M. Okamoto et al. / Journal of Organometallic Chemistry 664 (2002) 59Á
/
65
61
However, the same replacement of P to Si gave a better
selectivity in PdH2SiW6Mo6O40 (entry 3).
Table 3
Effect of the weight ratio of H2O against AcOH solvent
a
The increase of Mo/Pd ratio to 0.05Á0.1 decreased
/
b
Entry AcOH (g) H2O (g) Product (mmol)
1/(1ꢂ2ꢂ3) (%)
biphenyl productivity and selectivity (entries 6 and 7).
The change of all protons by TBA cations in
H3PW6Mo6O40 also decreased the selectivity (entry 10).
The most striking result is that PdHPMo12O40 itself
acted as a catalyst to give 95% selectivity of biphenyl
with a lower yield of biphenyl (entry 9). The productiv-
ity ratio of entries 9/entry 8 for biphenyl is 0.23. On the
contrary Pd usage ratio of entries 9/entry 8 is 0.1. This
means that the complex of the type of PdHPMo12O40
can produce as much as 2.3-fold of biphenyl in the case
of entry 9. This suggests that when a proton type HPA
cocatalyst is added to the Pd catalyst it may form a kind
1
2
3
1
2
3
4
3
2
1
0
0
1
2
3
0.80 0.03 0.02 94
2.42 0.08 0.07 94
1.40 0.01 0.02 98
0.00 0.00 0.00
Á/
a
Conditions: volume of autoclave (100 ml), Pd(OAc)2 (0.13 mmol),
H3PMo12O40 (5.5 mmol), PhH (38 mmol), AcOH, H2O, O2 (10 atm),
130 8C, 4 h.
we tried to add H2O to the solvent to increase the
solubility of a HPA in the solvent. The results are shown
in Table 3. Best yield of biphenyl was obtained in a
solvent with AcOH/H2O ratio of 2. Entries 2 and 3 also
showed these effects giving higher yields than the case of
no water in the solvent. When H2O only was used as a
solvent, no biphenyl was formed.
of PdÁHPA complex to act as an active catalyst to
/
facilitate the production of biphenyl.
2.3. Effect of composition of H3PW12ꢁnMonO40 against
TONs of Pd
It is interesting to see the effect of composition of
heteropolyacid on the productivity of Pd. Fig. 2 shows
such results. One can see that the TON ratio in the
presence of H3PW12ꢁnMonO40 and in the absence of it
increases with the increase of n from 1 to 4 and stays
constant until 6, then increases again to 12. However,
the ratio does not exceed 1 in the case of these types of
heteropolyacids up to n 10. This means that selectivity
increase depresses the productivity of biphenyl in case of
a HPA of these types indicating the difference of type of
catalytic mode with or without HPA in the reaction.
H3PMo12O40 acted very effectively.
2.5. Effect of the amount of H3PMo12O40 in AcOHÁ
H2O solvent
/
By adding more HPA in a more soluble solvent, the
productivity of biphenyl may increase. We changed the
amount of HPA added to the solvent of AcOHÁH2O
/
(weight ratio 2:1). These results are shown in Table 4.
When 0.06 mmol amount of HPA is added to the system
it gave all biphenyl and no phenols at all (entry 2). It is a
very surprising fact. When the amount was increased up
to 0.28 mmol the selectivity stayed 100% and the yield
also stayed almost constant. However, the yield de-
creased after that. It may be due to the formation of an
undesired active complex. Fig. 3 shows the yield of
biphenyl versus the amount of the HPA. From this
figure one can see that around 0.2 mmol of HPA is
appropriate for this reaction.
2.4. Effect of weight ratio of H2O against AcOH in a
solvent
We found that a complex formation of Pd with a
HPA might accelerate the reaction. By this information
Table 4
a
Effect of the amount of H3PMo12O40 in AcOHÁH2O solvent
/
b
Entry H3PMo12O40 (mmol) Product (mmol)
1/(1ꢂ2ꢂ3) (%)
1
2
3
1
2
3
4
5
0.01
0.06
0.17
0.28
0.56
0.80 0.03 0.02
94
3.13 0.00 0.00 100
3.63 0.01 0.00 100
3.46 0.01 0.00 100
1.78 0.02 0.00
99
Fig. 2. Effect of composition of H3PW12ꢁn Mon O40. Conditions:
volume of autoclave (100 ml), Pd(OAc)2 (0.13 mmol), H3PW12ꢁn Mo-
a
Conditions: volume of autoclave (100 cm3), Pd(OAc)2 (0.13
n O40 (Mo/Pdꢃ
130 8C, 4 h. TON with a HPA/TON without a HPA.
/
0.5), PhH (38 mmol), AcOH (3.0 g), O2 (10 atm),
mmol), H3PMo12O40, PhH (38 mmol), AcOH (2.0 g), H2O (1.0 g),
O2 (10 kg cmꢁ2), 130 8C, 4 h.
a