Resonance Contributor Stability

[dttom]

If considering the following reaction:

 

CH2=CH-CH=CH2 + Br2

 

the following intermediate would be formed:

 

(A) CH2Br-C*H-CH=CH2 <--> (B) CH2Br-CH=CH-C*H2

 

From (A), there forms:

CH2Br-CHBr-CH=CH2 (in lesser amount)

 

From (B), there forms:

CH2Br-CH=CH-CH2Br (in greater amount)

 

I wonder why the latter is in greater amount as (A) is a more stable secondary carbocation while (B) is a primary one.

 

Maybe in resonance there is a constant interconversion between contributor structures but here the two have been written out and the lesser product could only be form when Br- bombards into (A) and the greater product when Br- bombards into (B). The more significant amount of (B)-derived product suggests that the (B) resonance contributor is the more frequent forms during resonance, and the overall structure of the intermediate in resonance should bias towards (B) forms. I am really interested in why it is so.

[paradox57eag]

The molecule (b) is symmetrical about the double bond.So it directly affects the stability......More the symmetry more the stability.........So more Of (b) than (a)

Refer Markonikov's rule if u want to......

Reference: Morrison N Boyd, Organic Chemistry.

[dttom]

You are right in the product from B being more stable than product from A, either because it is more symmetric or through hyperconjugation and so according to Sayzeff's rule it is more stable. But if we consider the stability of intermediate it seems that intermediate A is more stable than intermediate B as A is a secondary carbonium ion better stabilized by hyperconjugation. So is there any paradox here?

[vedmecum]

May be this will help you .......... look for the both thermodynamically and kinetically derived product ........... in A there is a formation of allylic intermediate which is more stable than primary carbocation as in case of B .......... so A intermediate will form at a faster rate than carbocation ........ results in formation of 1,2 addition product in major amount ...........so 1,2 product is kinetically favored ........ now look on the final product ......... 1,4 product is more stable than 1,2 as 1,4 product is more substituted alkene so thermodynamically favored ......... so as soon as 1,2 product form there is enough collision energy ( even at room temp. ) to pass over the energy barrier from 1,2 product to 1,4 product........