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Catalytic Hydrogenation of Olefins in the
presence of Wilkinsonís Catalyst
The first homogeneous catalytic hydrogenation of unsaturated organic compounds was achieved using (P(C 6 H 5 ) 3 ) 3 RhCl (Wilkinsonís catalyst) in 1965.^1 Detailed studies have resulted in the proposed mechanism shown below in figure 1. Molecular hydrogen is activated through oxidative addition to the unsaturated rhodium complex shown in step 2. The resulting hydrido complex allows for addition of an olefin or other unsaturated species with a concomitant loss of phosphine. The rate limiting insertion reaction in step 5 is followed by a reductive elimination step which completes the catalytic cycle.
Figure 1. Proposed mechanism for hydrogenation of alkenes by Wilkinsonís catalyst.^2
Synthesis of RhCl(PPh 3 ) 3 , Wilkinsonís Catalyst
RhCl 3 ∑ 3H 2 O + P(C 6 H 5 ) 3 → RhCl(P(C 6 H 5 ) 3 ) 3
Place 20 mL of absolute ethanol in a 25-mL round bottom flask equipped with a magnetic stir bar. Attach a reflux condenser and place the apparatus in a heating mantel on a magnetic stir plate. Heat the ethanol to just below the boiling point. Remove the condenser momentarily, and add ~600 mg (2. mmol, a large excess) of triphenylphosphine to the hot ethanol and stir until dissolved. Remove the condenser momentarily once again, and add 100 mg
(0.48 mmol) of hydrated rhodium (III) chloride to the solution and continue to stir. Heat the solution to a gentle reflux. Initially, a deep red-brown solution is obtained, which during additional heating under reflux will slowly form yellow crystals. After ~20- minutes of reflux, the yellow crystals are converted into shiny burgundy-red crystals. Collect the product crystals by suction filtration on a Hirsch funnel while the solution is hot. Wash the crystals with three 1-mL portions of hot ethanol followed by three 1 mL portions of diethyl ether. Dry the crystals on the filter by continuous suction. Calculate the percentage yield and determine the melting point of the product.
Absorption of Hydrogen by Wilkinsonís Catalyst (Oxidative Addition)
RhCl(P(C 6 H 5 ) 3 ) 3 + H 2 → RhCl(P(C 6 H 5 ) 3 ) 3 H 2
Dissolve ~50 mg of RhCl(P(C 6 H 5 ) 3 ) 3 in 1 mL of deoxygenated chloroform (chloroform can be deoxygenated by bubbling with an inert gas such as argon or nitrogen for 2 minutes) small flask fitted with a septum. Bubble H 2 gas through the solution for a few minutes, whereupon the red solution will turn pale yellow (this change may be subtle). Concentrate the solution under the flow of H 2 gas. When the solution is sufficiently concentrated (~0. mL), add deoxygenated diethyl ether drop-wise until precipitation occurs. Cool the flask in an ice-water bath and collect the yellow crystals by suction filtration using a Hirsch funnel. Calculate the percent yield. Obtain the IR spectrum of each compound. Assign the peaks in the infrared spectrum.
Catalytic Hydrogenation
Flush a flask equipped with a magnetic stir bar and a rubber septum with H 2 through a long needle. Add 10 mL of toluene to this flask and saturate it with hydrogen gas by bubbling for 10 minutes. Quickly remove the septum and with stirring dissolve 25 mg of Wilkinsonís catalyst in the solvent. Note: The catalyst is relatively insoluble in toluene, but in the presence of H 2 gas it dissolves fairly rapidly due to the formation of the more soluble dihydrido species. The resulting solution is pale yellow. Discontinue the stirring, remove the septum, and with a Pasteur pipet, drop-wise, add 1 mL of freshly distilled cyclohexene. As soon as the alkene is added the solution turns deep red-brown in color. Reattach the septum. Flush the flask with hydrogen gas. Upon stirring in an atmosphere of H 2 , the color
of the solution lightens to pale yellow. Note: If the H 2 gas flow is stopped and stirring is discontinued even for only 30 seconds, the solution again turns deep red-brown. Remove a small sample of the solution for GC analysis. The presence of cyclohexane formed as a result of the hydrogenation of cyclohexene should be evident in the chromatogram. (Conditions: Silicone column, 8 ft. length, flow rate: 50 mL/min He, temperature 80-85∞ C) Calculate the percent conversion of cyclohexene to cyclohexane from the relative peak areas.
Kinetics of Hydrogenation ñ Single Turnover
Figure 2. Proposed mechanism for hydrogenation of cyclohexene by (PPh 3 ) 3 RhH 2 Cl.
In the presence of excess phosphine ligand the loss of hydrogen from species A (refer to figure
- is suppressed and can be ignored under the conditions of this experiment (k--2 ~ 10-4^ s-1). The proposed mechanism begins with A reacting with an alkene ( E ) to form the intermediate species B with the loss of one phosphine ligand ( P ). B then reacts in a rate determining step (R.D.S.) to give the hydrogen insertion product C which rapidly eliminates the alkane and in the presence of phosphine and absence of hydrogen reforms the original (PPh 3 ) 3 RhCl starting material. Inspection of the above mechanism with regards to species A , B and C results in the following differential equations.
4 [^ ][^ ]^4 [^ ][^ ]
dA k A E k B P dt −
4 [^ ][^ ]^4 [^ ][^ ]^ [^ ]
dB k B P k A E k B dt −
5 [^ ]
dC k B dt
The goal is to manipulate these equations so that only one experimentally observable variable remains. In the experimental conditions described below an excess of triphenylphosphine and cyclohexene are added in single turnover conditions. A single turnover in a catalytic cycle means that the conditions are such that the reaction stops after one cycle. Because the reaction is stopped after a single turnover the concentrations of phosphine and cyclohexene do not change during the experiment and can be treated as constants. In equation 1 there are two concentration variables A and B. We will attempt to rewrite this equation so that the B term is eliminated. Any equilibria or mass action expressions can also be used in working with these equations. The equilibrium expression for the first step in the mechanism is shown below.
4 4 4
[ ][ ]
[ ][ ]
B P k K A B k −
At any point during the reaction the total amount of rhodium containing species in solution ( T ) is the sum of A , B and C and the amount of reactive rhodium ( R ) is equal to A + B since once C is formed it can no longer react to produce A or B. If this is true then the following relations hold.
[ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ]
T A B C
R A B T C
Manipulation of the equilibrium expression leads to the following :
4 4
4
1 [ ][ ] [ ] [ ]
[ ][ ] [ ] [ ]
[ ] [ ] [ ]
[ ] [ ]
A E P A
K B P K E B
P A B
K E B
Note that in the second transformation, [B]/[B] was added to each side of the equation. Substitution of eq. 5 into eq. 6 and rearrangement gives the following expression for [B] in terms of R , P and E.
4 4 4 4
[ ] [ ] [ ]
1 [ ]
[ ] [ ] [^ ]
[ ]
[ ] [ ]
[ ]
[ ] [ ]
P R R
B
K E B P
K E
R K E
B
P K E
Thus far no assumptions have been made during these derivations. Letís take a closer look at [ B ] during the course of the reaction. Since the
may use that type of analysis to obtain values for kobs. However, there are simple methods to linearize exponential data. An example dataset is shown below in figure 3.
Figure 3. Plot of absorbance vs. time for a single
he starting absorbance value (A 0 ) and the
turnover experiment (red line). The blue points are the experimental data that is linearized in figure 2 below.
T
absorbance value at ìinfiniteî time (A∞) is required to linearize this data. A plot of ln[(A 0 -A∞)/(A-A∞)] vs. time should be linear with a slope equal to kobs as shown in figure 4.
Figure 4. Linearized plot of the data shown in figure
ote that not all of the data shown in figure 3 is used
lternate Fitting Procedure on-linear curve fitting
Create a spreadsheet similar to the one pictured
N
in figure 4. At long times the absorbance values approach the ìinfiniteî value and indeterminate results are obtained. In this case only the data from 0 to 650 seconds are used, shown in blue. The rate constant obtained for this experiment was determined to be 6.2 × 10 -3^ s-1.
A
Using ìSolverî in Excel for n
below. Import your kinetics data into columns A and B. Column A should be the time values (in seconds) and B the absorbance values.
A B C D E F G 1 0 0.17 3 0.187 1. 86E- 04 A 0.1 85 2 5 0.182 0.193 1.23E-04 k 0. 3 10 0.190 0.198 7.16E-05 c 0. 4 15 0.197 0.204 4.39E-05 X^ 2 1.2 2 E- 5 20 0.205 0.209 1.57E- 6 25 0.212 0.214 3.03E- 7 30 0.218 0.219 5.41E- 8 35 0.224 0.224 2.60E- 9 40 0.230 0.228 2.69E-
he equation for an exponential rise to a maximum
olumn C is the predicted curve based upon the
$G$3+$G$1(1-EXP(-$G$2A1))
olumn D is the square of the difference between the
(B1-C1)^
is the sum of the chi squares values,
T
value is given in eq. 11 above. Where A is the amplitude of the curve, c is the offset from zero and k obs is the observed rate constant. A and c are easily estimated from inspection of the data. The value for c is estimated as the absorbance at time zero, and the amplitude is estimated as the difference between the absorbance at time zero and the absorbance at infinite time (see figure 3 above). The rate constant can be estimated as 1/t1/2 (t1/2 = half-life).
C guess values of A, c and k. The following formula should be entered into cell C1 and copied into all of column C
=
C real data (column B) and the predicted data (column C), called chi squared( χ^2 ). The following formula should be entered into cell D1 and copied into all of column D.
=
Cell G4 (^) i^2 i
enter the following in cell G
SUM(D:D).
the predicted curve is very close to the
If experimental curve then the value for (^) i^2 i
∑ χ will be
minimum value of (^) i^2
small. The best fit curve is then expected to have the
i
∑ χ.^ We can use the ìSolverî
add-in in Excel to fi e values of A, c and k that
i i
nd th
result in the minimum value for ∑ χ^2 (cell G4).
rocedure to Fit the Data
. A new pop-up window will appear.^3
nction since we are trying to minimize
ows the solver to vary the
values to fit the data.
revert to your
values make sense,
inetics Analysis, Determination of K 4 and k 5
ne and e observed rate constant for each reaction your
P
- Under the ìToolsî menu select ìSolverî
- In the box labeled ìSet Target Cellî type in $G$4.
- Below this select the ìEqual Toî section to the ìminî fu the value in cell G4. In the box labeled ìBy Changing Cellsî type $G$1:$G$3. This all values for A, c and k to minimize the sum of chi squared. Now click on ìSolveî. The program will alter your initial
- A new pop-up window will appear asking if you want to keep the new values or original values. Select ìKeep Solver Solutionî and click the ìOKî button. The best fit values for A, c and k will now be in cells G1:G3. To be sure the plot the columns B and C vs. column A. The two curves should match very closely. If they do not, then you need to make better guesses for A, c, and k to start with. Enter new guess values and repeat the procedure.
K
Record the mass of PPh 3 , volume of cyclohexe th group performs. Turn in these values to your TA before you leave the lab. Look up the class experimental data on the course website and plot 1/ k obs (y-axis) vs. [PPh 3 ]/[C 6 H 10 ] as in equation 10 above. In this analysis we are plotting a ratio of concentrations. Since the triphenylphosphine and cyclohexene are in the same volume of solution one does not need to know the total volume of solution in order to calculate the concentration ratio.
[PPh ] 3 =moles of PPh 3
[C H 6 10 ] moles of C H 6 10
Report the values for K 4 and k 5. Compare the class alues to those reported in the literature.
eferences
v
R
rn, J.A.; Wilkinson, G.; Young, J.F., Chem. Comm., 1965 , 17 (b) Osborn, J.A.; Jardine,
hem. Soc.,Chem. omm. , 1973 , 629-30. (b)Halpern, J.; Okamoto, T.;
he tools enu, the add-in needs to be installed. To do this
- (a) Osbo
F.H.; Young, J.F.; Wilkinson, G., J. Chem. Soc. A. Inorg. Phys. Theor. , 1965 , 1711-32.
(a) Halpern, J.; Wong, C.S., J. C C Zakhariev, A., J. Mol. Cat. , 1976 , 2 , 65-68.
If you do not see this as an option in t m select ìAdd-Insî under the ìToolsî menu and check the solver add-in box. Note: You may need the Excel installation disk to add this feature.
