Question

The decreasing order of $p{K}_a$ values of the labelled nitrogen atoms in the above compound is :

• A. 3 > 2 > 1 > 4 > 5
• B. 4 > 3 >2 > 1 > 5
• C. 3 > 4 > 2 > 1 > 5
• D. 5 > 1 > 2 > 4 > 3

Answer 1

The correct option is C 3 > 4 > 2 > 1 > 5

We have to draw conjugate acids to check basic character:

Conjugate acids for $\left(3\right)$ and $\left(4\right)$ are shown in the above structure.

Basic character is least of $\left(5\right)$ as it is between two $-R$ showing group (Electron withdrawing group) which decreases basic strength. After that least basic is one as it has one $-C=O$ group adjacent to it $(-R)$.

$\left(3\right)$ is most basic as electronegativity of $N$ is less than that of oxygen and it will easily let hold of its lone pair to go into resonance, while ${}^{\prime }{O}^{\prime }$ being more electronegative will not allow it's lone pair easily. So, a conjugate base of $\left(3\right)$ is more stable than $\left(4\right)$. Therefore $\left(3\right)$ is more basic than $\left(4\right)$

$\left(2\right)$ is a primary amine whose lone pair is in resonance. it is more basic than $\left(1\right)$ and $\left(5\right)$

Decreasing $p{K}_a$ order: $\left(3\right)>\left(4\right)>\left(2\right)>\left(1\right)>\left(5\right)$