The $p K_{b}$ values of ammonia, methenamine and benzenamine (aniline) are $4.75, 3.38$ & $9.38$ respectively. Arrange them in the increasing order of their basic strength.

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Solution

Lower the value of $p K_{b}$ , higher is the basicity of amine.
The increasing order of their basic strength corresponds to decreasing order of $p K_{b}$ values.
$b e n z e n a m i n e (a n i l i n e) (9.38) < a m m o n i a (4.75) < m e t h a n a m i n e (3.38)$

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Q. Arrange the following
(i) In decreasing order of the

p K_{b}

values:

C_{2} H_{5} N H_{2}, C_{6} H_{5} N H C H_{3}, (C_{2} H_{5})_{2} N H

and

C_{6} H_{5} N H_{2}

(ii) In increasing order of basic strength:

C_{6} H_{5} N H_{2}, C_{6} H_{5} N (C H_{3})_{2}, (C_{2} H_{5})_{2} N H

and

C H_{3} N H_{2}

(iii) In increasing order of basic strength:
(a) Aniline, p-nitroaniline and p-toluidine

Q. Arrange the following in increasing order of their basic strength;
Aniline, p-nitroaniline, p-toulidine

Arrange the following:

(i) In the decreasing order of the $p K_{b}$ values
$C_{2} H_{5} N H_{2}, C_{6} H_{5} N H C H_{3}, (C_{2} H_{5})_{2} N H$ and $C_{6} H_{5} N H_{2}$

(ii) In increasing order of basic strength:
$C_{6} H_{5} N H_{2}, C_{6} H_{5} N (C H_{3})_{2}, (C_{6} H_{5})_{2} N H a n d C H_{3} N H_{2}$

(iii) In increasing order of basic strength:
(a) Aniline, p-nitroaniline and p-toluidine
(b) $C_{6} H_{5} N H_{2}, C_{6} H_{5} N H C H_{3}, C_{6} H_{5} C H_{2} N H_{2}$

(iv) In decreasing order of basic strength in gas phase :

$C_{2} H_{5} N H_{2}, (C_{2} H_{5})_{2} N H, (C_{2} H_{5})_{3} N a n d N H_{3}$

(v) In increasing order of boiling point:

$C_{2} H_{5} O H, (C H_{3})_{2} N H, C_{2} H_{5} N H_{2} .$

(vi) In increasing order of solubility in water :

$C_{6} H_{5} N H_{2}, (C_{2} H_{5})_{2} N H, C_{2} H_{5} N H_{2} .$

Q. When nitration of aniline occurs, ortho, meta and para nitro aniline is formed. Arrange them in the increasing order of their

%

yield.

Q. Arrange the compounds in an increasing order of their basic strengths:

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The pKb values of ammonia, methenamine and benzenamine (aniline) are 4.75,3.38 & 9.38 respectively. Arrange them in the increasing order of their basic strength.

Lower the value of pKb, higher is the basicity of amine. The increasing order of their basic strength corresponds to decreasing order of pKb values.benzenamine (aniline) (9.38)<ammonia (4.75)<methanamine (3.38)

The $p K_{b}$ values of ammonia, methenamine and benzenamine (aniline) are $4.75, 3.38$ & $9.38$ respectively. Arrange them in the increasing order of their basic strength.

Lower the value of $p K_{b}$ , higher is the basicity of amine.
The increasing order of their basic strength corresponds to decreasing order of $p K_{b}$ values.
$b e n z e n a m i n e (a n i l i n e) (9.38) < a m m o n i a (4.75) < m e t h a n a m i n e (3.38)$