Question

Choose the correct order of decreasing basic strength of the following compounds in aqueous solution.
(i) $C_6{H}_{5}N{H}_2$
(ii) $C_2{H}_{5}N{H}_2$
(iii) $N{H}_3$
(iv) $(C{H}_3{)}_{2}NH$

• A. $\left(i\right)>\left(ii\right)>\left(iii\right)>\left(iv\right)$
• B. $\left(iv\right)>\left(ii\right)>\left(iii\right)>\left(i\right)$
• C. $\left(ii\right)>\left(i\right)>\left(iii\right)>\left(iv\right)$
• D. $\left(iv\right)>\left(iii\right)>\left(ii\right)>\left(i\right)$
• E. $\left(ii\right)>\left(iv\right)>\left(iii\right)>\left(i\right)$

Answer 1

Answer: (B)

$\left(iv\right)>\left(ii\right)>\left(iii\right)>\left(i\right)$

More the availability of $N-atom$ to donate its lone pair, more will be basicity of amine.
Thus, presence of electron releasing groups (like $-C{H}_3,-C_2{H}_5)$ increases the basicity but presence of electron withdrawing group (Like $-C_6{H}_5)$ decreases the basicity. Ammonia $\left(N{H}_3\right)$ because of the absence of any group is less basic than the amines having electron releasing group. Thus, the order of basic strength of the given compounds is
$\underset{\underset{(\text{because of two electron releasing}-C{H}_{3}group)}{\left(iv\right)}}{(C{H}_3{)}_{2}NH}>\underset{\left(iii\right)}{C_2{H}_{5}N{H}_2}>\underset{\left(iii\right)}{N{H}_3}>\underset{\left(i\right)}{C_2{H}_{5}N{H}_2}$