IChO 2017 · Theory problem 2

Kinetic isotope effect (KIE) and zero-point vibrational energy (ZPE)

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Calculation of ZPE and KIE

Kinetic isotope effect (KIE) is a phenomenon associated with a change in rate constant of the reaction when one of the atoms is replaced by its isotope. KIE can be used to confirm whether a particular bond to hydrogen is broken in the reaction. Harmonic oscillator model is used to estimate the difference in the rate between \(\ce{C-H}\) and \(\ce{C-D}\) bond activation (D – deuterium).

The vibrational frequency (\(\nu\)) represented by harmonic oscillator model is

\[ \nu=\frac{1}{2\pi}\sqrt{\frac{k}{\mu}} \]

where \(k\) is the force constant and \(\mu\) is the reduced mass.

The vibrational energies of the molecule are given by

\[ E_n=\left(n+\frac{1}{2}\right)h\nu \]

where \(n\) is vibrational quantum number with possible values of 0, 1, 2, ... The energy of the lowest vibrational energy level (\(E_n\) at \(n=0\)) is called zero-point vibrational energy (ZPE).

2.1 Calculate the reduced mass of \(\ce{C-H}\) (\(\mu_{CH}\)) and \(\ce{C-D}\) (\(\mu_{CD}\)) in atomic mass unit.

Assume that the mass of deuterium is twice that of hydrogen.

Note: If a student is unable to calculate the values for \(\mu_{CH}\) and \(\mu_{CD}\) he/she can use \(\mu_{CH}=1.008\) and \(\mu_{CD}=2.016\) for the subsequent parts of the question. The given values are not necessarily be close to the correct values.

2.2 Given that the force constant (\(k\)) for \(\ce{C-H}\) stretching is the same as that for the \(\ce{C-D}\) stretching and the \(\ce{C-H}\) stretching frequency is \(2900\ \mathrm{cm^{-1}}\), find the corresponding \(\ce{C-D}\) stretching frequency (in \(\mathrm{cm^{-1}}\)).

2.3 According to the \(\ce{C-H}\) and \(\ce{C-D}\) stretching frequencies in question 2.2, calculate the zero-point vibrational energies (ZPE) of \(\ce{C-H}\) and \(\ce{C-D}\) stretching in \(\mathrm{kJ\,mol^{-1}}\).

Note: If a student is unable to calculate the values for ZPE in 2.3, he/she can use use \(ZPE_{CH}=7.23\ \mathrm{kJ\,mol^{-1}}\) and \(ZPE_{CD}=2.15\ \mathrm{kJ\,mol^{-1}}\) for the subsequent parts of the question. The given values are not necessarily be close to the correct values.

Kinetic isotope effect (KIE)

Due to the difference in zero-point vibrational energies, a protonated compound and its corresponding deuterated compounds are expected to react at different rates.

For the \(\ce{C-H}\) and \(\ce{C-D}\) bond dissociation reactions, the energies of both transition states and both products are identical. Then, the isotope effect is controlled by the difference in the ZPE's of the \(\ce{C-H}\) and \(\ce{C-D}\) bonds.

2.4 Calculate the difference in the bond dissociation energy (BDE) between \(\ce{C-D}\) bond and \(\ce{C-H}\) bond (\(BDE_{CD}-BDE_{CH}\)) in \(\mathrm{kJ\,mol^{-1}}\).

2.5 Assume that the activation energy (\(E_a\)) for the \(\ce{C-H}/\ce{C-D}\) bond cleavage is approximately equal to the bond dissociation energy and the Arrhenius factor is the same for both \(\ce{C-H}\) and \(\ce{C-D}\) bond cleavage. Find the relative rate constant for the \(\ce{C-H}/\ce{C-D}\) bond cleavage (\(k_{CH}/k_{CD}\)) at \(25\,^\circ\mathrm{C}\).

Using KIE to study reaction mechanism

The oxidation of nondeuterated and deuterated diphenylmethanol using an excess of chromic acid was studied.

Figure 1

2.6 Let \(C_0\) be the initial concentration of either nondeuterated diphenylmethanol or deuterated diphenylmethanol and \(C_t\) its concentration at time \(t\). The experiment led to two plots (Figure 2a and Figure 2b), from which the first-order rate constant can be determined.

Figure 2

Figure 2a

Figure 3

Figure 2b

Which plot should be for the oxidation of nondeuterated diphenylmethanol and which one is for the oxidation of deuterated diphenylmethanol?

For each statement, select your answer by marking “✓” in one of the square.

The oxidation of nondeuterated diphenylmethanol:

☐ Figure 2a
☐ Figure 2b

The oxidation of deuterated

☐ Figure 2a
☐ Figure 2b

2.7 Determine \(k_{CH}\), \(k_{CD}\) (in \(\mathrm{min^{-1}}\)), and the \(k_{CH}/k_{CD}\) of this reaction from the plots in question 2.6.

2.8 The mechanism has been proposed as follows:

Figure 4

According to the information in 2.6 and 2.7, which step should be the rate determining step?

Select your answer by marking “✓” in one of the corresponding squares:

☐ Step (1)
☐ Step (2)
☐ Step (3)

Official English text of the problem.

Answers

  • 2.1: \(\boxed{\mu_{CH}=0.9299\ \mathrm{u},\quad \mu_{CD}=1.726\ \mathrm{u}}\).
  • 2.2: \(\boxed{\tilde{\nu}_{CD}=2.13\times10^3\ \mathrm{cm^{-1}}}\).
  • 2.3: \(\boxed{ZPE_{CH}=17.35\ \mathrm{kJ\,mol^{-1}},\quad ZPE_{CD}=12.73\ \mathrm{kJ\,mol^{-1}}}\).
  • 2.4: \(\boxed{BDE_{CD}-BDE_{CH}=+4.61\ \mathrm{kJ\,mol^{-1}}}\).
  • 2.5: \(\boxed{k_{CH}/k_{CD}=6.43}\) at \(25\,^\circ\mathrm{C}\).
  • 2.6: Nondeuterated diphenylmethanol: Figure 2b. Deuterated diphenylmethanol: Figure 2a.
  • 2.7: \(\boxed{k_{CH}\approx1.2\times10^{-2}\ \mathrm{min^{-1}},\quad k_{CD}\approx1.8\times10^{-3}\ \mathrm{min^{-1}},\quad k_{CH}/k_{CD}\approx6.7}\).
  • 2.8: Step (3) is the rate determining step.

Solution

The calculations use the standard atomic masses

\[ m_C=12.01\ \mathrm{u},\qquad m_H=1.008\ \mathrm{u},\qquad m_D=2m_H=2.016\ \mathrm{u}, \]

and the constants

\[ \begin{aligned} h&=6.62607015\times10^{-34}\ \mathrm{J\,s},\\ c&=2.99792458\times10^{10}\ \mathrm{cm\,s^{-1}},\\ N_A&=6.02214076\times10^{23}\ \mathrm{mol^{-1}},\\ R&=8.314462618\ \mathrm{J\,mol^{-1}\,K^{-1}}. \end{aligned} \]

Here \(\tilde{\nu}=\nu/c\) denotes the spectroscopic wavenumber in \(\mathrm{cm^{-1}}\).

2.1 — Reduced masses

For two bonded atoms,

\[ \mu=\frac{m_1m_2}{m_1+m_2}. \]

Therefore,

\[ \mu_{CH} =\frac{(12.01)(1.008)}{12.01+1.008}\ \mathrm{u} =\boxed{0.9299\ \mathrm{u}}, \]

\[ \mu_{CD} =\frac{(12.01)(2.016)}{12.01+2.016}\ \mathrm{u} =\boxed{1.726\ \mathrm{u}}. \]

2.2 — \(\ce{C-D}\) stretching wavenumber

For equal force constants, both frequency and wavenumber are proportional to \(\mu^{-1/2}\). Thus,

\[ \frac{\tilde{\nu}_{CD}}{\tilde{\nu}_{CH}} =\sqrt{\frac{\mu_{CH}}{\mu_{CD}}}. \]

Using unrounded reduced masses,

\[ \tilde{\nu}_{CD} =(2900\ \mathrm{cm^{-1}}) \sqrt{\frac{0.9299493}{1.7262341}} =2128.52\ \mathrm{cm^{-1}}. \]

Hence,

\[ \boxed{\tilde{\nu}_{CD}\approx2.13\times10^3\ \mathrm{cm^{-1}}}. \]

The greater reduced mass lowers the stretching frequency.

2.3 — Zero-point vibrational energies

At \(n=0\), the energy per oscillator is \(\tfrac12h\nu\). The molar zero-point energy is therefore

\[ ZPE=\frac12N_Ah\nu=\frac12N_Ahc\tilde{\nu}. \]

With \(c\) expressed in \(\mathrm{cm\,s^{-1}}\),

\[ \frac12N_Ahc =5.9813283\times10^{-3}\ \mathrm{kJ\,cm\,mol^{-1}}. \]

Consequently,

\[ ZPE_{CH} =(5.9813283\times10^{-3})(2900) =\boxed{17.35\ \mathrm{kJ\,mol^{-1}}}, \]

\[ ZPE_{CD} =(5.9813283\times10^{-3})(2128.52) =\boxed{12.73\ \mathrm{kJ\,mol^{-1}}}. \]

The unrounded values, \(17.34585\) and \(12.73138\ \mathrm{kJ\,mol^{-1}}\), are retained for the following calculations.

2.4 — Difference in bond dissociation energies

Let \(D_e\) be the energy difference between the minimum of the bond potential and the dissociated products. Under the stated assumptions, \(D_e\) is the same for both isotopes. Dissociation begins from the vibrational ground state, so

\[ BDE_{CH}=D_e-ZPE_{CH}, \qquad BDE_{CD}=D_e-ZPE_{CD}. \]

Subtracting gives

\[ \begin{aligned} BDE_{CD}-BDE_{CH} &=ZPE_{CH}-ZPE_{CD}\\ &=17.34585-12.73138\\ &=\boxed{+4.61\ \mathrm{kJ\,mol^{-1}}}. \end{aligned} \]

The \(\ce{C-D}\) bond requires more energy to dissociate because its vibrational ground state lies lower.

2.5 — Relative rate constant at \(25\,^\circ\mathrm{C}\)

The Arrhenius equation is

\[ k=A\exp\left(-\frac{E_a}{RT}\right). \]

With equal pre-exponential factors,

\[ \frac{k_{CH}}{k_{CD}} =\exp\left(\frac{E_{a,CD}-E_{a,CH}}{RT}\right). \]

Using \(E_a\approx BDE\), the activation-energy difference is \(4.61447\ \mathrm{kJ\,mol^{-1}}\). At \(T=298.15\ \mathrm{K}\),

\[ \frac{k_{CH}}{k_{CD}} =\exp\left[ \frac{4.61447\times10^3\ \mathrm{J\,mol^{-1}}} {(8.314462618\ \mathrm{J\,mol^{-1}\,K^{-1}})(298.15\ \mathrm{K})} \right] =\boxed{6.43}. \]

Thus, this model predicts that \(\ce{C-H}\) cleavage is about \(6.4\) times faster.

2.6 — Identification of the plots

The organic compounds are:

Compound SMILES
Nondeuterated diphenylmethanol, also called benzhydrol OC(c1ccccc1)c1ccccc1
Diphenylmethanol deuterated at the methine carbon OC([2H])(c1ccccc1)c1ccccc1
Benzophenone, the oxidation product O=C(c1ccccc1)c1ccccc1

The deuterium is attached to carbon, as shown in the reaction scheme. Neither substrate has a stereogenic center because the two phenyl groups are identical.

For first-order disappearance,

\[ -\frac{dC}{dt}=kC, \qquad \ln\left(\frac{C_0}{C_t}\right)=kt. \]

Therefore, the slope of each graph is its rate constant. With the oxidant in excess, these are observed pseudo-first-order rate constants.

Figure 2b reaches the same value of \(\ln(C_0/C_t)\) in a much shorter time than Figure 2a. It therefore represents the faster, nondeuterated substrate.

The answer-sheet selections are:

  • Nondeuterated diphenylmethanol: mark Figure 2b.
  • Deuterated diphenylmethanol: mark Figure 2a.

2.7 — Rate constants from the plots

Using the origin and the clearly readable points near \(\ln(C_0/C_t)=0.70\):

For Figure 2b,

\[ k_{CH}\approx \frac{0.70}{60\ \mathrm{min}} =1.17\times10^{-2}\ \mathrm{min^{-1}} \approx\boxed{1.2\times10^{-2}\ \mathrm{min^{-1}}}. \]

For Figure 2a,

\[ k_{CD}\approx \frac{0.70}{400\ \mathrm{min}} =1.75\times10^{-3}\ \mathrm{min^{-1}} \approx\boxed{1.8\times10^{-3}\ \mathrm{min^{-1}}}. \]

Taking the ratio before rounding,

\[ \frac{k_{CH}}{k_{CD}} \approx \frac{0.70/60}{0.70/400} =\frac{400}{60} =\boxed{6.7}. \]

These are graphical estimates for the oxidation experiment at \(20\,^\circ\mathrm{C}\).

2.8 — Rate determining step

The large normal isotope effect, \(k_{CH}/k_{CD}\approx6.7\), indicates that cleavage of the carbon–hydrogen or carbon–deuterium bond contributes directly to the rate-controlling transition state.

In the proposed mechanism:

  • Step (1) is an inorganic acid–base equilibrium and involves no substrate \(\ce{C-H}\) bond.
  • Step (2) forms the chromate ester while retaining the methine \(\ce{C-H}\) bond.
  • Step (3) breaks that \(\ce{C-H}\) bond as the carbonyl group forms.

The chromate ester is diphenylmethyl hydrogen chromate(VI), O=[Cr](=O)(O)OC(c1ccccc1)c1ccccc1. Its elimination is represented by the balanced equation

\[ \ce{(C6H5)2CH-O-Cr(=O)2-OH + H2O -> (C6H5)2C=O + H3O+ + HCrO3-}. \]

The additional species are water, O; hydronium, [OH3+]; and hydrogen trioxidochromate(IV), \(\ce{HCrO3^-}\), [O-][Cr](=O)O.

In the proposed elimination, water accepts the methine hydrogen, the electrons of the \(\ce{C-H}\) bond form the carbonyl \(\pi\) bond, and the \(\ce{O-Cr}\) bond breaks with its electron pair transferred toward chromium. The chromium SMILES specify formal connectivity; they do not encode this electron flow, coordination geometry, or spin state.

Accordingly, within the proposed mechanism, mark

\[ \boxed{\text{Step (3)}}. \]

The solution was checked against the official solution.