Question

What is the order of basicity of the following compounds?

${CH}_3{NH}_2,{\left({CH}_3\right)}_{2}NH,{\left({CH}_3\right)}_{3}N$ (in protic solvent)

• A. ${\left({CH}_3\right)}_{2}NH>{CH}_3{NH}_2>{\left({CH}_3\right)}_{3}N$
• B. ${\left({CH}_3\right)}_{2}NH<{CH}_3{NH}_2<{\left({CH}_3\right)}_{3}N$
• C. ${\left({CH}_3\right)}_{2}NH>{\left({CH}_3\right)}_3>{CH}_3{NH}_{2}N$
• D. None of these

Answer 1

The correct option is A ${\left({CH}_3\right)}_{2}NH>{CH}_3{NH}_2>{\left({CH}_3\right)}_{3}N$

Greater is the stability of the substituted ammonium cation, stronger should be the corresponding amine as a base. Thus, the order of the basicity of aliphatic amines should be primary > secondary > tertiary, which is opposite to the inductive effect based order. Secondly, when the alkyl group is small, like $–C{H}_3$ group, there is no steric hindrance to H-bonding. In case the alkyl group is bigger than $C{H}_3$ group, there will be steric hindrance to H-bonding. Therefore, the change of nature of the alkyl group, e.g., from $–C{H}_{3}to–C_2{H}_5$ results in change of the order of basic strength. Thus, there is a subtle interplay of the inductive effect, solvation effect and steric hinderance of the alkyl group which decides the basic strength of alkyl amines in the aqueous state.

The order of basic strength in case of methyl substituted amines in aqueous solution is as follows:

$(C{H}_3{)}_{2}NH>C{H}_{3}N{H}_2>(C{H}_3{)}_{3}N>N{H}_3$