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I'm still learning huckel rule and aromaticity in different compound. For the 1-methoxy-1,3,5,7-octatetraene, is it correct that it has 8 electrons inside the cycle, but since they will be 8, then it is antiaromatic?

And for the Imidazole, is it 6 electron inside because the N: doesn't contribute to the pi, or is it 8?

enter image description here

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  • $\begingroup$ Why would someone downvote a question? I'm new here in chemistry.stackexchange and I'm still learning...can someone explain? $\endgroup$
    – user99277
    Nov 1, 2020 at 18:43
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    $\begingroup$ chemistry.stackexchange.com/questions/27099/… chemistry.stackexchange.com/questions/33332/… $\endgroup$
    – Mithoron
    Nov 1, 2020 at 18:50
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    $\begingroup$ I’m voting to close this question because there are two parts to the question, and both are duplicates (see Mithoron's comment). If this were just one question, we could close it as a duplicate, but because there are two, we can't. $\endgroup$
    – Karsten ♦
    Nov 14, 2020 at 21:18
  • $\begingroup$ I'm voting otherwise. $\endgroup$
    – user99277
    Nov 15, 2020 at 4:26

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Advice

Take a closer look at the lone pairs and double bonds that participate only and only in the delocalization of electrons.

1-methoxy-1,3,5,7-octatetraene

As you can see, 1-methoxy-1,3,5,7-octatetraene has four double bonds in an octagonal cycle. The lone pairs of the adjacent oxygen atoms, even though they directly influence the delocalization by means of the resonance effect, aren't actually part of the cycle and cannot be included amongst the pi electrons of the double bonds. Therefore, your molecule is antiaromatic. Its structure is not planar, in fact it looks something like this (the picture below includes only the cyclic backbone): enter image description here

For aromatic molecules, planarity is absolutely necessary. But Huckel's rule which I'm sure you're familiar with, provides the reasoning behind the planarity of the molecule.

The imidazole

As for the imidazole, your assumption is absolutely correct. Here we get to know a scenario similar to pyridine: enter image description here

The lone pair does not participate in the aromatic cycle. This can be easily checked by taking a look at the hybridization of the nitrogen atom. If the lone pair were to be part of the aromatic system, it should have been perpendicular to the other orbitals of the cycle. However, this is not the case. The 2p orbitals of the nitrogen are actually involved in the aromatic cycle, while the aforementioned paired electrons get to be embedded in an sp2 orbital, placed perpendicular to the plane of the cycle. This actually has significant consequences on the basicity of pyridine. On the other hand, in order not to interrupt the aromatic cycle, the lone pairs of the other nitrogen atos must come into play when counting pi electrons. Thus, the molecule is aromatic.

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    $\begingroup$ @user432797 Yes you're right. $\endgroup$
    – Mithoron
    Nov 1, 2020 at 23:38
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    $\begingroup$ Apologies, I didn't see the comments you had posted after my answer. If a molecule has an uninterrupted network of pi electrons and 4n pi electrons, it is usually antiaromatic. Because they lack the stabilization that would otherwise be present in the aromatic molecules thanks to the pi orbital that delocalizes over the whole molecule (see Huckel's rule), the molecule stabilizes itself by adopting non-planar configurations. $\endgroup$
    – TheRelentlessNucleophile
    Nov 2, 2020 at 17:34
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    $\begingroup$ Rule of thumb: uninterrupted network of pi electrons, planar molecule, 4n+2 electrons, extended pi orbital that delocalizes over the whole molecule: aromatic. Violate one rule and the molecule is rendered either antiaromatic (4n pi electrons) or non-aromatic. Now, I lack the appropriate knowledge to tell you exactly why do we bother with antiaromatic molecules in the first place, but let's just say that, as far as I know, if you were to draw the Frost circles in a Huckel diagram, their HOMOs would be at the ground energy level, rendering the orbitals non-bonding. $\endgroup$
    – TheRelentlessNucleophile
    Nov 2, 2020 at 17:39
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    $\begingroup$ Also, keep in mind that several of these molecules prefer to exist as diradicals, instead of completely stable molecules $\endgroup$
    – TheRelentlessNucleophile
    Nov 2, 2020 at 17:39
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    $\begingroup$ Your question was downvoted becausw it looked like homework and also because of lack of details involving YOUR work in answering the question ans sharing YOUR knowledge. Please refer to the rules of ChemSE. $\endgroup$
    – TheRelentlessNucleophile
    Nov 3, 2020 at 4:27