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How do I arrange these chemical bonds in an increasing/decreasing order based upon their lengths?

Propa-1,2-dien-1-ylidenecyclopropane

I understand that the bond (a) will be longer than (b) because (a) is a bond between sp² hybrid and sp hybrid carbon atoms, as opposed to (b) which is a bond between two sp hybrid carbon atoms. But how do we compare bond (c)?

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    $\begingroup$ I know this is only meant to be a theoretical exercise, but I ran a search and this molecule does not exist (as far as I can tell). Therefore, all explanations should only be taken at face value - i.e. they are merely a prediction of the relative bond lengths, as opposed to a rationalisation of observed bond lengths. $\endgroup$
    – orthocresol
    Apr 17, 2017 at 10:11

2 Answers 2

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Label the double bonds (a), (b), and (c), in correspondence with the image provided in the question. Label the carbons in the cyclopropane ring $\ce{A}$, $\ce{B}$, and $\ce{C}$, with $\ce{A}$ being the carbon with the double bond (c).

At $\ce{A}$, the usual $\ce{sp^2}$ hybridization would lead to bond angles of $120^\circ$, whereas a lack of hybridization (i.e. simple $\ce{p}$ orbitals) would lead to bond angles of $90^\circ$. The latter is not usually favored, but the high ring strain in cyclopropane leads to carbon-carbon bonds $\ce{A-B}$ and $\ce{A-C}$ having greater $\ce{p}$ character than is usual. The double bond (c) at $\ce{A}$ thus receives more $\ce{s}$ character, and is stronger than a usual $\ce{sp^2{-}sp}$ bond would be (i.e. (a)).

We conclude, therefore, that the bond strengths are (b) > (c) > (a), and hence that the bond lengths are (a) > (c) > (b).

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    $\begingroup$ A stronger bond does not necessarily mean it is a shorter bond. It works in this case, but it's not a correlation-causation consistent notion. One should be very, very careful using this. $\endgroup$
    – Martin - マーチン ♦
    Apr 17, 2017 at 1:31
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    $\begingroup$ @Martin-マーチン, thank you for the comment; I didn't know this. Would correcting my argument by saying that bonds with greater $s$ character are shorter be equally (in)valid? $\endgroup$
    – a-cyclohexane-molecule
    Apr 17, 2017 at 1:57
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    $\begingroup$ In this case this works mainly because you are comparing carbon-carbon bonds, which is the same reason why the strength argument works. (I have been sloppy about that in my own answer, which I need to revise.) Higher s character indeed means shorter bond here since the optimal overlap (balancing Pauli repulsion, etc.) is closer to the nuclei. And this is assuming that the pi bonds are essentially the same in all bonds (they are not). But for a rudimentary analysis, this gives the correct result. $\endgroup$
    – Martin - マーチン ♦
    Apr 17, 2017 at 2:05
  • $\begingroup$ I have removed the references to my answer since I deleted it as it is not correct. I currently don't have the time to revise it thoroughly and I think your analysis is good enough from an intuitive point of view. $\endgroup$
    – Martin - マーチン ♦
    Apr 17, 2017 at 9:32
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You can draw Hyper conjugating Structures to compare bond length. More the α Hydrogen of alkene, more are the Hyper conjugating structures, more is the delocalization of π-bond and more is the bond length (since double bond character decreases, bond length increases).

Thus bond length C would be greater than A.

As there are 4 α hydrogens around C, whereas '0' α hydrogens around A.

Therefore the order of bond length would be C>A>B

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