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 :

Percentage composition of compound $$X$$ is C = 61.01 %, H = 15.25 %, N = 23.74 %.

First, determine the empirical formula.
Moles of each element (using atomic masses C = 12, H = 1, N = 14):
$$n_C = \frac{61.01}{12} = 5.084$$
$$n_H = \frac{15.25}{1} = 15.25$$
$$n_N = \frac{23.74}{14} = 1.695$$

Divide by the smallest value (1.695) to obtain the simplest ratio:
$$\frac{n_C}{1.695} = 3.00,\; \frac{n_H}{1.695} = 9.00,\; \frac{n_N}{1.695} = 1.00$$

Empirical formula = $$\mathrm{C_3H_9N}$$.

All isomers of $$\mathrm{C_3H_9N}$$ are primary amines: $$\mathrm{CH_3CH_2CH_2NH_2}$$ (1-aminopropane) and $$\mathrm{(CH_3)_2CHNH_2}$$ (2-aminopropane, isopropylamine).

Reaction of $$X$$ with $$HNO_2/H_2O$$: a primary aliphatic amine is converted into the corresponding alcohol with loss of nitrogen.
$$\mathrm{RCH_2NH_2 + HNO_2 \rightarrow RCH_2OH + N_2 \uparrow + H_2O}$$

Therefore compound $$Y$$ is the alcohol having the same carbon skeleton as $$X$$.

Case 1: If $$X = \mathrm{CH_3CH_2CH_2NH_2}$$ → $$Y = \mathrm{CH_3CH_2CH_2OH}$$ (1-propanol).
Controlled oxidation of 1-propanol gives propanal ($$\mathrm{CH_3CH_2CHO}$$), which does not give the iodoform test. Case 2: If $$X = \mathrm{(CH_3)_2CHNH_2}$$ → $$Y = \mathrm{(CH_3)_2CHOH}$$ (2-propanol, isopropyl alcohol).
Controlled oxidation of 2-propanol gives acetone ($$\mathrm{CH_3COCH_3}$$). Both 2-propanol (because it contains the $$\mathrm{CH_3CH(OH)}$$ group) and acetone (a methyl ketone) give a positive iodoform test.

The experiment shows that the oxidation product $$Z$$ does give the iodoform test; hence Case 2 is correct.

Therefore $$X$$ must be 2-aminopropane (isopropylamine): $$\mathrm{(CH_3)_2CHNH_2}$$.

Option B which is: isopropylamine.