46E - Organic Chemistry (Mcmurry)

Question

The ammonium ion ( ${{\mathbf{NH}}_4}^{+}$, pK_a=9.25) has a lower pKa than the

methylammoniumion (${\mathbf{CH}}_3{{\mathbf{NH}}_3}^{+}$ , pK_a=10.66). Which is the stronger

base, ammonia $\left({\mathbf{NH}}_3\right)$ or methylamine $\left({\mathbf{CH}}_3{\mathbf{NH}}_2\right)$ ? Explain.

Step-by-Step Solution

Answer

Methylamine $\left(C{H}_{3}N{H}_2\right)$  is a stronger base than ammonia $\left(N{H}_3\right)$ . Therefore the conjugate base of the methylammonium ion is stronger. Therefore, methylamine is a stronger base than ammonia.

Acid-Base Strength

Acids differ in their ability to donate H⁺. Stronger acids, such as HCl, react almost completely with water, whereas weaker acids, such as acetic acid (CH₃COOH), react only slightly. The exact strength of a given acid HA in water solution is described using the acidity constant ($K_a$ ) for the acid-dissociation equilibrium.

$\begin{array}{l}{\text{HA + H}}_2\text{O }\rightleftarrows {\text{ A}}^{-}{\text{ + H}}_{\text{3}}{\text{O}}^{-}\\ K_a\underset{}{}\text{=}\underset{}{}\frac{{\text{[H}}_{\text{3}}{\text{O}}^{+}{\text{][A}}^{-}\text{]}}{\left[HA\right]}\end{array}$

 

Identify the stronger base with the help of value

Acid strengths are normally using  $\text{p}{K}_a$values rather than $K_a$  values, where the $\text{p}{K}_a$ is the negative common logarithm of $K_a$:

 $\text{p}K\text{a = -log }{K}_a$

The stronger base has a larger $\text{p}{K}_a$, and the weaker base has a smaller  $\text{p}{K}_a$value.

Predict the stronger base between ammonia & methylamine

The stronger the acid (smaller  ), the weaker its conjugate base. Since  is a stronger acid than  , methylamine  is a stronger base than ammonia .

Note: Since methylamine has an electron-donating group (methyl). Therefore electron density on nitrogen atom increase. Hence it is a stronger base, whereas, in ammonia, an electron-donating group is absent. So it is a weak base when compared to methyl amine.