Among the compounds given below,the order of | vedclass

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(A) The correct order of basicity is $I > III > II > IV$.In compounds $I$ (piperidine) and $III$ (morpholine),the nitrogen atom is $sp^3$-hybridized,m

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$I > III > II > IV$

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(A) The correct order of basicity is $I > III > II > IV$.In compounds $I$ (piperidine) and $III$ (morpholine),the nitrogen atom is $sp^3$-hybridized,making them more basic than compounds $II$ and $IV$.Between $I$ and $III$,compound $III$ is less basic than $I$ due to the electron-withdrawing effect of the oxygen atom present in the ring.In compounds $II$ (pyridine) and $IV$ (pyrrole),the nitrogen atom is $sp^2$-hybridized.In compound $II$,the lone pair on the nitrogen atom is in an $sp^2$ orbital and is not involved in resonance,making it available for protonation.In compound $IV$,the lone pair on the nitrogen atom is involved in the aromatic sextet (resonance),making it the least basic.Thus,the order is $I > III > II > IV$.

Among the compounds given below,the order of basicity is:
$(I)$ Piperidine
$(II)$ Pyridine
$(III)$ Morpholine
$(IV)$ Pyrrole

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Among the compounds given below,the order of basicity is:$(I)$ Piperidine$(II)$ Pyridine$(III)$ Morpholine$(IV)$ Pyrrole