Is Option 2 the Correct Answer for Huckel's Law?

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The discussion revolves around the application of Huckel's Law, specifically the formula 4n+2 for determining aromaticity based on pi electrons. Participants express confusion about the interpretation of electron counts, particularly regarding lone pairs and their inclusion in the calculations. Option 2 is proposed as the correct answer, indicating an aromatic structure with n=0, while options 1 and 3 also yield 6 pi electrons, suggesting they could be aromatic as well. There is uncertainty about the validity of option 3 as a possible structure, prompting further analysis of electron distribution on the carbon atoms. Clarification on which electrons to consider in Huckel's rule is emphasized as crucial for accurate conclusions.
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Homework Statement

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Homework Equations


Huckel's Law: 4n+2=pi electrons

The Attempt at a Solution


I'm not sure what the dash and two dots in the brackets mean, are those lone pair of electrons?
Not including them as pi electrons:
option 1: 4= 4n+2
n=1/2, so no aromatic
option 2: 2= 4n+2
n= 0, so aromatic
option 3: 4=4n+2, n=1/2
so the answer should be option 2?
 
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I don't know if you were told which electrons to consider when you take the value of n in huckels rule
Does option 1 have just 4 electrons involved in the resonance?
 
Suraj M said:
I don't know if you were told which electrons to consider when you take the value of n in huckels rule
Does option 1 have just 4 electrons involved in the resonance?
Doesn't Huckels rule involve pi electrons only? Does it involve all the electrons taking part in the resonating structures?
 
Do you think you've considered ALL the pi electrons?
 
Suraj M said:
Do you think you've considered ALL the pi electrons?
Oh, but then both option 1 and 3 work out to be aromatic as both have 6 pi electrons.
 
Is option 3 even a possible structure? Just look at it
Count the number of electrons on each carbon!
 
Suraj M said:
Is option 3 even a possible structure? Just look at it
Count the number of electrons on each carbon!
Ah, I see! Thanks.
 
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