Question

Basic strength of \[C{H_3}C{H_2}N{H_2}\].......... to that of \[C{H_2} = CHN{H_2}\]
A. Stronger
B. Weaker
C. Equal
D. None of the above

Answer 1

Hint: Basicity of the compound can be identified by the lone pair atoms present in that compound. Basicity increases with an increase in a negative charge. The nitrogen atom in amine has one lone pair of electrons. Saturated amines are carbon-carbon single-bonded amines whereas unsaturated amines are carbon=carbon double-bonded amines.

Complete step by step answer:
It is known that the Nitrogen atom in amine has one lone pair of electrons. Ethylamine and vinyl amine are coming under primary amine compounds.
Generally, Basicity increases with an increase in a negative charge.
The inductive effect plays an important role to determine the basicity of the compound. The inductive effect deals with the transmission of unequal sharing of the bonding electron. The inductive effect has two groups such as +I and –I groups. +I group is also called electron releasing groups and it tends to donate its electron pair to the nearer atom. The examples for electron releasing groups are \[ - OH, - N{H_2}, - Me, - F\].
$ - I$ the group is also called an electron-withdrawing group and it tends to pull the electrons of neighboring atoms towards itself. The examples for electron-withdrawing groups are \[ - CO, - N{O_2}, - COOR,CN\].
When compared to ethylamine and vinyl amine, ethyl amine possesses higher basicity than the vinyl amine. Because, in ethylamine, the carbon nearer to amine is a methyl group and they are electron releasing groups. So, the methyl group helps to increase electron charge density and it makes the compound more basic.
In vinyl amine, the double bond is present nearer to an amine. Generally double bonds tend to withdraw electrons from neighboring atoms. Thus, the amine group present in vinyl amine is less basic than ethylamine. This is because the double bond present in vinyl amine decreases the electron charge density and it makes the compound less basic.

Thus, the correct answer is option A.

Note: Basicity of the compound can also be identified in terms of hybridization. If s- character increases, the acidity of the compound increases. This eventually leads to a decrease in basicity. The hybridization of carbon atoms in ethylamine is \[s{p^3}\] whereas, in vinyl amine, its hybridization is \[s{p^2}\]. When compared to \[s{p^3}\] and \[s{p^2}\] hybridization, \[s{p^3}\] hybridization has less s- character than \[s{p^2}\] hybridization. Thus, ethylamine is more basic than the vinyl amine.