OChem Markovnikov question

[Intent]

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.

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[biomolecmed]

NBS donates a bromine is a radical halogenation reaction. Since propene has only 3 carbons and bromine goes only to the most stable radical, it will go to primary allylic radical, he se the methyl group. Now, lets say we had propane reacting with NBS, then most stable radical will be secondary allylic, the ch2. Therefore the reaction would go markovnikov and halogen would go to the most substituted carbon. Book is right describing a general reaction, it just used a bad example to illustrate it. Hope this helps.

 

[Intent]

Thanks. Can you explain to me why the H3C-CH is less substituted than the C=CH? I'm still struggling with the fact that the Br attaches to the CH2 instead of the Carbon with the methyl group attached. Sorry if I sound like a triple *****.

 

[MrSummerlin]

NBS will react with alkenes 1 in aqueous solvents to give bromohydrins 2. The preferred conditions are the portionwise addition of NBS to a solution of the alkene in 50% aqueous DMSO, DME, THF, or tert-butanol at 0°C.[2] Formation of a bromonium ion and immediate attack by water gives strong Markovnikov addition and anti stereochemical selectivities.

It not a radical reaction...has nothing to do with it being allyic...is all about the aqueous environment. You see later NBS without the aqueous environment would to that propene into an alkyl halide
(Wikipedia) -have to love it

 

[Intent]

Thanks but I'm still not understanding why the carbon atom the bromine attached to is considered the less highly substituted one. This is what I need to know, why is the CHCH3 less substituted than CH2? I can't understand why is on that carbon and not on the other one. It seems like this reaction is the opposite of markovnikovs rule. Anyone? Why did the Br attach to carbon 1 instead of carbon 2? I'm either missing a concept somewhere or something else. I understand how to make the carbocation intermediary and I can follow the drawing, I just can't understand why that is markovnikov.

 

[MrSummerlin]

The br in an aqueous environment can only attach to one of the carbons involved in the double bond so stop bringing up ch3(it has nothing to do with the reaction). CH2 is less substituted because it only attached to one carbon attached and that one carbon is the one it is double bonded to. CH in the double is attached to CH3 and CH2, so more of its Hydrogen have been replaced(also called substituted), so it is more highly substituted.

 

[Intent]

I GOT IT!!! I found it in the markovnikov addition link you included in prior post. This is what I needed to know, "The same is true when an alkene reacts with water in an addition reaction to form alcohol. The hydroxyl group (OH) bonds to the carbon that has the greater number of carbon-carbon bonds, while the hydrogen bonds to the carbon on the other end of the double bond, that has more carbon-hydrogen bonds."

THANK YOU SO MUCH!!! YOUR THE MAN!

 

[MrSummerlin]

go on utorrent and download organic chemistry author: mcmurry edition: 7th ed and look at the chapter summaries for 7,8,10,11 and it has all these reactions summarized really good.

Glad I could help.