See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Concept: SN1 reactivity depends on the stability of the carbocation intermediate formed after the departure of the leaving group. More stable the carbocation, faster the SN1 reaction. Step 1: Identify the carbocation formed from each compound upon ionization. - Compound (1): H3C-C(=O)-CH2(+) — a primary carbocation adjacent to a carbonyl group (acyl group). The carbonyl group is electron-withdrawing by both induction and resonance (no resonance stabilization of adjacent carbocation; in fact, resonance destabilizes it because the carbonyl oxygen withdraws electrons). This is actually a primary carbocation that is strongly destabilized by the adjacent electron-withdrawing carbonyl group. - Compound (2): H3C-CH2-CH2(+) — a primary carbocation with no special stabilization. - Compound (3): (H3C)3C(+) — a tertiary carbocation, stabilized by hyperconjugation and inductive donation from three methyl groups. Step 2: Rank carbocation stabilities. - (3) tert-butyl cation: tertiary, most stable due to three methyl groups providing electron density via hyperconjugation and induction. - (2) n-propyl cation: primary, moderate stability. - (1) chloromethyl ketone cation: primary AND adjacent to electron-withdrawing carbonyl, least stable (most destabilized). Step 3: Therefore, SN1 reactivity order follows carbocation stability: 3 > 2 > 1 Step 4: Why other options fail: - (a) 1 > 2 > 3: Incorrect; tertiary carbocation is most stable, not least. - (b) 1 > 3 > 2: Incorrect; compound 1 forms the least stable carbocation. - (d) 3 > 1 > 2: Incorrect; compound 1 is worse than compound 2 for SN1 due to carbonyl destabilization. Therefore, the correct answer is C.