What is the product of a reaction involving 1,3-cyclohexadiene and DBr?
Answer:
There are two main addition products in this case.
When you protonate (with your D+) you do so on the END of the diene system (C1) so that you will have an allylic cation (C2-C3-C4). The Br(-) then adds to either end (C2 or C4) of that allylic cation. If it adds to C2, then you've got a C3=C4 double bond, a bromine on C2, and a D on C1. If bromide adds to C4 then you've got a C2=C3 double bond, a bromine on C4, and a D on C1. You will have a mixture in both cases of Br trans to D and Br cis to D.
i don't really know what you mean by DBr,but if you mean that it is some positive ion the the reaction will proceed like this-
the best example to take is HBr.
1st 1 of the double bonds will be cleaved,being
asymmetrical,the bromine will add on to the carbon atom with less hydrogens(Markovnikov's rule) and the positive part will go to the carbon with more hydrogen atoms bonded to it.
then the same thing will repeat itself for the second bond hence we will get 1,3-cyclodibromohexane.
I'm not absolutely sure though i think this should happen, please forgive me if im wrong.
the double bond will break and the Br will be added to one side and the D will be added to the other side, a trans- addition
below is a view of the reaction site after addition
...H...Br
.|...|
C-C-C
...|...|
..D..H
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