Alkene ring expansion

morgan8222

I need to react 1,3-methylcyclopentene with H-Br.

Do I do a ring expansion here?

What I came up with is 1-bromo-3-methylcyclohexane.

Am I completely off? Is there just one major product?

I would appreciate the help!

aroc91

Besides the fact that your naming of your intended reactant is a bit wonky, by what mechanism would HBr expand a ring? Acids don't break carbon-carbon bonds like that.

If you meant this as your reactant-



you would make 4 diastereomers, depending on which end of the double bond and from which side the bromide attaches.

jd12345

I dont really understand the name of your reactant. Are there two -methyl groups?
But assuming there is a methyl group attached to one of the sp2 carbon - then if H+(from HBr) reacts with the double bond the positive charge will go to the tertiary carbon.

The ring will then try to expand as cyclohexane is much less strained than cyclopentane.
So you would have ring expansion.

aroc91

http://chemwiki.ucdavis.edu/Organic_...rearrangements

It seems ring expansion could only happen with a carbocation that isn't on the ring, which would not happen with methylpentene.

morgan8222

Sorry about the naming. I am just starting organic chemistry. I will try to explain the structure...

A 5-membered ring with one double bond and two methyl groups off of it.
Say the double bond is between 1 and 2. Number around the ring clockwise 1-5. One methyl is on 1, and the other methyl is on 3.
This is supposed to react with H-Br.

I know that the H will attach to one of the double bond C's so the pi bond will disappear. I know the 5 and 6-membered rings are most favorable. Does this shift to a 6 membered? If so, how does this happen? Could someone take me through it? I think I understand the process with a non-ringed alkenes, but I am confused about this.

Yanick

Have you learned about relative stability of carbocations? Just think about what type of carbocation will form with the addition of the H⁺ to the double bond. There are two possibilities, pick the most stable one-assuming you have learned this, otherwise just read the couple of pages in your text, its not THAT complicated.

IMO this problem is more a question of carbocation stability than ring stability/strain (5 membered rings are pretty stable and are abundant in nature).

The tricky part to this problem is the stereochemistry of the product(s). Do it mechanistically with all possibilities and see what you get.


Hypothetically speaking, a hydride shift may lead to a ring expansion but see my comments above WRT carbocation stability and ring strain.

I'm being vague because this is probably one of the easiest problems you can face in an O-Chem class. Its pretty straightforward, except for the stereochem.

morgan8222

The carbocation is most stable when it is on a tertiary carbon, so if possible, it will shift to be more stable. So the H would go on the secondary carbon, and the tertiary carbon is the carbocation. So, the Br would be on the same C as the methyl.
I think this is correct, but what about a ring expansion? I think 6 membered rings are a little more favored than 5.

Yanick

Think about what would need to happen to get that ring expansion, aroc19 gave a good hint.

Ask yourself:

Where would the carbocation need to be for the ring expansion to occur?

How can you get there from the intermediate which we have already established (the tertiary carbocation)?

Would this be a favorable intermediate?

morgan8222

Would the carbocation need to be on a C other than the C's that make up the ring? It would need to be on the methyl? Could do a hydride shift to achieve this. But that isn't favorable because you would be shifting from a tertiary to a primary. Am I correct?
So no ring expansion?

Yanick

AFAIK this is correct. Carbocation rearrangements require anion shifts; meaning the adjacent group shifts with its electrons. This effectively "shifts" the positive charge. The thing is, this type of shift typically occurs to create a MORE stable intermediate, not less. Although a six membered ring may be more favorable in the end, the intermediates/T-states are highly unfavorable; thus the reaction will likely not proceed through this pathway.

This is easy to visualize if you know how to draw reaction coordinates (reaction progress on the X-axis and relative potential energy on the Y-axis). You will most likely, later in the course, learn about the phenomenon of Thermodynamic vs. Kinetic control where you will have to think about the relative energetics of the final products and the intermediates on the pathway(s) to getting to these products. Don't make yourself crazy yet about this type of thing, but its important to think from an energetics point of view.

Now go and work out the stereochemistry of the products, this is a bit of a tricky part which usually catches many students off guard.

morgan8222

It's not just one product? Is the product then a 5-membered ring with a methyl and Br on 1 and the other methyl on 3? We are not doing stereochem yet with these. Wouldn't the most stable and therefore major product be what I came up with?

Yanick

Yes you are correct AFAIK. I apologize for confusing you with the stereochemistry thing, its just that usually stereochemistry stuff is done before getting into any kind of reactions and/or mechanisms etc (typically in the first 3 or so chapters of an intro to O-chem book, along with acids/bases, nomenclature and all the other fundamental stuff). If you are not required to know about stereochemistry yet, then I guess you're done but just understand that there will be more than 1 major product due to the generation of a stereo center in the product.

morgan8222

Okay. Thanks for your help!