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We know that intermolecular hydrogen bonding is seen in hydrogen peroxide $(\ce{H2O2})$. But is intramolecular H-bonding, i.e. hydrogen bond between an oxygen atom and a hydrogen atom attached to the adjacent oxygen atom possible? How strong will it be, compared to intermolecular H-bonding and why?

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    $\begingroup$ If there is intramolecular h-bonding occuring (which, off the cuff, I strongly doubt), my guess is that it would be very weak with respect to intermolecular h-bonding due to the angle, A(OOH), being far from optimal in a normal h-bond. I'll see what an NBO analysis says. It could give us a clue if there is any sort of overlap that could give rise to a hydrogen-bond like intramolecular interaction. $\endgroup$ – LordStryker Oct 29 '13 at 18:32
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Preamble:

I will conveniently avoid entering into a debate as to what a hydrogen bond is, given the numerous theories which attempt to do so from various perspectives. I will, however, attempt to use the relatively cheap and easy natural-bond orbital (NBO) analysis method to examine the possible existence of an intramolecular hydrogen bond (HBond) in the ground-state, global minimum configuration of H$_2$O$_2$. In doing so, we will assume that the mere existence of an $n\rightarrow\sigma^{*}$ overlap indicates that an intramolecular HBond is indeed present (a commonly accepted metric). For comparison, I will look at the planar form of H$_2$O$_2$ (containing a trans-type bond configuration) in order to minimize any overlap between the oxygen lone pair and the corresponding $\sigma^{*}$ orbital. I note here that the NBO analysis is purely a qualitative method, and caution should be used when attempting to make quantitative comparisons with this method.

Methods:

The H$_2$O$_2$ molecule was constructed in $C_2$ symmetry (Structure A). A full optimization and corresponding harmonic vibrational frequency computation was employed at the MP2/heavy-aug-cc-pVTZ level of theory, where non-hydrogen (i.e. heavy) atoms are augmented with diffuse functions. The basis set is denoted henceforth as haTZ. The magnitudes of the residual Cartesian gradient components of the optimized structure were less than $4.0\times10^{-5}$ $E_{\textrm{h}}$ $a_0^{-1}$. The frozen-core approximation was invoked for all computations. A subsequent NBO analysis of the fully optimized H$_2$O$_2$ molecule was performed at the MP2/haTZ level of theory using the SCF density. This procedure was also applied to a H$_2$O$_2$ molecule constructed in $C_{2h}$ symmetry (Structure B). All computations were performed with the G09 (Rev. D01) software package.

Results:

Structure A is a minimum on the MP2 potential energy surface with zero imaginary modes of vibration whereas structure B is a transition state with one imaginary mode of vibration.

The NBO analysis of structure A reveals two $n\rightarrow\sigma^{*}$ interactions (where $n$ is an electron lone pair on oxygen and $\sigma^{*}$ is the $\sigma$ antibonding orbital of the neighboring O$-$H entity. The existence of exactly two of these interactions is due to the symmetry of the molecule, and only one will be reported here for simplicity.

The following figure illustrates the overlap of a lone electron pair on O with the adjacent $\sigma^{*}$ orbital in structure A. The occupation number for this antibond, 0.00351, indicates a very small overlap. The corresponding second-order perturbation energy E$^{(2)}$(donor$\rightarrow$acceptor) is a modest 1.74 kcal mol$^{-1}$. We note that this is merely an estimated measure of the strength of the interaction.

H2O2 nsigma* overlap

Interestingly, upon rotation of the torsion angle, $\tau$(H-O-O-H), to 180$^\circ$ (structure B), the $n\rightarrow\sigma^{*}$ interaction disappears. The intramolecular hydrogen bonding may play an important role in stabilizing the lowest-energy conformer in H$_2$O$_2$ (structure A) as the transition state, structure B, does not contain this interaction.

Comparing the strength of an intramolecular HBond to an intermolecular HBond is not straightforward. Intermolecular interactions are frequenly and efficiently computed by using the supermolecular approach by taking difference of the electronic energy of a complex and the energies of each non-interacting body which make up the complex. Intramolecular interactions cannot be determined in this way without breaking bonds, thereby skewing the resulting number. To facilitate comparison to the intramolecular HBond in H$_2$O$_2$, an NBO analysis was performed on the water dimer (H$_2$O)$_2$ (in $C_S$ symmetry) due to the presence of a strong, intermolecular hydrogen bond. The corresponding E$^{(2)}$ for an $n\rightarrow\sigma^{*}$ interaction is a much larger 6.82 kcal mol$^{-1}$, nearly 5 kcal mol$^{-1}$ stronger than that seen in H$_2$O$_2$

Conclusions:

An NBO analysis of the lowest energy conformer H$_2$O$_2$ indicates the possibility of the presence of weak, intramolecular hydrogen bonds. These intramolecular hydrogen bond-like interactions could play a role in stabilizing the H$_2$O$_2$ complex. The strength of the interaction is much smaller than that computed for the water dimer by about 5 kcal mol$^{-1}$.

(Note: I strongly recommend performing a Bader analysis on H$_2$O$_2$ to compare with the NBO analysis before reaching any solid conclusions.)


ADDENDUM:

I have tried to avoid calling this intramolecular interaction a hydrogen bond. Based on the metric imposed in the preamble, it has a property of a hydrogen bond, however, the strength of this intramolecular interaction is so weak, it really does not reflect a true hydrogen bond (e.g. A typical hydrogen bond is about 10 kcal mol$^{-1}$ and a weak hydrogen bond is usually considered to be on the order of 2-3 kcal mol$^{-1}$. If we take the NBO results at face value, the interaction is just too weak to be considered as a hydrogen bond. Coupled with the fact that the geometry itself does not facilitate reasonable HBonding configurations.

Upon reflection of these issues, I conclude that there is, in fact, no intramolecular hydrogen bonding in H$_2$O$_2$.

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  • $\begingroup$ I went through your thorough answer,and though i could not understand much of the middle part,I understood that there cannot be an intramolecular hydrogen bond.How did you do this analysis?Is it from a research paper? $\endgroup$ – scienceauror Oct 30 '13 at 19:13
  • $\begingroup$ I was just thinking about this and water,not referring to experimental data,can you help me guess whether the strength of(intermolecular) hydrogen bond in $H_2O_2$ is more/less than in $H_2O$?I think the presence of two oxygens may make the hydrogen atom more electrophilic,hence making a stronger H-bond.Am I right? $\endgroup$ – scienceauror Oct 30 '13 at 19:18
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    $\begingroup$ @scienceauror I just decided to do this off the cuff as a mini-project. I wouldn't dream of lifting this kind of work from a paper without citing it. It was fun and I learned something. Also, when you are talking about intermolecular HBonding in H$_2$O$_2$, you are referring to something like a dimer now? If so, it would be trivial to figure out a reasonable HBond strength. $\endgroup$ – LordStryker Oct 30 '13 at 19:58
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    $\begingroup$ @scienceauror I went back and looked at the R(HO) bond lengths in H$_2$O$_2$ and H$_2$O. They were 0.97 Ang and 0.96 Ang respectively. From this alone I would say that the OH bond in water is stronger than in hydrogen peroxide. This sort of makes sense as the neighboring oxygen in H2O2 is 'much larger' than a hydrogen atom (in water), thereby 'pushing' electron density away from the adjacent oxygen, causing a lengthened OH bond. $\endgroup$ – LordStryker Oct 31 '13 at 15:31
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    $\begingroup$ The effect you are describing could be Hyperconjugation? $\endgroup$ – Martin - マーチン Apr 9 '14 at 4:49
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My molecular models set has led me to believe that hydrogen peroxide can conform to form 2 bent hydrogen bonds and one bent covalent bond between the $\ce{OH}$ groups.

This means that there can be a triple bond between the 2 oxygen atoms. The triple bent bond (i.e., 3 banana bonds) of which 2 are of the hydrogen bond type and one is the Linus Pauling bent-covalent bond type, like in ethylene ($\ce{CH2CH2}$), tau bonds, banana bonds, bent bonds, and indirect angular bonds.

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    $\begingroup$ You're wrong - hydrogen bonds can't be bent. Even if they could they wouldn't be able to offset ring strain. $\endgroup$ – Mithoron Aug 19 '15 at 23:54
  • $\begingroup$ Can you provide any evidence for these claims? They seem pretty unlikely to me. $\endgroup$ – bon Aug 20 '15 at 9:01
  • $\begingroup$ The normal direct covalent bond between the 2 OH groups becomes bent due to 1 repulsion between the non-bonding electron pairs, 1 on each oxygen atom and 2 attractions of the 2 hydrogen atoms for the 2 Non-bonding electron pairs. These attractions (2) and 1 repulsion are all in the same direction causing the direct bond to become bent, now forming a triple bent bonds. This lends greater stability to the gaseous form of H2 O2so in the conformation where $\endgroup$ – Chuck Boldwyn Aug 23 '15 at 17:45
  • $\begingroup$ The normal direct covalent bond between the 2 OH groups becomes bent due to 1 repulsion between the non-bonding electron pairs, 1 on each oxygen atom and 2 attractions of the 2 hydrogen atoms for the 2 Non-bonding electron pairs. These attractions (2 bent hydrogen bonds and 1 repulsion ( non-bonding pair repulsion) are all in the same direction causing the direct bond to become bent, now forming a set of triple bonds, all bent. This lends to far greater stability to the gaseous form of H2O2. In the liquid, H2O2 forms intermolecular hydrogen bonding with itself and with H2O & would predominate $\endgroup$ – Chuck Boldwyn Aug 23 '15 at 18:16
  • $\begingroup$ You will need a molecular model set that can form bent bonds of the covalent type (Linus Pauling type bent bond) and to show the bent hydrogen bonds. My models are unique and can easily illustrate non-bonding electron pairs and their interactions with their surroundings & can easily illustrate interactions to form bent bonds. I have no way to show a photo of the model here but being creative you can draw this molecule with pencil & paper, if you follow these directions above $\endgroup$ – Chuck Boldwyn Aug 23 '15 at 18:17

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