Question

The correct order of increasing basicity is

$C{H}_3\underset{I}{C{H}_2}N{H}_{2}C{H}_3\stackrel{\stackrel{NH}{||}}{\underset{II}{C}}C{H}_3\stackrel{\stackrel{O}{||}}{\underset{III}{C}}N{H}_2$

• A. II < III < I
• B. $I\approx III<II$
• C. I < II < III
• D. III < I < II

Answer 1

The correct option is D

III < I < II

II is most basic because delocalisation of electron pair leads to negative charge on N making it electron -rich and hence liable to be attacked by proten very easily. Moreover, the corresponding protonated species is very much stable because of equivalent contributing structures.

Species III is least basic because of delocalisation of electron pair on N, making it less available for protonation. Species I lies in mid-way, thus