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Question
Arrange the following in the increasing order of their pKb values.
C6H5NH2,C2H5NH2,C6H5NHCH3.
C6H5NH2,C2H5NH2,C6H5NHCH3.
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Solution
The increasing order for pKb values is C2H5NH2(3.29)<C6H5NHCH3(9.30)<C6H5NH2(9.38). The numbers in bracket indicates pKb. Higher is the pKb value, lower is the basic strength of amine. Thus, the increasing order for pKb values indicates
decreasing order of the basic strength of amine. Aniline is least basic and ethanamine is most basic.
Aniline is least basic as unshared pair of electrons on N atom is involved in resonance with benzene ring and cannot be easily donated. When an alkyl group such as methyl or ethyl is present, the electron releasing inductive effect (+I effect) of alkyl group increases the electron density on N atom and unshared pair of electrons on N atom can be easily donated.
The increasing order for pKb values is C2H5NH2(3.29)<C6H5NHCH3(9.30)<C6H5NH2(9.38). The numbers in bracket indicates pKb. Higher is the pKb value, lower is the basic strength of amine. Thus, the increasing order for pKb values indicates
decreasing order of the basic strength of amine. Aniline is least basic and ethanamine is most basic.
Aniline is least basic as unshared pair of electrons on N atom is involved in resonance with benzene ring and cannot be easily donated. When an alkyl group such as methyl or ethyl is present, the electron releasing inductive effect (+I effect) of alkyl group increases the electron density on N atom and unshared pair of electrons on N atom can be easily donated.
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