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प्रश्न
Account for the following:
pKb of aniline is more than that of methylamine.
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उत्तर १
The one which has less basicity has more pKb. The pKb of aniline is more than that of methylamine because aniline has five resonance structures, due to which the unshared electron pair of the nitrogen atom present in it gets delocalized on the benzene ring. This reduces the electron density and aniline does not donate electrons; i.e., the basicity of aniline is less.
On the other hand, the presence of a methyl group in methylamine increases the +I effect, due to which the electron density on the nitrogen atom increases and it readily donates electrons, i.e., methylamine is more basic. Therefore, the pKb of aniline is more than that of methylamine.
उत्तर २
In aniline, the lone pair of electrons on the N atom is delocalized over the benzene ring. As a result, the electron density of nitrogen decreases. In contrast, in CH3NH2 (+I) effect of CH3 increases the electron density on the N atom. Therefore, aniline is a weaker base than methylamine, and hence its pKb value is higher than that of methylamine.
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संबंधित प्रश्न
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\[\ce{C6H5NH2 + H2SO4 (conc.)}\]
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\[\ce{C6H5NH2 + Br2 (aq) ->}\]
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\[\begin{array}{cc}
\ce{CH3CN to CH3 - C - CH3}\\
\phantom{...........}||\\
\phantom{...........}\ce{O}
\end{array}\]
Do the following conversions in not more than two steps:
