# Homework Help: Reaction with Br2

1. May 6, 2014

### utkarshakash

1. The problem statement, all variables and given/known data
Give all possible products of the following reaction. How many stereoisomers of each product could be obained

2. Relevant equations
See attached image

3. 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.

1 - 2 products
2 - 4 products
3 - 2 products

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2. May 6, 2014

### Saitama

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?

3. May 6, 2014

### utkarshakash

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.

4. May 6, 2014

### 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.

5. May 6, 2014

### utkarshakash

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.

6. May 6, 2014

### AGNuke

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

7. May 6, 2014

### Saitama

AGNuke, do you still remember this stuff? :rofl:

8. May 6, 2014

9. May 7, 2014

### AGNuke

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

10. May 7, 2014

### 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)

Last edited: May 7, 2014
11. May 7, 2014

### utkarshakash

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 ?

Last edited: May 7, 2014
12. May 7, 2014

### 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.

13. May 7, 2014

### utkarshakash

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

14. May 7, 2014

### 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.

15. May 7, 2014

### utkarshakash

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

16. May 7, 2014

### 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.