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(C) $1$. Empirical formula calculation: $C = 61.01/12 = 5.08$,$H = 15.25/1 = 15.25$,$N = 23.74/14 = 1.69$. Dividing by the smallest value $(1.69)$: $C

Vedclass · Accepted Answer

$(CH_3)_2CH-NH_2$

Vedclass · Answer

(C) $1$. Empirical formula calculation: $C = 61.01/12 = 5.08$,$H = 15.25/1 = 15.25$,$N = 23.74/14 = 1.69$. Dividing by the smallest value $(1.69)$: $C = 3$,$H = 9$,$N = 1$. The empirical formula is $C_3H_9N$. $2$. Reaction with $HNO_2$: Aliphatic primary amines react with $HNO_2$ to form unstable diazonium salts,which decompose to form alcohols $(Y)$ and evolve $N_2$ gas. For $C_3H_9N$ (isopropylamine),the reaction is: $(CH_3)_2CH-NH_2 + HNO_2 \rightarrow (CH_3)_2CH-OH (Y) + N_2 + H_2O$. $3$. Oxidation and Iodoform test: Compound '$Y$' is isopropyl alcohol (propan$-2-$ol). Controlled oxidation of propan$-2-$ol gives acetone $(Z)$,which is $CH_3COCH_3$. Acetone contains the $CH_3CO-$ group and thus gives a positive iodoform test. Therefore,'$X$' is isopropylamine,$(CH_3)_2CH-NH_2$.

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