Solving Alkene Reactions Problems for Chem Exam

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SUMMARY

This discussion focuses on solving alkene reaction problems in preparation for a chemistry exam. Key reactions include the addition of bromine (Br2) to 1,3-cyclohexadiene and 1,4-cyclohexadiene, where the former undergoes both 1,2- and 1,4-addition, while the latter is treated as an isolated diene. The mechanism of bromination of propylene in methanol (CH3OH) is explored, leading to products C3H6Br2 and C4H9BrO, with the latter involving a reaction between methanol and the brominated intermediate. Additionally, the hydrolysis of 1,2-dimethylcyclopentene using BH3 and H2O2/NaOH is confirmed to yield the same product as direct hydrolysis due to the symmetry of the alkene.

PREREQUISITES
  • Understanding of alkene reactions and mechanisms
  • Knowledge of bromination and halogenation processes
  • Familiarity with the concepts of conjugated and isolated dienes
  • Basic principles of hydrolysis and alcohol reactions
NEXT STEPS
  • Study the mechanisms of electrophilic addition reactions in alkenes
  • Learn about the differences between conjugated and isolated dienes
  • Investigate the role of solvents like methanol in organic reactions
  • Explore the hydroboration-oxidation reaction and its applications
USEFUL FOR

Chemistry students preparing for exams, organic chemistry enthusiasts, and educators seeking to clarify alkene reaction mechanisms.

ChemDoodle
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Hey I'm studying for my chem exam & I am having difficulty solving some problems.

1. 1,3cyclohexadiene + Br2= ?
Here,should i treat it like a conjugated diene? Meaning that the Br molecules will have 1,2 and 1,4 additions on the diene and 2 products will be there?

2. 1,4cyclohexadiene + Br2 = ?
Here should i t eat it like an isolated diene? & if so..where should the Br add? On which bond? Or will there be 2 products? With Br adding on the 1 double bond in the first & 4 double bond in the second?

3. When propylene is treated with bromine in methanol (CH3OH),two products are formed C3H6Br2 & C4H9BrO.Explain meachanistically how each are formed.
Here,the first product is simple..its obtained by halogenation of alkenes..simple addition of the Br molecules on the pi-bond.But i wasn't able to determine how the second product is obtained? Obviously methanol has something to do with it..But what kind of reaction is it? We studied hyrohalogenation but nothing about alcohols & halogens together.

4. 1,2-dimethylcyclopentene + BH3 + H2O2/NaOH =
This gives us the same product as direct hydrolysis since the alkene is symmetrical,correct?

Any help in these questions would be greatly appreciated.They are just doubts that need clarifying :) & my exam is tomorrow.
 
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Anyone,please? :(
 
I guess I'm too late to help you for your exam. How did you do on the test? You were right about some of these.
 
ChemDoodle said:
Hey I'm studying for my chem exam & I am having difficulty solving some problems.

1. 1,3cyclohexadiene + Br2= ?
Here,should i treat it like a conjugated diene? Meaning that the Br molecules will have 1,2 and 1,4 additions on the diene and 2 products will be there?

2. 1,4cyclohexadiene + Br2 = ?
Here should i t eat it like an isolated diene? & if so..where should the Br add? On which bond? Or will there be 2 products? With Br adding on the 1 double bond in the first & 4 double bond in the second?

3. When propylene is treated with bromine in methanol (CH3OH),two products are formed C3H6Br2 & C4H9BrO.Explain meachanistically how each are formed.
Here,the first product is simple..its obtained by halogenation of alkenes..simple addition of the Br molecules on the pi-bond.But i wasn't able to determine how the second product is obtained? Obviously methanol has something to do with it..But what kind of reaction is it? We studied hyrohalogenation but nothing about alcohols & halogens together.

4. 1,2-dimethylcyclopentene + BH3 + H2O2/NaOH =
This gives us the same product as direct hydrolysis since the alkene is symmetrical,correct?

Any help in these questions would be greatly appreciated.They are just doubts that need clarifying :) & my exam is tomorrow.

1) Probably condition dependent. I have seen both 1,4 and 1,2 addition in similar cases.

2) Try drawing the mechanism, and see if you can get addition to both double bonds with one equivalent.

3) What is the intermediate in halogenation?

4) As per your previous thread, if we assume the lack of locant for the ene means it's a 1-ene. I'm not sure you get hydrolysis.
 
chemisttree said:
I guess I'm too late to help you for your exam. How did you do on the test? You were right about some of these.

86,highest grade :P I went early the next day & asked my prof about them hehe.
Thanx tho :D
 

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