Question

Among the following compounds, the increasing order of their basic strength is :

• A. $\left(II\right)<\left(I\right)<\left(IV\right)<\left(III\right)$
• B. $\left(II\right)<\left(I\right)<\left(III\right)<\left(IV\right)$
• C. $\left(I\right)<\left(II\right)<\left(IV\right)<\left(III\right)$
• D. $\left(I\right)<\left(II\right)<\left(III\right)<\left(IV\right)$

Answer 1

The correct option is A $\left(II\right)<\left(I\right)<\left(IV\right)<\left(III\right)$

Among the following compounds, the increasing order of their basic strength is $\left(II\right)<\left(I\right)<\left(IV\right)<\left(III\right)$

$\left(III\right)$ is most basic due to $+I$ effect of $-C{H}_3$ group attached to $N$. This increases electron density on $N$ so that $N$ can easily donate its lone pair of electrons to suitable acid.

$\left(II\right)$ is least basic as the lone pair of electrons on $N$ is part of pi electrons of aromatic ring. Donation of this lone pair of electrons will break the stability of the aromatic ring.

$\left(I\right)$ is less basic than $\left(IV\right)$ as the lone pair of electrons on $N$ is in conjugation with aromatic ring.