Give reasons for the following : (a) Haloalkanes react with KCN to form alkyl cyanides, while with AgCN they form isocyanides. (b) is more reactive towards $\mathrm{S_N2}$ displacement as compared to . (c) Grignard reagents should be prepared under anhydrous conditions.
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(a) ionic vs covalent → C- vs N-attack; (b) benzylic sp³ vs aromatic sp²; (c) water destroys the C−Mg bond.
Step 1: read the clue about (A) plus Grignard. The formula $\mathrm{C_3H_5N}$ has very few hydrogens for three carbons, which hints at a triple bond. Since it reacts with a Grignard reagent and then with water to give a carbonyl product, (A) is almost certainly a nitrile. The fit is propanenitrile, $\mathrm{C_2H_5CN}$. A nitrile plus a Grignard followed by hydrolysis is the classic way to build a ketone.
Step 2: use the tests to nail down (B). Compound (B) gives an orange-red colour with 2,4-DNP, so it has a C to O double bond. But it fails Tollens' and Fehling's, so it is not an aldehyde, it is a ketone. It also fails iodoform, so it is not a methyl ketone, and it does not decolourise bromine water, so it has no C to C double bond. The phenyl ethyl ketone that ticks every box is propiophenone, $\mathrm{C_6H_5COC_2H_5}$.
Step 3: identify (C) from the oxidation. Harsh oxidation with chromic acid burns off the side chain attached to the benzene ring and leaves a single $\mathrm{-COOH}$ on it. That gives benzoic acid, $\mathrm{C_6H_5COOH}$, whose formula $\mathrm{C_7H_6O_2}$ matches (C) exactly.
Step 4: write the reaction of (A) with phenylmagnesium bromide. The Grignard adds across the C to N triple bond to give a magnesium-bound imine salt, which on acidic hydrolysis turns into the ketone and ammonia: \[ C_2H_5C{\equiv}N \xrightarrow{C_6H_5MgBr} C_2H_5C(=N{-}MgBr)C_6H_5 \xrightarrow{H_3O^+} C_2H_5COC_6H_5 + NH_3 \] So (A) is propanenitrile, (B) is propiophenone, and (C) is benzoic acid.
