(I) | (II) |
(III) | (IV) |
A) IV > I > III > II
B) III > I > IV > II
C) II > I > III > IV
D) I > III > II > IV
Correct Answer: D
Solution :
[d] In I the unshared pair of electrons on N always available for protonation in III due to presence of electronegative O atom the electron density on N is decreased. In II and IV resonance suppresses the basic character.You need to login to perform this action.
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