[noromon119]

I have serious issues with inductive effects and their effect on halogenation.

What will be the major product when a mole of HBr is reacted with one mole of penta(1,3)diene?
This is similar to my previous post.

Obviously, due to markovnikov's rule, if the double bond at 1 is broken, Br will be substituted at 2. But, if double bond at 3 is broken, will the postive charge come on 3, as it will be resonance stabilised?? And I also know I should consider 1,4  addition. Im just confused which double bond will be affected and why.

Please help me out. I hope you guys understand where im lacking info.

[orgopete]

These questions are prototypal kinetic v thermodynamic reactions. However, in this case, they both yield the same product. Protonation gives two allylic carbocations which are identical by symmetry. Thus addition at either end will give the same product, 4-bromo-2-pentene. If DBr were used instead of HBr, then a mixture could be noted.

[noromon119]

The ones in boxes are due to shift of pi bonds

1, 3 and 4 will be our choices and 2, 5 will be due to shift of pi electrons (for 1,4 addition),

1 2 are identical. Hence, obviously, its the product. And 3 is not predominant because it is not resonance stabilized. What about 4,5??

[orgopete]

Rationally, 4 and 5 could also be present in relationship to their stability compared to 1 and 2. By cation stability, we would expect them to be minor as a prary carbocations is less stable. I don't know actual product distribution for this reaction.