A student performed analysis of aliphatic organic compound ‘X’ which on analysis gave C = 61.01%, H = 15.25%, N = 23.74%.
This compound, on treatment with HNO₂/H₂O produced another compound ‘Y’ which did not contain any nitrogen atom. However, the compound ‘Y’ upon controlled oxidation produced another compound ‘Z’ that responded to iodoform test.
The structure of ‘X’ is :

Solution

Chemistry ma’am · 3 steps

Find the formula from the percentages first, then let the iodoform test pick the right isomer.

1. Empirical formula. Divide each percentage by its atomic mass: C = 61.01/12 = 5.08, H = 15.25/1 = 15.25, N = 23.74/14 = 1.70. Dividing by the smallest gives C : H : N = 3 : 9 : 1, so X is C₃H₉N (molar mass 59).
2. Follow the reactions. A primary amine with HNO₂ loses its nitrogen as N₂ and gives an alcohol, which is the nitrogen-free Y. Z gives the iodoform test, so Z must contain a CH₃CO⁻ group. That means Y was a CH₃CH(OH)⁻ alcohol.

   Option	Formula	Y (alcohol)	Z on oxidation	Iodoform
   (a)	C₄H₁₁N, C = 65.75%	butan-2-ol	butanone	yes, but wrong formula
   (b)	C₃H₉N	CH₃CH₂CH₂OH	CH₃CH₂CHO	no
   (c)	C₈H₁₁N, aromatic	1-phenylethanol	acetophenone	yes, but not aliphatic
   (d)	C₃H₉N	CH₃CH(OH)CH₃	CH₃COCH₃	yes

3. So. Only isopropylamine, (CH₃)₂CHNH₂, fits both the formula C₃H₉N and the iodoform result. The answer is (d).

The idea to keep. Primary amine plus HNO₂ gives an alcohol. If that alcohol oxidises to an iodoform-positive compound, it was CH₃CH(OH)⁻, so the NH₂ sat on a carbon carrying a CH₃.

Answer from the NTA final answer key. Solution by PrepWiser.