Question

As we will see in Chapter 23, C-H bonds are sometimes more acidic than O-H bonds. Explain why the ${\mathbf{pK}}_{\mathbf{a}}$ of${\mathbf{CH}}_{\mathbf{2}}{\left(\mathrm{CHO}\right)}_{\mathbf{2}}$ is lower than the${\mathbf{pK}}_{\mathbf{a}}$ of$\mathbf{OH}{\left({\mathrm{CH}}_2\right)}_{\mathbf{3}}\mathbf{OH}$(9 vs. 16).

Short answer

The conjugate base of the compound${\mathrm{CH}}_2{\left(\mathrm{CHO}\right)}_2$is much more stabilized due to resonance than the conjugate base of the compound $\mathrm{OH}{\left({\mathrm{CH}}_2\right)}_3\mathrm{OH}$.

The conjugate base which is formed by the loss of a proton from $\mathrm{OH}{\left({\mathrm{CH}}_2\right)}_3\mathrm{OH}$has only one Lewis structure and is therefore less stable, as shown below:

Conjugate base of $\mathrm{OH}{\left({\mathrm{CH}}_2\right)}_3\mathrm{OH}$

The conjugate base formed by the loss of a proton from the compound$\mathrm{OH}{\left({\mathrm{CH}}_2\right)}_3\mathrm{OH}$is resonance stabilized and has the following resonating structures.

Among the resonating structures, two structures have a negative charge on the oxygen atom, which makes the conjugate base even more stable.

Resonance stabilized conjugate base of$\mathrm{OH}{\left({\mathrm{CH}}_2\right)}_3\mathrm{OH}$

Step-by-step solution

The meaning of pKa and the conjugate base of an acid

Some compounds act as proton donors and are termed the Bronsted–Lowry acids. The strength of the Bronsted–Lowry acids can be determined by the loss of a proton from the polar -OH bond.

The deprotonation of an acid can be done by a base whose conjugate acid has a higher value of${\mathrm{pK}}_{\mathrm{a}}$ . The conjugate base of a strong acid is always weak.

The strength of an acid is determined by the value of${\mathrm{pK}}_{\mathrm{a}}$ .

The comparison of the pKa of the given compounds

The ${\mathrm{pK}}_{\mathrm{a}}$of the compound ${\mathrm{CH}}_2{\left(\mathrm{CHO}\right)}_2$is lower than the of ${\mathrm{pK}}_{\mathrm{a}}$of $\mathrm{OH}{\left({\mathrm{CH}}_2\right)}_3\mathrm{OH}$. This can be explained in terms of the strength of the conjugate base of the given compounds. The compound ${\mathrm{CH}}_2{\left(\mathrm{CHO}\right)}_2$has the following structure:

Structure of ${\mathrm{CH}}_2{\left(\mathrm{CHO}\right)}_2$

The conjugate base which is formed by the loss of a proton from this molecule has three resonating structures,as shown below:

Resonance stabilized conjugate base of ${\mathrm{CH}}_2{\left(\mathrm{CHO}\right)}_2$

The stronger conjugate base of a compound corresponds to the weaker acid. On the other hand, the weaker conjugate base corresponds to the stronger acid.

The conjugate base formed by the loss of a proton from this compound is resonance stabilized and has the following resonating structures.

Among the resonating structures, two structures have a negative charge on the oxygen atom which makes the conjugate base even more stable.

The compound $\mathrm{OH}{\left({\mathrm{CH}}_2\right)}_3\mathrm{OH}$has the following structure:

Structure of $\mathrm{OH}{\left({\mathrm{CH}}_2\right)}_3\mathrm{OH}$

The conjugate formed by the loss of a proton from the above compound has only one Lewis structure as shown below:

Conjugate base of $\mathrm{OH}{\left({\mathrm{CH}}_2\right)}_3\mathrm{OH}$

The conjugate base of the compound${\mathrm{CH}}_2{\left(\mathrm{CHO}\right)}_2$ is much more stabilized due to resonance than the conjugate base of the compound $\mathrm{OH}{\left({\mathrm{CH}}_2\right)}_3\mathrm{OH}$. Therefore, the compoundis${\mathrm{CH}}_2{\left(\mathrm{CHO}\right)}_2$ more acidic than the compound $\mathrm{OH}{\left({\mathrm{CH}}_2\right)}_3\mathrm{OH}$.