See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Concept: This reaction involves a cyclopropylmethyl tosylate system undergoing E2 or solvolysis/elimination under basic conditions with heat. The substrate is 1-(1-methylcyclopropyl)cyclohexyl tosylate, where the OTs and the spiro cyclopropane bearing a methyl group are on the same carbon (C1) of the cyclohexane ring. Step 1 - Identify the substrate: The molecule has a quaternary carbon (C1 of cyclohexane) bearing: (1) the cyclohexane ring, (2) OTs (leaving group), (3) a cyclopropane ring with a CH3 group (spiro junction). The OTs is on a wedge/dashed bond indicating stereochemistry. Step 2 - Reaction pathway: Under HO- (strong base) and heat (delta), the tosylate acts as a leaving group. The cyclopropylcarbinyl system (cyclopropane adjacent to the carbon bearing OTs) is prone to ring opening. When OTs leaves, the cyclopropane ring opens to give a homoallylic/allylic carbocation or the base assists in an E2-type process with cyclopropane ring opening. Step 3 - Ring opening of cyclopropane: The ionization of OTs gives a cyclopropylcarbinyl cation at C1 of cyclohexane. The cyclopropyl ring opens to give a homoallylic cation. Specifically, the bond of the cyclopropane ring between C1 (spiro carbon) and one of the cyclopropane carbons breaks, placing the positive charge on the ring-opened carbon. This generates an allylic system within the cyclohexane ring. Step 4 - Product formation: When the cyclopropane (bearing CH3) fused at C1 of cyclohexane opens, the methyl-bearing carbon of the cyclopropane becomes part of an endocyclic double bond. The ring opening incorporates the cyclopropane carbons into the cyclohexane ring, expanding it, or alternatively, the cyclopropane opens to place the double bond endocyclically. Given the spiro cyclopropane at C1: opening of the cyclopropane bond between C1(cyclohexyl) and the CH2 of cyclopropane (not the CH-CH3) gives a carbocation stabilized by the adjacent CH3-bearing carbon, leading after deprotonation to 1-methylcyclohex-2-ene (option b). Step 5 - Why option (b) is major: The cyclopropylcarbinyl cation rearranges/opens to give the most stable alkene. The endocyclic trisubstituted alkene (1-methylcyclohex-2-ene) is more stable than the exocyclic methylenecyclohexane (option c) or other less substituted alkenes. Option (a) would require a different connectivity. Option (d) is a less substituted product. The homoallylic rearrangement from the cyclopropylcarbinyl cation followed by loss of proton gives 1-methylcyclohex-2-ene as the major, more stable (Zaitsev) product. Why other options fail: (a) 1-methylcyclohex-1-ene would require the methyl to be directly on the double bond carbon in a specific way not favored here. (c) Methylenecyclohexane is an exocyclic less-substituted alkene, less stable. (d) 4-methylcyclohex-1-ene would require methyl migration far from the reaction center. Therefore, the correct answer is B.