JEE Chemistry 2026 Question with Solution

Given: The compound X is an aliphatic organic compound with composition C = $61.01\%$, H = $15.25\%$, N = $23.74\%$.

Find: The structure of X.

Step 1: Empirical formula check The given percentages match isopropylamine, that is $CH_3-CH(NH_2)-CH_3$.

Step 2: Reaction with $HNO_2 / H_2O$ Primary aliphatic amines react with nitrous acid to give alcohols.

$$CH_3-CH(NH_2)-CH_3 \rightarrow CH_3-CH(OH)-CH_3$$

So compound Y is isopropanol, which contains no nitrogen atom.

Step 3: Controlled oxidation of Y Isopropanol on oxidation gives acetone.

$$CH_3-CH(OH)-CH_3 \rightarrow CH_3-CO-CH_3$$

So compound Z is acetone.

Step 4: Iodoform test Acetone gives a positive iodoform test because it contains the $COCH_3$ group.

Therefore, X = isopropylamine, and the correct option is B.

Given: X forms a nitrogen-free compound Y with $HNO_2/H_2O$, and Y on controlled oxidation gives Z which responds to iodoform test.

Find: Which option matches X.

From the nitrous acid reaction:

• Primary aliphatic amines give alcohols.
• Secondary amines give nitroso compounds.
• Tertiary amines do not give this conversion.

Hence X must be a primary aliphatic amine.

Now Y gives, on oxidation, a compound positive to iodoform test. This means Y must be a secondary alcohol of the type that oxidizes to a methyl ketone.

$$CH_3-CH(OH)-CH_3 \xrightarrow{[O]} CH_3-CO-CH_3$$

Thus Y is isopropanol.

So the starting amine X must be the corresponding amine:

$$CH_3-CH(NH_2)-CH_3$$

This is isopropylamine.

Therefore, the correct option is B.