[[V]]

I have these two problems.




The answers are :


I understand "B" because the Bromine wants to replace a hydrogen atom on 2° Carbon instead of the 1° Carbon because of Markovnicks rule.

However, in example "A" the bromine replaces the hydrogen on the 1° Carbon. Can someone please explain exactly what this is in this case?

Thanks
-V

[Schrödinger]

The first reaction is a free radical reaction. Stability of free radical is to be taken into account.

The second one is an electrophilic substitution reaction. CH₃ is an ortho/para directing group.

[NahiEw]

For (a), there is no doubt that there will be a mixture of the product you pose and a bromomethyl benzene through radical halogenation. The question is why the first one is favored. What I can think of so far is that if a bromine radical were to react with a hydrogen in the benzene, then it leaves an extra electron in the already electron-rich ring. Now, remember that the benzene ring is a conjugated system with p-orbitals aligning all parallelly? If you had a residual electron in the conjugated system then you would somewhat distort the conjugated system since the carbon baring  that electron would have sp3 like character whereas all other carbons in the ring has sp2. You can also use molecular orbital theory to prove that bromination on the benzene ring is unfavorable. If you had an extra electron in the system, then that electron would be positioned in one of the anti-bonding orbital.
Bromination on the methyl group, on the other hand, would not have much effect on the electronic structure of the molecule since -CH3 is sp3 and that the transition state CH2* is also sp3.

For (b), it's not because of the order of the carbon but because of the reactants you are dealing with. Br2 will "react" with FeBr3 to from an electrophile (Br2FeBr3). Because the benzene can also act as a nucleophile with its double bonds, it will attack one end of the Br (Electrophilic-> Br-BrFeBr3) forming a tetrahedral intermediate carbocation. To restore aromaticity, a base would extract the proton in the same carbon which reforms the bezene ring.

Hope this helps.

[orgopete]

For the radical halogenation of toluene (Example a), bromine is not sufficiently reactive for any notable reaction without catalysis. With FeBr3, the electrophilic reaction can take place, Example b. For a reaction to take place in Example a, bromine must be homolyzed by heat or light. There are two possible reactions, at the benzylic position or on the aromatic ring. The bond energy of an aromatic CH bond is quite strong (113 kcal/mole) compared with the benzylic CH bond (90 kcal/mole). Therefore the benzylic halogenation is the most energetically favorable reaction.

[[V]]

Thanks I think I understand.

So is it safe to say, the lesson I take home is; markovnick's rule only applies in the presence of a catalyst?

[Borek]

Correct spelling: Markovnikov's rule.

[Schrödinger]

Markovnikov's rule applies only when there is electrophilic addition across a double bond. The 2 reactions that you have posted are just substitution reactions. I don't think the rule applies here.

[[V]]

but there is a double bond in there, its a cycloalkene ring.
Are you sure markovnick's rule has nothing to do with this?

Assuming it does not,
in the case of the free radical I think I understand why the Br replaces the H on the CH3. Is it because all the other H's are on the ring which contain double bonds therefore are held much closer & have a much higher bond energy?

In the reaction with FeBr3 however, I still don't understand why Br wants to replace an H on the ring instead of one of the H's on the methyl groups.

Can someone please try and explain this to me? Thanks.

[Schrödinger]

but there is a double bond in there, its a cycloalkene ring.
Are you sure markovnick's rule has nothing to do with this?

Yes there is a double bond. But the reaction is different.
Consider propene. Adding HBr to it will result in mixture of products in which the major product is the one with Br on the 2nd Carbon atom, not on the 1st. This is Markovnikov's rule.
The product is a mixture of 1-bromopropane and 2-bromopropane. A double bond disappears, and two atoms add themselves on to the molecule.

But in the case of addition of Bromine to the benzene ring as in your question, the double bond remains intact. Br just replaces the H in the ring. Hence Markovnikov's rule is not applicable here. It is only for addition reactions, not substitution reactions.

[[V]]

Thanks after some research I think I understand the first one.
Basically, UV light breaks Br2 into two free radicals. One of them steals a Hydrogen from the methyl group, and the other free radical Br will stick itself to whats left of the methyl group. Makes sense. & because there is no catalyst, the H's on the methyl are easier to break off than the  hydrogens on the benzene ring, yes?


But for part B, can you please explain the mechanisms & intermediate steps that make Br replace the H on the ring? What exactly is it about the Catalyst FeBr3 that makes the Hydrogens ON the ring more ready to break off than the Hydrogens on the methyl?

Thanks. Its real important that I know exactly how its doing what its doing.

[NahiEw]

For (a), correct, the benzene ring is practically inert when expose to Br2 unless a catalyst is involved.
For (b), refer to this mechanism (http://faculty.leeu.edu/~ebrown/PhH2PhBr.swf) to see how Br2+FeBr3 complex becomes an electrophilic site followed by nucleophilic attack of the electro-rich benzene ring on the Br end.

[[V]]

Thanks that was very helpful.
So FeBr3 is an electron PAIR acceptor. So this is what makes it different from becoming a free radical right? If Br was missing just ONE electron it would be called a Free Radical; and would end up attacking the H on the methyl instead.

Since Br is 2 electrons short, it makes sense that it attacks the double bond on the aromatic ring, and then the rest of the rxn follows as showed in that flash.

Is my reasoning correct so far?
Why does it refer to Br as Br+ instead of Br 2+ in that case? Does this have something to do with the fact that still has shared pairs on the FeBr3 molecule?

I think I am pretty close to understanding this reaction, thanks for the *delete me*

[NahiEw]

This reaction is NOT a radical reaction. It's simply electrophilic addition.

Why would the Br be 2+ in the complex? Refer to formal charges on the Br and you'll find that it's only 1+.

[[V]]

I still don't understand why it couldnt have done the very same thing on the methyl substituant.
Why does it do this on the ring instead?

[NahiEw]

I still don't understand why it couldnt have done the very same thing on the methyl substituant.
Why does it do this on the ring instead?

Think about what kind of mechanism (b) is and then identify the nucleophile and electrophile. Do you think methyl group can be a nucleophile?