See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Concept: The reactivity of carbonyl compounds towards nucleophilic addition (such as Grignard reagents) depends on two factors: (1) steric hindrance at the carbonyl carbon and (2) electronic effects (electrophilicity of the carbonyl carbon). Step 1 – Identify the carbonyl compounds: - Compound (I): Acetaldehyde (CH3CHO) — aldehyde with one methyl group and one H on the carbonyl carbon. - Compound (II): Formaldehyde (HCHO) — aldehyde with two H atoms on the carbonyl carbon (least hindered). - Compound (III): Acetophenone (CH3COC6H5) — ketone with a methyl group and a phenyl group on the carbonyl carbon. Step 2 – Steric considerations: - Formaldehyde (II) has no substituents other than H on both sides of the carbonyl carbon → least steric hindrance → most reactive. - Acetaldehyde (I) has one CH3 group → moderate steric hindrance → moderately reactive. - Acetophenone (III) has one CH3 and one bulky C6H5 group → most steric hindrance → least reactive. Step 3 – Electronic considerations: - In ketones, two alkyl/aryl groups donate electron density to the carbonyl carbon, reducing its electrophilicity. - In acetophenone, the phenyl ring provides additional resonance stabilization (conjugation with C=O), further reducing the electrophilicity of the carbonyl carbon compared to simple aldehydes. - Formaldehyde has no electron-donating groups, making its carbonyl carbon the most electrophilic. - Acetaldehyde has one electron-donating methyl group, making it less electrophilic than formaldehyde but more than acetophenone. Step 4 – Order of reactivity: II (formaldehyde) > I (acetaldehyde) > III (acetophenone) Step 5 – Why other options fail: - (a) I > II > III: Incorrect because formaldehyde is less hindered and more electrophilic than acetaldehyde, so II should be more reactive than I. - (c) II > III > I: Incorrect because acetophenone (III) is a ketone with greater steric and electronic deactivation than acetaldehyde (I); hence I > III, not III > I. - (d) I > III > II: Incorrect for the same reasons as (a) and (c). Therefore, the correct answer is B.