Question

An organic compound (A) with molecular formula C$_3$H$_5$N on reaction with C$_6$H$_5$MgBr followed by hydrolysis, gives a compound (B). Compound (B) forms an orange-red precipitate with 2,4-DNP reagent and does not give iodoform test. It neither reduces Tollens' or Fehling's reagent nor does it decolourise bromine water. On drastic oxidation with chromic acid it gives a carboxylic acid (C) having molecular formula C$_7$H$_6$O$_2$. Identify the compounds (A), (B) and (C). Write the reactions of compound (A) with C$_6$H$_5$MgBr followed by hydrolysis to give compound (B).

Hint:

Nitrile + Grignard reagent followed by hydrolysis always gives a ketone.

Answer 1

Step 1: Identify C from its molecular formula and drastic oxidation.
C is a carboxylic acid with formula $C_7H_6O_2$. This matches benzoic acid: $C = C_6H_5COOH$. Since C comes from strong oxidation of B, B must contain a phenyl group with a side chain.
Step 2: Identify B using chemical test observations.
B gives 2,4-DNP test (carbonyl present), does not give Tollens'/Fehling's test (B is a ketone, not aldehyde), and does not give iodoform test (no $CH_3CO{-}$ unit). Since oxidation gives $C_6H_5COOH$, B must be propiophenone: $B = C_6H_5COCH_2CH_3$.
Step 3: Identify A and write the reaction with $C_6H_5MgBr$.
$C_3H_5N$ with one degree of unsaturation is propanenitrile. Grignard addition then hydrolysis: \[ CH_3CH_2CN + C_6H_5MgBr \xrightarrow{H_3O^+} C_6H_5COCH_2CH_3 \] \[ \boxed{A = CH_3CH_2CN,\quad B = C_6H_5COCH_2CH_3,\quad C = C_6H_5COOH} \]