IChO 2023 · Theory problem 6

Direct Conversion of Methane to Methanol

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Methane is widely available as natural gas making it an attractive feedstock for the chemical industry, such as for the production of methanol. However, control of this process is challenging as methanol is more easily oxidized than methane. Overoxidation is avoided in a chemical looping process, where active sites of copper-exchanged zeolite catalysts provide only the single oxygen atom required for oxidation to methanol and are consumed. In a second step, the catalyst is regenerated with oxygen in the absence of methane. The scheme below shows two potential catalytic copper sites. During the reaction, Cu(II) is reduced to Cu(I).

Figure 1

6.1 Give the number of S1 sites and the number of S2 sites required to oxidize one methane molecule to methanol.

In the absence of oxygen, the methanol formed does not desorb from zeolite. If the reaction is performed in a container with constant volume and temperature (an autoclave), a pressure-drop results only from the consumption of methane, which can be considered as an ideal gas. In a 1 L autoclave containing 200 mg of zeolite loaded with 4.3 % wt. copper, the initial methane pressure \(p_0 = 933\ \mathrm{Pa}\) dropped to \(p_\infty = 925\ \mathrm{Pa}\) after completion of the reaction at 528 K.

6.2 Calculate the percentage of copper that reacted.

6.3 Experimental data are plotted in Figure 1. Based on this, decide on the (pseudo) order of the oxidation of \(\ce{CH4}\).

(a) The reaction is of (pseudo) zeroth order.

(b) The reaction is of (pseudo) first order.

(c) The reaction is of (pseudo) second order.

Figure 2

Figure 1. Semi-logarithmic graph of normalized methane pressure versus time for the reaction with copper-loaded zeolite containing sites S1 and S2 in an autoclave at various temperatures. The symbols denote experimental data points. The solid lines are linear fits to appropriate time ranges. The dotted lines are guides for the eyes

6.4 Write down the (pseudo) rate law for the oxidation of \(\ce{CH4}\) that is consistent with the experimental data under the given conditions. Note that it may depend on the concentrations of \(\ce{CH4}\) as well as of sites S1 and S2 and on the rate constants.

6.5 Choose the correct statement(s).

(a) At least two types of copper sites react, each with a different rate constant.

(b) The overall methane oxidation by zeolite is faster at higher temperature.

(c) At higher temperature, a larger fraction of the copper sites will have reacted with methane after completion of the reaction.

(d) One of the reactions becomes slower at higher temperature.

Paramagnetic sites S1 can be observed by electron paramagnetic resonance (EPR) spectroscopy, whereas diamagnetic sites S2 do not give an EPR signal. EPR spectroscopy measures the number of electron spins. Thus, the number of S1 sites is proportional to the double integral \(I_2\) of the EPR spectrum, i.e. \([\mathbf{S1}] \propto I_2\). Spectra were measured at different temperatures \(T\) and at each temperature at different times, \(t\), after initiating the reaction.

6.6 Derive the equation, which is linear in time, that relates \(I_2(t)\) to the rate constant for the loss of S1 sites.

6.7 Tick the boxes on the answer sheet for each measurement that needs to be calibrated with a known Cu(II) standard.

(a) Total number of paramagnetic Cu(II) sites in the sample.

(b) Concentration of paramagnetic Cu(II) sites in the sample.

(c) Rate constant.

(d) Types of different paramagnetic Cu(II) species in the sample.

From EPR measurements, it is known that the rate constant for the reaction with S1 sites at 528 K is \(2.604 \times 10^{-3}\ \mathrm{s^{-1}}\).

6.8 Considering Figure 1 and based on a calculation, decide if methane reacts faster or slower with S2 sites than with S1 sites.

(a) Methane reacts faster with S1.

(b) Both reaction rates are the same.

(c) Methane reacts faster with S2.

Methanol can be further converted into valuable olefins with different zeolite catalysts. In this process, one observes an intermediate product with molar mass \(86.09\ \mathrm{g\,mol^{-1}}\), elemental analysis (55.8 % wt. \(\ce{C}\), 7.0 % wt. \(\ce{H}\)) and an \({}^{1}\mathrm{H}\) NMR spectrum consisting of signals at four different chemical shifts:

a: 12.2 ppm (1H, s), broad, disappears when \(\ce{D2O}\) is added

b: 6.3 ppm (1H, d)

c: 5.7 ppm (1H, d)

d: 2.0 ppm (3H, s)

6.9 Draw the structure of the intermediate product and assign protons a and d.

The United States Department of Energy assigned twelve chemical compounds only containing \(\ce{C}\), \(\ce{H}\) and \(\ce{O}\) as platform chemicals. These are the most promising candidates, easy to prepare from renewable resources and with multiple target derivatives to be prepared from them. One of them is compound A, that can either be further derivatized or used for example in medicinal applications or in detergents.

  • \({}^{1}\mathrm{H}\) NMR in DMSO: 7.81 ppm (a, s), 13.0 ppm (b, s, broad, disappears when \(\ce{D2O}\) is added), both signals have the same integral.
  • \({}^{13}\mathrm{C}\) NMR: 165.1 ppm (1), 150.6 ppm (2) and 120.6 ppm (3).
  • \(M = 156.03\ \mathrm{g\,mol^{-1}}\). Elemental analysis: 46.15 % wt. \(\ce{C}\), 2.56 % wt. \(\ce{H}\).

6.10 Give a possible structure of A and assign all protons and carbon 1.

Official English text of the problem.

Answers

  • 6.1: 2 S1 sites or 1 S2 site per methane molecule.
  • 6.2: \(2.7\%\) of the copper reacted.
  • 6.3: (b) — (pseudo) first order.
  • 6.4: With \(k_i\) defined as the first-order disappearance constant of site population \(i\), \[ -\frac{d[\ce{CH4}]}{dt} =\frac{k_1}{2}[\mathbf{S1}]+k_2[\mathbf{S2}]. \]
  • 6.5: (a) and (b).
  • 6.6: \[ \boxed{\ln\!\left(\frac{I_2(t)}{I_2(0)}\right)=-k_1t}. \]
  • 6.7: (a) and (b).
  • 6.8: (c) — methane reacts faster with S2. The slow and fast graphical decay constants are approximately \(2.6\times10^{-3}\) and \(8.0\times10^{-3}\ \mathrm{s^{-1}}\), respectively.
  • 6.9: Methacrylic acid (2-methylprop-2-enoic acid), C=C(C)C(=O)O. Proton a is the carboxylic acid proton; d denotes the three methyl protons.
  • 6.10: Furan-2,5-dicarboxylic acid, O=C(O)c1ccc(C(=O)O)o1. Protons a are the two ring hydrogens at positions 3 and 4; b are the two carboxylic acid protons. Carbon signal 1 represents both carboxyl carbonyl carbons.

Solution

6.1

Methane, C, is oxidized to methanol, CO. The carbon oxidation number changes from \(-4\) in \(\ce{CH4}\) to \(-2\) in \(\ce{CH3OH}\), corresponding to the loss of two electrons per methane molecule.

Each copper atom accepts one electron when Cu(II) is reduced to Cu(I). Therefore, two copper atoms must be reduced per methane molecule:

  • S1 contains one copper atom, so two S1 sites are required.
  • S2 contains two copper atoms, so one S2 site is required.

After including catalyst regeneration, the balanced overall conversion is

\[ \ce{2CH4 + O2 -> 2CH3OH}, \]

where molecular oxygen has SMILES O=O.

6.2

Because methanol remains adsorbed, the decrease in gas amount equals the amount of methane consumed. Using \(R=8.314\ \mathrm{Pa\,m^3\,mol^{-1}\,K^{-1}}\),

\[ \begin{aligned} n_{\ce{CH4},\mathrm{reacted}} &=\frac{(p_0-p_\infty)V}{RT}\\ &=\frac{(933-925)\ \mathrm{Pa}\times1.00\times10^{-3}\ \mathrm{m^3}} {8.314\ \mathrm{Pa\,m^3\,mol^{-1}\,K^{-1}}\times528\ \mathrm{K}}\\ &=1.82\times10^{-6}\ \mathrm{mol}. \end{aligned} \]

The initial copper mass is

\[ m_{\ce{Cu}}=200\ \mathrm{mg}\times0.043 =8.6\ \mathrm{mg}. \]

Using \(M(\ce{Cu})=63.55\ \mathrm{g\,mol^{-1}}\),

\[ n_{\ce{Cu},0} =\frac{8.6\times10^{-3}\ \mathrm{g}} {63.55\ \mathrm{g\,mol^{-1}}} =1.35\times10^{-4}\ \mathrm{mol}. \]

Two copper atoms react per methane molecule, regardless of which site type is involved:

\[ n_{\ce{Cu},\mathrm{reacted}} =2n_{\ce{CH4},\mathrm{reacted}} =3.64\times10^{-6}\ \mathrm{mol}. \]

Thus,

\[ \boxed{ \text{Copper reacted} =\frac{3.64\times10^{-6}}{1.35\times10^{-4}}\times100\% \approx2.7\%. } \]

6.3

Define the normalized remaining pressure change as

\[ q(t)=\frac{p(t)-p_\infty}{p_0-p_\infty}. \]

For a single first-order contribution,

\[ q(t)=e^{-kt}, \qquad \ln q(t)=-kt. \]

Thus, first-order behavior produces a straight line on the given semilogarithmic plot. The two slope regimes indicate two kinetic contributions, each consistent with pseudo-first-order decay.

The correct choice is (b).

6.4

The methane concentration changes by only

\[ \frac{933-925}{933}\times100\%=0.86\%. \]

It can therefore be treated as approximately constant while the reactive copper sites are consumed.

Define \(k_1\) and \(k_2\) by the disappearance of the respective site populations:

\[ -\frac{d[\mathbf{S1}]}{dt}=k_1[\mathbf{S1}], \qquad -\frac{d[\mathbf{S2}]}{dt}=k_2[\mathbf{S2}]. \]

Expressing the amounts per common fixed volume, the stoichiometry from 6.1 gives

\[ -\frac{d[\ce{CH4}]}{dt} =-\frac12\frac{d[\mathbf{S1}]}{dt} -\frac{d[\mathbf{S2}]}{dt}. \]

Hence,

\[ \boxed{ -\frac{d[\ce{CH4}]}{dt} =\frac12 k_1[\mathbf{S1}]+k_2[\mathbf{S2}]. } \]

The factor \(1/2\) accounts for the consumption of two S1 sites per methane molecule. Equivalently, if the constants are defined directly for methane consumption,

\[ v=k'_1[\mathbf{S1}]+k'_2[\mathbf{S2}], \qquad k'_1=\frac{k_1}{2},\quad k'_2=k_2. \]

Any dependence on the approximately constant methane concentration is incorporated into these pseudo-first-order constants. For example, an underlying site-disappearance law \(K_i[\ce{CH4}][\mathbf{S}i]\) would give \(k_i=K_i[\ce{CH4}]_0\). The present experiment does not independently establish the intrinsic order in methane.

To connect this law to Figure 1, let \(S_{1,0}\) and \(S_{2,0}\) denote the initial concentrations of reactive sites. The remaining methane uptake capacity is

\[ [\ce{CH4}](t)-[\ce{CH4}]_\infty =\frac12S_{1,0}e^{-k_1t}+S_{2,0}e^{-k_2t}. \]

Consequently,

\[ \boxed{ q(t)= \frac{\frac12S_{1,0}e^{-k_1t}+S_{2,0}e^{-k_2t}} {\frac12S_{1,0}+S_{2,0}}. } \]

This sum of two exponentials explains the changing slope and the eventual dominance of the slower population.

6.5

  • (a) Correct. The two slope regimes are consistent with two site populations having different rate constants.
  • (b) Correct. The fitted slopes become more negative at higher temperature, indicating faster consumption of the reactive sites and methane.
  • (c) Not supported. The pressure change is normalized by its final magnitude. This removes the information needed to compare the absolute fractions of copper that eventually react at different temperatures.
  • (d) Incorrect. Both fitted kinetic regimes become faster as temperature increases.

Therefore, select (a) and (b).

6.6

At a fixed temperature and under fixed measurement conditions,

\[ I_2(t)=\alpha[\mathbf{S1}](t), \]

where \(\alpha\) is a constant instrumental proportionality factor.

Integrating the first-order site-loss equation gives

\[ \int_{[\mathbf{S1}]_0}^{[\mathbf{S1}](t)} \frac{d[\mathbf{S1}]}{[\mathbf{S1}]} =-k_1\int_0^t dt, \]

so

\[ \ln\!\left( \frac{[\mathbf{S1}](t)}{[\mathbf{S1}]_0} \right)=-k_1t. \]

Substituting the EPR intensities cancels \(\alpha\):

\[ \boxed{ \ln\!\left(\frac{I_2(t)}{I_2(0)}\right)=-k_1t. } \]

Plotting this logarithmic ratio against time gives a straight line with slope \(-k_1\).

6.7

A known Cu(II) standard determines the conversion between integrated EPR signal intensity and the absolute number of spins.

Therefore, calibration is needed for:

  • (a) the total number of paramagnetic Cu(II) sites;
  • (b) their concentration.

For (c), the unknown intensity calibration factor cancels in \(I_2(t)/I_2(0)\), so the rate constant can be obtained without absolute calibration.

For (d), different paramagnetic species are distinguished through spectral features such as resonance positions and hyperfine patterns; absolute spin-count calibration is unnecessary.

Select (a) and (b).

6.8

At 528 K, the late-time fitted segment passes approximately through

\[ (t,y)=(1000\ \mathrm{s},-9.1) \quad\text{and}\quad (2000\ \mathrm{s},-11.7), \]

where \(y=\ln q\). Its decay constant is

\[ \begin{aligned} k_{\mathrm{slow}} &\simeq-\frac{(-11.7)-(-9.1)} {(2000-1000)\ \mathrm{s}}\\ &=2.6\times10^{-3}\ \mathrm{s^{-1}}. \end{aligned} \]

This agrees with the EPR value

\[ k_1=2.604\times10^{-3}\ \mathrm{s^{-1}}, \]

so the slow population is S1.

For the early, steeper segment, the dotted guides give approximately

\[ (t,y)=(200\ \mathrm{s},-5.4) \quad\text{and}\quad (400\ \mathrm{s},-7.0). \]

Thus,

\[ \begin{aligned} k_{\mathrm{fast,app}} &\simeq-\frac{(-7.0)-(-5.4)} {(400-200)\ \mathrm{s}}\\ &=8.0\times10^{-3}\ \mathrm{s^{-1}}. \end{aligned} \]

The second population must therefore be faster than S1:

\[ \boxed{\text{Methane reacts faster with S2: choice (c).}} \]

If S2 dominates the early fitted segment, this gives \(k_2\approx8.0\times10^{-3}\ \mathrm{s^{-1}}\), approximately \(3.1k_1\). The exact separation of the two contributions is unnecessary for deciding which is faster.

6.9

The oxygen mass fraction is

\[ 100.0-55.8-7.0=37.2\%. \]

Combining the elemental analysis with the molar mass gives the approximate numbers of atoms per molecule:

\[ N_{\ce{C}} =\frac{0.558(86.09)}{12.01}\approx4, \]

\[ N_{\ce{H}} =\frac{0.070(86.09)}{1.008}\approx6, \]

\[ N_{\ce{O}} =\frac{0.372(86.09)}{16.00}\approx2. \]

The molecular formula is therefore

\[ \boxed{\ce{C4H6O2}}. \]

Its degree of unsaturation is

\[ \mathrm{DBE}=\frac{2(4)+2-6}{2}=2. \]

The broad, exchangeable signal at 12.2 ppm identifies a carboxylic acid proton. The carboxyl carbonyl accounts for one degree of unsaturation.

The two signals at 6.3 and 5.7 ppm, each integrating to one proton, indicate two nonequivalent terminal alkene hydrogens. The 2.0 ppm singlet integrating to three protons is a methyl group attached to the substituted alkene carbon.

These features give methacrylic acid (2-methylprop-2-enoic acid):

\[ \boxed{\ce{CH2=C(CH3)-C(=O)-OH}} \]

SMILES: C=C(C)C(=O)O.

To draw it, place a terminal methylene group double-bonded to a carbon bearing both a methyl group and a carboxylic acid group.

The assignments are:

  • a, 12.2 ppm: the carboxylic acid \(\ce{O-H}\) proton.
  • d, 2.0 ppm: the three protons of the methyl group attached to the alkene carbon.

The two terminal methylene hydrogens are chemically nonequivalent and can split each other through geminal coupling, accounting for signals b and c. The terminal alkene has no \(E/Z\) configuration.

6.10

The oxygen mass fraction is

\[ 100.00-46.15-2.56=51.29\%. \]

The approximate atom counts are

\[ N_{\ce{C}} =\frac{0.4615(156.03)}{12.01}\approx6, \]

\[ N_{\ce{H}} =\frac{0.0256(156.03)}{1.008}\approx4, \]

\[ N_{\ce{O}} =\frac{0.5129(156.03)}{16.00}\approx5. \]

Thus,

\[ \boxed{\ce{C6H4O5}}, \qquad \mathrm{DBE}=\frac{2(6)+2-4}{2}=5. \]

The two proton signals have equal integrals. With four hydrogens in total, each signal represents two hydrogens.

The broad signal at 13.0 ppm indicates two equivalent carboxylic acid protons. Two carboxyl groups account for four oxygen atoms, two carbon atoms, and two degrees of unsaturation. The remaining fragment contains four carbons and one oxygen, with three degrees of unsaturation: a ring and two double bonds.

A symmetric furan ring bearing two equivalent carboxyl groups satisfies these requirements. A suitable structure is furan-2,5-dicarboxylic acid:

SMILES: O=C(O)c1ccc(C(=O)O)o1.

Draw a five-membered aromatic furan ring, numbering its oxygen as position 1. Attach a carboxylic acid group to each adjacent carbon, positions 2 and 5. The remaining ring carbons, positions 3 and 4, each bear one hydrogen.

The requested assignments are:

  • a, 7.81 ppm: the two equivalent ring hydrogens at positions 3 and 4. Equivalent hydrogens do not split each other, consistent with the singlet.
  • b, 13.0 ppm: the two equivalent carboxylic acid \(\ce{O-H}\) protons.
  • Carbon signal 1, 165.1 ppm: the two equivalent carboxyl carbonyl carbons.

The remaining carbon signals are also consistent:

  • 150.6 ppm, signal 2: ring carbons 2 and 5, directly bonded to the ring oxygen and bearing the carboxyl substituents.
  • 120.6 ppm, signal 3: proton-bearing ring carbons 3 and 4.

The solution was checked against the official solution.