See image — Reaction Mechanism Chemistry Question

💡 Solution & Explanation

Step 1 - Concept: Grignard reagents (PhMgBr, phenylmagnesium bromide) react with carbonyl compounds followed by acid workup (H+) to give alcohols. The class of alcohol produced depends on the carbonyl compound used: - Formaldehyde (HCHO) → primary alcohol (the carbon bearing OH comes from formaldehyde, and only one carbon is attached to it via the Grignard) - Aldehyde (RCHO, R ≠ H) → secondary alcohol (the carbon bearing OH has two carbon substituents: R and Ph) - Ketone (R2C=O) → tertiary alcohol (the carbon bearing OH has three carbon substituents: R, R, and Ph) - A compound with an acidic proton (e.g., active methylene compound like acetylacetone) acts as a proton source and simply protonates PhMgBr, giving benzene (PhH) instead of an alcohol product Step 2 - Matching each reaction: (a) PhMgBr + (A) → 1° alcohol: A 1° alcohol results when PhMgBr attacks formaldehyde (H-C(=O)-H). The product is Ph-CH2-OH (benzyl alcohol), a primary alcohol. Therefore A = (s) H-C(=O)-H (formaldehyde). (b) PhMgBr + (B) → 2° alcohol: A 2° alcohol results when PhMgBr attacks an aldehyde (RCHO where R ≠ H). CH3-C(=O)-H (acetaldehyde) gives Ph-CH(OH)-CH3, a secondary alcohol. Therefore B = (r) CH3-C(=O)-H (acetaldehyde). (c) PhMgBr + (C) → 3° alcohol: A 3° alcohol results when PhMgBr attacks a ketone. CH3-C(=O)-CH3 (acetone) gives Ph-C(OH)(CH3)2, a tertiary alcohol. Therefore C = (q) CH3-C(=O)-CH3 (acetone). (d) PhMgBr + (D) → benzene: When PhMgBr encounters a compound with sufficiently acidic protons, protonation of the Grignard occurs rather than nucleophilic addition, releasing Ph-H (benzene). Acetylacetone (CH3-C(=O)-CH2-C(=O)-CH3) has acidic alpha-CH2 protons (pKa ~9) that can protonate PhMgBr to give benzene. Therefore D = (p) CH3-C(=O)-CH2-C(=O)-CH3 (acetylacetone). Step 3 - Why other options fail: - Acetone (q) with PhMgBr cannot give a 1° or 2° alcohol; it gives 3°. - Formaldehyde (s) with PhMgBr cannot give a 2° or 3° alcohol; it gives 1°. - Acetylacetone (p) does not give an alcohol product with PhMgBr because its active methylene protons protonate the Grignard before nucleophilic addition occurs. Therefore, the correct answer is {"a": ["S"], "b": ["R"], "c": ["Q"], "d": ["P"]}.