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Topic: Alkene addition - really simple  (Read 4294 times)

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Offline cabaal

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Alkene addition - really simple
« on: October 21, 2010, 02:48:08 PM »


I'm having trouble figuring out where the H and Cl go. I know it's either carbon 3 or 4, but I'm not sure which and my book doesn't have any similar examples.

My guess is the methyl group affects the carbocation stability (i.e., the farther the carbocation is from the methyl group, the more stable it is.) I would say that the hydrogen ion goes on carbon 3 and the chloride ion goes on carbon 4. Is this correct?

Offline syndakitz

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Re: Alkene addition - really simple
« Reply #1 on: October 21, 2010, 02:51:24 PM »
You are most likely going to have a mixture between both products. But in essence, the more yielding product should be the one you suggested.

Offline karlosshughes

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Re: Alkene addition - really simple
« Reply #2 on: October 27, 2010, 12:58:49 AM »
No. The methyl group actually helps to stabilise the carbocation as it can donate some electron density from one of the C-H sigma bonds into the empty 2p orbital of the carbocation. The proton will add to the end of the double bond with the ethyl group forming the carbocation on the carbon next to the i- pr group.

The product will be 2-methyl-3-chlorohexane.
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Offline Cherriyan

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Re: Alkene addition - really simple
« Reply #3 on: October 28, 2010, 10:02:15 AM »
In any such questions, if there's a confusion, first recognize all the possibilities of the attack of the electophile and hence draw the structure of carbocations.
In this case, both the carbocations will be primary. So the factor which decides the stability of the carbocation is hyperconjugation. Check for the number of alpha hydrogens in each case. If the positive charge is on C number 3, then there are 3 alpha Hydrogens. On the other hand if the electromeric shift is such that the positive charge is on C number 4, then there are 4 alpha Hydrogens. This implies, the latter intermediate is more stable as it will be having more number of hyperconjugative forms. So the attack of Cl- will take place at C number 4. Thus, the product formed in major amount will be 4-chloro-2-methylhexane.

Offline orgopete

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Re: Alkene addition - really simple
« Reply #4 on: October 28, 2010, 11:00:55 PM »
This is a really good question. Will alkyl substitution stabilize the carbocation as suggested by karlos or will hyperconjugation win as suggested by Cherriyan?

Can someone look up this reaction (if it exists) and post the actual data? My instinct suggests the 3-chloro will be greater as suggested by karlos, but I think as a mixture. I would also expect some rearranged product.
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Offline karlosshughes

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Re: Alkene addition - really simple
« Reply #5 on: October 28, 2010, 11:14:07 PM »
Alkyl groups stabilise carbocations by induction of e- density- this is quite a fundamental concept in organic chemistry and my explanation was taken from a textbook (just to make sure)- consultation of any organic chemistry textbook will confirm this.

However, I should mention that I was incorrect in saying electrons will be donated from the sigma bond of a C- H orbital on one of the methyl groups- it will actually be donated from the C- H sigma orbital on the carbon adjacent to the carbocation centre. But nevertheless, the more alkyl groups are, the more electron density is donated by induction which further stabilises the cation.

Think of Markovnikov's rule- the proton is attatched to the carbon of the alkene which had the most protons to start with, i.e. the cation is formed on the carbon with the most alkyl (or aryl) groups. Although this doesn't help specifically in this case, since both carbons of the alkene have one proton, the theory behind this is the same.

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Offline karlosshughes

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Re: Alkene addition - really simple
« Reply #6 on: October 28, 2010, 11:14:57 PM »
P.s. the phenomenon I was describing is hyperconjugation.
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Offline karlosshughes

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Re: Alkene addition - really simple
« Reply #7 on: October 28, 2010, 11:19:24 PM »
At the risk of spamming, this is explained in chapter 17 of Organic Chemistry by Clayden et al (one of the authors of which happens to be one of my lecturers), page 409.

Also, I agree there will probably be some of the tertiary chloride formed as the tertiary cation can be formed by proton migration (didn't think of that one!).
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