Question

In the following compounds, the decreasing order of basic strength will be:
(A) ${{({{C}_{2}}{{H}_{5}})}_{2}}NH>{{C}_{2}}{{H}_{5}}N{{H}_{2}}>N{{H}_{3}}$
(B) ${{({{C}_{2}}{{H}_{5}})}_{2}}NH>N{{H}_{3}}>{{C}_{2}}{{H}_{5}}N{{H}_{2}}$
(C) $N{{H}_{3}}>{{C}_{2}}{{H}_{5}}N{{H}_{2}}>{{({{C}_{2}}{{H}_{5}})}_{2}}NH$
(D) ${{C}_{2}}{{H}_{5}}N{{H}_{2}}>N{{H}_{3}}>{{({{C}_{2}}{{H}_{5}})}_{2}}NH$

Answer 1

Ethyl and Methyl are electron donating groups. These groups increasing electron density on the nearby attached more electronegative atom. Ammonia has a lone pair attached to it which is available for donation.

Complete step by step answer: The ammonia reacts as a base because of the active lone pair on the nitrogen. Nitrogen is more electronegative than hydrogen and so attracts the bonding electrons in the ammonia molecule towards itself. That means that in addition to the lone pair, there is a build-up of negative charge around the nitrogen atom.
-Methyl and ethyl groups are electron donating groups, they increase electron density on their nearby electronegative atoms. Because alkyl groups donate electrons to the more electronegative nitrogen. The inductive effect makes the electron density on the alkyl amine's nitrogen greater than the nitrogen of ammonium. Correspondingly, primary, secondary, and tertiary alkyl amines are more basic than ammonia.
-Therefore, according to this, diethyl amine will be more basic than that of ethyl amine, and in turn, ethyl amine will be more basic than ammonia.

Hence the correct option is A option.

Note: Ammonia is a colourless gas with a characteristically pungent smell. It is lighter than air, its density being 0.589 times that of air. It is easily liquefied due to the strong hydrogen bonding between molecules; the liquid boils at $-33.3~{}^\circ C$ ($-27.94~{}^\circ F$ ), and freezes to white crystals at $-77.7~{}^\circ C$ .