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: Concept
These are three distinct organic transformations: (a) Grignard reaction with nitrile, (b) Etard-type oxidation of aromatic methyl group, (c) formation of cyclic imide.

Step 2: Analysis (a)
Propanenitrile ($CH_3CH_2CN$) reacts with phenyl Grignard reagent ($C_6H_5MgBr$) to give an imine salt intermediate. On hydrolysis, this yields a ketone. The aryl group from Grignard adds to the carbon of $-CN$: Product is phenyl ethyl ketone (1-phenyl-1-propanone, propiophenone): $C_6H_5-CO-CH_2CH_3$.

Step 3: Analysis (b)
p-Fluorotoluene ($F-C_6H_4-CH_3$) treated with $CrO_3$/acetic anhydride (Etard conditions for aromatic methyls) oxidizes the $-CH_3$ group to $-CHO$ while the ring and the $-F$ substituent remain intact. Product: p-Fluorobenzaldehyde ($F-C_6H_4-CHO$).

Step 4: Analysis (c)
Phthalic acid (benzene-1,2-dicarboxylic acid) treated with $NH_3$ forms phthalamide (diamide). On subsequent strong heating, the two amide groups lose water to form a cyclic imide. Product: Phthalimide ($C_6H_4(CO)_2NH$).

Final Answer:
(a) $C_6H_5-CO-CH_2CH_3$ (Propiophenone)
(b) $F-C_6H_4-CHO$ (p-Fluorobenzaldehyde)
(c) Phthalimide $[C_6H_4(CO)_2NH]$ -- a cyclic imide