An aldol condensation was carried out on equal amounts of propionaldehy de and butyraldehyde. The reaction gave rise to four distinct products. What were their structures, and how did each arise?

In the base-catalyzed aldol condensation reaction, there will be the formation of enolate ions from propionaldehyde and butyraldehyde. Let's denote the enolate ion of propionaldehyde as Enolate 1 and the enolate ion of butyraldehyde as Enolate 2. Enolate 1 (propionaldehyde): \[ H_2C=CH-CH_2 O^-\] Enolate 2 (butyraldehyde): \[ H_2C=CH-CH_2 -CH_2 O^-\]

Reacting each enolate with their corresponding aldehyde will give us the self-aldol product. Product 1 (propionaldehyde self-aldol): \[ 2 \times CH_3-CH_2-CHO \rightarrow CH_3- CH_2 - CH(OH) -CH_2 -CHO \] Product 2 (butyraldehyde self-aldol): \[ 2 \times CH_3-CH_2 -CH_2 -CHO \rightarrow CH_3-CH_2 -CH_2 -CH(OH) - CH_2 - CHO \]

Mixing enolates 1 and 2 with the opposite aldehyde will yield the cross-aldol products. Product 3 (crossed-aldol, propionaldehyde enolate with butyraldehyde): \[ CH_3 - CH_2 - CHO + H_2C=CH-CH_2 O^- \rightarrow CH_3- CH_2 - CH(OH) - CH_2 - CH_2 - CHO\] Product 4 (crossed-aldol, butyraldehyde enolate with propionaldehyde): \[ CH_3 - CH_2 - CH_2 - CHO + H_2C=CH-CH_2 -CH_2 O^- \rightarrow CH_3 - CH_2 - CH_2 - CH(OH) - CH_2 - CH_2 - CHO \] So, the four distinct products formed due to aldol condensation between propionaldehyde and butyraldehyde are, 1. Product 1 (propionaldehyde self-aldol): \[ CH_3 - CH_2 - CH(OH) - CH_2 - CHO \] 2. Product 2 (butyraldehyde self-aldol): \[ CH_3 - CH_2 - CH_2 - CH(OH) - CH_2 - CHO \] 3. Product 3 (crossed-aldol, propionaldehyde enolate with butyraldehyde): \[ CH_3 - CH_2 - CH(OH) - CH_2 - CH_2 - CHO \] 4. Product 4 (crossed-aldol, butyraldehyde enolate with propionaldehyde): \[ CH_3 - CH_2 - CH_2 - CH(OH) - CH_2 - CH_2 - CHO \]