IChO 1985 · Theory problem 4

Kinetics of the racemisation of alpha-pinene

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At a temperature around \(200\,^\circ\mathrm{C}\) the racemisation of pinene can be followed in the gaseous phase by measuring the optical rotation.

If, for example, you take the (+)-enantiomer of α-pinene

Figure 1

an equilibrium is gradually established between the two enantiomers (optical isomers). The two opposing reactions are both of the first order.

In 1927 D. F. Smith obtained the following data in his study of racemisation of α-pinene:

\(T/\mathrm{K}\) \(\alpha_1\) \(\alpha_2\) \(t/\mathrm{min}\)
490.9 32.75 18.01 579
490.9 29.51 15.59 587
503.9 30.64 8.74 371
505.4 12.95 8.05 120
510.1 23.22 6.15 216

\(\alpha_1\) and \(\alpha_2\) are the values for optical rotation in terms of the dimensions of the polarimeter scale; \(t\) is the time which has elapsed between the two measurements.

Problems:

4.1 What is the value for the equilibrium constant for the racemisation?

What is the corresponding value of \(\Delta_{\mathrm r}G^\circ\) (racemisation)?

What is the relationship between the forward and backward rate constants, \(k_1\) and \(k_{-1}\), in a state of dynamic equilibrium?

4.2 State the rate equation for the racemisation of pinene.

Derive a relationship which could be used to calculate the rate constant for the conversion of the (+)-enantiomer into the (-)-enantiomer using the data given in the table.

4.3 Calculate the rate constant for this reaction at the four temperatures given in the table.

4.4 Calculate the average value of the activation energy for this reaction. You should take the average of the values at a minimum of three temperatures or use a graphical method.

HINT:

If the loss of concentration of a substance obeys the rate equation:

\[ -\frac{dc}{dt}=k(2c-\text{constant}) \]

Then the dependence of concentration on time is given by:

\[ \ln\frac{2c_0-\text{constant}}{2c-\text{constant}}=2kt \]

where \(c_0\) is the initial concentration at time \(t=0\).

Official English text of the problem.

Answers

  • 4.1: \(K=1\), \(\Delta_{\mathrm r}G^\circ=0\), and \(k_1=k_{-1}\equiv k\).
  • 4.2: With \(c_+\) and \(c_-\) the enantiomer concentrations and \(C=c_++c_-\), \[ -\frac{dc_+}{dt}=k(c_+-c_-)=k(2c_+-C), \qquad \boxed{k=\frac{1}{2t}\ln\frac{\alpha_1}{\alpha_2}}. \]
  • 4.3: The one-way conversion constants are:
\(T/\mathrm{K}\) \(k/\mathrm{min^{-1}}\) \(k/\mathrm{s^{-1}}\)
490.9 \(5.30\times10^{-4}\) \(8.83\times10^{-6}\)
503.9 \(1.69\times10^{-3}\) \(2.82\times10^{-5}\)
505.4 \(1.98\times10^{-3}\) \(3.30\times10^{-5}\)
510.1 \(3.08\times10^{-3}\) \(5.13\times10^{-5}\)

The value at \(490.9\ \mathrm K\) is the mean of the two determinations.

  • 4.4: An Arrhenius straight-line fit using all four temperatures gives \[ \boxed{E_{\mathrm a}\approx189\ \mathrm{kJ\,mol^{-1}}}. \]

Solution

4.1

The two species are:

  • (+)-(1R,5R)-α-pinene, denoted \(P_+\): CC1=CC[C@H]2C[C@@H]1C2(C)C.
  • (-)-(1S,5S)-α-pinene, denoted \(P_-\): CC1=CC[C@@H]2C[C@H]1C2(C)C.

The SMILES specify absolute configuration; the \(+\) and \(-\) signs specify measured optical rotation, whose sign is not encoded in SMILES.

Their balanced interconversion is

\[ \ce{(+)-C10H16 <=>[k_1][k_{-1}] (-)-C10H16}. \]

Enantiomers have equal standard molar Gibbs energies in an achiral environment. Consequently,

\[ \Delta_{\mathrm r}G^\circ =\mu_-^\circ-\mu_+^\circ=0. \]

Using

\[ \Delta_{\mathrm r}G^\circ=-RT\ln K \]

gives

\[ \boxed{K=1}. \]

Thus the equilibrium mixture contains equal amounts of the two enantiomers:

\[ c_{+,\mathrm{eq}}=c_{-,\mathrm{eq}}. \]

At dynamic equilibrium the opposing reaction rates are equal:

\[ k_1c_{+,\mathrm{eq}}=k_{-1}c_{-,\mathrm{eq}}. \]

Therefore,

\[ \boxed{k_1=k_{-1}\equiv k}. \]

4.2

The net rate of disappearance of the (+)-enantiomer is

\[ -\frac{dc_+}{dt} =k_1c_+-k_{-1}c_-. \]

Since both rate constants equal \(k\),

\[ -\frac{dc_+}{dt}=k(c_+-c_-). \]

The total concentration is conserved:

\[ C=c_++c_-, \qquad c_-=C-c_+. \]

Hence the required concentration rate equation is

\[ \boxed{-\frac{dc_+}{dt}=k(2c_+-C)}. \]

At a fixed temperature and optical path length, the two enantiomers contribute equal and opposite rotations. Therefore, for a proportionality constant \(\beta\),

\[ \alpha=\beta(c_+-c_-)=\beta(2c_+-C). \]

Differentiating and substituting the rate equation,

\[ \frac{d\alpha}{dt} =2\beta\frac{dc_+}{dt} =-2k\beta(2c_+-C) =-2k\alpha. \]

Integration between two measurements separated by time \(t\) gives

\[ \ln\frac{\alpha_2}{\alpha_1}=-2kt, \]

or

\[ \boxed{\alpha_2=\alpha_1e^{-2kt}}, \qquad \boxed{k=\frac{1}{2t}\ln\frac{\alpha_1}{\alpha_2}}. \]

The factor of two is essential: the optical rotation decays with constant \(k_1+k_{-1}=2k\), whereas the requested one-way conversion constant is \(k\). Neither measurement needs to correspond to an initially pure enantiomer.

4.3

For example, the first experiment gives

\[ k= \frac{\ln(32.75/18.01)}{2(579\ \mathrm{min})} =5.16387\times10^{-4}\ \mathrm{min^{-1}}. \]

Applying the same expression to every row gives the following intermediate values:

\(T/\mathrm{K}\) \(t/\mathrm{min}\) \(\ln(\alpha_1/\alpha_2)\) \(k/\mathrm{min^{-1}}\)
490.9 579 0.597976 \(5.16387\times10^{-4}\)
490.9 587 0.638100 \(5.43526\times10^{-4}\)
503.9 371 1.254396 \(1.69056\times10^{-3}\)
505.4 120 0.475424 \(1.98093\times10^{-3}\)
510.1 216 1.328562 \(3.07537\times10^{-3}\)

The arithmetic mean of the two measurements at \(490.9\ \mathrm K\) is

\[ \begin{aligned} \overline{k} &=\frac{5.16387\times10^{-4} +5.43526\times10^{-4}}{2}\ \mathrm{min^{-1}}\\ &=5.29956\times10^{-4}\ \mathrm{min^{-1}}\\ &\approx\boxed{5.30\times10^{-4}\ \mathrm{min^{-1}}}. \end{aligned} \]

The other three values, rounded to three significant figures, are

\[ \boxed{ \begin{aligned} k(503.9\ \mathrm K)&=1.69\times10^{-3}\ \mathrm{min^{-1}},\\ k(505.4\ \mathrm K)&=1.98\times10^{-3}\ \mathrm{min^{-1}},\\ k(510.1\ \mathrm K)&=3.08\times10^{-3}\ \mathrm{min^{-1}}. \end{aligned}} \]

Division by \(60\) converts these values to \(\mathrm{s^{-1}}\).

4.4

The Arrhenius equation is

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

Thus a plot of

\[ y=\ln\left(\frac{k}{1\ \mathrm{min^{-1}}}\right) \quad\text{against}\quad x=\frac1T \]

should be a straight line with slope

\[ m=-\frac{E_{\mathrm a}}R. \]

Using the mean rate constant at \(490.9\ \mathrm K\), the four plotting points are:

\(1/T\;(\mathrm{K^{-1}})\) \(\ln[k/(1\ \mathrm{min^{-1}})]\)
0.002037075 −7.542716
0.001984521 −6.382695
0.001978631 −6.224188
0.001960400 −5.784329

The points lie close to a descending straight line. An unweighted least-squares fit, implementing the graphical method numerically, gives

\[ y=38.8473-(2.27775\times10^4\ \mathrm K)x. \]

Using \(R=8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}\),

\[ \begin{aligned} E_{\mathrm a} &=-Rm\\ &=(8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}) (2.27775\times10^4\ \mathrm K)\\ &=1.89372\times10^5\ \mathrm{J\,mol^{-1}}. \end{aligned} \]

Therefore,

\[ \boxed{E_{\mathrm a}\approx189\ \mathrm{kJ\,mol^{-1}}}. \]

The solution was checked against the official solution.