Question

Question: Which compound in each pair is more basic:

a. ${\left({\mathrm{CH}}_3\right)}_2\mathrm{NH}\mathrm{and}{\mathrm{NH}}_3$

b. ${\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{NH}}_2\mathrm{and}{\mathrm{ClCH}}_2{\mathrm{CH}}_2{\mathrm{NH}}_2$ ?

Short answer

Answer

a.${\left({\mathrm{CH}}_3\right)}_2\mathrm{NH}$

b.${\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{NH}}_2$

Step-by-step solution

Basicity of amines

Primary $1°,2°\mathrm{and}3°$alkylamines are more basic than ${\mathrm{NH}}_3$ because of the electron-donating inductive effect of the R groups. The alkyl groups increase the electron density at the nitrogen.Therefore, the electrons at the nitrogen become freer to donate.

Explanation

a.${\left({\mathrm{CH}}_3\right)}_2\mathrm{NH}$is a more basic compound because it has two methyl groups bonded to the nitrogen atom.

${\mathrm{NH}}_3$is not bonded with an alkyl group.

b. ${\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{NH}}_2$is a more basic compound because it has an ethyl group bonded to the nitrogen atom.

${\mathrm{ClCH}}_2{\mathrm{CH}}_2{\mathrm{NH}}_2$is bonded with the Cl group, which is an electronegative atom. Therefore, the${\mathrm{ClCH}}_2{\mathrm{CH}}_2{\mathrm{NH}}_2$compound is less basic.