Question

The reaction involves acetone (propan-2-one) with sodium nitrite (NaNO2) and hydrochloric acid (HCl). This combination of reagents generates nitrous acid (HNO2), which is the active species for nitrosation. Nitrous acid exists in equilibrium with the nitrosonium ion (NO+) in acidic conditions, which acts as an electrophile.

Ketones with alpha-hydrogens undergo nitrosation at the alpha-carbon. The mechanism involves the following steps:

1. Enolization: Acetone, an unsymmetrical ketone, exists in equilibrium with its enol form in the presence of acid.

   CH3-CO-CH3 <=> CH3-C(OH)=CH2

2. Electrophilic Attack: The enol form acts as a nucleophile and attacks the electrophilic nitrosonium ion (NO+).

   CH3-C(OH)=CH2 + NO+ -> [CH3-C(OH+)-CH2-NO]

3. Deprotonation: The intermediate carbocation loses a proton from the hydroxyl group to restore the carbonyl, forming an alpha-nitrosoketone.

   [CH3-C(OH+)-CH2-NO] - H+ -> CH3-CO-CH2-NO

4. Tautomerization: Alpha-nitrosoketones are generally unstable and rapidly tautomerize to their more stable alpha-oximino ketone form.

   CH3-CO-CH2-NO <=> CH3-CO-CH=N-OH

Answer 1

The product of the reaction is 1-(hydroxyimino)propan-2-one. Structure: CH3-CO-CH=N-OH

Answer 2

Acetone (CH3COCH3) reacts with NaNO2/HCl (which generates HNO2) via an alpha-nitrosation reaction. The ketone first enolizes, and the enol then undergoes electrophilic attack by NO+ (nitrosonium ion). The resulting alpha-nitrosoketone (CH3COCH2NO) tautomerizes to the more stable alpha-oximino ketone.