9
$\begingroup$

I can not understand why a peroxide $\ce{R-O-O-R}$ is considered reactive and unstable.

Going down one row on the periodic table, a disulfide bridge ($\ce{R-S-S-R}$) is apparently super stable and super important to proteins 3d structure.

At the same time, a thioester is considered to be as unstable/reactive as ATP and the ester is the stable one?

What am I conceptually missing here to explain this contradiction?

Improve this question
$\endgroup$
8
  • 3
    $\begingroup$ Why do you think the trend should be the same, though? $\endgroup$
    – orthocresol
    Commented Jul 19, 2017 at 15:28
  • 6
    $\begingroup$ You're trying to compare trends between -O-O- and -S-S- bonds, and -C-O-C- and -C-S-C- bonds. $\endgroup$
    – Eashaan Godbole
    Commented Jul 19, 2017 at 15:47
  • 3
    $\begingroup$ chemistry.stackexchange.com/questions/431/… $\endgroup$
    – Mithoron
    Commented Jul 19, 2017 at 16:29
  • 2
    $\begingroup$ From @Mithoron's link, I also found this chemistry.stackexchange.com/questions/47056/… which explains why s-s bonds are stronger than o-o bonds. $\endgroup$
    – Tyberius ♦
    Commented Jul 28, 2017 at 17:57
  • 2
    $\begingroup$ The whole point is that you're comparing apples and oranges. The O-O bond is weaker than the S-S bond, so peroxides are less stable than the corresponding disulfides. But there aren't any O-O or S-S bonds in esters/thioesters, so there's no reason why the same argument should apply, and there's no "contradiction" in the fact that thioesters are more reactive than esters (which is due to poor overlap of sulfur's 3p orbital with the 2p orbitals of C=O). $\endgroup$
    – orthocresol
    Commented Jul 31, 2017 at 17:44

2 Answers 2

Reset to default
10
+25
$\begingroup$

Some hard data: bond enthalpies (in $\pu{kJ mol-1}$)

$$\begin{array}{c|c|c|c} \text{Bond} & \text{Enthalpy} & \text{Bond} & \text{Enthalpy} \\ \hline \ce{C-C} & 350 & \ce{Si-Si} & 226\\ \ce{N-N} & 163 & \ce{P-P} & 201\\ \ce{O-O} & 146 & \ce{S-S} & 226\\ \ce{F-F} & 155 & \ce{Cl-Cl} & 240 \\ \end{array}$$

We observe a decrease in bond energy from carbon to nitrogen and from silicon to phosphorus. It is clear and probably have same reason: appearance of lone electron pair of the atom. There is a repulsion between lone pairs, that is in partially compensated by stronger bonds formed by smaller atoms (the atomic radii falls down to the end of the row). When moving from second to third row, this repulsion is reduced thanks to larger size of atoms of the elements of the third row.

The general decrease in energy from carbon to silicon is thanks to increased size of the orbitals of valence level, making them more diffuse with less efficient overlap and longer bond length.

The stability of esters vs thioesters is ruled by different reasons. Sulfur is not as electronegative as oxygen, but when you consider an anion, sulfur anions have larger radius and are less prone to grabbing first positive charge they find. So sulfur anions are more stable intermediates (kinetically), and thus polar dissociation of thioester bonds is easier.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Regarding the last paragraph; it is correct if you are considering a dissociative mechanism of the form $\ce{RC(=O)SR' -> RC#O+ + ^-SR'}$. I'd suspect, though, that the usual mechanism for thioester cleavage is associative. So I think it would be to do with the electrophilicity of C=O, which is in this case dependent on resonance with the lone pair on sulfur (poor overlap = less donation into π* orbital = more electrophilic). $\endgroup$
    – orthocresol
    Commented Aug 5, 2017 at 4:32
  • $\begingroup$ @orthocresol The C-S bond is still easier to break than C-O bond in this case as well. $\endgroup$
    – permeakra
    Commented Aug 5, 2017 at 17:45
-2
$\begingroup$

It has to do with electronegativity (EN). Remember the concept of "formal charge"? In a bond between molecule 1 and 2, if molecule 1 is more electronegative than molecule 2, then the electrons will be drawn closer to molecule one.

That is general, but to get to your specific question, Oxygen's EN is 3.5. Sulfur's EN is aprox 2.5. Carbon's EN is also 2.5.

So in a Carbon oxygen bond, the electrons will shift to oxygen. In a carbon-sulfur bond, the electrons will be relatively in the middle, thus one can expect a carbon-sulfur, and sulfur-sulfur bonds to be relatively equal in strenght.

On the other hand, the peroxides (when bonded to carbons) will always pull the electrons towards them more than the carbons wil, and at some point, the bond will eventually snap.

Improve this answer
$\endgroup$
2
  • 1
    $\begingroup$ Electronegativity is only one factor, and rarely the deciding factor, in bond strength. I highly doubt that C-C, C-S, and S-S bonds are all of the same strength. Your last paragraph is also not exactly on topic. The issue is not the C-O bond breaking, it is the O-O bond. The argument is also incorrect, as C-O bonds are generally not easy to break at all. $\endgroup$
    – orthocresol
    Commented Aug 4, 2017 at 13:32
  • $\begingroup$ And another factor is.... $\endgroup$
    – El chimisto
    Commented Aug 4, 2017 at 13:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.