 | | | What the difference between anti markovnikov rule and markovnikov rule
Asked Apr 30, 2011, 01:56 AM
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3 Answers What the difference between anti markovnikov rule and markovnikov rule, and when the anti markovnikov happens . ? Is it only when , there is BH3,THF/H2O2, OH- ?
I can understand if you explain it deeply, I'm studying organic chemistry course .. Thread Summary |
3 Answers
 | Full Member | |
Apr 30, 2011, 07:59 AM
| | | Markovnikov'v rule explains the regiochemistry in the addition of unsymmetrical reactants to alkenes. Most commonly this is seen in the addition of water (H-OH) and hydrogen bromide (H-Br). For Markovnikov's rule to hold sway, the reaction must proceed through a carbocation mechanism, The Markovnikov product is, of course, formed from the more stable carbocation. As an example of this, look at the addition of water to propene under acid catalysis: CH3CH=CH2 + H-OH ---H+--> CH3CHOHCH3. 2-Propanol is the "Markovnikov Product"
The hydroboration/oxidation sequence gives, as you tell, the other isomeric propanol (1-propanol). This is the "AntiMarkovnikov Product" We also see this reversal of regiochemistry with the addition of HBr in the absence (Markovnikov) and presence (AntiMarkovnikov) of peroxides.
I think it should always be pointed out that in antiMarkovnikov additions the RULE still holds true - it is the reactants that are altered. When H-OH is added with acid catalysis the major product is formed from the more stable carbocation. Stoichiometrically, you add H+ then OH-. If you examine the mechanism of hydroboration/oxidation you see that the sequence results in the addition of H- then OH+ -- you have reversed the polarity of water! (BH3 is s borohydrIDE) In HBr addition, the difference is the ionic vs the radical mechanisms.
I hope this works for you. A picture is worth a thousand words but I have a new computer and no Chemdraw yet. | | |  | Junior Member | |
Apr 30, 2011, 08:57 AM
| | | Comment on DrBob1's post Quote:
Originally Posted by DrBob1 Markovnikov'v rule explains the regiochemistry in the addition of unsymmetrical reactants to alkenes. Most commonly this is seen in the addition of water (H-OH) and hydrogen bromide (H-Br). For Markovnikov's rule to hold sway, the reaction must proceed through a carbocation mechanism, The Markovnikov product is, of course, formed from the more stable carbocation. As an example of this, look at the addition of water to propene under acid catalysis: CH3CH=CH2 + H-OH ---H+--> CH3CHOHCH3. 2-Propanol is the "Markovnikov Product"
The hydroboration/oxidation sequence gives, as you tell, the other isomeric propanol (1-propanol). This is the "AntiMarkovnikov Product" We also see this reversal of regiochemistry with the addition of HBr in the absence (Markovnikov) and presence (AntiMarkovnikov) of peroxides.
I think it should always be pointed out that in antiMarkovnikov additions the RULE still holds true - it is the reactants that are altered. When H-OH is added with acid catalysis the major product is formed from the more stable carbocation. Stoichiometrically, you add H+ then OH-. If you examine the mechanism of hydroboration/oxidation you see that the sequence results in the addition of H- then OH+ -- you have reversed the polarity of water! (BH3 is s borohydrIDE) In HBr addition, the difference is the ionic vs the radical mechanisms.
I hope this works for you. A picture is worth a thousand words but I have a new computer and no Chemdraw yet. | i got it thanks :)
i was confused because i related it , with the syn and the anti !
haha i thoght the antimark.. is an addition to some side rather than other side .. something like that.. | | |  | Full Member | |
Apr 30, 2011, 10:42 AM
| | | Comment on DrBob1's post Quote:
Originally Posted by DrBob1 Markovnikov'v rule explains the regiochemistry in the addition of unsymmetrical reactants to alkenes. Most commonly this is seen in the addition of water (H-OH) and hydrogen bromide (H-Br). For Markovnikov's rule to hold sway, the reaction must proceed through a carbocation mechanism, The Markovnikov product is, of course, formed from the more stable carbocation. As an example of this, look at the addition of water to propene under acid catalysis: CH3CH=CH2 + H-OH ---H+--> CH3CHOHCH3. 2-Propanol is the "Markovnikov Product"
The hydroboration/oxidation sequence gives, as you tell, the other isomeric propanol (1-propanol). This is the "AntiMarkovnikov Product" We also see this reversal of regiochemistry with the addition of HBr in the absence (Markovnikov) and presence (AntiMarkovnikov) of peroxides.
I think it should always be pointed out that in antiMarkovnikov additions the RULE still holds true - it is the reactants that are altered. When H-OH is added with acid catalysis the major product is formed from the more stable carbocation. Stoichiometrically, you add H+ then OH-. If you examine the mechanism of hydroboration/oxidation you see that the sequence results in the addition of H- then OH+ -- you have reversed the polarity of water! (BH3 is s borohydrIDE) In HBr addition, the difference is the ionic vs the radical mechanisms.
I hope this works for you. A picture is worth a thousand words but I have a new computer and no Chemdraw yet. | You were confusing regiochemistry with stereochemistry. | | | | Thread Tools | Search this Thread | | | | Add your answer here.
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