Label the three most acidic hydrogen atoms in lactic acid, ${\mathbf{\text{CH}}}_{\mathbf{\text{3}}}{\mathbf{\text{CH(OH)CO}}}_{\mathbf{\text{2}}}\mathbf{\text{H}}$, and rank them in order of decreasing acidity. Explain your reasoning.

Hydrogen atoms that tend to get liberated as ${\mathbf{\text{H}}}^{\mathbf{+}}$ions are referred to as acidic hydrogens. If a hydrogen atom is connected to an electronegative atom, then that hydrogen atom is regarded as acidic hydrogen.

The effects such as resonance and inductive effects are used to determine the most acidic protons. The hydrogens found in the ${\text{CH}}_3$group are considered the least acidic since they are connected to the ${\text{sp}}_3$hybridized carbon atom.

The conjugate base that is formed due to the removal of that particular hydrogen is not resonance stabilized.

Acidic hydrogens in lactic acid

The O-H protons (b and c) are more acidic when compared to the C-H proton (a) because of the different element effects. The acidic hydrogens are ranked in order$\text{c>b>a}$.

The most acidic proton will have resonance stabilization when it is removed. Hence, the conjugate base of this particular proton is most stable.

The conjugate base created by the proton loss possesses two resonance structures, and one of the resonance structures has a negative charge on carbon.

Conjugate base formed by loss of proton a

The conjugate base that is formed by loss of b has no resonance stabilization. However, a negative charge is present on the oxygen atom, which is an electronegative atom.

Conjugate base formed by loss of proton b

The conjugate base that is formed by the loss of c possesses resonance stabilization. A negative charge is present on the oxygen atom in both the resonance structures. This makes the proton c to be acidic.

Conjugate base formed by loss of proton c