Oxygenation and Hydrochlorination of Vaska’s Complex Ir(Cl)[P(C6H5)3]2(CO)

Author: Adam Capriola

CHM 2521 Section 151, Department of Chemistry, Saint Joseph’s University

Date Submitted:April 22, 2010

Abstract

Vaska’s complex, Ir(Cl)[P(C6H5)3]2(CO), reacts with O2 at room temperature to form Ir(Cl)[P(C6H5)3]2(CO)(O2) at a ratio of 3.11:1 reactant to product and with HCl to form Ir(Cl)2[P(C6H5)3]2(CO)H at 47.7% yield. The CO stretch on the IR spectra of these compounds is found at a lower frequency than that of the CO stretch from Vaska’s complex because the additional ligands lead to increased π back-bonding. In comparison to the 31P NMR spectrum of Ir(Cl)[P(C6H5)3]2(CO), the signal given by the 31P NMR spectrum of Ir(Cl)2[P(C6H5)3]2(CO)H is shifted upfield because of additional electronegative ligands added to the metal, which draw in electron density and deshield the phosphorus molecules. Finally, the 1H NMR spectra of Ir(Cl)[P(C6H5)3]2(CO) and Ir(Cl)2[P(C6H5)3]2(CO)H are nearly identical aside from a signal given off by Ir(Cl)2[P(C6H5)3]2(CO)H at δ -15.36 representative of the Ir-H addition. This signal is seen as a triplet of quartets due to coupling with Ir and P.

Introduction

Vaska’s complex, Ir(Cl)[P(C6H5)3]2(CO), is a versatile complex because of its ability to bind other additional ligands.1 Whereas most complexes contain 18 electrons and are considered to be saturated, Vaska’s complex contains only 16 electrons and is thus able to add certain two-electron donors to reach 18 valence electrons. It can also undergo oxidative addition in which the Ir1 center inserts into the σ bond of certain molecules, and the oxidation state of Ir is increased while reaching 18 electrons. Oxygenation and hydrochlorination of Ir(Cl)[P(C6H5)3]2(CO) proceed in the following manners:

Scheme 1

Scheme 2

These oxidative additions are of interest because the relative ease of identifying their products. The additional ligands will both increase π back-bonding to the metal and draw in electron density. The effects of these two characteristics are apparent through IR and 31P NMR spectroscopy, as a weakened CO bond will cause a lowering of CO stretch frequency in an IR spectrum, and deshielded phosphorus will be shifted upfield in a 31P NMR spectrum.

Experimental

All syntheses were carried out in air and the reagents and solvents were purchased from commercial sources and used as received unless otherwise noted. The synthesis of Ir(Cl)[P(C6H5)3]2(CO)(O2)(2)andIr(Cl)2[P(C6H5)3]2(CO)H (3)were based on reports published previously.1

Ir(Cl)[P(C6H5)3]2(CO) (1). The1H NMR, 31P NMR, and IR spectra of (1)were taken by Dr. Graham. This compound was not synthesized but purchased commercially. 1H NMR (CDCl3): δ 7.25-7.97 (3 H, series of signals, -C6H5). 31P NMR (CDCl3): δ 24.5 (s, -P(C6H5)3). FTIR (ATR) ν(CO)1951 cm-1 (s, C-O linkage).

Ir(Cl)[P(C6H5)]2(CO)(O2) (2). A stir bar, 1 (0.010 g, 1.28 x 10-5 mol), and toluene (10 mL) were subsequently added to a 25 mL single neck round bottom flask. The flask was covered with a septum and the vessel was degassed with O2 for 3 minutes. The solution was then stirred at room temperature at moderate speed for 1 h. The septum was taken of the flask and the solvent was removed via rotary evaporation so that only a few mL of solution remained. One drop of this solution was placed onto the ATR and allowed to dry before taking the IR spectrum of the complex. FTIR (ATR) ν(CO) 1952 cm-1 (m, C-O linkage), ν(CO)2000 cm-1 (m, C-O linkage).

Ir(Cl)2[P(C6H5)3]2(CO)H (3). 1 (0.032 g, 4.10 x 10-5 mol), THF (10 mL), HCl (concentrated, 5 drops), and Et2O (50 mL) were subsequently added to a 250 mL Erlenmeyer flask. The solution was swirled around for about 5 minutes to allow a whitish precipitate to form. The solution was filtered using a small frit and the precipitate was vacuum dried. The product was determined to be 3 (0.016 g, 1.96 x 10-5 mol, 47.7% yield based on the amount of 1 used). An extension of the 1H NMR spectra of this substance was given out by Dr. Graham. 1H NMR (CDCl3): δ -15.36 (tq, JIr, JP,Ir-H), δ 7.24-7.91 (3 H, series of signals, -C6H5). 31P NMR (CDCl3): δ -1.74 (s, -P(C6H5)3). FTIR (ATR) ν(CO) 1951 cm-1 (s, C-O linkage), ν(CO) 2021 cm-1 (s, C-O linkage).

Results

The reaction of 1 with O2 was not directly measured for percent yield, but could indirectly be measured via the IR spectrum of the product 2. The stretch at 1952 cm-1 was telling of 1 and the stretch at 2000 cm-1 was indicative of 2. The ratio of the area of these peaks was 3.11:1, reactant to product. The reaction of 2 with HCl resulted 0.016 g of product, which was determined to be 3. This translated to 1.96 x 10-5 mol and a 47.7% yield based on the amount of 1 used. The reactants and products reacted in 1:1 ratios in both instances. The IR spectrum of 1 was used to differentiate the two stretches in the 1950 to 2000 cm-1 region on the IR spectra of 2 and 3. The stretch around 1950 cm-1 could be identified as the CO stretch from 1, and the stretches around 2000 cm-1 were from the CO stretches of 2 and 3.

The 1H NMR spectrum of 1 showed a series of signals from δ 7.25-7.97 which was representative of protons attached to the phenyl rings. The 1H NMR spectrum of 3had these same phenyl signals, but additionally contained a signal from the hydride at δ -15.36 that was coupled with Ir and P into a triplet of quartets. The 31P NMR spectrum of 1 gave a singlet peak at δ 24.5 representative of the two identical phosphorus molecules on the compound. The 31P NMR spectrum of 3 displayed its singlet peak shifted upfield to δ -1.74.

Discussion

The ratio of formation of 2 seems reasonable, but it is unable to be determined whether or not that is a high or low ratio given the reaction time. If the solution was given more time to react, perhaps more product would have formed. If the flask was degassed with O2 for a longer amount of time, it is likely that would also favor the formation of more product, as additional oxygen would increase the interactions with 1 to form 2. In retrospect, the solution could have been stirred at more rigorous speed, as that would also likely increase the interactions between 1 and O2. The percent yield of 3 was close to 50%, which seems fairly good, though not all of the product that was weighed out actually was 3. It can be seen on the IR spectrum of the product that a significant amount of1 remained, due to the stretch visible around 1950 cm-1.

The IR spectra for 1, 2, and 3 are all very similar. The IR spectrum of 1 shows a single CO stretch at 1951 cm-1. The IR spectra of 2 and 3 also show stretches around 1950 cm-1, which suggests that those products obtained contained unreacted 1. The IR spectra of 2 and 3 also contain a second CO stretch around 2000 cm-1. These stretches are representative of desired product. The lowered frequency is due to increased π back-bonding from the addition of O2 and HCl to the Ir in1 in each case. This strengthens the Ir-C bond and weakens the C-O bond causing the shifts in frequency.1

The 1H NMR spectra of 1 and 3 are nearly identical save for the signal given off by the Ir-H bond by 3 at δ -15.36. This signal is first coupled with phosphorus, which has a spin of 1/2.1 There are two phosphorus, so using the equation 2nI + 1, the value of 3 is obtained. Because the spin is 1/2, this means that 3 peaks will be observed in a 1:2:1 ratio. The hydrogen is then coupled to iridium, which has a spin of 3/2.2 There is only one iridium, so using the equation 2nI + 1, the value of 4 is obtained. Because the spin is 3/2, this means that the 4 peaks will be observed in a 1:1:1:1 ratio.3 This explains the appearance of the hydride signal.

The 31P NMR spectrum of 3 has a singlet peak shifted upfield from that of the 31P NMR spectrum of 1. This is due to the electronegativity of thehydrogen and chlorine added to the metal, which draw electron density away from the phosphorus molecules leaving them less shielded. The oxidation state of Ir in complex 1 is +1 while being +3 in complexes 2 and 3. Its electron count in 1 is 16 electrons, while its electron count in2 and 3 is 18 electrons.

Conclusion

The main purposes of the experiments were to synthesize 2 and 3, confirm their structures by comparing their various spectra to those of 1, and to determine their percent yields or reactant to product ratios. The ratio of reactant to product for2was 3.11:1 which was obtained from the ratio of the areas of the two CO stretches on its IR spectrum representative of reactant and product. The percent yield of 3 was calculated to 47.7%, but was in reality lower because of CO stretches on its IR spectrum representative of 1. The visibility of this stretch means the reaction did not go to completion.

The structures of 2 and 3 were somewhat validated by their IR spectra, which gave CO stretches around 2000 cm-1. It was expected to see CO stretches for those compounds in this area because the additional ligands would cause increased π back-bonding, making the C-O bond weaker and thus lowering the frequency.1 A series of peaks seen on both the 1H NMR spectra of1 and 3 around δ 7.24-7.91 is suggestive of phenyl groups. The only notable difference between the 1H NMR spectra of 1 and 3 is the addition of a triplet of quartets at δ -15.36 for compound 3. The splitting of the peaks is due to coupling of the proton from the Ir-H bond to Ir and to the two P. The P split the signal into triplets of 1:2:1 ratio and the Ir split those signals into quartets of 1:1:1:1 ratio. Finally, the comparison of the 1 and 331P NMR spectra seem to confirm the identity of 3 as the singlet seen in the spectrum of 1 is shifted upfield in the spectrum of 3. This is due to the addition of H and Cl to the compound, which are electronegative and draw electron density away from the P, leaving it deshielded.

References

(1) Angelici, R. J.; Girolami, G. S.; Rachufuss T. B. Synthesis and Technique in Inorganic Chemistry: A Laboratory Manual; University Science Books: Sausilito, CA, 1999; pp 189-195, 259.

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