A) \[I>II>III\]
B) \[III>II>I\]
C) \[I>II\approx III\]
D) \[I=II\equiv III\]
Correct Answer: A
Solution :
[a] Note the point of difference in the given compounds which here lies at \[\beta \]-carbon. In I, II, III, the \[\beta \]-carbon atoms are \[s{{p}^{3}},s{{p}^{2}}\] and sp hybridised respectively which in turn cause the difference in their s-character. We know that more is the s character of an atom, greater will be its electron-withdrawing nature. Thus sp (50% s character) hybridised carbon is most electron-withdrawing, while \[s{{p}^{3}}\] (25% s-character) is least electron-withdrawing. Further, we know that presence of an electron-withdrawing group decreases basicity of an amine. ThusYou need to login to perform this action.
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