See image — Alcohols Phenols and Ethers Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the substrate: 1,2-dimethyl propyl p-toluene sulfonate. The '1,2-dimethyl propyl' group refers to a 3-carbon chain with methyl groups at C1 and C2, making the carbon bearing the leaving group (OTs) a tertiary carbon (C1 has two methyls and is connected to an isopropyl-like framework). Actually, 1,2-dimethylpropyl means the carbon attached to OTs has: one methyl, one H attached to C2 which itself has one methyl and connects further - this is a secondary substrate. Re-examining: 1,2-dimethylpropyl = CH3-CH(CH3)-CH(CH3)- wait. 1,2-dimethylpropyl p-toluenesulfonate: the propyl backbone is C1-C2-C3, with methyl at C1 and C2. So C1 bears: OTs, CH3, H, and -CH(CH3)-CH3 group making it a tertiary carbon (C1 has OTs, one methyl, connected to C2 which has a methyl). Actually C1: OTs, CH3, H, C2 = secondary. But with a secondary/tertiary substrate and acetic acid (a polar, nucleophilic solvent of moderate nucleophilicity) at 75°C (elevated temperature favors ionization), the conditions strongly favor unimolecular mechanisms. Step 2 - Reaction conditions analysis: Acetic acid is a polar protic solvent with weak nucleophilicity. Polar protic solvents stabilize carbocations through solvation, strongly favoring SN1/E1 pathways. The substrate is a secondary (or tertiary after rearrangement) alkyl tosylate. The elevated temperature (75°C) further promotes ionization. Step 3 - Mechanism determination: Under these conditions, the rate-determining step is ionization of the C-OTs bond to form a carbocation intermediate (unimolecular). This carbocation can then: (a) be captured by the solvent (acetic acid acting as nucleophile) → substitution via SN1, or (b) lose a proton to a base → elimination via E1. Both SN1 and E1 share the same carbocation intermediate. Step 4 - Why not SN2/E2: SN2 requires a strong nucleophile and backside attack, which is disfavored in polar protic solvents with weak nucleophiles like acetic acid. E2 requires a strong base and is bimolecular, also disfavored here. Step 5 - Conclusion: Both substitution (SN1) and elimination (E1) products arise from the same carbocation intermediate through unimolecular mechanisms. Therefore, the correct answer is D.