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Topic: Stablization of cyclic carbocations  (Read 6308 times)

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Offline asascl1992

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Stablization of cyclic carbocations
« on: November 07, 2009, 12:22:45 PM »
The question asks "If the most stable carbocation, utilizing all possible modes of stabilization, is formed, what is the protonation site in the conjugated alkene shown below?"

Now this is an non-aromatic system but it has a aromatic resonance structure that I assume it will switch to albeit not equivalently. The resonance structure would give cyclopropene a net positive charge and would give cyclopentadiene a net negative charge. Thus, I thought that only 4 and 5 would be options since the cyclopropene would have a net positive charge and that is where a protonation would be favored. I then chose 4 instead of 5 since 4 is a tertiary carbon while 5 is a secondary carbon. However, the answer key states that option 2 is where the protonation site will occur, which makes no sense to me!

Offline orgopete

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Re: Stablization of cyclic carbocations
« Reply #1 on: November 07, 2009, 12:34:58 PM »
Let's try this again. Your analysis is correct. The resonance structures can be two aromatic systems, a cyclopropenyl cation and a cyclopentadienyl anion. In that case, why would protonation occur on the cyclopentadienyl cation? Draw the result of the protonation product from C-4. Is it aromatic, non-aromatic, or antiaromatic?

If the cyclopentadiene is more electron rich from electron donation of the cyclopropenyl methylene group, on which carbon will protonation occur? Try each of the carbons, 1, 2, and 3, and determine which carbocation is more stable? Interestingly for me, I thought it was going to be C-3.
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