Dehydration Problem: Solving for the Location of a Double Bond in a Carbocation

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SUMMARY

The discussion centers on the formation of a double bond in a carbocation during a dehydration reaction, specifically referencing a problem from a homework assignment. Participants highlight the application of Zaitsev's rule to determine the location of the double bond, noting that both potential sites (positions a and b) appear equivalent. The solution involves considering the enol form of the molecule rather than simply protonating the hydroxyl group at the bridgehead carbon. This approach clarifies the mechanism leading to the correct double bond formation.

PREREQUISITES
  • Understanding of carbocation stability and formation
  • Familiarity with Zaitsev's rule in elimination reactions
  • Knowledge of enol and keto forms of organic compounds
  • Basic principles of dehydration reactions in organic chemistry
NEXT STEPS
  • Study the mechanism of dehydration reactions in organic chemistry
  • Learn about the implications of Zaitsev's rule on elimination reactions
  • Explore the concept of enolization and its significance in reaction mechanisms
  • Investigate the stability of different carbocation types and their formation pathways
USEFUL FOR

Organic chemistry students, educators, and researchers interested in reaction mechanisms, particularly those involving carbocations and elimination reactions.

vijayramakrishnan
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Homework Statement


Please see question number 31

http://cms.fiitjee.co/Resources/DownloadCentre/Document_Pdf_183.pdf

Homework Equations


none

The Attempt at a Solution



i got up to second step in the solution(scroll downwards) but in the last step a carbocation is formed and double bond can be formed in two places b and a. they seem to be equivalent places,i don't know where the double bond will be formed.
 
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Look up Zaitsev's rule.
 
TeethWhitener said:
Look up Zaitsev's rule.
but here both are equally substituted sir
 
Ah, I see now. You're wondering why the double bond doesn't form between the two bridgehead carbons? This might be because you're envisioning protonating the OH group at the bridgehead carbon and eliminating a water to give a tertiary carbocation. In this case however, you have to do something a little different. Start by drawing the molecule in the enol form. Can you get it from here?
 

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