The correct order of basicity of the given compounds is

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${CH}_{3} {CH}_{2} {NH}_{2} > HO {({CH}_{2})}_{3} {NH}_{2} > HO {({CH}_{2})}_{2} {NH}_{2}$

${CH}_{3} {CH}_{2} {NH}_{2} < HO {({CH}_{2})}_{3} {NH}_{2} > HO {({CH}_{2})}_{2} {NH}_{2}$

${CH}_{3} {CH}_{2} {NH}_{2} > HO {({CH}_{2})}_{3} {NH}_{2} < HO {({CH}_{2})}_{2} {NH}_{2}$

${CH}_{3} {CH}_{2} {NH}_{2} < HO {({CH}_{2})}_{3} {NH}_{2} < HO {({CH}_{2})}_{2} {NH}_{2}$

answer is C.

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Detailed Solution

The correct order of basicity of the given compounds is as follows:

${CH}_{3 } {CH}_{2} {NH}_{2} > HO {({CH}_{2} )}_{3} {NH}_{2} > HO {({CH}_{2} )}_{2} {NH}_{2} $

$OH$ group is an electron withdrawing group and hence, it increases the partial positive charge on $N$ atom. Therefore, lone pair of $N$ cannot be easily donated and also, the basicity decreases.

As the number of carbon atoms between $- OH$ group and amino group increases, the electron withdrawing capacity of $- OH$ group on amine N decreases and basicity increases.

The $- OH$ group causes decrease in the basicity of $- {NH}_{2}$ group

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