Question

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 HNO\(_2\)/H\(_2\)O 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:

• A. Ph—CH—NH\(_2\) (with CH\(_3\) substituent)
• B. CH\(_3\)—CH—NH\(_2\) (with CH\(_3\) substituent)
• C. CH\(_3\)—CH\(_2\)—CH—CH\(_3\) (with NH\(_2\) substituent)
• D. CH\(_3\)—CH\(_2\)—CH\(_2\)—NH\(_2\)

Hint:

For nitrous acid reactions: - Primary aliphatic amines → alcohols. - Secondary amines → nitroso compounds. - Tertiary amines → no reaction.

Answer 1

The Correct Option is B

Step 1: Empirical formula check.
Given percentages:
C = 61.01%, H = 15.25%, N = 23.74%
This matches with isopropylamine (CH3-CH(NH2)-CH3).

Step 2: Reaction with HNO2 / H2O.
Primary aliphatic amines react with nitrous acid to give alcohols.
CH3-CH(NH2)-CH3 → (HNO2 / H2O) → CH3-CH(OH)-CH3
Thus, compound Y is isopropanol.

Step 3: Controlled oxidation of Y.
Isopropanol oxidizes to acetone.
CH3-CH(OH)-CH3 → (O) → CH3-CO-CH3
Compound Z is acetone.

Step 4: Iodoform test.
Acetone gives a positive iodoform test due to the presence of the COCH3 group.

Therefore:
X = isopropylamine (CH3-CH(NH2)-CH3)