Which aldol addition product is formed when two molecules of ${\mathrm{CH}}_3\mathrm{CHO}$ (acetaldehyde) undergo base-catalyzed aldol condensation?

• A. ${\mathrm{CH}}_3\mathrm{CH}={\mathrm{CH}}_2\mathrm{CHO}$
• B. ${\mathrm{CH}}_3\mathrm{COOH}$
• C. ${\mathrm{CH}}_3\mathrm{CH}\left(\mathrm{OH}\right){\mathrm{CH}}_2\mathrm{CHO}$
• D. ${\mathrm{CH}}_3\mathrm{COCH}{\mathrm{CH}}_3$

Answer: (C)

Verified step by step guidance

1. Identify the reactants and conditions: Two molecules of acetaldehyde (CH\_3CHO) undergo a base-catalyzed aldol reaction. The base deprotonates the alpha hydrogen of one acetaldehyde molecule to form an enolate ion.
2. Form the enolate ion: The alpha hydrogen adjacent to the carbonyl group in acetaldehyde is acidic. Under basic conditions, this hydrogen is removed, generating a resonance-stabilized enolate ion, which acts as a nucleophile.
3. Nucleophilic attack on the carbonyl carbon: The enolate ion attacks the electrophilic carbonyl carbon of a second acetaldehyde molecule, forming a new carbon-carbon bond and generating an alkoxide intermediate.
4. Protonation of the alkoxide intermediate: The alkoxide ion formed is protonated by water or the solvent, resulting in the aldol addition product, which is a β-hydroxy aldehyde. In this case, the product is 3-hydroxybutanal (CH\_3CH(OH)CH\_2CHO).
5. Recognize that this aldol addition product can further undergo dehydration under the reaction conditions to form an α,β-unsaturated aldehyde, but since the question asks for the aldol addition product, focus on the β-hydroxy aldehyde structure.