A student performed analysis of an 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 :
- CH3-CH2CH(NH2)-CH3
- CH3CH2CH2NH2
- Ph-CH(CH3)-NH2
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Answer
(4) [figure]
Solution
Empirical formula of X:
C : 61.0112 = 5.08; H : 15.251 = 15.25; N : 23.7414 = 1.696
C : H : N = 5.081.696 : 15.251.696 : 1.6961.696 = 3 : 9 : 1
X is C3H9N (an aliphatic amine).
X with HNO2 gives Y without nitrogen (N2 is evolved), X is a primary amine and Y is an alcohol.
Controlled oxidation of Y gives Z which gives the iodoform test, Z is a methyl ketone, so Y is the secondary alcohol propan-2-ol.
(CH3)2CH-NH2 HNO2/H2O (CH3)2CH-OH + N2
(CH3)2CH-OH [O] CH3COCH3 (gives iodoform test)
X = propan-2-amine.
Option (1) is C4H11N (C = 65.75%), option (2) gives propan-1-ol propanal (no iodoform test) and option (3) is aromatic.
Answer: (4)