JEE Main 2026 — Jan 28, Shift 2 — Chemistry solution

JEE Main2026Jan 28, Shift 2Chemistry

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:

Solution

Step 1 — empirical formula from %: $\frac{61.01}{12} : \frac{15.25}{1} : \frac{23.74}{14} = 5.08 : 15.25 : 1.70 \approx 3 : 9 : 1$ . Empirical formula C $_{3}$ H $_{9}$ N. Both options (1) and (3) have this formula.

Step 2 — chemical clues: X is a primary aliphatic amine (reacts with HNO $_{2}$ to lose N $_{2}$ and give an alcohol Y). Mild oxidation of Y produces Z that gives a positive iodoform test, so Z must contain CH $_{3}$ CO- (or CH $_{3}$ CH(OH)- as precursor).

If X = n-propylamine (option 1): HNO $_{2}$ gives n-propanol, mild oxidation gives propanal (no methyl-keto group, no iodoform).

If X = isopropylamine (option 3): HNO $_{2}$ gives isopropanol $(CH_{3})_{2} CHOH$ ; controlled oxidation gives acetone $CH_{3} COCH_{3}$ which gives a positive iodoform test.

$∴$ X $= (CH_{3})_{2} CHNH_{2}$ .

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