Answer 1

The final product is Cyclohexane-1,2-dione.

Answer 2

The reaction proceeds in two main steps:

Step 1: Nitrosation of Cyclohexanone

1. Generation of Nitrous Acid: Sodium nitrite ($\text{NaNO}_2$) reacts with hydrochloric acid ($\text{HCl}$) to generate nitrous acid ($\text{HNO}_2$). $\text{NaNO}_2 + \text{HCl} \rightarrow \text{HNO}_2 + \text{NaCl}$

2. Formation of Nitrosonium Ion: Nitrous acid is protonated and then loses water to form the nitrosonium ion ($\text{NO}^+$), which is a strong electrophile. $\text{HNO}_2 + \text{H}^+ \rightleftharpoons \text{H}_2\text{O}^+-\text{N}=\text{O} \rightleftharpoons \text{H}_2\text{O} + \text{NO}^+$

3. Enolization of Ketone: Cyclohexanone, being a ketone with alpha-hydrogens, exists in equilibrium with its enol form.

4. Electrophilic Attack: The nucleophilic alpha-carbon of the enol form attacks the electrophilic nitrosonium ion ($\text{NO}^+$), forming an alpha-nitroso ketone.

5. Tautomerization: The alpha-nitroso ketone is generally unstable and rapidly tautomerizes to the more stable alpha-oxime ketone (2-(hydroxyimino)cyclohexan-1-one).

Step 2: Hydrolysis of the Oxime

1. Acid-catalyzed Hydrolysis: The alpha-oxime ketone (2-(hydroxyimino)cyclohexan-1-one) is subjected to acidic hydrolysis ($\text{H}_3\text{O}^+$). Under these conditions, the oxime group ($=\text{NOH}$) is hydrolyzed to a carbonyl group ($=\text{O}$), similar to the hydrolysis of an imine. This reaction releases hydroxylamine ($\text{NH}_2\text{OH}$).

The final product is cyclohexane-1,2-dione.