul 31, 2014 at 1:46

Sterics can affect basicity not only through solvation but also through inhibition of resonance.

Steric inhibition of resonance is present only in benzene rings. Presence of any group at the ortho position of benzoin acid,it throws the carboxylic acid group out of the plane so it's mesomeric connection w tin benzene ring vanishes thus ortho substituted benzoic acids are stronger than meta and para substituted benzoic acids.

In the case of meta-xylene with an $\ce{-NH2}$ group between its methyl groups,

enter image description here

resonance is inhibited because the methyl groups prevent the amine group from adopting a perfectly planar conformation. This inhibition of resonance decreases the electron-withdrawing capabilities of the aromatic ring. As a result, electron density is concentrated on the amine group, and meta-xylene with such a configuration is unusually basic.

Further reading:



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