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enter image description here


Here is how I approached this:

As we know, alc. KOH results in dehydrohalogenation of the compound which takes place as an anti elimination. And then, we're simply conducting halogenation of the compound which would result in a dibromo substituded product.


Where I'm facing problems is deciding the final orientation of the product. How will we represent the product in the fischer representation as the initial reactant is given in its fischer representation as well.


For reference, here's the product (B) given in my book:

enter image description here

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    $\begingroup$ A quick way to solve this particular question would be to notice that the first reaction is anti elimination, and the second is anti addition, so you can simply replace the atoms being eliminated (H, Br) with those being added (Br, Br). $\endgroup$ – Ashish Ahuja May 17 at 6:37
  • $\begingroup$ @AshishAhuja Oh. So in a way we can say that the first anti elimination then again an anti addition sort of balance out and it appears as if it was a syn addition? $\endgroup$ – Prajwal Tiwari May 17 at 8:11
  • $\begingroup$ Yes, something like that. Also, there are tricks for figuring out the final product (since there are only four cases, cis,trans + syn,anti) without actually drawing the newman projections which you can use in objective examinations, which would make solving such a question a very fast process, unlike the newman approach. $\endgroup$ – Ashish Ahuja May 17 at 11:31
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This is a fun one, and a common question in this course. First, as you say, H and Br are lost by anti elimination, since they helpfully tell you base and E2 are involved. That means the H and Br are lost from opposite faces of the future double bond, leaving us with a trans intermediate that has a mirror plane (the plane of the molecule)

enter image description here

Now we need to add Br+ to make a bromonium intermediate. Notice that both carbons end up as (R) or else both end up as (S), so two enantiomers are possible here.

![enter image description here

But, whichever carbon is attacked by Br- ends up with an inversion of symmetry, so you now have two carbons of opposite chirality. The result, as shown in your answer, is a molecule with a mirror plane. In other words, adding bromine to opposite sides of a trans double bond makes both sides come out as mirror images, once you turn the bromines around to line up. This compound is a meso compound. But ... what about the other isomer if the Br+ originally goes to the other side of the double bond? Well, just turn your Fisher structure upside down. :)

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