Final answer:
The reaction of propanal with 2-methylpropanal in the presence of NaOH is an aldol condensation, resulting in an α,ß-unsaturated ketone. This reaction is classified as a synthesis reaction.
Step-by-step explanation:
The reaction between propanal and 2-methylpropanal in the presence of NaOH is an example of an aldol condensation reaction. In aldol condensation, an alpha hydrogen of one aldehyde molecule is removed by the base, NaOH, to form an enolate ion, which then attacks the carbonyl carbon of a second aldehyde molecule to form a ß-hydroxy aldehyde.
Under the reaction conditions, the ß-hydroxy aldehyde often undergoes a subsequent dehydration to form an α,ß-unsaturated aldehyde or ketone. In this particular case, the product formed would be a α,ß-unsaturated ketone.
This reaction is classified as a synthesis reaction because it involves the combining of two different molecules (propanal and 2-methylpropanal) to form a new, more complex molecule.