See image — Alcohols Phenols and Ethers Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the substrate: The alkene shown is trans-2-butene (E-2-butene), where H3C and CH3 are on opposite sides of the double bond, with H atoms on each sp2 carbon. Step 2 - Understand OsO4 dihydroxylation mechanism: OsO4 performs a syn (cis) addition of two OH groups across the double bond. Both hydroxyl groups are added to the same face of the alkene simultaneously via a cyclic osmate ester intermediate. Step 3 - Apply syn addition to trans-2-butene: In trans-2-butene, the two methyl groups are on opposite faces. When OsO4 adds both OH groups syn (same face), the two OH groups end up on the same face. For trans-2-butene, syn addition from either face gives the same product due to the geometry: both OH groups add to the same face, but because the methyl groups are trans to each other, the resulting diol has OH groups on the same side in the transition state. This gives the (R,S) = meso compound (compound 1 in Fischer projection: both OH groups on the same side, i.e., H-OH and H-OH both pointing right). Step 4 - Confirm: For trans-2-butene + OsO4 (syn dihydroxylation): syn addition from top face gives (2R,3S) and syn addition from bottom face gives (2S,3R), but for trans-2-butene these two are identical - they are the meso compound (since the molecule has an internal plane of symmetry). Therefore only the meso diol (compound 1) is formed. Step 5 - Why other options fail: - Options 2 and 3 represent the (R,R) and (S,S) enantiomers (the threo/anti diol), which would result from anti addition, not syn addition by OsO4. - Option (b) 1:1 mixture of 2 and 3 would be the result of anti addition to trans-2-butene (like epoxidation followed by ring opening). - Option (d) is incorrect because OsO4 gives exclusively syn addition. - Option (c) Only 2 is incorrect. Therefore, the correct answer is A.