IChO 2013 · Theory problem 3

Meerwein-Schmidt-Ponndorf-Verley reaction

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In the table for 3.3, A–F are editorial labels identifying the six structures in reading order.

PROBLEM 3

Meerwein-Schmidt-Ponndorf-Verley reaction

Meerwein-Schmidt-Ponndorf-Verley (MSPV) reaction is a useful tool for reduction of carbonyl compounds to alcohols. The reaction is the reduction of carbonyl compounds by low molecular weight alcohols in the presence of alkoxides of aluminium or other metals:

Figure 1

The mechanism of the reaction includes coordination of carbonyl compound by aluminium alkoxide, hydride transfer in the inner sphere of the complex and subsequent transalkoxylation. It can be schematically represented as follows (transalkoxylation is shown as a one-step process for brevity):

Figure 2

The reaction is reversible and shifting the equilibrium to the desired product requires some excess of the reductant. In some cases (e. g. in the case of reduction of aromatic aldehydes and ketones) the equilibrium constant is so large that the reverse reaction can be neglected.

The table below contains standard entropies and standard enthalpies of formation of liquid substances at 298 K. The boiling points of the substances at 1 bar are also given.

Substance \(\Delta_{\mathrm f}H^\circ_{298}\), kJ mol\(^{-1}\) \(S^\circ_{298}\), J mol\(^{-1}\) K\(^{-1}\) \(t_{\mathrm{vap}}\), °C
Acetone −248.4 200.4 56
Isopropanol −318.1 180.6 82
Cyclohexanone −271.2 229.0 156
Cyclohexanol −348.2 203.4 161

3.1 a Calculate the minimum isopropanol to cyclohexanone mass ratio which is required to reach a 99 % yield of reaction at 298 K. Assume that:

i) the reaction mixture eventually gets at equilibrium,

ii) no products are initially present.

3.1 b Choose the appropriate way(s) to increase the cyclohexanol yield. Tick the relevant boxes.

Increase the temperature up to 50 °C using a reflux
Increase the temperature up to 60 °C, evaporating (distilling) the acetone
Add some ethanol to the reaction mixture
Add some ethanal to the reaction mixture

3.2 Often the rate-limiting step in the MSPV reaction is the hydride transfer or the alcoholysis of the alkoxide after hydride transfer. For these two cases, using the above mechanism (2), derive an expression for the rate of reaction as a function of current concentrations of a carbonyl compound, isopropanol and a catalyst. In both cases determine the rate orders in the reactants and the catalyst. Assume that all reaction steps before the limiting step are fast and reversible. Use equilibrium approximation, if necessary. For brevity use the following notation: A for carbonyl compound, B for isopropanol, C for catalyst. Denote intermediates as you wish.

MSPV reaction can be used to obtain chiral alcohols, if the chiral catalyst is employed. For instance, Campbell et al. used the catalyst based on the chiral 2,2’-dihydroxy-1,1’-binaphtyl (BINOL), which is synthesized in situ from binaphtol and trimethylaluminium:

Figure 3

The chirality of BINOL is due to the sterically hindered rotation around the \(\ce{C-C}\) bond. Though perfectly stable at room temperature, BINOL may racemize when heated.

3.3 Which of the phenols below can form stable (at room temperature) enantiomers so that they can be used in the same fashion to produce a chiral catalyst? Tick the relevant boxes.

Substance Can be used Substance Can be used
A B
C D
E F

A:

Figure 4

B:

Figure 5

C:

Figure 6

D:

Figure 7

E:

Figure 8

F:

Figure 9

Enantiomeric excess, \(ee\), is used to characterize the enantiomeric purity of the substance. This quantity equals to the ratio of the difference of concentrations of enantiomers \(R\) and \(S\) to their sum:

\[ ee=\frac{[R]-[S]}{[R]+[S]} \]

Enantiomeric excess of the pure \(R\) isomer is unity, \(ee\) of the racemic mixture is zero.

3.4 When using the enantiomerically pure \(\ce{(BINOL)Al(OiPr)}\) as a catalyst for reduction of \(\alpha\)-bromoacetophenone, the \(ee\) of the product equals 81 %. What is the \(ee\) of the product if the catalyst \(ee\) equals 50 %? Provide your calculation with an illustration or derivation of the final formula.

Official English text of the problem.

Answers

  • 3.1 a: The minimum initial mass ratio is \[ \boxed{\frac{m(\text{isopropanol})}{m(\text{cyclohexanone})}\approx6.94:1}. \]

  • 3.1 b: Tick distilling off acetone at 60 °C and adding ethanol. Leave the other two boxes unticked.

  • 3.2: With \([C]\) denoting free catalyst:

  • Hydride transfer limiting: \[ \boxed{v=k_{\mathrm H}K_1[A][C]},\qquad \boxed{(n_A,n_B,n_C)=(1,0,1)}. \]

  • Alcoholysis limiting: \[ \boxed{v=k_{\mathrm{alc}}K_1K_2[A][B][C]},\qquad \boxed{(n_A,n_B,n_C)=(1,1,1)}. \]

  • 3.3: Tick A, C and D: the top-left, middle-left and middle-right structures.

  • 3.4: \[ \boxed{ee_{\mathrm{product}}=0.81(0.50)=0.405=40.5\%}. \]

Solution

3.1 a

The reacting substances are:

  • Cyclohexanone: O=C1CCCCC1
  • Isopropanol, or propan-2-ol: CC(O)C
  • Cyclohexanol: OC1CCCCC1
  • Acetone, or propan-2-one: CC(=O)C

The balanced reaction is

\[ \ce{C6H10O(l) + C3H8O(l) <=> C6H12O(l) + C3H6O(l)}. \]

From the supplied formation enthalpies,

\[ \begin{aligned} \Delta_{\mathrm r}H^\circ &=(-348.2-248.4)-(-271.2-318.1)\\ &=-7.3\ \mathrm{kJ\,mol^{-1}}. \end{aligned} \]

Similarly,

\[ \begin{aligned} \Delta_{\mathrm r}S^\circ &=(203.4+200.4)-(229.0+180.6)\\ &=-5.8\ \mathrm{J\,mol^{-1}\,K^{-1}}. \end{aligned} \]

Consequently, at 298 K,

\[ \begin{aligned} \Delta_{\mathrm r}G^\circ &=\Delta_{\mathrm r}H^\circ-T\Delta_{\mathrm r}S^\circ\\ &=-7.300-298(-0.0058)\\ &=-5.5716\ \mathrm{kJ\,mol^{-1}}. \end{aligned} \]

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

\[ K=\exp\left(-\frac{\Delta_{\mathrm r}G^\circ}{RT}\right) =\exp\left(\frac{5571.6}{8.314(298)}\right) =9.4765. \]

Assume ideal liquid-mixture behaviour, so activities can be represented by mole fractions. Because the reaction has equal total stoichiometric coefficients on both sides, the total-mole factors cancel.

Take initially \(1.000\) mol cyclohexanone and \(b\) mol isopropanol. At 99 % conversion:

Substance Initial amount / mol Equilibrium amount / mol
Cyclohexanone 1.000 0.010
Isopropanol \(b\) \(b-0.990\)
Cyclohexanol 0 0.990
Acetone 0 0.990

Thus,

\[ K=\frac{(0.990)^2}{0.010(b-0.990)}. \]

Solving,

\[ b=0.990+\frac{0.990^2}{0.010(9.4765)} =11.3324. \]

Using molar masses \(60.10\ \mathrm{g\,mol^{-1}}\) for isopropanol and \(98.15\ \mathrm{g\,mol^{-1}}\) for cyclohexanone,

\[ \frac{m(\text{isopropanol})}{m(\text{cyclohexanone})} =\frac{11.3324(60.10)}{1.000(98.15)} =6.939. \]

Therefore,

\[ \boxed{m(\text{isopropanol}):m(\text{cyclohexanone})\approx6.94:1}. \]

This is the minimum because any smaller initial isopropanol charge gives an equilibrium conversion below 99 %.

3.1 b

Proposed change Tick? Reason
Heat to 50 °C under reflux No The forward reaction is exothermic, so heating decreases its equilibrium constant. Reflux returns volatile substances to the mixture.
Heat to 60 °C while distilling acetone Yes Removing acetone, a reaction product, shifts the reaction towards cyclohexanol. Acetone has the lowest listed boiling point, 56 °C.
Add ethanol Yes Ethanol supplies an additional reducing alcohol and is oxidized to ethanal.
Add ethanal No Ethanal is a hydrogen acceptor and promotes oxidation of alcohols, opposing the desired reduction.

The additional substances are ethanol CCO and ethanal, or acetaldehyde, CC=O.

Their coupled transfer-hydrogenation reaction is

\[ \ce{C6H10O + CH3CH2OH <=> C6H11OH + CH3CHO}. \]

Adding ethanol favours the forward direction. Adding ethanal favours the reverse direction, consuming cyclohexanol.

For heating alone, the sign also follows from the van ’t Hoff equation:

\[ \frac{\mathrm d\ln K}{\mathrm dT} =\frac{\Delta_{\mathrm r}H^\circ}{RT^2}<0. \]

3.2

Let \(P\) denote the product alcohol and \(Q\) acetone. The mechanism can be written schematically as

\[ \ce{A + C <=> I_1}, \]

\[ \ce{I_1 <=> I_2}, \]

\[ \ce{I_2 + B <=> C + P + Q}. \]

Here:

  • \(A\) is the carbonyl substrate, generic SMILES [*:1]C(=O)[*:2].
  • \(B\) is isopropanol, CC(O)C.
  • \(C\) is aluminium triisopropoxide, CC(C)O[Al](OC(C)C)OC(C)C.
  • \(P\) is the corresponding alcohol, generic SMILES [*:1]C(O)[*:2].
  • \(Q\) is acetone, CC(=O)C.
  • \(I_1\) is the carbonyl–aluminium coordination complex, represented schematically by CC(C)O[Al](OC(C)C)(OC(C)C)<-O=C([*:1])[*:2].
  • \(I_2\) is the mixed aluminium alkoxide with coordinated acetone after hydride transfer, represented by CC(C)O[Al](OC(C)C)(OC([*:1])[*:2])<-O=C(C)C.

The wildcard groups denote the substrate substituents. These SMILES describe the monomeric connectivity used in the problem; <- denotes oxygen-to-aluminium coordination. They do not show the cyclic arrangement during hydride transfer.

In that transfer, the coordinated isopropoxide donates its carbon-bound hydrogen to the substrate carbonyl carbon while becoming acetone. The substrate carbonyl oxygen becomes an alkoxide oxygen. Free hydride need not be released.

Define the pre-equilibrium constants

\[ K_1=\frac{[I_1]}{[A][C]}, \qquad K_2=\frac{[I_2]}{[I_1]}. \]

Hydride transfer is rate-limiting.

The first coordination step is at equilibrium:

\[ [I_1]=K_1[A][C]. \]

Hydride transfer occurs within \(I_1\), so its forward rate is

\[ v=k_{\mathrm H}[I_1] =k_{\mathrm H}K_1[A][C]. \]

Hence the orders are

\[ \boxed{n_A=1,\qquad n_B=0,\qquad n_C=1}. \]

Free isopropanol does not participate in this limiting step: the hydrogen donor is already an isopropoxide ligand on the catalyst.

Alcoholysis is rate-limiting.

Both preceding steps are at equilibrium:

\[ [I_2]=K_2[I_1]=K_1K_2[A][C]. \]

The limiting step involves \(I_2\) and one molecule of isopropanol:

\[ v=k_{\mathrm{alc}}[I_2][B] =k_{\mathrm{alc}}K_1K_2[A][B][C]. \]

Therefore,

\[ \boxed{n_A=1,\qquad n_B=1,\qquad n_C=1}. \]

These expressions refer to free catalyst concentration and to forward rates, or net rates when the reverse reaction is negligible.

If catalyst concentration instead means total catalyst loading \(C_T\), binding can produce saturation. For hydride transfer limiting, neglecting the rapidly consumed downstream intermediate,

\[ C_T\simeq[C]+[I_1], \qquad v=\frac{k_{\mathrm H}K_1[A]C_T}{1+K_1[A]}. \]

For alcoholysis limiting,

\[ C_T=[C]+[I_1]+[I_2], \qquad v=\frac{k_{\mathrm{alc}}K_1K_2[A][B]C_T} {1+K_1[A](1+K_2)}. \]

Thus the reaction remains first order in total catalyst, while its apparent substrate order can decrease from one towards zero as binding saturates.

3.3

Stable biaryl enantiomers require both:

  1. A stereogenic axis: the two paths around the aromatic framework at each end of the axis must be distinguishable.
  2. Sufficiently hindered rotation to prevent rapid interconversion at room temperature.

The following SMILES specify connectivity. Ordinary SMILES do not reliably distinguish these biaryl axial configurations; the two stable atropisomers must additionally be designated \((R_a)\) and \((S_a)\). The flat drawings do not select one of them.

  • A — top left: tick.
    2′,3′-Dimethoxy-[1,1′-binaphthalene]-2,3-diol, c1(-c2c(OC)c(OC)cc3ccccc23)c(O)c(O)cc2ccccc12.
    Both ends of the binaphthyl axis have distinguishable ring paths. The crowded ortho and peri environments strongly hinder rotation, allowing stable atropisomers.

  • B — top right: leave unticked.
    1-(Anthracen-9-yl)naphthalene-2,3-diol, c1(-c2c3ccccc3cc3ccccc23)c(O)c(O)cc2ccccc12.
    The two paths through the unsubstituted anthracen-9-yl group are symmetry-equivalent. Thus the connecting bond does not constitute a stereogenic axis, despite its steric crowding.

  • C — middle left: tick.
    9-(2,3-Dimethoxynaphthalen-1-yl)anthracene-2,3-diol, c1(-c2c(OC)c(OC)cc3ccccc23)c2cc(O)c(O)cc2cc2ccccc12.
    Unlike B, the anthracene fragment has hydroxyl substituents on only one terminal ring, making its two paths different. The naphthalene end also has distinct paths, and rotation is strongly hindered.

  • D — middle right: tick.
    6,6′-Dimethoxy-[1,1′-biphenyl]-2,2′-diol, COc1cccc(O)c1-c1c(O)cccc1OC.
    Each ring has different ortho substituents, hydroxyl and methoxy, so the axis is stereogenic. Occupation of all four ortho positions provides the rotational hindrance needed for stable enantiomers.

  • E — bottom left: leave unticked.
    [1,1′-Biphenyl]-2,2′-diol, Oc1ccccc1-c1ccccc1O.
    Twisted conformations can be chiral, but the two hydroxyl groups alone do not sufficiently hinder rotation. The mirror-image conformations interconvert too readily at room temperature.

  • F — bottom right: leave unticked.
    [1,1′-Biphenyl]-2,3-diol, c1(-c2ccccc2)c(O)c(O)ccc1.
    The unsubstituted phenyl end has equivalent ring paths, so the connecting bond is not a stereogenic axis.

The completed answer-sheet pattern is therefore

Row Left box Right box
Top ✓
Middle ✓ ✓
Bottom

3.4

The substrate is 2-bromo-1-phenylethan-1-one, commonly called \(\alpha\)-bromoacetophenone or phenacyl bromide: O=C(CBr)c1ccccc1.

Reduction gives 2-bromo-1-phenylethan-1-ol:

  • \((R)\)-enantiomer: O[C@@H](CBr)c1ccccc1
  • \((S)\)-enantiomer: O[C@H](CBr)c1ccccc1

The balanced reaction is

\[ \ce{C6H5COCH2Br + (CH3)2CHOH <=> C6H5CH(OH)CH2Br + (CH3)2CO}. \]

BINOL is [1,1′-binaphthalene]-2,2′-diol, Oc1ccc2ccccc2c1-c1c(O)ccc2ccccc12.

Its catalyst is aluminium isopropoxide chelated by binaphthalene-2,2′-diolate, represented by CC(C)O[Al]1Oc2ccc3ccccc3c2-c2c(O1)ccc3ccccc23. Its axial configuration must additionally be specified as discussed in 3.3.

Assume the two catalyst enantiomers act independently and have equal total activity. They then give equal and opposite product selectivities.

Call the catalyst enantiomer that favours \(R\) \(C_+\), and its mirror image \(C_-\). With pure \(C_+\), the product fractions satisfy

\[ f_R+f_S=1,\qquad f_R-f_S=0.81. \]

Therefore,

\[ f_R=\frac{1+0.81}{2}=0.905, \qquad f_S=\frac{1-0.81}{2}=0.095. \]

Pure \(C_-\) gives the reversed proportions.

A catalyst \(ee\) of 50 % means

\[ f_{C_+}=\frac{1+0.50}{2}=0.75, \qquad f_{C_-}=\frac{1-0.50}{2}=0.25. \]

The product fractions are consequently

\[ f_R=0.75(0.905)+0.25(0.095)=0.7025, \]

\[ f_S=0.75(0.095)+0.25(0.905)=0.2975. \]

Thus,

\[ ee_{\mathrm{product}} =\frac{0.7025-0.2975}{0.7025+0.2975} =0.405. \]

More generally, if \(e_0\) is the product \(ee\) obtained with pure catalyst and \(e_C\) is the catalyst \(ee\),

\[ \begin{aligned} f_R&=\frac{1+e_C}{2}\frac{1+e_0}{2} +\frac{1-e_C}{2}\frac{1-e_0}{2} =\frac{1+e_Ce_0}{2},\\ f_S&=\frac{1-e_Ce_0}{2}. \end{aligned} \]

Hence

\[ \boxed{ee_{\mathrm{product}}=e_Ce_0=40.5\%}. \]

The solution was checked against the official solution.