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. 2021 Jul 5;10(9):867–876. doi: 10.1002/open.202100047

Table 3.

Monomer sequence distributions of poly(ethylene‐co‐DC)s and poly(ethylene‐co‐DD)s prepared by Cp*TiCl2(O‐2,6‐iPr2‐4‐R‐C6H2) [R=H (1), tBu (2), CHPh2 (4), CPh3 (5), SiMe3 (6), SiEt3 (7), 3,5‐Me2C6H3 (9)] −MAO catalyst systems. (ethylene 6 atm, toluene).[a]

run

cat.

comonomer

content[b]/

triad sequence distribution[c] [%]

dyads[d] [%]

r E [e]

r C [e]

rEr C [f]

[mol %]

EEE

EEC+CEE

CEC

ECE

CCE+ECC

CCC

EE

EC+CE

CC

1

1

DC

21.4

39.6

31.5

6.73

18.0

4.25

55.3

42.6

2.12

3.12

0.12

0.37

12

5

DC

21.1

36.0

34.5

7.62

15.6

6.20

53.3

43.6

3.10

2.94

0.17

0.50

14

5

DC

21.4

42.8

28.9

5.69

17.6

4.98

57.3

40.3

2.49

3.41

0.15

0.51

15

6

DC

20.1

43.9

29.9

4.75

17.5

4.07

58.8

39.2

2.04

3.60

0.12

0.45

18

6

DC

21.3

39.2

32.4

5.71

18.3

4.48

55.4

42.4

2.24

3.14

0.13

0.40

22

7

DC

21.4

36.4

34.4

6.94

17.6

4.67

53.6

44.1

2.33

2.92

0.13

0.37

27

9

DC

21.1

41.9

26.3

9.31

16.6

5.98

55.0

42.0

2.99

3.14

0.17

0.54

28

1

DD

17.6

47.8

30.6

3.67

14.9

3.03

63.1

35.3

1.52

3.68

0.09

0.33

30

2

DD

18.3

50.1

27.4

4.81

14.2

3.46

0.76

63.8

34.4

2.49

3.82

0.15

0.57

36

4

DD

17.5

47.2

31.8

3.42

13.5

4.04

63.1

34.9

2.02

3.73

0.12

0.45

40

6

DD

16.3

50.4

29.6

3.28

14.1

2.62

65.2

33.5

1.31

4.02

0.08

0.32

42

7

DD

15.3

52.4

29.0

3.04

13.2

2.29

66.9

31.9

1.14

4.32

0.07

0.32

44

7

DD

16.7

46.0

33.0

3.43

14.5

3.08

62.5

36.0

1.54

3.58

0.09

0.32

48

9

DD

17.6

49.3

28.9

4.16

14.9

2.80

63.7

34.9

1.40

3.76

0.08

0.31

49

9

DD

17.7

45.9

30.7

5.30

14.0

4.13

61.2

36.7

2.06

3.44

0.12

0.40

[a] Detailed polymerization conditions, see Tables 1 and 2, C=comonomer [1‐decene (DC), 1‐dodecene (DD)]. [b] Comonomer contents in copolymer estimated by 13C NMR spectra. [c] Calculated by 13C NMR spectra, E=ethylene, C=comonomer [DC, DD]. [d] [EE]=[EEE]+1/2[EEC+CEE], [EC]=[CEC]+[ECE]+1/2{[EEC+CEE]+[CCE+ECC]}, [CC]=[CCC]+1/2[CCE+ECC]. [e] r E= [C]0/[E]0×2[EE]/[EC+CE], r C= [E]0/[C]0×2[CC]/[EC+CE]. [f] r Er C = 4[EE][CC]/[EC+CE]2.