halogenation of an alkene question

[jlee1986]

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hey guys,

thanks in advance for responding to this message..i really appreciate it! after the halonium ion is made (ex: Br+ bonded to two carbons), why is the more substituted carbon attacked by the Br-? Wouldn't the lesser substituted carbon be attacked by Br- since it is less stable?

 

[levyusupov]

your problem concerns with the stability of the carbocation.
the more substituted the carbocation the better. therfore when you form carbocation the most substituted will be attacked by the nucleophile.

for further info, check your orgo textbook about carbocations and its applications.

 

[Shunwei]

hey guys,

thanks in advance for responding to this message..i really appreciate it! after the halonium ion is made (ex: Br+ bonded to two carbons), why is the more substituted carbon attacked by the Br-? Wouldn't the lesser substituted carbon be attacked by Br- since it is less stable?

The bromonium ion is positively charged, and translocates much of this charge on the carbon atoms forming the triangular complex. These carbon atoms therefore are like carbocations, and the most stable one gets attacked on the backside by the remaining bromide anion. This is just like oxymercuration, or the opening of epoxide rings by acid-catalysis.

 

[jlee1986]

yeah, but why would the more substituted carbon be more likely to be attacked? is it because of the fact that there are more carbons around it and they spread the charge of the electrons amongst those other carbon atoms? or is it because when the Br+ breaks its bond with the most substituted carbon, it is the more stable than the other carbon to become a carbocation? thanks 🙂

your problem concerns with the stability of the carbocation.
the more substituted the carbocation the better. therfore when you form carbocation the most substituted will be attacked by the nucleophile.

for further info, check your orgo textbook about carbocations and its applications.

 

[Shunwei]

yeah, but why would the more substituted carbon be more likely to be attacked? is it because of the fact that there are more carbons around it and they spread the charge of the electrons amongst those other carbon atoms? or is it because when the Br+ breaks its bond with the most substituted carbon, it is the more stable than the other carbon to become a carbocation? thanks 🙂

alkyl groups are considered electron-releasing, so next to a carbocation it acts as a stabilizing influence which makes it a more stable intermediate that is preferentially attacked.

 

[Mstoothlady2012]

The bromonium ion is positively charged, and translocates much of this charge on the carbon atoms forming the triangular complex. These carbon atoms therefore are like carbocations, and the most stable one gets attacked on the backside by the remaining bromide anion. This is just like oxymercuration, or the opening of epoxide rings by acid-catalysis.

I thought bromonium ion is negatively charged

 

[joonkimdds]

hey guys,

thanks in advance for responding to this message..i really appreciate it! after the halonium ion is made (ex: Br+ bonded to two carbons), why is the more substituted carbon attacked by the Br-? Wouldn't the lesser substituted carbon be attacked by Br- since it is less stable?

ok here is the answer 🙂
if Br attack H with less stable carbon, it will form less stable carbocation.
if Br attack H with stable carbon, it will form stable carbocation.

our goal is to make a stable carbocation 🙂
was it a good explanation ? I think it was...even I am surprised with my own words,

 

[IsRaEl21]

It should follow Markonikoff's rule the more substituted will form the Carob cation on + charge. If u really want to know the reason why u should open either ur orgo book, or read the Destroyer's solution, there is a small lecture about it. Good Luck!!!