Diene + NBS

[skyisblue]

What products(s) would you expect from the reaction of 1,4-hexadiene with NBS? What is the structure of the most stable radical intermediate?

There are a total of 5 products. One intermediate produces 2 and the other produces 3.

I just don't understand how they distributed the products accordingly to each radical intermediate.
How do you know which goes to which?

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[Lonely Sol]

The 5 different intermiediates that can be formed are as follows:
Radical on:
1. C-1
2. C-2
3. C-4
4. C-5
5. C-3

Well, Maybe someone can correct me if I am wrong, but it make sense to me
that C-3 will the most favored position for the radical to form, since it is an allylic position with respect to both double bonds, therefore, C-3 radical is fovored!

Final answer for the rxn should be: 3-bromo-1,4hexadiene!

 

[skyisblue]

The 5 different intermiediates that can be formed are as follows:
Radical on:
1. C-1
2. C-2
3. C-4
4. C-5
5. C-3

Well, Maybe someone can correct me if I am wrong, but it make sense to me
that C-3 will the most favored position for the radical to form, since it is an allylic position with respect to both double bonds, therefore, C-3 radical is fovored!

Final answer for the rxn should be: 3-bromo-1,4hexadiene!

Because there are 5 products, I would consider that there are 5 radicals as well. BUT, the solution states that there are 2 radical intermediates only. The one that is most stable is the one with 3 radicals.

I provided attachments. Please explain almighty lonelysol.

 

[Lonely Sol]

Because there are 5 products, I would consider that there are 5 radicals as well. BUT, the solution states that there are 2 radical intermediates only. The one that is most stable is the one with 3 radicals.

I provided attachments. Please explain almighty lonelysol.

I guess I dont understand your question, I think, the only thing is that the most stable position for a radical is allylic position and if the allylic position is at secondary carbon then that is even better. Therefore, since the compound is 1,4-hexadiene, then the C-3 position is very favored because it is the allylic position for both double bonds plus it is not a primary carbon, therefore, I would say the answer is 3-bromo-1,4-hexadiene

Also, dont worry about intermiediates, it will never be on the test, but i guess its good to know if you are going to take Org. II

I might be wrong! This just makes sense to me!

 

[BodybldgDoc]

he is right. allylic takes precendence in this case because it would be the most stable. However also keep in mind if it came down to benzylic vs allylic then benzylic radical > allylic radical because of delocalization and stability.
I would agree with 3-bromo-1,4-hexadiene being the major product.
Focus on the major products of the reactions and less so on the intermediates and minor products which are things that wont show up on the test.

 

[Lonely Sol]

he is right. allylic takes precendence in this case because it would be the most stable. However also keep in mind if it came down to benzylic vs allylic then benzylic radical > allylic radical because of delocalization and stability.
I would agree with 3-bromo-1,4-hexadiene being the major product.
Focus on the major products of the reactions and less so on the intermediates and minor products which are things that wont show up on the test.

Yea, stability order should be benzylic>allylic>3>2>1