When 2 moles of acetone react with one mole of benzaldehyde, they can undergo a condensation reaction known as the crossed aldol condensation. In this reaction, one mole of benzaldehyde serves as an electrophile while the acetone acts as a nucleophile.
Reaction Steps:
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Enolate Ion Formation: One mole of acetone (propan-2-one) undergoes deprotonation to form an enolate ion.
\[ \text{CH}_3\text{COCH}_3 \xrightarrow{\text{Base}} \text{CH}_2\text{COCH}_3^- \]
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Nucleophilic Addition: The enolate ion formed from acetone acts as a nucleophile and attacks the carbonyl carbon of benzaldehyde.
\[ \text{CH}_2\text{COCH}_3^- + \text{C}_6\text{H}_5\text{CHO} \rightarrow \text{C}_6\text{H}_5\text{C}(OH)(\text{CH}_3)_2\text{CO} \]
This forms a β-hydroxy ketone intermediate.
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Dehydration: The β-hydroxy ketone can then undergo dehydration to form an α,β-unsaturated ketone.
\[ \text{C}_6\text{H}_5\text{C}(OH)(\text{CH}_3)_2\text{CO} \xrightarrow{\text{Dehydration}} \text{C}_6\text{H}_5\text{C}=\text{C}(\text{CH}_3)_2\text{CO} + \text{H}_2\text{O} \]
Major Product:
The final product after dehydration will be 4-phenyl-3-buten-2-one (or mesityl oxide) if only the condensation occurs once. The reaction simplifies as follows:
\[ 2 \text{CH}_3\text{COCH}_3 + \text{C}_6\text{H}_5\text{CHO} \rightarrow \text{C}_6\text{H}_5\text{C}(=O)(\text{CH}_3)_2 + \text{H}_2\text{O} \]
Overall Reaction:
Summarizing the overall reaction, we can write it as:
\[ 2 \text{CH}_3\text{COCH}_3 + \text{C}_6\text{H}_5\text{CHO} \rightarrow \text{C}_6\text{H}_5\text{C}(=O)(\text{CH}_3)_2 + \text{H}_2\text{O} \]
Therefore, the major product from this reaction is mesityl oxide (4-phenyl-3-buten-2-one).