An organic compound X having molecular mass 60 is found to contain C = 20%, H = 6.67% and N = 46.67%, while rest is oxygen. On heating, it gives ammonia along with a solid residue. The solid residue gave violet colour with alkaline copper sulphate solution. The compound X is

Options

A. `CH_3CH_2CONH_2`
B. `CH_3NCO`
C. `CH_3CONH_2`
D. `(NH_2)_2CO`

Answer

D

Explanation

Percentage of Oxygen = 100 - (20.0 + 6.67 + 46.67) = 26.66
C : H : N : O = `20/12 : (6.67)/(1.0) : (46.67)/(14) : (26.66)/(16) = 1.67 : 6.67 : 3.33 . 1.67 = 1: 4: 2:1`
`:. E.F. = CH_4 N_2O`
and mol w.t. = 60 (given)
`:. M.F. xx ("Mol.wt")/("E.F.wt") = CH_4 N_2O xx60/60 = CH_4N_2O`
MF of `(NH_(2))_(2)` CO matches with that of `CH_4N_2O` therefore, the given compound is urea, `NH_2CONH_2`. Since urea on heating loses NH, and gives a residue called biuret which gives a violet colouration with alk. `CuSO_4` solution, therefore, the given organic compound is `NH_2 CONH_2`
`NH_2 CONH_2+HNHCONH_2 overset(Delta)underset(-NH_3)rarr`
`underset("Biuret")(NH_2CNHCONH_2)overset(Alk. CuSO_3)rarr` Violet colouration