Reaction of Br2 with Possible Products & Stereoisomers

[utkarshakash]

Homework Statement

Give all possible products of the following reaction. How many stereoisomers of each product could be obained

Homework Equations

See attached image

The Attempt at a Solution

For product 1, there is 1 chiral centre and 1 double bond. ∴No. of products = 2²
For 2, it will be same as 1.
For 3, two chiral centres. Thus, 2² products.

Correct answer according to solution:-
1 - 2 products
2 - 4 products
3 - 2 products

 

[Saitama]

Homework Statement

Give all possible products of the following reaction. How many stereoisomers of each product could be obained

Homework Equations

See attached image

The Attempt at a Solution

For product 1, there is 1 chiral centre and 1 double bond. ∴No. of products = 2²
For 2, it will be same as 1.
For 3, two chiral centres. Thus, 2² products.

Correct answer according to solution:-
1 - 2 products
2 - 4 products
3 - 2 products

I am not sure if I understood the problem statement and your attempt. The question asked to find the possible products but then what does the following mean:

Correct answer according to solution:-
1 - 2 products
2 - 4 products
3 - 2 products

What does 1,2 and 3 represent? Do they represent the structures you have drawn and are they correct according to the solution key?

 

[utkarshakash]

I am not sure if I understood the problem statement and your attempt. The question asked to find the possible products but then what does the following mean:

What does 1,2 and 3 represent? Do they represent the structures you have drawn and are they correct according to the solution key?

They represent the structures in my diagram and are correct according to the solution key.
The total possible products are 3 but then the problem also asks the number of stereoisomers of each possible products.

 

[AGNuke]

Ask yourself how does Bromine attach to the given molecule? Does it involve the formation of planar intermediate at the site of bonding or is it something else? Just a food for thought.

 

[utkarshakash]

Ask yourself how does Bromine attach to the given molecule? Does it involve the formation of planar intermediate at the site of bonding or is it something else? Just a food for thought.

The bromonium ion attaches itself to the double bond and then, the bromide ion attacks resulting in the formation of product. I guess the intermediate must be non-planar.

 

[AGNuke]

Really? The triangular C-Br-C intermediate forms and that is non-planar?

 

[Saitama]

AGNuke, do you still remember this stuff?

 

[utkarshakash]

Really? The triangular C-Br-C intermediate forms and that is non-planar?

Have a look at this.

http://www.masterorganicchemistry.c...ation-of-alkenes-with-br2-to-give-dibromides/

The diagram clearly shows that the intermediate is non-planar.

 

[AGNuke]

AGNuke, do you still remember this stuff?

No, I don't. I was talking just about the C-Br-C bond.

 

[AGNuke]

Please note that having a double bond doesn't confirm stereoisomerism unless there is a flexibility to have more than one orientation (hint: Double bond in the ring, product 1)

Second one - I'm sure you can do it.

Third one - Again, go through the mechanism and witness how Br^- attacks that "intermediate". You'll reach the conclusion. (Hint: The stereocentres are not independent of each other)

 

[utkarshakash]

Please note that having a double bond doesn't confirm stereoisomerism unless there is a flexibility to have more than one orientation (hint: Double bond in the ring, product 1)

Second one - I'm sure you can do it.

Third one - Again, go through the mechanism and witness how Br^- attacks that "intermediate". You'll reach the conclusion. (Hint: The stereocentres are not independent of each other)

This means whenever a double bond is present in the ring, only one product is obtained, right? For third one, are cis and trans products formed ?

 

[AGNuke]

Nope. Try to remember how Bromide ion attack. The Bromidium ion exclusively runs away from the direction of attack of the Bromide ion, so the product will always be Trans.

But as I said, their chirality is not independent, use this fact to arrive at the conclusion that you'll get a racemic mixture of two possible trans product.

 

[utkarshakash]

Nope. Try to remember how Bromide ion attack. The Bromidium ion exclusively runs away from the direction of attack of the Bromide ion, so the product will always be Trans.

But as I said, their chirality is not independent, use this fact to arrive at the conclusion that you'll get a racemic mixture of two possible trans product.

I still can't understand how the two stereocentres are dependent.

 

[AGNuke]

Since the bromide ion and brominium ion are always opposite to each other (as far as I can remember), they are always in Trans orientation. The two stereocentres each have one bromine attached to them. But since they are always in trans state, you can't independently permute the two stereocentres, or else there'll be two cases for cis orientation, which is not possible. Hence, the stereocentres are obliged to be dependent.

 

[utkarshakash]

Since the bromide ion and brominium ion are always opposite to each other (as far as I can remember), they are always in Trans orientation. The two stereocentres each have one bromine attached to them. But since they are always in trans state, you can't independently permute the two stereocentres, or else there'll be two cases for cis orientation, which is not possible. Hence, the stereocentres are obliged to be dependent.

Then only one product should exist, and that too in trans form. But why is the answer 2?

 

[AGNuke]

Umm... if you change the orientation at one point, the orientation at another point will also change. Hence the two products. Now draw the products in the wedge-dash diagram and appreciate the fact that you'll be getting a racemic mixture.