E2 elimination in β-Lindane

  • #1
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Homework Statement


*How many halogens will be removed in the following E2 reactions combined?*
IMG_20180711_164003.JPG

The Attempt at a Solution


The question's language isn't error free. So, looking at the options, assuming it's *How many halogens will be removed in the following E2 reactions combined?*, the chair comformation of the following compound is:
IMG_20180711_170208.JPG

I know that, for elimination to take place, the chlorines and hydrogens must be anti-periplanar (only possible when both of them are axial). But as you see here, none of the hydrogens turn out to be axial. How does elimination take place then?
 

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Answers and Replies

  • #2
TeethWhitener
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I really want to help with this but I can’t make heads or tails of the question. My best guess is that at some point, you have to do a stereoinverting substitution which gives you the antiperiplanar geometry needed for an E2. Maybe that’s why they say “halogens” instead of “chlorines.”
 
  • #3
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How about substitution with PBr3? It's an SN2 reaction.:olduhh:
But would the reaction stop at mono-substitution?
 
  • #4
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Tell me, is PBr3 ok?
Would it stop at mono-substitution?
 
  • #5
TeethWhitener
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Tell me, is PBr3 ok?
Would it stop at mono-substitution?
In this problem, PBr3 is functionally the same as sodium iodide. I doubt it would stop at monosubstitution, but I imagine there’d be a ridiculous mixture of products and I have no idea what the major one would be. But again, you’d have to substitute before you did the elimination.
 
  • #6
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Come on, dude. Think...
Which kind of substitution would work?
 
  • #7
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Hey @TeethWhitener ! I have one more doubt. It's not directly related to this but still...
I have seen reactions where NaI + acetone gives substitution by -I (Finkelstein reaction) and also where it gives elimination. So how do I decide what happens when?
 
  • #8
TeethWhitener
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Hey @TeethWhitener ! I have one more doubt. It's not directly related to this but still...
I have seen reactions where NaI + acetone gives substitution by -I (Finkelstein reaction) and also where it gives elimination. So how do I decide what happens when?
Can you give a specific example of NaI giving elimination? NaI is not a strong base, so unless there’s a highly stabilized leaving group or sterically hindered carbocation (so that the nucleophilic addition rate is dramatically slowed), I can’t imagine it’s a very likely scenario.
 
  • #9
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Can you give a specific example of NaI giving elimination? NaI is not a strong base, so unless there’s a highly stabilized leaving group or sterically hindered carbocation (so that the nucleophilic addition rate is dramatically slowed), I can’t imagine it’s a very likely scenario.
Not with NaI alone, but with NaI + acetone, the above example
 
  • #10
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I'll dig for more specific examples, have seen them for sure. Keep this thread watched.
 
  • #11
DrDu
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I have this all very rusty, but I wonder whether in the first reaction, the product won't do a Markovnikov elimination so as to yield a trichlorobenzene.
 
  • #12
TeethWhitener
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I have this all very rusty, but I wonder whether in the first reaction, the product won't do a Markovnikov elimination so as to yield a trichlorobenzene.
Generally an E2 mechanism relies on the H and the leaving group being antiperiplanar to each other. In this case, they're all synperiplanar gauche to one another. However, I imagine if this reaction were actually performed, you'd get an intractable mixture of products.
 
  • #13
DrDu
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Generally an E2 mechanism relies on the H and the leaving group being antiperiplanar to each other.
More specifically, an E2 elimination. I would have rather expected an E2 substitution, followed by an elimination. Here, I- is a better nucleofuge than OH-.
 
  • #14
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I would have rather expected an E2 substitution, followed by an elimination. Here, I- is a better nucleofuge than OH-.
No, I think they just mean to ask two different reactions from the same reactant. We need to treat these as two different cases. Look at options (c) 8 & (d) 10. They need the halogens removed in the E2 reactions combined.
Substitutions would be unlikely due to the alc. KOH unless you add some other reagent first, like 3 mol equiv. PBr3 in a polar aprotic solvent
 
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