#1
TOP-1
0.98 ± 0.00
C[C]12[C]3(C)[C]4(C)[C]5(C)[C]1(C)[Rh+2]2345
#1
TOP-1
0.98 ± 0.00
C[C]12[C]3(C)[C]4(C)[C]5(C)[C]1(C)[Rh+2]2345
#1
TOP-1
0.98 ± 0.00
C[C]12[C]3(C)[C]4(C)[C]5(C)[C]1(C)[Rh+2]2345
#1
TOP-1
0.98 ± 0.00
C[C]12[C]3(C)[C]4(C)[C]5(C)[C]1(C)[Rh+2]2345
#1
TOP-1
0.98 ± 0.00
C[C]12[C]3(C)[C]4(C)[C]5(C)[C]1(C)[Rh+2]2345
#1
TOP-1
0.98 ± 0.00
C[C]12[C]3(C)[C]4(C)[C]5(C)[C]1(C)[Rh+2]2345
#1
TOP-1
0.98 ± 0.00
C[C]12[C]3(C)[C]4(C)[C]5(C)[C]1(C)[Rh+2]2345
#1
TOP-1
0.98 ± 0.00
C[C]12[C]3(C)[C]4(C)[C]5(C)[C]1(C)[Rh+2]2345
#1
TOP-1
0.98 ± 0.00
C[C]12[C]3(C)[C]4(C)[C]5(C)[C]1(C)[Rh+2]2345
#2
0.00 ± 0.00
Clc1ccc(C2=[N]3->[Rh+3]<-[n]4c(N3C(c3ccccc3)C2)sc2ccccc24)cc1
#2
0.00 ± 0.00
Clc1ccc(C2=[N]3->[Rh+3]<-[n]4c(N3C(c3ccccc3)C2)sc2ccccc24)cc1
#2
0.00 ± 0.00
Clc1ccc(C2=[N]3->[Rh+3]<-[n]4c(N3C(c3ccccc3)C2)sc2ccccc24)cc1
#2
0.00 ± 0.00
Clc1ccc(C2=[N]3->[Rh+3]<-[n]4c(N3C(c3ccccc3)C2)sc2ccccc24)cc1
#2
0.00 ± 0.00
Clc1ccc(C2=[N]3->[Rh+3]<-[n]4c(N3C(c3ccccc3)C2)sc2ccccc24)cc1
#2
0.00 ± 0.00
Clc1ccc(C2=[N]3->[Rh+3]<-[n]4c(N3C(c3ccccc3)C2)sc2ccccc24)cc1
#2
0.00 ± 0.00
Clc1ccc(C2=[N]3->[Rh+3]<-[n]4c(N3C(c3ccccc3)C2)sc2ccccc24)cc1
#2
0.00 ± 0.00
Clc1ccc(C2=[N]3->[Rh+3]<-[n]4c(N3C(c3ccccc3)C2)sc2ccccc24)cc1
#2
0.00 ± 0.00
Clc1ccc(C2=[N]3->[Rh+3]<-[n]4c(N3C(c3ccccc3)C2)sc2ccccc24)cc1
#3
0.00 ± 0.00
Clc1ccc(C2=[N](->[Rh+3])N(c3nc4ccccc4s3)C(c3ccccc3)C2)cc1
#3
0.00 ± 0.00
Clc1ccc(C2=[N](->[Rh+3])N(c3nc4ccccc4s3)C(c3ccccc3)C2)cc1
#3
0.00 ± 0.00
Clc1ccc(C2=[N](->[Rh+3])N(c3nc4ccccc4s3)C(c3ccccc3)C2)cc1
#3
0.00 ± 0.00
Clc1ccc(C2=[N](->[Rh+3])N(c3nc4ccccc4s3)C(c3ccccc3)C2)cc1
#3
0.00 ± 0.00
Clc1ccc(C2=[N](->[Rh+3])N(c3nc4ccccc4s3)C(c3ccccc3)C2)cc1
#3
0.00 ± 0.00
Clc1ccc(C2=[N](->[Rh+3])N(c3nc4ccccc4s3)C(c3ccccc3)C2)cc1
#3
0.00 ± 0.00
Clc1ccc(C2=[N](->[Rh+3])N(c3nc4ccccc4s3)C(c3ccccc3)C2)cc1
#3
0.00 ± 0.00
Clc1ccc(C2=[N](->[Rh+3])N(c3nc4ccccc4s3)C(c3ccccc3)C2)cc1
#3
0.00 ± 0.00
Clc1ccc(C2=[N](->[Rh+3])N(c3nc4ccccc4s3)C(c3ccccc3)C2)cc1
#4
0.00 ± 0.00
Clc1ccc(C2=NN(c3sc4ccccc4[n]3->[Rh+3])C(c3ccccc3)C2)cc1
#4
0.00 ± 0.00
Clc1ccc(C2=NN(c3sc4ccccc4[n]3->[Rh+3])C(c3ccccc3)C2)cc1
#4
0.00 ± 0.00
Clc1ccc(C2=NN(c3sc4ccccc4[n]3->[Rh+3])C(c3ccccc3)C2)cc1