What is the correct decreasing order of basic | vedclass

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(D) The basic strength depends on the availability of the lone pair of electrons on the nitrogen atom.$I$ (Pyrrole): The lone pair on nitrogen is invo

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$II > III > IV > I$

Vedclass · Answer

(D) The basic strength depends on the availability of the lone pair of electrons on the nitrogen atom.$I$ (Pyrrole): The lone pair on nitrogen is involved in aromaticity (delocalized),making it the least basic.$II$ (Piperidine): The lone pair is on an $sp^3$ hybridized nitrogen and is localized,making it the most basic.$III$ (Pyridine): The lone pair is on an $sp^2$ hybridized nitrogen and is localized,but $sp^2$ nitrogen is more electronegative than $sp^3$ nitrogen,making it less basic than $II$.$IV$ (Aniline): The lone pair is delocalized into the benzene ring through resonance,making it more basic than $I$ but less basic than $III$.Therefore,the correct decreasing order of basic strength is $II > III > IV > I$.

What is the correct decreasing order of basic strength?

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