I found out from drawing it out that there are 2 products: 4-bromo-3,4-dimethyl-2-hexene and 2-bromo-3,4-dimethyl-3-hexene. I want to know which forms most of the time.

I used the fact in allyl carbons, the double bonds will travel to the carbon-carbon single bonds to form a double bond. I also used the fact that tertiary carbocations are more stable due to sigma bonds that can participate in hyperconjugation - that is delocalization of electrons by the overlap of sigma bond orbital with an empty p-orbital on an adjacent carbon.

I would like to see some of these reactions in a lab setting, but for now I'm learning organic chemistry reactions like as if it were math.

  • $\begingroup$ The second product you propose has four substituents, including one double bond, at C-3, but anyway HBr does not brominate alkenes or dienes just like that. Are you learning OC by yourself? $\endgroup$
    – Karl
    May 7 '17 at 13:36
  • $\begingroup$ Yes I am. I'm stationed at a ship and my only resource is a book and me thinking. $\endgroup$ May 7 '17 at 18:30
  • $\begingroup$ Ah. Navy_Colors. Wow. (Of course alkenes react with HBr, what was I thinking? The second product should be 2-Bromo-..., right?) Can you describe the first step of the reaction? Markownikow would be the reasonable thing to think. $\endgroup$
    – Karl
    May 7 '17 at 18:59
  • $\begingroup$ Same question on slightly different substrate chemistry.stackexchange.com/q/59964/16683 $\endgroup$
    – orthocresol
    May 8 '17 at 8:56
  • $\begingroup$ @orthocresol Similar indeed, only C-3 here is too crowded to really allow the 1,2-addition, isn't it? $\endgroup$
    – Karl
    May 9 '17 at 11:16

Conjugated dienes can undergo two types of additions with HX, they can undergo 1,2-additions and 1,4-additions. A 1,2-addition occurs when HX adds to only one of the double bonds, not interacting with the other. a 1,4-addition occurs when HX adds X to one of the double bonds, and because the system is conjugated, the hydrogen will be added three carbons away from X. The 1,2-adduct and 1,4-adduct are either thermodynamically or kinetically stable. The kinetic product will form quicker, but is less stable, it will dominate at low temperatures and short reaction times. The thermodynamic product is more stable but takes more time to form. It will be the major product at high temperature and long reaction times. In this case, the kinetic product would be 2-bromo-3,4-dimethyl-3-hexene, the thermodynamic product would be 4-bromo-3,4-dimethyl-2-hexene, because the thermodynamic product has the bromine on the most substituted carbon. Depending on the reaction conditions, either could form. At low temps. and short reaction times, 2-bromo-3,4-dimethyl-3-hexene will form, at high temps. and long reaction times, 4-bromo-3,4-dimethyl-2-hexene will form.


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