Destroyer Ochem #122

[lex1489]

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This question asks which reaction is correct. I don't understand the reaction in letter D.

1-butene reacts with Br2 in methanol. It forms (not sure how to name this) Br-CH2-CH(OCH3)-CH2-CH3. Br is anti to the OCH3 (methoxy, right?) group How does this mechanism work? And how do you name the product?

 

[Classic59]

Hey, I believe this mechanism involves the initial attack of the pi electrons on the Br2 atom, forming a cyclic bromonium bridge, which is shaped like triangle with partial positive character. Then, the methanol attacks the more substituted carbon C2, ANTI to the bridge, breaking the C2-Br bond. A proton transfer restores neutrality, and you got your product!
I would call it 1-bromo-2-methoxybutane. 🙂

 

[lex1489]

Ahh that's right. Without the methoxy available, the Br- left over from Br2 would instead anti-attack the most substituted carbon, correct?

 

[edcampbe]

That's right. Nucleophilic solvents compete with the bromide ion to attack the intermediate, so a mixture would occur. Either the bromides or the bromide/methoxy will be anti.