Question

Arrange the following in the increasing order of the ${pK}_b$ values:

i) $C_2{H}_{5}N{H}_2$
ii)$C_6{H}_5{NHCH}_3$
iii)$(C_2{H}_5{)}_{2}NH$
iv) $C_6{H}_5{NH}_2$

• A. i < ii < iii < iv
• B. iii < i < ii < iv
• C. i < iii < ii < iv
• D. iv < ii < i < iii

Answer 1

The correct option is D iv < ii < i < iii

$p{K}_b$ value gives a measure of the basicity of a compound. Higher the value of $K_b$ of a base, stronger is the base. In aniline, $C_6{H}_{5}N{H}_2$, lone pair on nitrogen atom are withdrawn into the benzene ring. Benzene ring is an electron withdrawing group and hence, the lone pair is not easily available on nitrogen atom, so it cannot attract a proton easily.

N-Methylaniline, $C_6{H}_{5}NHC{H}_3$ has benzene ring attached to it as well as a methyl group attached to the nitrogen atom. The benzene ring withdraws lone pair from nitrogen atom . Methyl group attached to the nitrogen atom is an electron donating group and it donates electron to nitrogen atom increasing electron density .

Thus the $K_b$ value for N-Methylaniline is higher than aniline and it is a stronger base than aniline.

In $C_2{H}_{5}N{H}_2$, ethyl groups are electron donating group. It donates electrons to nitrogen atom increasing the electron density over nitrogen atom. Thus Ethylamine is a stronger base than N-Methylaniline.

In diethylamine, $(C_2{H}_5{)}_{2}NH$, two Ethyl groups attached to nitrogen atom are electron donating group. increasing the electron density on nitrogen atom. hence, diethylamine is a stronger base than ethylamine because of the presence of two electron donating groups.

Thus the correct option is D.