Explain why the ${\mathbf{\text{C-H}}}_{\mathbf{\text{a}}}$ bond is much more acidic than the ${\mathbf{\text{C-H}}}_{\mathbf{\text{b}}}$bond in pentan-2-one.

The power of a particular acid can be understood by checking its reaction with a particular base. Strong acids are regarded to possess small or negative ${\mathbf{\text{pK}}}_{\mathbf{\text{a}}}$values.

The formation of the conjugate base by the removal of ${\text{H}}_{\text{a}}$proton from pentan-2-one can be given as follows:

Removal of proton from pentan-2-one

The resonance stabilization of ${\text{H}}_{\text{a}}$proton can be given as follows:

Resonance stabilization of proton

The${\text{H}}_{\text{a}}$ proton is more acidic because of the formation of the resonance stabilized conjugate base that is shown above.

The formation of the conjugate base by the removal of ${\text{H}}_{\text{b}}$ proton from pentan-2-one

Removal of proton from pentan-2-one

The conjugate base formed by the removal of ${\text{H}}_{\text{b}}$ is not resonance stabilized and is hence, not regarded to be acidic.