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

In summary: 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.In summary, eight halogens will be removed in the following E2 reactions combined.
  • #1
baldbrain
236
21

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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  • #2
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
How about substitution with PBr3? It's an SN2 reaction.:olduhh:
But would the reaction stop at mono-substitution?
 
  • #4
Tell me, is PBr3 ok?
Would it stop at mono-substitution?
 
  • #5
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.
 
  • #6
Come on, dude. Think...
Which kind of substitution would work?
 
  • #7
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
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.
 
  • #9
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
 
Last edited:

What is E2 elimination in β-Lindane?

E2 elimination in β-Lindane is a chemical reaction that involves the removal of a hydrogen atom and a halogen atom (typically chlorine or bromine) from adjacent carbon atoms, resulting in the formation of a double bond. This reaction is also known as a dehydrohalogenation reaction.

What is the mechanism of E2 elimination in β-Lindane?

The mechanism of E2 elimination in β-Lindane involves a concerted reaction, meaning that the hydrogen and halogen atoms are removed simultaneously in a single step. The reaction is typically initiated by a strong base, such as potassium hydroxide, which abstracts the hydrogen atom, forming a carbanion intermediate. The halogen atom then leaves, resulting in the formation of the double bond.

What factors influence the rate of E2 elimination in β-Lindane?

The rate of E2 elimination in β-Lindane is influenced by several factors, including the strength of the base, the steric hindrance of the atoms involved, and the stability of the resulting double bond. A stronger base will result in a faster reaction, while steric hindrance can slow down the reaction. A more stable double bond will also favor the elimination reaction.

What are the potential side reactions of E2 elimination in β-Lindane?

One potential side reaction of E2 elimination in β-Lindane is an E1 elimination, where the halogen atom leaves first, forming a carbocation intermediate, followed by the removal of the hydrogen atom. Another possible side reaction is the formation of a cyclic product, where the double bond is formed within the same molecule instead of between two separate molecules.

What is the significance of E2 elimination in β-Lindane in organic synthesis?

E2 elimination in β-Lindane is a useful reaction in organic synthesis as it provides a way to form carbon-carbon double bonds, which are important structural motifs in many organic compounds. This reaction is also commonly used in the synthesis of pharmaceuticals and natural products. Understanding the factors that influence the reaction can help chemists design more efficient and selective synthetic routes.

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