O-Chem: Alkene Reaction with HBr and Methanol

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SUMMARY

The reaction of cis-2-butene with HBr and methanol (CH3OH) yields two final products: 2-bromo-2-methylbutane and 1-bromo-2-methylbutane. The HBr first attacks the double bond of cis-2-butene, leading to the formation of a carbocation intermediate. Methanol then acts as a nucleophile, attacking the carbocation to form the final products. The presence of two secondary carbons allows for the formation of bromine on either side of the structure, resulting in two distinct products.

PREREQUISITES
  • Understanding of alkene reactions and mechanisms
  • Knowledge of carbocation stability and rearrangements
  • Familiarity with nucleophilic substitution reactions
  • Basic grasp of organic chemistry nomenclature
NEXT STEPS
  • Study the mechanism of alkene reactions with hydrogen halides
  • Learn about carbocation rearrangements in organic reactions
  • Research nucleophilic substitution reactions involving alcohols
  • Explore the role of solvents in organic reactions, specifically methanol
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Organic chemistry students, educators, and anyone interested in understanding alkene reactions and their mechanisms.

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Homework Statement



What are the two final products when cis-2-butene reacts with HBr and CH3OH?

Homework Equations



Reactions of Alkenes.

The Attempt at a Solution



I'm not sure if the HBr will attack the double bond first or if the OH on methane will take the H off HBr and create a radical. This sample final I'm working on does not have a key, so there is no way I can check the correct answer and reverse engineer it.

Thank you :)
 
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Remember there are two products-- what product would each of the options you outlined yield?
 
Since there are two secondary carbons, I think I could end up with Bromine on either side of the structure and that the cis-2-butene would be an alkane. I don't know how the methanol plays a part in this though. It's not a peroxide.
 

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