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

Arrange the following in order of their decreasing acidic strength:
$(I)$ $CH_3NH_3^+$
$(II)$ $C_6H_5NH_3^+$
$(III)$ $[H_2N-C(=NH)-NH_2]H^+$
$(IV)$ $O_2N-C_6H_4-NH_3^+$

In the chemical reaction given above,$A$ and $B$ respectively are:

Arrange the following in decreasing order of their $pK_{b}$ values:

Identify $Z$ in the above sequence of reactions.

If aniline is treated with a $1:1$ mixture of conc. $HNO_{3}$ and conc. $H_{2}SO_{4}$,$p$-nitroaniline and $m$-nitroaniline are formed in nearly equal amounts. This is due to:

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