Why is this not a geometric isomer?

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Discussion Overview

The discussion revolves around the classification of certain compounds as geometric isomers, specifically addressing why one compound is not considered a geometric isomer despite initial assumptions. The scope includes conceptual clarification regarding isomerism in organic chemistry.

Discussion Character

  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant expresses uncertainty about why a specific compound (ii) is not classified as a geometric isomer, suggesting it could be trans-pent-2-ene due to the arrangement of ethyl and methyl groups around the double bond.
  • Another participant questions whether compounds (i) and (ii) are isomers at all.
  • A participant references external sources to argue that (i) and (ii) are not geometric isomers because they are mirror images of each other, suggesting they are the same compound.
  • There is a discussion about the implications of having three options for isomerism, with one participant stating that (iii) and (iv) do not make sense in the context of cis-trans isomerism.
  • One participant clarifies that being mirror images does not necessarily mean the compounds are identical, emphasizing the importance of the ability to coincide through bodily movement rather than mere reflection.

Areas of Agreement / Disagreement

Participants do not reach a consensus on whether (i) and (ii) are isomers or the classification of the compounds as geometric isomers. Multiple competing views remain regarding the nature of the compounds and their isomerism.

Contextual Notes

Some assumptions about the definitions of isomerism and the criteria for geometric isomers are not fully articulated, leading to potential misunderstandings. The discussion also touches on the complexity of isomer classification when multiple configurations are possible.

Declan McKeown
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ObccNpv.png


While I'm fairly sure this question is pretty straight forward, I can't remember why (ii) is not considered a geometric isomer. (The answer is (c)) http://imgur.com/ObccNpv

Looking at ii I would have thought it to be trans-pent-2-ene due to the ethyl and methyl group either side of the double bond, however it is not and I am not sure why. Would anyone happen to know why this is?

Thanks :)
 
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Are (i) and (ii) isomers at all?
 
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https://en.wikipedia.org/wiki/Cis–trans_isomerism

Well, (I) and (II) are not geometric isomers because they are the same compound just mirror images, besides both seem trans to me.
I II and III can not be geo. (cis-trans) isomers because that is three options.
III and IV don't make sense because IV is trans so III should be cis but it's weird because its not.

Sorry that's all I can help with my limited knowledge.
 
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IcyRealm said:
https://en.wikipedia.org/wiki/Cis–trans_isomerism

Well, (I) and (II) are not geometric isomers because they are the same compound just mirror images, besides both seem trans to me.
I II and III can not be geo. (cis-trans) isomers because that is three options.
III and IV don't make sense because IV is trans so III should be cis but it's weird because its not.

Sorry that's all I can help with my limited knowledge.

Being mirror images does not necessarily make them the same compound! You probably remember now. :oldsmile:

The essential is you can swivel it around a vertical axis and i coincides with ii - so it is the same thing. Any bodily movement leading one to coincide with another means it is the same thing. A reflection on the other hand is not a bodily movement.

I think that identity was the essential point of this exercise. Anything that has the same composition as another thing but is not the same thing is an isomer of it.
 
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