See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 – Identify the substrate and reaction. (E)-3-methyl-2-pentene: CH3-CH=C(CH3)-CH2CH3, where the double bond is between C2 and C3, with a methyl group on C3 and an ethyl group continuing the chain. The (E) geometry means the two larger groups (CH3 on C2 and ethyl-bearing C3 chain) are on opposite faces. Step 2 – Hydroboration-oxidation mechanism. BH3 adds in a concerted syn (cis) fashion across the double bond. The boron (and ultimately OH) goes to the less substituted carbon (anti-Markovnikov), and the H goes to the more substituted carbon. Both B and H add from the same face (syn addition). Oxidation with H2O2/HO- replaces B with OH with retention of configuration. Step 3 – Assign regiochemistry. Boron adds to C2 (less substituted, bears only one methyl), hydrogen adds to C3. So OH ends up at C2, and an extra H at C3. Step 4 – Assign stereochemistry using the (E) alkene geometry. In (E)-3-methyl-2-pentene, viewing C2=C3: on C2 the groups are H and CH3; on C3 the groups are CH3 and CH2CH3 (ethyl). (E) means CH3 on C2 and CH3 on C3 are on opposite sides (the higher-priority groups are trans). Syn addition of BH3 from one face gives a specific relative stereochemistry. The syn addition from the bottom face (less hindered, same face for B and H) of the (E)-alkene places OH and H at C2 and C3 in an anti relationship between the OH and the ethyl group — producing the threo/anti product. Step 5 – Analyze Newman projections. In option (a), the Newman projection along C2-C3 shows: front carbon (C2) has OH (up/top), Et (left), H (right); back carbon (C3) has H (left), Me (right), Me (bottom). This corresponds to syn addition onto the (E)-alkene giving the product where OH and H are syn-added, with the correct relative stereochemistry: anti relationship between OH and the methyl on C3, consistent with syn hydroboration on the (E)-alkene. Options (b), (c), and (d) represent either wrong regiochemistry, wrong stereochemistry (anti addition products), or the enantiomer/diastereomer arising from syn addition on the opposite face giving a minor or incorrect product relative to the major syn-addition outcome on (E)-alkene. Option (a) correctly represents the major product of syn hydroboration-oxidation of (E)-3-methyl-2-pentene with OH at C2 and the correct relative configuration. Therefore, the correct answer is A.