In the following compounds,the order of basic | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) $(I)$ Piperidine is the most basic. The $N$ lone pair is in an $sp^3$ hybrid orbital and there is no resonance.$(II)$ In pyridine,the $N$ lone pai

Vedclass · Accepted Answer

$I > III > II > IV$

Vedclass · Answer

(D) $(I)$ Piperidine is the most basic. The $N$ lone pair is in an $sp^3$ hybrid orbital and there is no resonance.$(II)$ In pyridine,the $N$ lone pair is in an $sp^2$ hybrid orbital. The $sp^2$ hybrid is smaller than the $sp^3$ hybrid,so there is a stronger attraction to the nucleus,making it less basic than $I$.$(III)$ In morpholine,the $N$ lone pair is in an $sp^3$ hybrid orbital. However,the electron-withdrawing effect of the $O$ atom in the ring makes the lone pair less available for donation compared to piperidine.$(IV)$ Pyrrole is a very weak base. The $N$ lone pair is involved in the aromatic sextet. Protonation would destroy the aromaticity,which is an energetically unfavourable process.Therefore,the correct order of basicity is $I > III > II > IV$.

In the following compounds,the order of basicity is:
$I$: Piperidine
$II$: Pyridine
$III$: Morpholine
$IV$: Pyrrole

Similar Questions

Which one of the following gives the most stable diazonium salt?

Which of the following compounds does $NOT$ react with bromine in an alkaline medium?

The major products $A$ and $B$ from the following reactions are:

Which of the following reactions will not take place?

The major product formed in the following reaction is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module

In the following compounds,the order of basicity is:$I$: Piperidine$II$: Pyridine$III$: Morpholine$IV$: Pyrrole