A student performed analysis of aliphatic organic compound 'X' which on analysis gave $\mathrm{C}=61.01\%$, $\mathrm{H}=15.25\%$, $\mathrm{N}=23.74\%$. This compound, on treatment with $\mathrm{HNO}_{2}/\mathrm{H}_{2}\mathrm{O}$ 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) $\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{CH}_{2}\mathrm{NH}_{2}$

JEE Main · 2026 · 28 Jan 2026 · Shift 2 · Compounds Containing Nitrogen

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) $CH_{3} CH_{2} CH_{2} NH_{2}$
(B)(see structure)
(C)(see structure)
✓ Correct
(D)(see structure)

Answer: (C)

Solution

The mass percentages give mole ratios

C : H : N = \frac{61.01}{12} : \frac{15.25}{1} : \frac{23.74}{14} = 3 : 9 : 1

, so X is

C_{3} H_{9} N

. A single nitrogen lost entirely on treatment with

HN O_{2} / H_{2} O

marks X as a primary amine, which is converted to the corresponding alcohol Y. For the controlled oxidation of Y to give a compound Z that answers the iodoform test, Y must be a secondary alcohol of the

C H_{3} - CHOH -

type; here propan-2-ol is oxidised to acetone, a methyl ketone. Hence X is isopropylamine,

(C H_{3})_{2} CHN H_{2}

JEE Main · 2026 · 28 Jan 2026 · Shift 2 · Single correct option

Practise more: all JEE Main PYQs on Compounds Containing Nitrogen →

JEE Main · 2026 · 28 Jan 2026 · Shift 2 · Compounds Containing Nitrogen

Solution

More JEE Main questions on Compounds Containing Nitrogen