Question about Markovnikov Addition

[tony montana]

I am trying to understand how and why these reactions differ.

 

 

 

I know that the nucleophile should always attach to the less substituted carbon of a double bond in an addition alkene reaction. However, in my textbook, there are two example questions that are confusing me. One example shows a hydrogen as the nucleophile and the other example shows Bromide as the nucleophile.

 

 

 

Would this have anything to do with the fact that the Bromide came from a free-radical?

[hypervalent_iodine]

Can you tell me why they are confusing you?

[tony montana]

Why in some situations the Hydrogen is the nucleophile and in others the Bromide?

 

In other words, how can you tell what the nucleophile is in the problem?

[hypervalent_iodine]

Hydrogen only acts as a nucleophile when it is a hydride. In the form of HBr, H is not a nucleophile. It might be easier for me to help you if you could put the question on here, is that possible?

[spin-1/2-nuclei]

Hello,

 

The Markonikov vs. anti Markonikov products are examples of what is called regiochemistry. The Markonikov vs. anti-Markonikov product is formed depending on where the most stable carbocation or radicle will be formed during the intermediate of the reaction.

 

Thus,

 

The Markonikov vs anti-Markonikov product depends on which mechanism the reaction is undergoing... In the case of radical addition or hydroboration you have a situation where the opposite regiochemistry results from the inversion of the order of steps in the mechanism.

 

 

In the case of the anti-Markonikov product...

After the initiation step to form the X radical with whatever random initiator, the X radical - in your case Br, will (via the fishhook arrows for pictorial representation) propogate via formation of a bond with one electron from the double bond of the alkene and the other electron from the double bond of the alkene will go to form the most stable radical...

 

That is to say that, the radical will be on the carbon with the most substituents. To obtain your product the radical will then propogate with another H-Br [H-X] by again forming the most stable radical. Thus the bond between the H and Br, consisting of two electrons, will send one of it's electron with the hydrogen to form the H-C bond and quench the radical (producing your anti-Markonikov product). Now, the new radical just formed, the Br radical (being the most stable), will go on to another propogation step with another alkene.

 

termination can occur between any two radicals, but no new radicals will be formed during that step. Do not confuse the propagation with termination step.

 

*Note that the rate between propogation between the alkene radical and an H-Br molecule will be faster than the rate for propogation between an alkene radical and an alkene. This is because of sterics, etc.

 

So the reason you get the different regiochemistry -i.e. anti-Markonikov - arises from stability of the intermediate..

So review a good source on radical mechanisms and this will make more sense.. remember your three steps in a radical reaction are 1.) initiation, 2.) propogation, and 3.) termination. Also review radical stability.

 

In normal Markonikov addition - you are typically forming a carbocation - which will always form on the carbon with the most substituents. Then you will have attack of a nucleophile (I can go over what makes something a good nucleophile with you in another post if you wish)

 

So, in the mechanism to produce Markonikov regiochem, the alkene will use one of it's bonds to deprotonate H-Br. The reason the hydrogen is "grabbed" first is because it is less stable as a H- than Br is as a Br- (if you don't understand why I can go over what makes something a good nucleophile and likely to hold a negative charge and other such concepts that will make this more clear) but, for now - if you don't understand Hard Soft Acid Base theory - let's just accept that the hydrogen is grabbed because it is less able to exist unbonded/with a negative charge than the Br is..

 

Now that you have the carbon hydrogen bond formed, formal charges tell us that the carbon opposite the carbon that received the hydrogen is now missing a bond and therefore has a positive charge..

 

primary carbocations are less stable than secondary carbocations which are less stable than tertiary carbocations. (I can go over all of this with you in another post too if you need but for now we have to accept at face value that, during reactions the most stable carbocation will typically form).

 

Thus, the hydrogen will go to the least substituted carbon of what used to be our alkene and as a result a positive charge must be assigned to the carbon best able to support it - i.e. - the carbon with the most substituents. Then, the Br- (being the nucleophile in this situation) will attack the positively charged carbon and the Markonikov product is formed with the Br on the most substituted carbon.

 

So, in the radical reaction that gives the anti-Markonikov product - the Br is delivered to the least substituted carbon because during propogation it needs to make a bond and it cannot make a bond with the most substituted carbon - because that carbon needs to take the other electron from the alkene bond and thus place the radical on the most substituted carbon..

 

In the Markonikov product the mechanism undergoes a carbocation - which forms on the most substituted carbon and it is the carbocation that is attacked by the Br in this case.

 

Hopefully this was helpful..

Cheers..

Edited August 29, 2011 by spin-1/2-nuclei