See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Concept: A mixed aldol addition reaction is efficient (i.e., gives predominantly one product) when one carbonyl component has no alpha-hydrogens (so it can only act as the electrophilic acceptor) and the other component acts as the nucleophilic enolate donor. This avoids the formation of multiple self-aldol products. Step 1 – Analyze option (b): Ph-CO-CH(CH3)-CH2OH. Retrosynthetically, this beta-hydroxy ketone can be disconnected at the beta-carbon–alpha-carbon bond (the C–C bond formed in the aldol). This gives: the enolate of acetaldehyde (CH3CHO, acting as donor) adding to benzaldehyde (PhCHO, acting as acceptor). Benzaldehyde has NO alpha-hydrogens, so it cannot form a self-aldol. Acetaldehyde provides the enolate. The crossed aldol between benzaldehyde (no alpha-H, acceptor only) and acetaldehyde (donor) is efficient and well-known. The product would be Ph-CH(OH)-CH2-CHO (the direct aldol adduct), which upon examination matches option (b) when viewed correctly as PhCO-CH(CH3)-CH2OH — wait, let us re-examine: Ph-C(=O)-CH(CH3)-CH2OH means a phenyl ketone. Retrosynthetic cut between the alpha-carbon and the carbonyl carbon bearing CH2OH: donor = enolate of PhCOCH3 (acetophenone, alpha to C=O), acceptor = formaldehyde (HCHO, no alpha-H). Acetophenone + formaldehyde (no alpha-H) → Ph-CO-CH(CH2OH)-... Actually re-reading structure (b): Ph-C(=O)-C(CHCH3)(CH2OH), the alpha carbon has both CH3 and CH2OH substituents. Retrosynthetic disconnection: acceptor = HCHO (formaldehyde, no alpha-H), donor = enolate of Ph-CO-CH2-CH3 (propiophenone). Formaldehyde has no alpha-hydrogens, so it acts solely as the electrophile. Propiophenone acts as the enolate donor. This mixed aldol is efficient because formaldehyde cannot self-condense via aldol (no alpha-H). The product is the beta-hydroxy ketone Ph-CO-CH(CH3)-CH2OH, which is exactly option (b). Step 2 – Eliminate other options: - Option (a): Ph-C(OH)(CH3)-CH2-CHO. This would require two components both capable of self-aldol reactions, making the mixed reaction inefficient with multiple products. - Option (c): Ph-CH2-CO-CH(CH3)-C(OH)(CH3)2. The retrosynthetic precursors would both have alpha-hydrogens, leading to a mixture of self-aldol and cross-aldol products; inefficient. - Option (d): Ph-CO-CH2-CH2-CHO. This is a 1,4-keto aldehyde (delta position), not a beta-hydroxy carbonyl compound at all; it is not a direct aldol addition product (aldol gives beta-hydroxy carbonyls, not 1,4-dicarbonyls without an OH). Step 3 – Conclusion: Option (b) is best prepared by an efficient mixed aldol addition because it uses formaldehyde (no alpha-H) as the acceptor and propiophenone as the donor, giving a single crossed aldol product. Therefore, the correct answer is B.