Question

The order of basicity of the compounds $C_2{H}_5{NH}_2,{\left(C_2{H}_5\right)}_{2}NH,{\left(C_2{H}_5\right)}_{3}N$ (in aprotic solvent) is:

• A. ${\left(C_2{H}_5\right)}_{3}N>{\left(C_2{H}_5\right)}_{2}NH>C_2{H}_5{NH}_2$
• B. ${\left(C_2{H}_5\right)}_{3}N<{\left(C_2{H}_5\right)}_{2}NH<C_2{H}_5{NH}_2$
• C. ${\left(C_2{H}_5\right)}_{3}N>C_2{H}_5{NH}_2>{\left(C_2{H}_5\right)}_{2}NH$
• D. none of these

Answer 1

Answer: (A)

${\left(C_2{H}_5\right)}_{3}N>{\left(C_2{H}_5\right)}_{2}NH>C_2{H}_5{NH}_2$

The order of basicity in aqueous solution when $RN{H}_2(R\ne C{H}_3)$ is $2^o>3^o>1^o>N{H}_3$

$\therefore (C_2{H}_5{)}_{2}NH>(C_2{H}_5{)}_{3}NH>C_2{H}_{5}N{H}_2>N{H}_3$