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Question: The increasing order of \(\text{ p}{{\text{K}}_{\text{b}}}\text{ }\) of the following compound is: ...

The increasing order of  pKb \text{ p}{{\text{K}}_{\text{b}}}\text{ } of the following compound is:

A)
B)
C)
D)

A)  A < C < D < B \text{ A }<\text{ C }<\text{ D }<\text{ B }
B)  B < D < A < C \text{ B }<\text{ D }<\text{ A }<\text{ C }
C)  C < A < D < B \text{ C }<\text{ A }<\text{ D }<\text{ B }
D)  B < D < C < A \text{ B }<\text{ D }<\text{ C }<\text{ A }

Explanation

Solution

. The basicity is the measure of the extent of the base to accept the proton or donate the electron pair. The basicity is expressed as a dissociation constant of the base  Kb\text{ }{{\text{K}}_{\text{b}}} . The basicity of an organic compound depends on the substituents. The electron releasing group increases the basicity of the base while the electron-withdrawing group decreases the basicity of the compound.

Complete step by step answer:
The basicity is determined by the  Kb\text{ }{{\text{K}}_{\text{b}}} value of a base. the base tends to abstract a proton or donate an electron pair. The value of the  Kb\text{ }{{\text{K}}_{\text{b}}} is equilibrium constant for the dissociation of the base.
The basicity depends on the substituents present in the compound.
The electron-donating group likes CH3 -\text{C}{{\text{H}}_{\text{3}}}\text{ } and OCH3 -\text{OC}{{\text{H}}_{\text{3}}}\text{ }releases the electron density towards the nitrogen of the compound. This stabilized the cation formed after donating an electron pair or accepting a proton. Thus from given compounds, compound B) and D) have high basicity.
Among CH3 -\text{C}{{\text{H}}_{\text{3}}}\text{ } and OCH3 -\text{OC}{{\text{H}}_{\text{3}}}\text{ } groups, the methyl group donates its electron via the hyperconjugation and inductive effect. However, the methoxy OCH3 -\text{OC}{{\text{H}}_{\text{3}}}\text{ }group donates the electron via the resonance. Thus compound B) has more basicity compared to D).
The electron-withdrawing group like -\text{F } and NO2 -\text{N}{{\text{O}}_{\text{2}}}\text{ } withdraws the electron density towards itself from the nitrogen of the compound. This destabilized the cation formed after donating an electron pair or accepting a proton. Thus from given compounds, compound A) and C) have low basicity.
Among -\text{F } and NO2 -\text{N}{{\text{O}}_{\text{2}}}\text{ } groups, the Florine group withdraws its electron via the inductive effect. However, the nitroNO2 -\text{N}{{\text{O}}_{\text{2}}}\text{ } group withdraws the electron via the resonance. Thus compound A) has more basicity compared to C).

Thus the order of basicity for the compounds is given as,
 B > D > A > C \text{ B }>\text{ D }>\text{ A }>\text{ C }
The  Kb\text{ }{{\text{K}}_{\text{b}}} is related to  pKb \text{ p}{{\text{K}}_{\text{b}}}\text{ } is given as,’
 pKb = logKb \text{ p}{{\text{K}}_{\text{b}}}\text{ }=\text{ }-\log {{\text{K}}_{\text{b}}}\text{ }
Thus, the order of increasing  pKb \text{ p}{{\text{K}}_{\text{b}}}\text{ } is given as,
 B < D < A < C \text{ B }<\text{ D }<\text{ A }<\text{ C }
So, the correct answer is “Option C”.

Note: Note that, we can summarise the basicity and  pKb \text{ p}{{\text{K}}_{\text{b}}}\text{ }as basicity is directly proportional to the resonance effect of electron releasing group and hyperconjugation effect of electron-donating group. However, the basicity is inversely related to the resonance effect of the electron-withdrawing group and hyperconjugation effect. Always remember that the resonance effect is the permanent effect but hyperconjugation is a temporary effect, thus resonance makes it more stable.
 Basicity  +R  1R +H  1H \text{ Basicity }\propto \text{ +R }\propto \text{ }\dfrac{1}{-\text{R}}\propto \text{ +H }\propto \text{ }\dfrac{1}{-\text{H}}\text{ }