Class of metallocenes and method of producing polyethylene
A polyethylene and metallocene technology, applied in the field of polyethylene, can solve the problems of compound complexity and high cost
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Embodiment 1
[0070] Embodiment 1, dichloro cyclopentadienyl (tetrahydroindenyl) zirconium Cp (H 4 Ind)ZrCl 2 preparation of
[0071] To a yellow solution of cyclopentadienyl(indenyl)zirconium dichloride (2.00 g, 5.84 mmol) in dichloromethane (40 mL) was added PtO 2 (0.20 g, 10 wt%), a stained yellow mixture was obtained. Stir the mixture with 80 psi H 2 Pressurize. Using 80psi H 2 After stirring for 2 hours, the solution was almost colorless. The hydrogen was then evacuated, and the reaction was filtered through Celite to obtain a black solid and a pale straw yellow solution. The solution was evaporated in vacuo to leave a white solid. Yield 1.69 g (84%). 1 H NMR (CD 2 Cl 2 ): δ1.60-1.81 (m, 4H, H 4 H on the ring of Ind, 2.50-2.59 (m, 2H, H 4 H on the ring of Ind, 2.76-2.85 (m, 2H, H 4 H on the ring of Ind, 5.80 (m, 2H, H 4 Cp-H of Ind), 6.37 (m, 1H, H 4 Cp-H) of Ind, 6.44 (s, 5H, Cp).
Embodiment 2
[0073] Example 2, dichlorocyclopentadienyl (1-methyltetrahydroindenyl) zirconium Cp (1-MeH 4 Ind)ZrCl 2 preparation of
[0074] To Cp(1-MeInd)ZrCl 2 (1.07g, 3.00mmol) bright yellow solution in dichloromethane (25mL) was added PtO 2 (0.10 g, 10 wt%), a stained yellow mixture was obtained. Stir the mixture with 80 psi H 2 Pressurize. Using 80psi H 2 After stirring for 45 minutes, the solution was gray-black. The hydrogen was then evacuated, and the reaction was filtered through celite to obtain a black solid and a light green solution. The solution was evaporated in vacuo to leave a white solid. Yield 0.95 g (88%). 1 H NMR (CD 2 Cl 2 ): δ1.60-1.78 (m, 4H, H 4 H on the ring of Ind, 1.96(s, 3H, Me), 2.34-2.56(m, 2H, H 4 H on the ring of Ind, 2.72-2.82 (m, 2H, H 4 H on the ring of Ind, 5.69 (d, 1H, H 4 Cp-H of Ind), 6.14(d, 1H, H 4 Cp-H) of Ind, 6.41 (s, 5H, Cp).
Embodiment 3
[0077] Example 3, dichloro (1,3-dimethylcyclopentadienyl) (tetrahydroindenyl) zirconium (1,3-Me 2 Cp)(H 4 Ind)ZrCl 2 preparation of
[0078] To (1,3-Me 2 Cp)IndZrCl 2 (0.80g, 2.16mmol) to a yellow solution in dichloromethane (20mL) was added PtO 2 (0.08 g, 10 wt%), a stained yellow mixture was obtained. Stir the mixture with 80 psi H 2 Pressurize. Using 80psi H 2 After stirring for 40 minutes, the solution was gray-black. The hydrogen was then evacuated and the reaction was filtered through celite to obtain a black solid and a colorless solution. The solution was evaporated in vacuo to leave a white solid. The solid was washed with pentane (3 x 10ml) and dried in vacuo. , yield 0.74 g (92%) of a white solid. 1 H NMR (CD 2 Cl 2 ): δ1.52-1.65 (m, 2H, H 4 H on the ring of Ind, 1.72-1.90 (m, 2H, H 4 H on the ring of Ind, 2.18(s, 3H, Me), 2.52-2.60(m, 2H, H 4 H on the ring of Ind, 2.77-2.86 (m, 2H, H 4 H on the ring of Ind, 5.73 (d, 2H), 6.31 (d, 2H), 6.07 (t, 1H)...
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