Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: Enola on June 15, 2016, 05:59:34 AM
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Hi. :)
First of all, sorry for my English. ;)
I have a question about reaction: pent-2-ene with concentrated HCl, and there's also KI as the reaction medium. I know that the product will be: 2-chloropropane and 3-chloropropane.
And... what is the main product, what is the mechanism of this reaction and why is there KI? ???
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You should be able to find the mechanism for an electrophilic addition to an alkene or as a Markovnikov addition reaction, the mechanism is the same.
Re: KI, this sound like an error or misinterpretation of the reaction. Was KI step two?
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Really, KI was there as the reaction medium. And I was very surprised and I was looking this reaction in a few books, and on many web pages but I haven't found anything like this so this is why I asked here.
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You should be able to find the mechanism for an electrophilic addition to an alkene or as a Markovnikov addition reaction, the mechanism is the same.
Re: KI, this sound like an error or misinterpretation of the reaction. Was KI step two?
Orgopete,
Is it possible they are trying to do a "one-pot" hydrochlorination/Finkelstein to give the alkyl iodide?
Ortho
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If one wanted the iodide, why attempt a two-step addition? Can't you just use HI directly? If you used a mixture of HCl/HI, what would you get? What was discussed in class?
It seems more plausible that this is a textbook problem and KI was a step two reaction. Another possibility is that this is an error. I know I have made plenty of them myself.
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This reaction is from an exam, and my teacher don't want to explain this because he wants to give this reaction again and he has very difficult character, he always gives weird reactions. He is known for that ... And I don't think it's an error, he was too confident about it.
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I know that the product will be: 2-chloropropane and 3-chloropropane
Chloropropanes are imposible product of this reaction. You probably mean chloropentanes.
For obtaining chloropentanes the presence of KI is rather unimportant.
If KI is used in the second step then iodides may be obtained, but the shorter way for them is using KI with anhydrous H3PO4.
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I know that the product will be: 2-chloropropane and 3-chloropropane
Chloropropanes are imposible product of this reaction. You probably mean chloropentanes.
Yes, of course. Sorry.
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Given the reaction, why would a mixture of chloropentanes be the result? That doesn't make sense.
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Is there possibility that in this reaction KI just is and that's all? Like, maybe my teacher is trying us and our knowledge and... flair?
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Would HCl react with KI, with I2 and H2 evolution, rather than with pentene at all?
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I don't think one should regard KI as such. The reaction as described should be mainly H3O(+), K(+), Cl(-), and I(-). That is why I questioned a chloride should be the major product.
Re: Deiodination reaction
Since I(-) can react with an aryl iodide to give an aryl-H + I2, then a similar reaction in this case would give pentane. Again, not a chloropentane. However, that would imply alkyl iodides react with iodide to give an alkane. Now think of all the reactions to give an iodide. If iodide reaction was that facile, they would all give alkanes instead. This doesn't seem likely.
Given the nature of the question, I'm guessing the product is a mixture of chlorides and iodides. The Finkelstein gives an iodide because NaCl precipitates from acetone driving the reaction forward. In this case, I'm guessing a product mixture similar to the the ratio of [Cl]/[ I]. Also, because this reaction is in water, I'm not excluding alcohol formation.
If this wasn't a test question, I'd ask someone to look up the reaction and tell us what actually happened. The poster could ask his or her professor, but I'm guessing the professor doesn't actually know either (besides refusing to answer). Practically speaking, this seems a dumb reaction to try, in my opinion.