Question

Among the following compounds\[{C_3}{H_7}N{H_2}\], \[N{H_3}\], \[C{H_3}N{H_2}\], \[{C_6}{H_5}N{H_2}\]and \[{C_2}{H_5}N{H_2}\], the least basic compound is?
A.\[{C_3}{H_7}N{H_2}\]\[\]
B.\[N{H_3}\]
C.\[C{H_3}N{H_2}\]
D.\[{C_6}{H_5}N{H_2}\]
E.\[{C_2}{H_5}N{H_2}\]

Answer 1

Hint: Basicity is the electron donor ability of compounds. Compounds having non-bonding electrons show basic character as they can donate their electrons. More donating power of electrons makes the compound more basic.

Complete Step by Step Solution:
All the compounds given in the question are basic because of the presence of nitrogen atoms in them which have one lone pair of electrons. But they have a difference in their basicity. The ones whose electrons are readily available for donation are more basic than the ones whose electrons are delocalised in the compound. Because if the electrons are delocalized, they are not readily available to donate.

From the given compounds, only \[{C_6}{H_5}N{H_2}\]i.e., aniline is the one where electrons are delocalised on the compound and its electron donation power decreases. Due to this, they are not readily available for donation and hence it is the least basic compound. After option D, Option E is less basic because of the pie bond of ethene on the adjacent position of amine. Here also, lone pairs of electrons of Nitrogen are delocalised in the compound. But, comparatively, it is less than aniline because of comparatively less delocalisation as it has less no. of pie bonds.

Image:Delocalisation of lone pair of electrons of Nitrogen in aniline compound.

So, option D is correct.

Note: Alkyl groups in electron-donating groups, increased the tendency of atoms to donate their electrons. So, they help in increasing the basicity of the compound. Basicity can be used to find compounds that can be easily protonated. So, the most basic compound is the one which gets protonated fast. Basic compounds donate their electrons and form bonds between them, in return, they become positively charged.