The α,β -unsaturated ketone shown below is formed as one of the...

To identify the reactants that produce the given $\alpha,\beta$-unsaturated ketone, we can perform a retro-aldol analysis by cleaving the carbon-carbon double bond.

The structure of the given compound is $CH_3-CH_2-C(=O)-CH=C(CH_3)_2$.

In an aldol condensation, the double bond is formed between the $\alpha$-carbon of the enolate (nucleophile) and the carbonyl carbon ($\beta$-carbon in the product) of the other reactant (electrophile).

Cleaving the $C_{\alpha}=C_{\beta}$ double bond:
1. The $\alpha$-carbon (the $-CH=$ group adjacent to the carbonyl) gains two hydrogen atoms to form the original ketone. This gives $CH_3-CH_2-C(=O)-CH_3$, which is butan-2-one.
2. The $\beta$-carbon (the $=C(CH_3)_2$ group) gains an oxygen atom to form the original carbonyl compound. This gives $O=C(CH_3)_2$, which is propanone (acetone).

Thus, the cross-aldol condensation between butan-2-one (acting as the enolate at the terminal methyl group) and propanone yields the given product.

Answer: Butan-2-one and propanone