Reverse of optical isomerism in a reaction

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Discussion Overview

The discussion revolves around the reversal of optical isomerism in the reaction of chiral 2-bromobutane with aqueous hydroxide ions. Participants explore the mechanisms involved, particularly focusing on the SN2 mechanism and the structural changes during the reaction.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Homework-related

Main Points Raised

  • One participant suggests that the hydroxide ion attacks the electron-deficient carbon to avoid repulsion from the bromine atom.
  • Another participant indicates that the reaction mechanism is likely an SN2 process.
  • A participant expresses confusion about how a dextro or leavo enantiomer could become the opposite, questioning the role of the bromine leaving group.
  • There is a mention of the formation of a trigonal bipyramidal structure during the reaction, where the substituents lie in one plane with the bromine and hydroxide perpendicular.
  • One participant likens the switching of substituents to an umbrella inverting, suggesting a visual analogy to understand the process.
  • Another participant describes the necessity for substituents to move to accommodate the hydroxide ion's attachment, leading to a change in symmetry when the bromine leaves.

Areas of Agreement / Disagreement

Participants generally agree on the involvement of the SN2 mechanism and the structural changes during the reaction, but there remains uncertainty about the specifics of how chirality is reversed and the exact timing of the switching process.

Contextual Notes

Some participants express uncertainty regarding the terminology used, such as whether it is appropriate to refer to the intermediate as a carbanion. The discussion also highlights the complexity of visualizing the reaction mechanism and the changes in molecular symmetry.

Who May Find This Useful

Students studying organic chemistry, particularly those interested in reaction mechanisms and stereochemistry, may find this discussion beneficial.

Kushal
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Homework Statement



when chiral 2-bromobutane reacts with aqueous hydroxide ions, the optical activity of the product is reversed. Suggest why the optical isomerism is reversed.


The Attempt at a Solution



i have no idea... but i don't know if the fact that the hydroxide ion will attack the electron deficient carbon (from C - Br) away from the Br atom to avoid repulsion may lead to the answer...

thanks!
 
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You are on the right track. Try to imagine how the reaction proceeds. Molecule model may come handy.
 
ermmm well i tried to imagine but i still can't see how a dextro or leavo enantiomer could become the opposite?!

because the Br- will go and the OH- will just attach to the electron deficient carbon.

does that have something to do with the SN2 mechanism?
 
Kushal said:
does that have something to do with the SN2 mechanism?

Yes.
 
i had a look at the mechanism but i can't find how there could be a reversal in chirality. a hint would be well appreciated.

thnks
 
You start with R1, R2, R3 and Br. You end with R1, R2, R3 and OH. During reaction R1, R2 and R3 are first "made flat" (when there exist intermediate, 5 substituded carbon) and then they are "switched" to the other side, as if they were reflected in the mirror.
 
so you mean that when the carbanion forms a trigonal bipyramidal structure, the R1, R2 and R3 lie on one plane with the Br and OH perpendicular to that plane.

but how does the switching occur?! does it occur after the Br- leaves?

thnks
 
I remember an old textbook (Ingold) likened it to an umbrella inverting. Imagine a 3-ribbed umbrella - not a perfect analogy but may help visualise. :smile:
 
Kushal said:
so you mean that when the carbanion forms a trigonal bipyramidal structure, the R1, R2 and R3 lie on one plane with the Br and OH perpendicular to that plane.

That's how it looks. Not sure if its OK to call it carbanion, I am called Mr.pH, not Mr.Organic :wink:

but how does the switching occur?! does it occur after the Br- leaves?

For the OH- to attach to carbon, R1, R2 and R3 must move to make place, then when Br- leaves, they have new place where they can move - at this moment they have to change symmetry. They behave like a hat turned inside out.
 
  • #10
epenguin said:
I remember an old textbook (Ingold) likened it to an umbrella inverting. Imagine a 3-ribbed umbrella - not a perfect analogy but may help visualise. :smile:

Pretty good example :smile: Somehow I missed the moment you posted the answer.
 
  • #11
niice analogies lol... thank you borek and epenguin...

i can get the picture now:)
 

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