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

Concept: Nitriles react with Grignard reagents to form ketones after hydrolysis. The molecular formula: \[ C_3H_5N \] suggests a nitrile.

Step 1: Identify compound (A). The nitrile having formula C$_3$H$_5$N is: \[ CH_3CH_2CN \] (Propanenitrile) \[ \boxed{A = CH_3CH_2CN} \]

Step 2: Reaction with phenylmagnesium bromide. \[ CH_3CH_2CN + C_6H_5MgBr \rightarrow \text{Intermediate} \] On hydrolysis: \[ CH_3CH_2COC_6H_5 \] is formed. This compound is propiophenone. \[ \boxed{B = C_6H_5COCH_2CH_3} \]

Step 3: Verify compound (B). It gives 2,4-DNP test because it contains a carbonyl group. It does not give Tollens' or Fehling's test because it is a ketone. It does not give iodoform test because it does not contain: \[ CH_3CO- \] group. Hence all observations are satisfied.

Step 4: Oxidation of compound (B). Strong oxidation of propiophenone converts the side chain into benzoic acid. \[ C_6H_5COCH_2CH_3 \xrightarrow{[O]} C_6H_5COOH \] Molecular formula: \[ C_7H_6O_2 \] Therefore: \[ \boxed{C = C_6H_5COOH} \] (Benzoic acid)

Reaction Sequence: \[ CH_3CH_2CN + C_6H_5MgBr \rightarrow CH_3CH_2C(MgBr)=NC_6H_5 \] \[ \xrightarrow{H_3O^+} CH_3CH_2COC_6H_5 \] \[ \boxed{ A = CH_3CH_2CN,\quad B = C_6H_5COCH_2CH_3,\quad C = C_6H_5COOH } \]