SolutionsAcidity

Identify the acid groups in the following molecules and write the expected equilibria with water, including the structures of the corresponding conjugate bases:

Ethane, benzene, 1,2-dichloroethane and tetrahydrofuran: no acid groups

These four compounds do not have any acid groups per se, because there is no hydrogen bonded to a heteroatom.

Therefore, an equilibrium with water does not make sense.

Carbanions that would form by removing a proton from ethane, benzene, 1,2-dichloroethane and tetrahydrofuran

Although an equilibrium with water cannot occur, we can imagine the structure of the hypothetical conjugate bases, as shown, by removing a proton. They all have an excess of electrons on a carbon, which is a very high-energy and therefore very unfavourable situation.

In these cases there is an acid function, because there are hydrogens bonded to heteroatoms.

Equilibrium of phenol with water and resonance forms of the phenoxide anion

Equilibrium of 3-pentanol with water to give its alkoxide

Equilibrium of 3-methylbutanoic acid with water and resonance forms of the carboxylate

Equilibrium of the thiol with water to give the thiolate anion

Equilibrium of 3-methylbutanamide with water and resonance forms of the anion; the form with the charge on nitrogen contributes less to the resonance hybrid

An equilibrium with water can be established, and the conjugate base has a negative charge on a heteroatom that is more electronegative than carbon, which is not as extremely unfavourable a situation as the previous ones. In some cases the charge can be delocalized by resonance, which further stabilizes the conjugate base.

The case of acetone is special. Its acidity is actually very low, but not as low as that of the compounds in the first group. Although there are no hydrogens attached to heteroatoms and the conjugate base has a negative charge on a carbon, acetone shows some acidity because of resonance and can establish an equilibrium with water.

Equilibrium of acetone with water and resonance forms of the enolate

The proximity of the C=O group allows the charge to be distributed towards the oxygen. Can you write resonance forms of the conjugate base to prove it? Surely you can... There you have it. Try it yourself!

Order the above compounds by groups from highest to lowest acidity.

The most acidic of all will be the one with the COOH group, whose very name is carboxylic "acid".

The conjugate base is the LEAST UNSTABLE of all because the charge is distributed between the two oxygens by resonance.

pKa = 4

Next in acidity could be the amide, very similar in structure to the carboxylic acid except that it has an NH₂ group instead of an OH group.

But the charge is not distributed between identical atoms, and therefore the conjugate base is much less stable than carboxylate.

pKa = 15 (amides are much less acidic than expected)

The next group of acids are phenols, in which the negative charge of the conjugate base lies on a heteroatom and can be delocalized by the aromatic ring.

pKa = 10

The next group of acids are alcohols and thiols, in which the negative charge of the conjugate base lies on a heteroatom but cannot be delocalized, because the O or S atoms are bonded to sp³ carbons.

pKa = 13-15

The second-to-last acid will be acetone. Although the negative charge of the conjugate base lies on a carbon, which is not very electronegative, it is next to the C=O group and its pi system. This allows a small delocalization of the electron density that slightly stabilizes the conjugate base.

pKa = 20

The last group of substances in this order of acidity are those that have no hydrogen bonded to a heteroatom.

pKa > 30