How many halogens will be removed in the E2 reactions combined?

AI Thread Summary
The discussion centers on the number of halogens removed in E2 reactions, with participants analyzing the conditions necessary for elimination to occur. Key points include the requirement for anti-periplanar geometry for E2 reactions, which is not met in the provided compound, leading to confusion about how elimination can proceed. There is debate over the role of PBr3 and NaI in substitution reactions, with suggestions that substitution must occur before elimination. Participants also express uncertainty about the resulting product mixtures and the likelihood of elimination versus substitution. Overall, the conversation highlights the complexities of predicting reaction outcomes in organic chemistry.
baldbrain
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Homework Statement


*How many halogens will be removed in the following E2 reactions combined?*
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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:
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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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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.”
 
How about substitution with PBr3? It's an SN2 reaction.:olduhh:
But would the reaction stop at mono-substitution?
 
Tell me, is PBr3 ok?
Would it stop at mono-substitution?
 
baldbrain said:
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.
 
Come on, dude. Think...
Which kind of substitution would work?
 
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?
 
baldbrain said:
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.
 
TeethWhitener said:
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
I'll dig for more specific examples, have seen them for sure. Keep this thread watched.
 
  • #11
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
DrDu said:
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
TeethWhitener said:
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
DrDu said:
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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