Question

Compare the basic strengths of the given compounds.

Answer 1

(a) As the number of $-Ph$ group increases, the availability of lone pair on nitrogen decreases. This is because the lone pair is attracted by the phenyl groups. Thus more the phenyl groups, less is the basic strength.

$\therefore$ Order of basic strength: $PhN{H}_2>P{h}_{2}NH>P{h}_{3}N$

(b) In the $3^{rd}$compound, the lone pair is involved in making the ring aromatic. Thus it will be least basic.

The $2^{nd}$ compound has less availability of lone pair on the nitrogen due to the presence of phenyl group. The availability of lone pair is maximum in $1^{st}$ one. Hence it will be most basic.

$\therefore$ Order of Basic strength:

(Refer to Image)

(c) The $2^{nd}$ compound has the least availability of lone pair due to the presence of phenyl group directly attached to nitrogen. Hence it is least basic.

As the distance of phenyl group increases, the lone pair is more available. This is because the electron withdrawing effect of the phenyl group decreases.

Hence, the order of basic strength is:-

$Ph-C{H}_2-C{H}_2-N{H}_2>C{H}_3-\underset{|}{CH}-N{H}_2>C{H}_3-C{H}_2-\underset{|}{NH}$

$Ph$ $Ph$