Question

What happens when: p-fluorotoluene is treated with \( CrO_{3} \) in the presence of acetic anhydride followed by hydrolysis with aqueous acid?

Hint:

This is a variation of the Etard reaction. Using acetic anhydride is key to stopping the oxidation at the aldehyde stage.

Answer 1

Step 1: Recognise why this particular oxidation is chosen.
Ordinary oxidation of a methyl group on a benzene ring runs straight past the aldehyde stage to the carboxylic acid, which is not what we want here. Using $CrO_3$ together with acetic anhydride, $(CH_3CO)_2O$, is a trick that traps the reaction at the aldehyde oxidation level, this is the Etard-type modification.
Step 2: Form the protected intermediate.
The methyl group of p-fluorotoluene is oxidised, and the resulting unstable species is immediately captured by the acetic anhydride as a gem-diacetate, $p\text{-}F-C_6H_4-CH(OCOCH_3)_2$. This diacetate is stable enough to isolate, unlike the aldehyde itself would be under the reaction conditions.
Step 3: Release the aldehyde by hydrolysis.
Treating the diacetate with aqueous acid, $H_3O^+$, hydrolyses both acetate groups off together and replaces them with a single oxygen, unmasking the aldehyde.
\[ p\text{-}F-C_6H_4-CH_3 \xrightarrow{CrO_3,\ (CH_3CO)_2O} p\text{-}F-C_6H_4-CH(OCOCH_3)_2 \xrightarrow{H_3O^+} p\text{-}F-C_6H_4-CHO \]
\[ \boxed{\text{4-fluorobenzaldehyde}} \]