Ring Expansion of 1,3-Methylcyclopentene with H-Br

In summary, a hydride shift may lead to a ring expansion, but this is not always the case and the stereochemistry of the product is a challenge.
  • #1
morgan8222
16
0
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!
 
Chemistry news on Phys.org
  • #2
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-

3-methylcyclopentene.GIF


you would make 4 diastereomers, depending on which end of the double bond and from which side the bromide attaches.
 
  • #3
I don't 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.
 
  • #4
http://chemwiki.ucdavis.edu/Organic_Chemistry/Organic_Chemistry_With_a_Biological_Emphasis/Chapter_15%3A_pi_electrons_as_nucleophiles/Section_15.7%3A_Carbocation_rearrangements [Broken]

It seems ring expansion could only happen with a carbocation that isn't on the ring, which would not happen with methylpentene.
 
Last edited by a moderator:
  • #5
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.
 
  • #6
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.
aroc91 said:
http://chemwiki.ucdavis.edu/Organic_Chemistry/Organic_Chemistry_With_a_Biological_Emphasis/Chapter_15%3A_pi_electrons_as_nucleophiles/Section_15.7%3A_Carbocation_rearrangements [Broken]

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

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.
 
Last edited by a moderator:
  • #7
Yanick said:
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.

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.
 
  • #8
morgan8222 said:
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.

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?
 
  • #9
Yanick said:
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?

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?
 
  • #10
morgan8222 said:
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?

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.
 
  • #11
Yanick said:
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.

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?
 
  • #12
morgan8222 said:
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?

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.
 
  • #13
Yanick said:
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.


Okay. Thanks for your help!
 

1. How does the ring expansion of 1,3-methylcyclopentene with H-Br occur?

The ring expansion of 1,3-methylcyclopentene with H-Br involves the addition of the H-Br molecule to the double bond of the 1,3-methylcyclopentene. This results in the formation of a new five-membered ring structure with a bromine atom attached to one of the carbon atoms.

2. What is the purpose of using H-Br in the ring expansion reaction?

H-Br serves as a source of a proton and a bromine atom in the ring expansion reaction. The proton adds to the double bond of 1,3-methylcyclopentene, while the bromine atom attaches to one of the carbon atoms, resulting in the formation of a new ring structure.

3. What type of reaction is the ring expansion of 1,3-methylcyclopentene with H-Br?

The ring expansion of 1,3-methylcyclopentene with H-Br is a type of electrophilic addition reaction. This means that the electrophilic H-Br molecule adds to the electron-rich double bond of 1,3-methylcyclopentene, leading to the formation of a new product.

4. What are the major products of the ring expansion reaction?

The major products of the ring expansion reaction are the new five-membered ring structure and the formation of a new carbon-carbon bond between the bromine atom and one of the carbon atoms in the original 1,3-methylcyclopentene molecule. The resulting product is known as 1-bromo-2-methylcyclopentane.

5. What factors can affect the rate of the ring expansion reaction?

The rate of the ring expansion reaction can be affected by several factors, including temperature, concentration of reactants, and the presence of any catalysts. Higher temperatures and higher concentrations of reactants can increase the rate of the reaction, while the presence of catalysts can lower the energy barrier for the reaction to occur, also increasing the rate.

Similar threads

  • Biology and Chemistry Homework Help
Replies
25
Views
2K
Replies
10
Views
997
  • Biology and Chemistry Homework Help
Replies
1
Views
1K
Replies
1
Views
854
Replies
9
Views
19K
  • Introductory Physics Homework Help
Replies
30
Views
2K
  • Introductory Physics Homework Help
Replies
7
Views
138
  • Linear and Abstract Algebra
2
Replies
55
Views
3K
Replies
2
Views
722
Back
Top