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Question: Arrange the following (i) In decreasing order of the pKb values: \[{C_2}{H_{5}}N{H_2},{C_{6}}{H_{...

Arrange the following
(i) In decreasing order of the pKb values: C2H5NH2,C6H5NHCH3,(C2H5)2NH    and  C6H5NH2{C_2}{H_{5}}N{H_2},{C_{6}}{H_{5}}NHC{H_3},{\left( {{C_2}{H_5}} \right)_2}NH\;\;and\;{C_{6}}{H_{5}}N{H_2}
(ii) In increasing order of basic strength: C6H5NH2,C6H5N(CH3)2,(C2H5)2NH  and  CH3NH2{C_{6}}{H_5}N{H_2},{C_6}{H_5}N{\left( {C{H_3}} \right)_2},{\left( {{C_2}{H_5}} \right)_2}NH\;and\;C{H_3}N{H_2}
(iii) In increasing order of basic strength: (a) Aniline, p-nitroaniline and p-toluidine

Explanation

Solution

The pKb value is the mathematical measurement of basicity or alkalinity and is associated with shifting of the acid-base equilibrium of a substance with water. Formula used for calculation of pKb is pKa+pKb=14pKa + pKb = 14 . If we know the value of pKa the value for pKb can be calculated.
Basic strength: it is defined as the ability of the species to accept the hydrogen ions from another species according to the Bronsted lowry concept. Higher the number of the accepted hydrogen ions greater will be the basic strength of the compound.

Complete step by step answer: Now we will discuss the orders of pKb value for each option:
i.The pKb value for the stronger bases have lower value:
Arranging the sequence in the increasing order of basicity:
C6H5NH2<C6H5NHCH3<C2H5NH2<(C2H5)2NH{C_6}{H_5}N{H_2} < {C_6}{H_5}NHC{H_3} < {C_{2}}{H_5}N{H_2} < {({C_2}{H_5})_{2}}NH
The ability of accepting increases from aniline to diethyl amine in these sequences so the trend of basicity also increases, diethyl having the most.
Arrangement of the sequence in the decreasing order of pKb will be: C6H5NH2>C6H5NHCH3>C2H5NH2>(C2H5)2NH{C_6}{H_5}N{H_2} > {C_6}{H_5}NHC{H_3} > {C_{2}}{H_5}N{H_2} > {({C_2}{H_5})_{2}}NH
Aniline having the lower value of basicity will have high value for pKb, as the stronger bases have the lower pKb value.
ii.The increasing order of the basicity will be as follows:
  C6H5NH2<C6H5NHCH3<CH3NH2<(C2H5)2NH\;{C_{6}}{H_5}N{H_2} < {C_{6}}{H_5}NHC{H_3} < C{H_3}N{H_2} < {({C_2}{H_5})_2}NH
The ability of accepting increases from aniline to diethyl amine in these sequences so the trend of basicity also increases, diethyl having the most.

iii.Basics strengths increasing order will be as follows:
p−nitroaniline < aniline < p−toluidinep - nitroaniline{\text{ }} < {\text{ }}aniline{\text{ }} < {\text{ }}p - toluidine
The ability to accept the hydrogen ions from nitro aniline increases to toluidine, so the trend of the basic strength also increases, toluidine having the highest.

Note: The lower basicity value of aniline is due to +R effect, due to which aniline donates its electrons to the benzene ring. The electron lone pair as a result of resonance on the N atom gets delocalized and is hardly available for protonation