Question

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.

Short answer

The resonance stabilization makes ${\text{C-H}}_{\text{a}}$to be more acidic than ${\text{C-H}}_{\text{b}}$.

Step-by-step solution

Acidity

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.

Formation of the conjugate base by the removal of Ha proton from pentan-2-one

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.

Formation of the conjugate base by the removal of Hb  proton from pentan-2-one

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.