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I am a not a chemistry student but a physics student. Nevertheless, I am quite familiar with molecular orbital theory and similar quantum chemistry concepts. However, I have problems understanding the ‚double bond‘ nomenclature, especially in conjugated molecules.

But let’s start simple: If we have a molecule like $C_2H_4$, there is a double bond between the two C atoms. This double bond is due to the overlap of the p orbitals and the overlap of the $sp_2$ orbitals. Until here, everything is clear for me. enter image description here $$$$

Now, let’s look at a conjugated molecule like 1,3 Butadiene: enter image description here

There are also double bonds between every second carbon atom (structural formular in the top left). In the picture above (b) we see, that all p-orbitals of all carbon atoms overlap, producing four different molecular orbitals. In (a) we also see that there are $sp_2$ bonds between all carbon atoms.

My question is:

Why do we draw double bonds exactly between the two first and the last two C atoms in the structural formula? How does this follow out of the MO-picture (where we have four different $\pi$ type MO)? Why isn’t it possible to draw the double bond between the second and third carbon atom instead of between the first two, or even add an electron and draw a double bond between all carbon atoms (since all p-orbitals overlap)? Does it have something to do with the fact that only for the HOMO-orbital there is a node between the central two carbon atoms (where there is a also single bond drawn).

Pictures are taken from LibreTexts

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    $\begingroup$ This isn't MO picture it's VBT, and you don't consider resonance structures. $\endgroup$
    – Mithoron
    Mar 9, 2023 at 17:47
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    $\begingroup$ In general, structural drawings are not intended to capture the full electronic structure of a molecule. They are simply agreed upon representations that allow a knowledgeable person to discern the structure. $\endgroup$
    – Andrew
    Mar 9, 2023 at 21:48
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    $\begingroup$ Oh these pictures are terrible! They try explaining one thing (the sigma framework) with Valence Bond Theory and another thing with Molecular Orbital Theory (the pi framework). While these two theories are equivalent in their outcome, you cannot just mix them in the interim stages. For example: there cannot be a HOMO in VBT as there are no MO. $\endgroup$ Mar 10, 2023 at 20:30

2 Answers 2

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As stated in the comments, the Lewis structures are representations of a localized bond picture, and you need so-called mesomeric contributors to show delocalized bonding (some of which do show the double bond in the center):

enter image description here

You might ask why the charges, and not single electrons. Single electrons on carbon 1 and carbon 4 don't fit the properties of the molecule (it is not a diradical).

An empirical way of distinguishing double and single bonds is to look at bond lengths.

For butadiene, the formal C=C bond length is 1.338 Å, and the formal C-C bond length is 1.454 Å.

Source: N. C. Craig et al.

So the structure in the middle is a better representation if a single contributor has to be picked.

[OP] Why isn’t it possible to draw the double bond between the second and third carbon atom instead of between the first two, or even add an electron and draw a double bond between all carbon atoms (since all p-orbitals overlap)?

You could add two electrons, and would think that you have a double bond between carbon two and three. However, this would mess with the pi-bonding of carbon 1 and 2 as well as carbon 3 and 4 (there are nodes there). In other examples, you do get "extra" bonding by adding electrons (going from cyclopentadiene to cyclopentadienyl anion, for example).

Craig, N. C.; Groner, P.; McKean D. C. Equilibrium Structures for Butadiene and Ethylene:  Compelling Evidence for Π-Electron Delocalization in Butadiene. J. Phys. Chem. A 2006, 110, 7461-7469. DOI 10.1021/jp060695b

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    $\begingroup$ Well, it is neither. For benzene, the bond order is 1.5 for all carbon-carbon bonds. For butadiene, the pi bond order is 0.45 for the central bond 0.89 for the other two bonds according to one source and another. $\endgroup$
    – Karsten
    Mar 9, 2023 at 20:25
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    $\begingroup$ @karsten if you do a Huckel-type calculation on the pi orbitals of benzene you discover that the total bond order in the ring is actually $1\frac23$ per linkage. Aromatic rings generally have a resonance stabilization equivalent to about one extra pi bond distributed around the ring. $\endgroup$ Mar 10, 2023 at 11:03
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    $\begingroup$ @OscarLanzi I was wondering why for butadiene, it did not add up. That’s unexpected from a VB view, and also not captured in the Lewis structure. $\endgroup$
    – Karsten
    Mar 10, 2023 at 12:04
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    $\begingroup$ Bond order is an entirely made up concept with arbitrary numbers. Without the proper reference frame the calculations are worse than guessing. Since sigma bonds are always stronger than pi bonds (choose your reference frame well) there could be an argument made, that the bond order of a sigma bond is actually 1.1 while an accompanying pi bond is 0.9; again be careful with the reference frame. The sigma bond in ethane is not the same as in ethene. And so it goes on and on and one… $\endgroup$ Mar 10, 2023 at 20:39
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    $\begingroup$ @Martin By extension, single bond and double bond is also an entirely made up concept. We have a lot of those in chemistry: oxidation states, partial charges on atoms, electronegativity, empty orbitals, electron pushing. Sometimes they are helpful, though. $\endgroup$
    – Karsten
    Mar 10, 2023 at 23:14
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The simple pictures chemists draw are used for convenience not to correctly represent the MO pictures of bonds

Chemistry is complicated. The simple pictures chemists draw to explain the structures of molecules are a trade-off between describing the structure unambiguously and misrepresenting the details of the bonds involved. We could draw pictures giving a better view of the bonding, but they would often be complex, ambiguous or make other things harder. So, often, we draw simple pictures to maximise clarity while assuming that other knowledge will avoid misinterpretations of the real nature of the bonding.

One of the things we want to be easy is unambiguously counting electrons. For example the commonest way to draw benzene is this:

benzene structure

This is clearly a misrepresentation of the actual structure of benzene (where all the bonds are equivalent and the reality involves a bunch of molecular orbitals with delocalised electrons). But this picture does allow a chemist to count the electrons (each line represents a notional two electron bond) which may be important and may help describe simple reaction mechanisms. Some chemists do the same drawing as a hexagon containing a circle to represent the fact that we know the electrons are delocalised, but this doesn't make it easy to count the number of bonding electrons accurately which can make explaining reaction mechanisms much harder.

A similar issue occurs with butadiene. We could choose a more "reliable" drawing of the molecule (or, as Karsten shows in his answer, several drawings) but this loses a lot in clarity without generating a lot more insight that helps in most situations. We assume that chemists have enough knowledge about bonding to recognise the situations where a better understanding of MO theory and its implications is actually needed to understand a molecule or its reactions.

For many structures and many reactions the "simple" drawing plus some chemical knowledge is good enough for the task at hand. And we don't need to use extra complexity in what is drawn unless it is vital for adding extra insight to the structure or the reaction mechanism.

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