Chemistry - What is the isomer distribution in monosubstituted fluorobullvalene?

Solution 1:

Extensive NMR studies of substituted bullvalenes were done in the 1960's and 70's, by Oth et al. Much of their original work was published in German. Thankfully, some of these topics were revisited (in English!) during the 1990's by Luz et al., and these included some rather nice low temperature ¹⁹F and ¹³C NMR studies in both solution and solid state (and make it much quicker and easier for me to read).

Let us assign the following isomers 1–4, as is done in the question:

To answer your question:

Oth et al.^[1] studied fluorobullvalene at −25 °C, and found no evidence by ¹⁹F of isomer 1. The equilibrium populations of isomers 4:3:2 was 78%:7%:15%. However, they could not distinguish between isomers 3 and 2, so it could be 78:15:7. In any case, predominantly isomer 4. They do state, however, that isomer 1 must exist ["Das C-Isomer muss also durchlaufen werden."], as the Cope rearrangement mechanism for isomer 4 to go to isomer 3 must go through isomer 1.

Luz et al.^[2] studied the solution state Cope rearrangement of a number of monosubstituted bullvalenes. At −30 °C, they identify the ratio of isomers 4:3:2 to be 87%:3.4%:9.6%. Again no evidence of isomer 1 by ¹⁹F at this temperature. However, at −55 °C, they observe a peak in the ¹⁹F spectrum that they ascribe to isomer 1, with an intensity of 0.2% relative to isomer 4. This peak broadens beyond detection at −40 °C. The ratios of isomers at −55 °C, then, was found to be 1.0:0.018:0.065:0.002 for isomers 4:3:2:1.

And just out of interest:

Of course, the other monsubstituted bullvalenes all have different preferred isomer distributions in solution. In the solid state, all monosubstituted bullvalenes studied by Luz crystallize as well-ordered single isomer crystals. Fluorobullvalene^[3] crystallizes exclusively as isomer 4. Cyanobullvalene, bullvalenecarboxylic acid^[4], and (ethylthio)bullvalene^[5] crystallize exclusively as isomer 3. Bromo- and iodobullvalene^[6] crystallize entirely as isomer 2.

Useful references for you:

1. Oth, J. F. M.; Merényi, R.; Röttele, H.; Schröder, G. Fluor-, chlor- und jodbullvalen. Tetrahedron Lett. 1968, 9 (36), 3941–3946. DOI: 10.1016/S0040-4039(00)72372-0.

2. Poupko, R.; Zimmermann, H.; Müller, K.; Luz, Z. Dynamic NMR Investigation of the Cope Rearrangement in Solutions of Monosubstituted Bullvalenes. J. Am. Chem. Soc. 1996, 118 (34), 7995–8005. DOI: 10.1021/ja954004t.

3. Müller, K.; Zimmermann, H.; Krieger, C.; Poupko, R.; Luz, Z. Reaction Pathways in Solid-State Processes. 1. Carbon-13 NMR and X-ray Crystallography of Fluorobullvalene. J. Am. Chem. Soc. 1996, 118 (34), 8006–8014. DOI: 10.1021/ja954005l.

4. Müller, K.; Zimmermann, H.; Krieger, C.; Poupko, R.; Luz, Z. Reaction Pathways in Solid-State Processes. 2. Carbon-13 NMR and X-ray Crystallography of Cyanobullvalene and Bullvalenecarboxylic acid. J. Am. Chem. Soc. 1996, 118 (34), 8015–8023. DOI: 10.1021/ja954006d.

5. Luger, P.; Roth, K. X-Ray, n.m.r., and theoretical studies of the structures of (ethylthio)bullvalene. J. Chem. Soc., Perkin Trans. 2 (1972-1999) 1989, 649–655. DOI: 10.1039/P29890000649.

6. Luz, Z.; Olivier, L.; Poupko, R.; Müller, K.; Krieger, C.; Zimmermann, H. Bond Shift Rearrangement of Chloro-, Bromo-, and Iodobullvalene in the Solid State and in Solution. A Carbon-13 and Proton NMR Study. J. Am. Chem. Soc. 1998, 120 (22), 5526–5538. DOI: 10.1021/ja9728029.

Solution 2:

Let me begin by thanking Martin and long for supplying the key reference (J. Am. Chem. Soc. 1996, 118 (34), 7995–8005). Importantly, long also provided the isomer ratio I was asking for; according to him, the ratio was estimated using the numbers at bottom of page 7997 (with the 0.002 added).

The free energy differences between the isomers of fluorobullvalene are reported in this paper as

$$\begin{align} \Delta G(1,4) &= \pu{+3.41 kcal/mol} \\ \Delta G(3,1) &= \pu{-2.21 kcal/mol} \\ \Delta G(2,3) &= \pu{-0.46 kcal/mol} \\ \end{align}$$

Here is a diagram showing the 4 possible isomers of fluorobullvalene in equilibrium, as well as a simple example of the symmetry-allowed, 6-pi electron Cope rearrangement that occurs in bullvalene and interconverts the isomers:

In this answer, the equilibrium concentrations of the four isomers were determined to be as follows. These numbers differ somewhat from long's numbers - which were an approximation long made based on nmr peak intensities - but appear to match those shown in Fig. 4 in the paper at 300 K:

\begin{align} [1]~ &=~~~ 0.23\% \\ [2]~ &=~ 20.27\% \\ [3]~ &=~~~ 9.37\% \\ [4]~ &=~ 70.13\% \\ \end{align}

The question left to be answered: why does the isomer distribution fall out like this, and, in particular, why such an avoidance of the cyclopropyl position by fluorine?

As we work towards an answer to this question we will discuss bonds that are $\ce{sp^{i}}$ hybridized. “i” is referred to as the hybridization index. The hybridization index is such that the ratio $1:i$ is the ratio of the s:p atomic orbitals mixed to form the hybrid orbital. For example in an $\mathrm{sp^3}$ bond, $i=3$, and the ratio of s:p atomic orbital content in the bond is $1:3$. Further, we can see from this that the percentage s character in a bond is given as

$$\text{s-character} = \frac{1}{1+i} \tag{1}$$

which is $1/(1+3) = 25\%$ in the $\mathrm{sp^3}$ case.

There are a number of ways to estimate the percent s-character in a bond. One way is through use of the formula

$$^1\!J_\ce{C-H} = \pu{500 Hz} \cdot \text{s-character in the bond} \tag{2}$$

where $^1\!J_\ce{C-H}$ is the one-bond C–H coupling constant.

Note: $\ce{^12C}$ does not have a magnetic moment, therefore it cannot interact with a proton spin and split it. $\ce{^13C}$ is a spin-½ nucleus, it can interact with a proton spin and cause signal splitting in the carbon and proton nmr spectra. $\ce{^13C}$ only occurs at 1.1% natural abundance, but one can still measure the resultant C–H splitting (e.g. the coupling constant).

Next, consider the polarization in a $\ce{C-F}$ bond. It will look something like this:

Because of the relative electronegativities of carbon and fluorine, electron density will shift away from carbon towards fluorine. To understand the effect of this electron shift, let's compare the hybridization in methane ($\ce{CH4}$) and methyl fluoride ($\ce{CH3-F}$). In methane, the carbon orbital used to form the C–H bond is exactly $\mathrm{sp^3}$ hybridized – this is demanded by the symmetry of the molecule. Using equation (1) we know that the percent s-character in the atomic orbital used by carbon to interact with and form a molecular orbital with hydrogen is 25% in this case. From equation (2) we would predict that $^1\!J_\ce{C-H}$ in methane would be (500*0.25) Hz, or 125 Hz. Indeed, the experimentally determined coupling constant for methane is measured as 125 Hz.

In methyl fluoride the experimentally observed $^1\!J_\ce{C-H}$ coupling constant is 149 Hz. Using equation (2) we find that there is 149/500 = 0.30 or 30% s-character in the carbon orbital used to form the C–H bond. Using equation 1, this is an $\mathrm{sp^{2.36}}$ bond. There are 3 of these C–H bonds, so a total of 90% of the available carbon 2s orbital is used in forming them. This leaves 10% of the carbon 2s orbital remaining and available for use in the C–F bond. Using equation 1 we find that i=9 and this carbon orbital is $\mathrm{sp^9}$ hybridized – very nearly pure p!

What a difference in bond hybridization between methane and methyl fluoride and it teaches an important lesson. Since the carbon 2s orbital is lower in energy than the carbon 2p orbital, the more s-character there is in a bond, the lower the energy of the electrons in that bond. If the electron density in an orbital is low (like in the carbon portion of the C–F bond in methyl fluoride), then nature doesn’t mix much s-character into that orbital and, instead, uses that s-character to stabilize orbitals that do have significant electron density. Hence, the carbon orbitals used to form the C–F bonds in methyl fluoride only have 10% s-character because the electron density remaining around carbon in that bond is so low. Instead, the s-character is used in the 3 C–H bonds which do contain significant electron density around carbon.

Before we apply this type of analysis to fluorobullvalene, let’s make some predictions based on what we’ve learned so far. Clearly, fluorine will preferentially bond with carbon atoms that have low s-character in the bonding orbital. The $^1\!J_\ce{C-H}$ coupling constant in cyclopropane is 161 Hz. Using equations (2) and (1), we find that the C–H bond in cyclopropane has 32% s-character and is $\mathrm{sp^{2.1}}$ hybridized. In bullvalene itself, carbon (4) in the tris-allylic position looks like a $\ce{sp^{3}}$ hybridized carbon whereas the olefinic carbons are probably $\ce{sp^{2}}$ (more or less). So right off the bat, we would expect more fluorine at position 4, than at either of the olefinic positions or the cyclopropyl position.

The relevant $^1\!J_\ce{C-H}$ coupling constants in the parent molecule bullvalene are presented in the table below, along with the percent s-character derived from equation (2), the hybridization index $i$ from equation (1), the value of $\mathrm{sp}^i$, and the percentage of the isomer.

$$\begin{array}{|c|c|c|c|c|c|}\hline \text{Carbon} & ^1\!J_\ce{C-H}\text{ / Hz} & \text{% s-character} & i & \mathrm{sp}^i & \text{% isomer} \\ \hline 1 & 169.0 & 33.8 & 1.96 & \mathrm{sp}^{1.96} & 0.23 \\ \hline 2 & 160.8 & 32.2 & 2.11 & \mathrm{sp}^{2.11} & 20.27 \\ \hline 3 & 166.1 & 33.2 & 2.01 & \mathrm{sp}^{2.01} & 9.37 \\ \hline 4 & 133.9 & 26.8 & 2.73 & \mathrm{sp}^{2.73} & 70.13 \\ \hline \end{array}$$

We immediately see that the amount of fluorine at a given position changes with the percentage s-character in the bond. Just as we expected, the higher the s-character, the lower the amount of that fluorinated isomer. In fact, if we measure the correlation coefficient between the percentage s-character in the bond and the percentage of the isomer with fluorine attached at that position, a correlation coefficient of –0.9982 is obtained (if we use long's suggested isomer distribution [0.18%, 5.99%, 1.66%, 92.17%] the correlation coefficient is –0.9875). A value of 1 would be a perfect correlation, a value of –1 would be a perfect inverse correlation. In the case at hand, the percentage s-character in the bond and the percentage of the isomer with fluorine attached at that position demonstrate a near-perfect inverse correlation. An inverse correlation means that as the percentage s-character increases, the isomer with fluorine attached at this position decreases and the high value for the coefficient indicates that the movement of the two values is highly correlated.

Bullvalene provided an interesting test case where we could measure fluorine’s affinity to bond to different types of carbon. From this example we can see that fluorine strongly prefers to form bonds to carbon that utilize carbon atomic orbitals with low s-character.

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Solution 3:

I had some free time and resources at hand and did some highly accurate quantum chemical calculations for this system. I took it as a chance to compute some benchmarks and play a little with density functional theory.

Originally I also wanted to compute transition barriers, but based on the previous findings I had to omit that, since I currently cannot afford ab initio calculations. However, I'd like to share these results.

For your reading convenience, here is the chart from longs post, in order to see the numeration ($\ce{X=F}$).

Electronic Energies $\Delta{}E^\circ$ are as usual in $\mathrm{kcal/mol}$. Thermodynamic data $\Delta{}E^\circ$ was estimated at the DF-BP86/def2-TZVP and DF-BP86/def2-SVP level of theory with $T=298.15~\mathrm{K}$ and $p=1~\mathrm{atm}$. Experimental data was taken from the cited paper. Product ratios were calculated from the electronic energies (except for BP86 where data was available) assuming that the contribution of thermodynamic corrections is neglectable, i.e. $\Delta{}E^\circ = \Delta{}G^\circ$. The level of degeneracy is for each conformer $g(\mathbf{1})=g(\mathbf{2})=g(\mathbf{3})=3$ and $g(\mathbf{4})=1$.
(I was somehow unable to reproduce the product ratio that ron calculated with his approach. I used How to obtain product ratio from energy differences via Boltzmann statistics)

\begin{array}{lrrrr}\hline \text{method} & \mathbf{1} & \mathbf{2} & \mathbf{3} & \mathbf{4} \\\hline \text{Exptl.}~\Delta G^\circ & 0.0 & -2.67 & -2.21 & -3.41 \\ \% & 0.4 & 38.0 & 17.5 & 44.2 \\\hline \text{BP86/SVP}~\Delta E^\circ & 0.0 & -3.3 & -3.5 & -2.6 \\ \phantom{\text{BP86/SVP}}~\Delta G^\circ & 0.0 & -3.0 & -3.0 & -2.6 \\ \% & 0.3 & 43.5 & 48.0 & 8.2 \\\hline \text{BP86/TZVP}~\Delta E^\circ & 0.0 & -3.2 & -3.0 & -3.1 \\ \phantom{\text{BP86/TZVP}}~\Delta G^\circ & 0.0 & -2.9 & -2.6 & -3.0 \\ \% & 0.4 & 48.6 & 30.6 & 20.5 \\\hline \text{PBE0/TZVP}~\Delta E^\circ & 0.0 & -3.0 & -2.8 & -3.3 \\ \% & 0.3 & 44.7 & 30.9 & 24.1 \\\hline \text{M06//BP86/TZVP}~\Delta E^\circ & 0.0 & -3.1 & -2.9 & -3.7 \\ \% & 0.2 & 38.5 & 27.2 & 34.1 \\\hline \text{MP2//BP86/TZVP}~\Delta E^\circ & 0.0 & -1.3 & -1.0 & -3.9 \\ \% & 0.4 & 3.7 & 2.1 & 93.8 \\\hline \text{CCSD(T)//BP86/TZVP}~\Delta E^\circ & 0.0 & -1.3 & -1.0 & -3.9 \\ \% & 0.4 & 3.7 & 2.1 & 93.8 \\\hline \end{array}

It is astonishing that MO6 almost exactly reproduces the experimental values, while the ab initio calculations CCSD(T) and MP2 conclude that there is almost only the $\mathbf{4}$ conformer. This was quite unexpected, maybe frequency calculations have to be done to obtain more fitting results.