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Hi, Im having difficulty understanding why a molecule is markovnikov instead of non-markovnikov. Here is the question:
When an unsymmetrical alkene such as propene is treated with NBS in aqueous DMSO, the major product has the bromine atom bonded to the less highly substituted carbon atom. Is this Markovnikov or non-Markovnikov orientation? Explain.
http://jpkc.zju.edu.cn/k/146/Organic_Chemistry_Level_1/Answer_for_mcmurry/Chapter7.pdf
heres the link it's problem 7.6
The answer is markovnikov according to the book.
Here is why I don't understand the answer. I thought that markovnikov's rule said that in the addition of HX to an alkene, the H attaches to the carbon with the fewer alkyl substituents and the X attaches to the carbon with more alkyl substituents. I think this is slightly different because it's being treated with NBS, and a halogen and an alcohol will attach to the alkene. My problem is, I don't know why the Br is attaching to the carbon bonded to the methyl group and not the carbon bonded to two H's. Doesn't the rule state that the halogen will bond to the carbon with the least amount of alkyl substituents (the one with the largest initial H's on it) ? Or is this the way the reaction works when NBS is involved? Any clarity would be greatly appreciated. Thanks.
When an unsymmetrical alkene such as propene is treated with NBS in aqueous DMSO, the major product has the bromine atom bonded to the less highly substituted carbon atom. Is this Markovnikov or non-Markovnikov orientation? Explain.
http://jpkc.zju.edu.cn/k/146/Organic_Chemistry_Level_1/Answer_for_mcmurry/Chapter7.pdf
heres the link it's problem 7.6
The answer is markovnikov according to the book.
Here is why I don't understand the answer. I thought that markovnikov's rule said that in the addition of HX to an alkene, the H attaches to the carbon with the fewer alkyl substituents and the X attaches to the carbon with more alkyl substituents. I think this is slightly different because it's being treated with NBS, and a halogen and an alcohol will attach to the alkene. My problem is, I don't know why the Br is attaching to the carbon bonded to the methyl group and not the carbon bonded to two H's. Doesn't the rule state that the halogen will bond to the carbon with the least amount of alkyl substituents (the one with the largest initial H's on it) ? Or is this the way the reaction works when NBS is involved? Any clarity would be greatly appreciated. Thanks.