An organic compound 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 `NH_(3)` along with a solid residue. The solid residue gives violet color with alkaline copper sulphate solution. The compounds is

Options

A. `CH_(3)CH_(2)CONH_(2)`
B. `CH_(3)NCO`
C. `CH_(3)CONH_(2)`
D. `(NH_(2))_(2)CO`

Answer

D

Explanation

`%O = 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_(2)O`
Now E.F. wt. = 12 + 4 + 28 + 16 = 60
and Mol. Wt = 60 (given)
`:. M.F. = E.F. xx ("Mol. wt.")/("E.F. wt") = CH_(4)N_(2)O xx (60)/(60)`
`= CH_(4)NH_(2)O`
Now out of the given four compounds `(CH_(3)CH_(2)CONH_(2) (a) = C_(3)H_(7)NO, CH_(3)CONH_(2) (c) = C_(2)H_(5)NO, CH_(3)NCO (b) = C_(2)H_(3)NO)`, the `CH_(4)N_(2)O`, there-fore, the given compound is urea, `NH_(2)CONH_(2)`.
Further since urea on heating loses `NH_(3)` 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)`, i.e., option (d) is correct.

`underset("Biuret")(NH_(2)CONHCONH_(2)) overset(Alk. CuSO_(4))rarr` Violet colouration