Question

How will you convert:
(i) Propanone to Propene,
(ii) Benzoic acid to Benzaldehyde,
(iii) Benzene to m-Nitroacetophenone?

Hint:

Aldehydes are generally more reactive than ketones towards nucleophilic attack due to less steric hindrance and only one electron-donating alkyl group.

Answer 1

Step 1: Convert propanone to propene.
First reduce the ketone to an alcohol using a mild hydride reducing agent, then dehydrate that alcohol under acid conditions to bring in the double bond. \[ CH_3COCH_3 \xrightarrow{NaBH_4} CH_3CH(OH)CH_3 \xrightarrow[\Delta]{conc.\,H_2SO_4} CH_3-CH=CH_2 \]
Step 2: Convert benzoic acid to benzaldehyde.
Benzoic acid is first turned into the more reactive acid chloride using thionyl chloride, and this acid chloride is then selectively reduced to the aldehyde stage using hydrogen over a poisoned palladium catalyst, stopping short of the alcohol. This partial reduction is known as the Rosenmund reduction. \[ C_6H_5COOH \xrightarrow{SOCl_2} C_6H_5COCl \xrightarrow[Pd/BaSO_4]{H_2} C_6H_5CHO \]
Step 3: Convert benzene to m-nitroacetophenone.
Benzene is first acylated under Friedel-Crafts conditions using acetyl chloride and anhydrous aluminium chloride to install the acetyl group, giving acetophenone. Since the acetyl group is a meta director, subsequent nitration with a nitrating mixture places the nitro group at the meta position. \[ C_6H_6 \xrightarrow[AlCl_3]{CH_3COCl} C_6H_5COCH_3 \xrightarrow[H_2SO_4]{conc.\,HNO_3} m\text{-}O_2N\text{-}C_6H_4\text{-}COCH_3 \]
Step 4: Note why the acetyl group directs the incoming nitro group to the meta position.
The carbonyl carbon of the acetyl group withdraws electron density from the ring through resonance, deactivating the ortho and para positions more strongly than the meta position, so the electrophile ends up meta to the acetyl group. \[ \boxed{\text{(i) Propene (ii) Benzaldehyde (iii) } m\text{-Nitroacetophenone, as shown above}} \]