Q10) Predict the major product of the following reaction. Br Br2 . Br سم و رسم Methanol Br Br Br ОН Br OMe Br OMe

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I picked d for this but I am not sure if it is correct. Can you please help me out? 

Thank you.

**Q10) Predict the major product of the following reaction.**

**Reaction:**  
A cyclohexane reacts with Br₂ in the presence of methanol.

**Possible Products:**  
a) Cyclohexane with one bromine (Br) attached  
b) Cyclohexane with two bromine atoms on adjacent carbon atoms  
c) Cyclohexane with two bromine atoms on opposite sides  
d) Cyclohexane with one bromine (Br) and one hydroxyl group (OH), where the bromine and hydroxyl are on adjacent carbon atoms (circled)  
e) Cyclohexane with one bromine (Br) and one methoxy group (OMe), where the bromine and methoxy are on adjacent carbon atoms  
f) Cyclohexane with one bromine (Br) and one methoxy group (OMe), where the bromine and methoxy are on opposite sides

**Explanation:**  
- Product d) is highlighted as the major product. This suggests that the reaction likely involves a regioselective or stereoselective process, leading to the formation of a bromohydrin due to the presence of methanol as a solvent, which facilitates the formation of the OH group.
Transcribed Image Text:**Q10) Predict the major product of the following reaction.** **Reaction:** A cyclohexane reacts with Br₂ in the presence of methanol. **Possible Products:** a) Cyclohexane with one bromine (Br) attached b) Cyclohexane with two bromine atoms on adjacent carbon atoms c) Cyclohexane with two bromine atoms on opposite sides d) Cyclohexane with one bromine (Br) and one hydroxyl group (OH), where the bromine and hydroxyl are on adjacent carbon atoms (circled) e) Cyclohexane with one bromine (Br) and one methoxy group (OMe), where the bromine and methoxy are on adjacent carbon atoms f) Cyclohexane with one bromine (Br) and one methoxy group (OMe), where the bromine and methoxy are on opposite sides **Explanation:** - Product d) is highlighted as the major product. This suggests that the reaction likely involves a regioselective or stereoselective process, leading to the formation of a bromohydrin due to the presence of methanol as a solvent, which facilitates the formation of the OH group.
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