Amines Part 3

Concept:Empirical formula calculation and reaction of primary amine with $\mathrm{HNO}_2$ to form alcohol. Iodoform test indicates methyl ketone or ethanol.Explanation:Given percentages: $61.01\%$ C, $15.25\%$ H, $23.74\%$ N.Compute moles: C: $61.01/12 \approx 5.08$, H: $15.25/1 = 15.25$, N: $23.74/14 \approx 1.696$.Divide by smallest: C: $5.08/1.696 \approx 3$, H: $15.25/1.696 \approx 9$, N: $1$.Empirical formula: $\mathrm{C}_3\mathrm{H}_9\mathrm{N}$. It is a saturated aliphatic primary amine.Only primary amines react with $\mathrm{HNO}_2/\mathrm{H}_2\mathrm{O}$ to give alcohols (Y) without nitrogen.Y on controlled oxidation gives Z that gives positive iodoform test.Iodoform test confirms presence of $\mathrm{CH}_3\mathrm{CO}-$ group or $\mathrm{CH}_3\mathrm{CH(OH)}-$.Thus Y must be a secondary alcohol $\mathrm{CH}_3\mathrm{CH(OH)R}$ (e.g., isopropanol).The only $\mathrm{C}_3\mathrm{H}_9\mathrm{N}$ primary amine that yields such Y is isopropylamine, $(\mathrm{CH}_3)_2\mathrm{CHNH}_2$.Oxidation: $(\mathrm{CH}_3)_2\mathrm{CHOH} \xrightarrow{[O]} \mathrm{CH}_3\mathrm{COCH}_3$ (acetone, gives iodoform).Among options, C is isopropylamine.Answer:Option C: $\underset{\mathrm{CH}_3}{\overset{\mathrm{CH}_3}{\mathrm{CH}}} - \mathrm{NH}_2$