See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: The reaction of an epoxide with triphenylphosphine (PPh3) proceeds via an SN2-type attack. PPh3 acts as a nucleophile, attacking the less hindered carbon of the epoxide in an SN2 fashion with inversion of configuration at that carbon. The oxygen is transferred to phosphorus forming POPh3, and the alkene is generated via an intramolecular elimination (betaine intermediate). Step 1 - Identify the epoxide stereochemistry: The starting material is cis-2,3-dimethyloxirane. The wedge/dash notation shows that on C2: H is on a dashed wedge (back) and Me is on a bold wedge (front); on C3: Me is on a dashed wedge (back) and H is on a bold wedge (front). This means the two methyl groups are on opposite faces - wait, let us re-read: H(dashes) on left carbon top, Me(dashes) on right carbon top, Me(bold) on left carbon bottom, H(bold) on right carbon bottom. The two methyls are trans to each other in the ring (one Me up, one Me down), but the ring connectivity makes this the cis epoxide (cis-2,3-dimethyloxirane means the two methyl groups are on the same side of the epoxide ring plane, i.e., both pointing in the same direction relative to the ring). Step 2 - Mechanism with PPh3: PPh3 attacks one of the epoxide carbons (C2 or C3, both equivalent by symmetry in this symmetric epoxide) via SN2, causing inversion at that carbon. This opens the epoxide to give a zwitterionic betaine: Ph3P(+)-C-C-O(-). The betaine then undergoes syn-elimination (the phosphorus and oxygen leave from the same face) to give the alkene and POPh3. Step 3 - Determine the alkene geometry: In cis-2,3-dimethyloxirane, both methyl groups are on the same face of the epoxide. PPh3 attacks with inversion at one carbon. After inversion at one carbon, the two methyl groups end up on opposite sides relative to each other in the betaine. The syn-elimination from this betaine gives cis-2-butene (the two methyl groups end up on the same side of the double bond). Detailed stereochemical analysis: For cis-epoxide + PPh3 (SN2 inversion at one center, then syn elimination): The net result is retention of relative stereochemistry going from epoxide to alkene when considering the overall transformation. The cis-epoxide gives cis-alkene as the major product. This is because the SN2 inversion at one carbon followed by syn elimination effectively gives the same relative arrangement as the original epoxide substituents. Step 4 - Why other options fail: (a) trans-2-butene would result from a trans-epoxide starting material under these conditions; (c) 1-butene requires rearrangement which is not favored; (d) isobutene requires carbon skeleton rearrangement. Therefore, the correct answer is B.