Question

What happens when: (a) Propanenitrile is treated with phenyl magnesium bromide followed by hydrolysis? (b) p-Fluorotoluene is treated with $CrO_3$ in presence of acetic anhydride followed by hydrolysis with aqueous acid? (c) Phthalic acid is treated with $NH_3$ followed by heating?

Hint:

Grignard + Nitrile ? Ketone (after hydrolysis). $CrO_3$/Ac$_2$O selectively converts $-CH_3$ to $-CHO$. Phthalic acid + $NH_3$ + heat ? Phthalimide.

Answer 1

Step 1: Reaction (a) — Grignard with nitrile.
Propanenitrile ($CH_3CH_2CN$) + phenylmagnesium bromide ($C_6H_5MgBr$): the Grignard reagent adds to the $C\equiv N$ triple bond, forming a metalated imine. On aqueous hydrolysis, the imine gives a ketone. Product: Propiophenone ($C_6H_5COCH_2CH_3$, 1-phenylpropan-1-one).
Step 2: Reaction (b) — $CrO_3$/acetic anhydride oxidation.
p-Fluorotoluene ($4-F-C_6H_4-CH_3$) with $CrO_3$/acetic anhydride oxidises $-CH_3$ to $-CHO$ selectively; on hydrolysis gives aldehyde. Product: p-Fluorobenzaldehyde ($4-F-C_6H_4-CHO$).
Step 3: Reaction (c) — Phthalic acid with $NH_3$ then heat.
Phthalic acid + $NH_3$ gives diamide (phthalamide); strong heating causes both amide groups to lose water and cyclise. Product: Phthalimide ($C_6H_4(CO)_2NH$).
Step 4: Summary.
(a) Propiophenone: $C_6H_5COCH_2CH_3$; (b) p-Fluorobenzaldehyde: $4-F-C_6H_4CHO$; (c) Phthalimide: $C_6H_4(CO)_2NH$.