The most basic 'N' atom in the following mole | vedclass

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

Question

(D) To determine the most basic nitrogen atom,we analyze the availability of the lone pair on each nitrogen atom:$1$. Nitrogen at position $A$ is part

Vedclass · Accepted Answer

$D$

Vedclass · Answer

(D) To determine the most basic nitrogen atom,we analyze the availability of the lone pair on each nitrogen atom:$1$. Nitrogen at position $A$ is part of an aromatic pyridine-like ring. Its lone pair is in an $sp^2$ orbital perpendicular to the $\pi$-system,making it available for protonation.$2$. Nitrogen at position $B$ is an aniline-type amine. Its lone pair is delocalized into the aromatic ring,reducing its basicity.$3$. Nitrogen at position $C$ is an amide nitrogen. Its lone pair is delocalized into the adjacent carbonyl group $(C=O)$,making it non-basic.$4$. Nitrogen at position $D$ is a secondary aliphatic amine in a saturated piperidine ring. Its lone pair is in an $sp^3$ orbital and is not involved in resonance or delocalization. Aliphatic amines are generally more basic than aromatic amines or pyridine-like nitrogens.Therefore,the nitrogen at position $D$ is the most basic.

The most basic $'N'$ atom in the following molecule is:

Similar Questions

If an organic compound is treated with $NaNO_2$ and $HCl$ in the presence of ice and nitrogen gas is evolved,what is the compound?

What is the product obtained upon the hydrolysis of a $1^o$ amine?

Complete reduction of $benzene-diazonium$ chloride with $Zn/HCl$ gives:

The correct order of basicity of amines in water is :

Identify the product of the following reaction:
$\text{Aniline} \xrightarrow[ii) \text{H}_2\text{O}, \Delta]{i) \text{NaNO}_2 + \text{HCl}, 273 \text{ K}} \text{Product}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module