[Cooper]

Hi,

I came across the question below. I figured that because carbon one on the butadiene is allylic, a carbocation would form there, due to its stability, after the alkene took a proton from HBr. Then HBr would attach to the carbon (product D).

They key says B is the answer. Is the answer B, because there pi electrons are conjugated and it doesn't want to break the conjugation so the terminal alkene is broken?

Thanks

[Dan]

They key says B is the answer. Is the answer B, because there pi electrons are conjugated and it doesn't want to break the conjugation so the terminal alkene is broken?

That's half the story - what you are arguing is that you get the more thermodynamically stable product in which the alkene product is conjugated to the phenyl ring.

The question states that this is the major product at all temperatures, which implies that it is also the kinetic product. Can you see why?

[Cooper]

They key says B is the answer. Is the answer B, because there pi electrons are conjugated and it doesn't want to break the conjugation so the terminal alkene is broken?

That's half the story - what you are arguing is that you get the more thermodynamically stable product in which the alkene product is conjugated to the phenyl ring.

The question states that this is the major product at all temperatures, which implies that it is also the kinetic product. Can you see why?

Thanks! Hm, is it the kinetic product as well because the energy barrier to break the conjugated double bond is higher than breaking the terminal alkene?