So I was "happily" doing organic chemistry homework when I came across this question:

For 1-methoxy-1,3-butadiene, which of the following resonating structure is the least stable?

$$\begin{align}(\mathbf{a})\ &\ \ce{H2\overset{+}{C}-\overset{+}{C}H-CH=CH-O-CH3} & (\mathbf{b})\ &\ \ce{H2\overset{+}{C}-CH=CH-CH=\overset{+}{O}-CH3} \\ (\mathbf{c})\ &\ \ce{H2\overset{+}{C}-\overset{-}{C}H-\overset{+}{C}H-\overset{+}{C}H-O-CH3} & (\mathbf{d})\ &\ \ce{H2C=CH-\overset{+}{C}H-CH=\overset{+}{O}-CH3}\end{align}$$

I decided to take a look through my textbook to see what I was missing. The rules to compare resonance structure energies (and therefore determine relative contributions) were in my textbook. It said:

The following structures are considered to contribute most:

  1. Structures having filled octet a for second row elements (C, N, O, F) are stable.
  2. Structures having minimum number of formal charges and maximum number of bonds.
  3. Structure in which negative charge appears on the most electronegative atom (C < N < O).
  4. Structure in which there is minimal charge separation while keeping the formal charges closer together.

Instead of answering my initial question, two more questions came to mind. They are (in order):

  1. Why do the octets of only the second row elements have to be filled in for greater relative contribution? I suppose it has something to do with the greater electronegativity values of the elements, but would still like some clarification.
  2. From option (a) and (c) in my question, how did the two positive charges get next to each other?

Resonance structures, as stated in my textbook, can be represented by two or more Lewis structures that differ only in the positions of the electrons. The options clearly do not just redistribute the electrons. Or at least I don't think so.

So if I had to guess, the answer would be (b) because the terminal carbocation is worse than an internal carbocation.

But it turns out, the correct answer is (c). And I think that it shouldn't even be considered a resonating structure.

  • 13
    $\begingroup$ Possibly there's an error in the question. None of the 4 structures represent 1-methoxy-1,3-butadiene; all of the structures represent the dication of 1-methoxy-1,3-butadiene. I suspect someone accidentally put a (+) sign instead of a (-) sign in each structure. Had they been drawn correctly (with an equal number of positive and negative charges), then (c) would be the least stable since it has 4 separated charges, all of the others only have 2 (rules 1 and 2 in your list). $\endgroup$
    – ron
    Commented Jun 19, 2015 at 16:47
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    $\begingroup$ I'm afraid your textbook sucks ;) $\endgroup$
    – Mithoron
    Commented Jun 19, 2015 at 17:06
  • 3
    $\begingroup$ Wow! I spent at least an hour on this question. I guess I should be more careful with my choice of text next time. @ron I can't believe I didn't see the mistake. Thanks. $\endgroup$
    – tkhanna42
    Commented Jun 19, 2015 at 17:09
  • 4
    $\begingroup$ Resonance structures are not real, they are only a description. Therefore there cannot be a most stable resonance structure. The premise of the question is wrong, regardless of the error in the formulae. See also here. $\endgroup$ Commented Jun 16, 2016 at 5:16
  • $\begingroup$ Can inductive effect due to nearby atoms not be considered as a point for stability? $\endgroup$
    – User
    Commented Jul 31, 2017 at 8:28

4 Answers 4


(C) is technically a valid resonance structure. You can derive it from (B) by moving the electron pair from the double bond between the carbons 2 and 3 (numbering from the left: this may not be correct from a nomenclature standpoint, but I'm using it for this answer) onto carbon 2 and moving the electron pair from the double bond between the oxygen and the carbon onto the oxygen.

(C) is least stable because it is, simply put, absolutely disgusting. Charge separation (carbons two and three) is gross, you have two cations in close proximity (carbons three and four), a terminal carbocation (carbon 1), and three atoms with empty valence shells (carbons 1, 3, and 4). So while the resonance structure is valid enough for (C) to be considered a resonance structure, that resonance structure happens to be the spawn of the devil.

That said, the compound in the answers isn't the same as the compound named in the question, like ron said.

As to the first part of your question, the filled valence shells are more strictly necessary for the second row elements because they don't have d orbitals to confound the picture.

  • $\begingroup$ I find the second paragraph not only offensive, but also wrong. There are no such things as most/least stable resonance structures; the whole premise of the question is void. While charge separated structures in organic molecules are often minor contributors, that changes quickly, if you go to other elements than carbon and co. They are very often essential contributors, which explain reactivity towards certain mechanisms. It may look odd to you, but it most certainly is not the spawn of the devil. And your last sentence hints at octet expansion, which has been disproved multiple times. $\endgroup$ Commented Jul 23, 2018 at 14:13

As ron stated in the question’s comments, none of the resonance structures display 1-methoxybuta-1,3-diene. In fact, they show an extremely unstable (if generateable) dication. Choose a random $+$, replace it with a $-$ (don’t choose the one on the oxygen) to get resonances for the structure the question asks for. That should also semi-answer your question about the two positive charges.

As for the question on octets: It is the more traditional viewpoint that elements of the third and higher periods such as sulphur could expand their octets (e.g. in $\ce{SO2}$). This was generally explained with contribution by d-orbitals (and Pauling’s dislike for charge separation).

Nowadays, however, these structures are more generally considered as abiding by the octet rule invoking charge separation where necessary and multicentred bonds where applicable. The reason for this is that d-orbitals are too far removed energetically from s and p of the same shells to actively take part in bonding.

Finally, there are sub-octet structures to consider (heptets in radicals or sextets in carbocations, carbenes and the like). These exist and are undebated. This violation of the octet rule is ‘less bad’ (by a mile) than a super-octet. Why? Well, you’re not filling electrons into high-energy contributed orbitals but rather removing them from stable ones.

That said, when considering the first of these octet paragraphs, it is a lot harder for me to think of compounds that have sub-octets for elements of higher periods — most examples I know are carbon or nitrogen centred.

On the topic of the ugliness of (c), Breaking Bioinformatics has answered that nicely. But you could have arrived at the same conclusion by checking your book’s rules:

  1. Does not apply here. No resonance has an octet on every carbon.

  2. (c) has four formal charges, all the others have two. Stop here, mark (c) and grab a piece of cake.


Following rules are 100%-accurate and can be used to determine stability:

  1. more the no. of π-bonds more stable the contributing structure is.

  2. if the no. of π-bonds are equal then … the molecule with no charge or with no charge separation is more stable.

  3. if the molecules are not containing any charges i.e neutral then … the molecule in which all atoms have a complete octet is more stable.

  4. if the atoms in a molecule have an incomplete octet or have charge separation then … the negative charge is more stable on a more electronegative atom (like N,O) and positive charge on a less electronegative atom.

  5. if above the condition is satisfied, too, in a molecule then … if in a molecule unlike charges are closer then it is more stable than the molecule which has unlike charges far away.

    And if like charges are closer to each other then it is more unstable than in which like charges are far away.


According to inductive effect OCH3 is electron withdrawing group and positive charge next to this carbon make it relatively least unstable .So c is correct answer.


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