13

Is the move from pi bonds to sigma bonds always favorable?

My professor claims so, except in the case of conjugated pi bonds. I can see how pi bonds going to sigma bonds might be favorable; I know that double bonds are actually not twice as strong as single bonds because of poorer overlap ... but are there any exceptions to this guideline that the reduction of pi bonds to sigma bonds is favorable except in the case of conjugated systems?

EDIT: what about in the case of organic chemistry? It certainly seems that hydrogenation (a way of transforming pi bonds to sigma bonds) is generally a favorable move.

enter image description here

LordStryker
  • 4,629
  • 1
  • 40
  • 61
Dissenter
  • 18,815
  • 49
  • 177
  • 341
  • 1
    I'm wondering if your question should ask "is hydrogenation always energetically favorable?" Kinetically it is not but thermodynamically it may be. – LordStryker Sep 26 '14 at 14:06
  • I think that should be the question too ... but my professor (falsely) equates hydrogenation with transformation of pi to sigma bonds. Hydrogenation is ONE way to transform pi bonds to sigma bonds, but there are other ways for one, and it seems to also be confusing a lot of answerers. @LordStryker. – Dissenter Sep 26 '14 at 14:07
  • 1
    By the way, in case it wasn't clear, I do like your question. – LordStryker Sep 26 '14 at 14:24
  • @LordStryker I also appreciate all the answers here; do you suggest opening a separate question though specifically about the favorability of hydrogenation? As you said there is a thermo and a kinetic component to favorability ... and the guy's just throwing around favorability rather loosely. – Dissenter Sep 26 '14 at 14:26
  • 1
    @Dissenter I've edited my answer to include some counterexamples from organic chemistry. – Philipp Sep 26 '14 at 14:28
  • 1
    @Dissenter "favorable" usually means "final energy is lower than initial" so that is how I am trying to use it in this context. I've updated my answer to take into account Phillip's comment indicating that with hydrogenation, we are not just considering the pi to sigma bond transformation of the CC bond but ALSO the creation of two single bonds with the hydrogens. These are two different things. For more info on hydrogenation and related energetics check out https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/addene2.htm. – LordStryker Sep 26 '14 at 14:30
  • @LordStryker and Philipp Let's make the comparison as clean, simple and direct as possible, let's consider isomerization. The heats of formation of 1-hexene and cyclohexane are -10.2 and -29.5 kcal/mole. Doesn't this suggest that a (if you will) carbon-carbon two-membered ring ("double bond) is less stable than a 6-membered ring? – ron Sep 26 '14 at 15:55
  • @ron Ah yes but we're accounting for the entire molecule with that energy comparison no? The individual double-bond itself is lower in energy than its corresponding single bond, however the MOLECULE itself is more stable (i.e. lower in energy) if the double bond is transformed to a single bond. I think that the initial question was a bit unclear whether we were only considering the actual bond or the entire molecular unit. Time has sort of clarified that but this is where this whole issue came from. – LordStryker Sep 26 '14 at 17:06
  • @Dissenter, could you clarify, as LordStryker phrased it, "whether we were only considering the actual bond or the entire molecular unit." My answer addressed the latter case, what was your intent? – ron Sep 26 '14 at 18:18

3 Answers3

8

When thinking about questions along these lines, I always try to remember the bent-bond description of ethylene. In all aspects related to physical reality (structure, electron distribution, etc.) the bent bond model pictured below on the right is equivalent to the pi model shown on the left (at least within the scope of all measurements made to date).

enter image description here

The "nice" thing about the bent bond model is that it casts ethylene as a highly-strained, two-membered ring. When wouldn't releasing this strain and forming sigma bonds be an energy lowering process. Aside from conjugated examples, I think that transformation of a strained, ethylenic system to a saturated, sigma-bonded system would always be energy lowering.

EDIT: Further Thoughts

Instead of analyzing hydrogenations or other reactions where the starting material and product have a different number of atoms, let's consider the following isomerization reaction.

enter image description here

Thermodynamically, cyclohexane is the most stable $\ce{C6H12}$ isomer in this series by a significant margin. All we've done is convert a C-C pi bond into a C-C sigma bond and convert 3 $\ce{C(sp^2)-H}$ bonds into 3 $\ce{C(sp^3)-H}$ bonds. Which (at least as I read it) was the OP's premise. So I would say that, in carbon-hydrogen systems, pi bonds generally prefer to transform into sigma bonds, as pi bonds are inherently destablilized due to poor overlap (pi model) or strain (bent bond model).

That said, I retract what I wrote above that carbon-carbon pi to sigma transformations would "always" be energy lowering. Consider the photolysis of benzene to produce dewar benzene, prismane and benzvalene. These conversions are examples where one or more pi bonds have been converted to one or more sigma bonds and the reaction is not energetically (thermodynamically) favorable.

enter image description here

The products are much less stable (due to the strain in the small rings that have been created) than the starting benzene. Because of these relative stabilities, the forward reaction does not occur thermally (in fact, thermally, each of those 3 products will revert to benzene), but by pumping in excess energy photochemically, the transformation can be brought about.

To summarize, I would maintain that generally, in model, carbon-hydrogen systems, ethylenic linkages are destabilized (due to poor overlap in the pi model, or strain in the bent bond model) and will thermodynamically prefer to exist as thermodynamically more-stable saturated linkages.

ron
  • 84,691
  • 13
  • 231
  • 320
  • "the bent bond model pictured below on the right is equivalent to the pi model shown on the left (at least within the scope of all measurements made to date)." how exactly are the two equivalent? – Dissenter Sep 26 '14 at 14:00
  • 1
    Bond angles, bond lengths and electron distribution are all unchanged. We just arbitrarily assign different hybridizations to carbon. – ron Sep 26 '14 at 14:02
  • What would the hybridization of carbon be in the bent bond model? @ron – Dissenter Sep 26 '14 at 14:03
  • 1
    The carbon orbital used in the C-H bond remains around 2.2, but the orbitals used in the two equivalent C-C bonds are around 4.3, instead of sp2 and p – ron Sep 26 '14 at 14:07
7

This is certainly not generally true. Have a look here for example: There you find that the bond energy for an $\ce{O-O}$ single bond is 142 kJ/mol while that of an $\ce{O=O}$ double bond is more than thrice as large with 494 kJ/mol. So, a reduction of $\ce{O=O}$ that would produce two $\ce{O-O}$ bonds would not be favorable. The same is true for $\ce{N-N}$ (167 kJ/mol) and $\ce{N=N}$ (418 kJ/mol) bonds. For nitrogen you even have the rare case that the $\ce{N#N}$ triple bond with 942 kJ/mol packs more bond strength than two isolated $\ce{N=N}$ double bonds.

In the context of organic chemistry you often have the situation that a reaction destroys a $\pi$ bond while in the process creating a new $\sigma$ bond. A double bond will always be stronger than a single bond, so going from double to single costs energy in any case. But usually when you reduce a double or triple bond you create new bonds at the expense of them. Only when those new bonds + the reduced bond are stronger than the old double/triple bond then the reaction is favorable. There are reactions where the opposite is the case. Phosphorous is infamous for its affinity to oxygen and very prone to form $\ce{P=O}$ double bonds. In the Wittig reaction (see for example here) you have an oxaphosphetane intermediate with a strained 4-membered ring with 4 $\sigma$ bonds that springs open to form one $\ce{C=C}$ and one $\ce{P=O}$ double bond.

Another case comes to mind. Hemiacetals (or 1,1-geminal diols), having two $\ce{C-O}$ $\sigma$ bonds, are usually unstable and prefer to destroy one of the $\ce{C-O}$ bonds by expelling the $\ce{R-O}$ group (or one of the $\ce{O-H}$ groups) thus forming a $\ce{C=O}$ bond. So, in this case one $\ce{C=O}$ bond beats two $\ce{C-O}$ bonds.

If you want to read more on the matter of the relation between single and multiple bonds concerning the strengths of the $\sigma$ and the $\pi$ interactions have a look at this paper by Kutzelnigg.

Community
  • 1
Philipp
  • 17,748
  • 4
  • 79
  • 118
  • what about in the context of just organic chemistry? Hydrogenation certainly seems favorable most of the time. – Dissenter Sep 26 '14 at 13:56
  • you should confront my professor with your knowledge. @Philipp – Dissenter Sep 26 '14 at 14:30
  • 1
    @Dissenter I smell a science battle :) – Philipp Sep 26 '14 at 14:30
  • 2
    @Philipp In the words of Manishearth... "FIGHT! FIGHT! FIGHT! FIGHT! FIGHT!" (http://meta.chemistry.stackexchange.com/questions/493/can-site-users-suggest-some-design-specifications-regarding-graduated-site-devel/495#495) – LordStryker Sep 26 '14 at 14:33
  • 2
    @LordStryker I will be a slaughter unlike any other witnessed in the course of history... only without all the corpses and the bloodshed – Philipp Sep 26 '14 at 14:37
  • 2
    @Philipp The NSA appreciates your comment and would like to inform you that their 'list' is now more rigorous. ;) – LordStryker Sep 26 '14 at 14:38
  • 1
    @LordStryker I always strive to make new friends ;) – Philipp Sep 26 '14 at 14:40
5

The answer to your question is no.

In terms of organic chemistry, consider the two hydrocarbons, $\ce{H_3C-CH_3}$ and $\ce{HC\equiv CH}$. The $\ce{C-C}$ bond dissociation energy is 368 kj/mol while the $\ce{C\equiv C}$ triple bond has a bond dissociation energy of 682 kj/mol, nearly twice that of the single bond.

The transformation of this triple to single bond may be thermodynamically favorable via hydrogenation but may not always be the case. How does temperature affect hydrogenation and can $\Delta H$ be positive under other conditions? Furthermore, if we want to talk about the 'strength' of the CC bond, then clearly the triple bond is stronger.

Phillip brought up a good point in the comments that merits attention. Hydrogenation contains an appreciable barrier for this process to occur. When we take into account the formation of the additional single bonds by hydrogen then the energy is lower. However, if we consider strictly a pi to sigma transformation (between two carbons in this example) and not the formation of other bonds, based on the energetics of the bond we are transforming, the energy will increase.

Reference for bond energies: http://web.chem.ucsb.edu/~zakariangroup/11---bonddissociationenergy.pdf

LordStryker
  • 4,629
  • 1
  • 40
  • 61
  • but isn't the heat of hydrogenation of acetylene still negative? – Dissenter Sep 26 '14 at 14:03
  • 1
    @Dissenter You asked for 'always'. Hydrogenation is just one way and we can consider a multitude of environmental conditions. – LordStryker Sep 26 '14 at 14:06
  • 2
    @Dissenter Yes it is, but that is not your frame of reference. A double bond will always be stronger than a single bond, so going from double to single costs energy in any case. But usually when you reduce a double or triple bond you create new bonds at the expense of them. Only when those new bonds + the reduced bond are stronger than the old double/triple bond then the reaction is favorable. – Philipp Sep 26 '14 at 14:09
  • 1
    @Philipp Your comment is spot on. – LordStryker Sep 26 '14 at 14:25
  • 1
    @LordStryker Thanks, I also could finally think of some organic reactions/compounds where double bonds win out over single bonds and included them in my answer. – Philipp Sep 26 '14 at 14:34
  • LordStryker and @Philipp "bond has a bond dissociation energy of 682 kj/mol, nearly twice that of the single bond." The point is that it is less than twice. There is some destabilization (strain energy is one way to describe it) in a C-C double bond compared to a single bond, and even more in a triple bond. Considering that the OP was addressing (roughly) homopolar, carbon-carbon systems, I think his professor's statement was correct. Hetero-polar (C-O) situations or breaking bonds without forming new ones aside, can you think of any C-C violations. – ron Sep 26 '14 at 14:57
  • @ron For C-C, no. But in my answer I showed that for some other homopolar bonds the bond dissociation energy of a double bond is higher than twice the energy of the single bond. At the bottom of my answer I linked a paper that analyses this relationship for the main group elements quite nicely. It is a little outdated but still useful. – Philipp Sep 26 '14 at 15:08
  • @Philipp Yes, but it perhaps it is the bonding scheme (hybridization) in these other homopolar cases that differentiates them from the carbon case. – ron Sep 26 '14 at 15:13
  • @ron It has to do with the different electronegativities of carbon, oxygen and nitrogen. This entails different relative positions of the s and p valence electron energy levels and so leads to different degrees of s-p-mixing in the bonding orbitals. For 3rd and later row elements some other factors come into play. With them the p-p-$\pi$-interaction is generally weaker due to the higher bond distance. Carbon is indeed a special case. The enormous variety of organic chemistry is largely due to the fact that C-C $\sigma$- and $\pi$-bonds are competitive with each other. – Philipp Sep 26 '14 at 15:25