Step 1: Identify compound A with molecular formula $C_3H_5N$.
Degrees of unsaturation for $C_3H_5N$ = $(2 \times 3 + 2 + 1 - 5)/2 = 4/2 = 2$. Two degrees of unsaturation corresponds to a triple bond ($C \equiv N$, i.e., a nitrile group). Compound A with $C_3H_5N$ and 2 DoU is propionitrile (ethyl cyanide): $CH_3CH_2CN$.
Step 2: Determine compound B from $A + C_6H_5MgBr$, then hydrolysis.
A nitrile reacts with a Grignard reagent followed by hydrolysis to give a ketone: \[ CH_3CH_2CN + C_6H_5MgBr \xrightarrow{1.\;\text{react}\;2.\;H_3O^+} CH_3CH_2-CO-C_6H_5 \] Compound B is propiophenone (phenyl ethyl ketone, 1-phenylpropan-1-one): $C_6H_5COCH_2CH_3$.
Step 3: Confirm compound B using all given chemical tests.
(i) Orange-red precipitate with 2,4-DNP: confirms a carbonyl group ($C=O$). B has a ketone $C=O$. (ii) No iodoform test: iodoform ($CHI_3$) forms only with $CH_3CO-$ or $-CHOH-CH_3$ groups. $C_6H_5CO-CH_2CH_3$ has $-CH_2-$ next to $C=O$, not $-CH_3$. No iodoform. (iii) No Tollens'/Fehling's test: aldehydes reduce these; ketones do not. B is confirmed as a ketone. (iv) No Br2 water decolourisation: no C=C bond in the aliphatic chain. All tests are consistent with propiophenone.
Step 4: Identify compound C from $B + NH_2OH$.
Ketones react with hydroxylamine ($NH_2OH$) in acid catalyst by nucleophilic addition-elimination to form oximes: \[ C_6H_5COCH_2CH_3 + NH_2OH \rightarrow C_6H_5C(=NOH)CH_2CH_3 + H_2O \] Compound C is the oxime of propiophenone: $C_6H_5C(=NOH)CH_2CH_3$.
Step 5: Summarise the structural identifications.
A = $CH_3CH_2CN$ (propionitrile). B = $C_6H_5COCH_2CH_3$ (propiophenone). C = $C_6H_5C(=NOH)CH_2CH_3$ (oxime of propiophenone).
Step 6: State the final boxed answer.
\[ \boxed{A = CH_3CH_2CN;\; B = C_6H_5COCH_2CH_3;\; C = C_6H_5C(=NOH)CH_2CH_3} \]