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Junior Member
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Apr 6, 2012, 01:42 AM
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Organic Chemistry
What are two possible structure for the carbonium ion C6H11 +?
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Full Member
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Apr 6, 2012, 07:26 AM
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There are SO MANY (dozens of) possibilities. Is there more to your question?
C6H12 (the conjugate acid of your carbocation) must have a double bond or be a cyclic molecule. Draw a candidate, lose one H atom and you have it!
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Junior Member
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Apr 6, 2012, 07:54 AM
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He says 2:
I gave cyclohexane with 1 H less and 1 C +vely charged,
The other one I don't know!
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Full Member
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Apr 6, 2012, 10:54 AM
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OK, now do the same thing to methylcyclopentane. Now you can eliminate any one of 5 different H atoms and make a differnet C+ ion.
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Junior Member
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Apr 6, 2012, 10:17 PM
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The question was this: the mechanism of the conversion of cyclohexene into cyclohexane involves the formation of a carbonium ion C6h11+. Draw the structural formulae of two conformations of this carbonium ion and show the mechanism for the conversion of the carbonium ion into the final product.
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Full Member
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Apr 7, 2012, 07:53 AM
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I'm not sure what the "two conformatons of the carbonium ion" would be. The carbocation is planar. A base (the conjugate base of your catalytic acid for instance) will remove a proton from one of the adjacent carbon atoms to yield the alkene.
Cycloalkane conformations are more commonly encountered in E2 eliminations where the leaving groups must be antiperiplanar to one another.
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Junior Member
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Apr 8, 2012, 06:03 AM
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WHile thanking you for trying to answer the above question and upon seeing that you are good in chemistry could you solve this:
at room temperature dinitrogen tetroxide is in equilibrium with nitrogen dioxide. Deduce an expression relating Kp for decomposition reaction with "a" the degree of dissociation and "P" the total pressure.
I have tried it out using 100 as an initial number of moles of dinitrogen tetroxide. I got this answer:
[(2aP)^2]/(1-a^2) = Kp
Forget the other answer. COuld you check please?
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Full Member
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Apr 8, 2012, 06:20 AM
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I'm an Organic chemist, and one of it's benefits is that I don't have to work with Phys Chem problems. Post this one as a new question.
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