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HRID513981

反应详情

EQUATION

反应方程式

HRID 513981 的结构方程式

REACTANTS

反应物

9
结构：C1CNCCNCCCNCCNC1
HCID64964

Cyclam

C10H24N4

结构：CN1CCCN(C)CCN(C)CCCN(C)CC1
HCID123502

1,4,8,11-Tetraazacyclotetradecane, 1,4,8,11-tetramethyl-

C14H32N4

结构：CC1CC(C)(C)NCCNC(C)CC(C)(C)NCCN1
HCID143337

5,5,7,12,12,14-Hexamethyl-1,4,8,11-tetraazacyclotetradecane

C16H36N4

结构：CN1CCCN2CCN(C)CCCN(CC1)CC2
HCID372702

4,11-Dimethyl-1,4,8,11-tetraazabicyclo[6.6.2]hexadecane

C14H30N4

结构：CN1CCCNCCN(C)CCCNCC1
HCID4659299

1,8-Dimethyl-1,4,8,11-tetraazacyclotetradecane

C12H28N4

结构：CCN1CCCN2CCN(CC)CCCN(CC1)CC2
HCID10158910

4,11-Diethyl-1,4,8,11-tetraazabicyclo[6.6.2]hexadecane

C16H34N4

结构：CN1CCNCCN(C)CCNCC1
HCID11063459

1,7-dimethyl-1,4,7,10-tetraazacyclododecane

C10H24N4

结构：CCCCCCCCN1CCCN2CCN(C)CCCN(CC1)CC2
HCID12011737

5-N-octyl-12-methyl-1,5,8,12-tetraazabicyclo[6.6.2]hexadecane

C21H44N4

结构：CCN1CCCNCCN(CC)CCCNCC1
HCID18451695

1,8-Diethyl-1,4,8,11-tetraazacyclotetradecane

C14H32N4

PRODUCTS

生成物

1
结构：C1CNCCNCCNCCN1
HCID64963

Cyclen

C8H20N4

PROCEDURE

实验过程

1,7-dimethyl-1,4,7,10-tetraazacyclododecane; 1,4,8,11-tetraazacyclotetradecane (cyclam); 1,4,8,11-tetramethyl-1,4,8,11-tetraazacyclotetradecane; 5,7,7,12,14,14-hexamethyl-1,4,8,11-tetraazacyclotetradecane; 1,8-dimethyl-1,4,8,11-tetraazacyclotetradecane; 1,8-diethyl-1,4,8,11-tetraazacyclotetradecane (diethylcyclam); 5,12-dimethyl-1,5,8,12-tetraazabicyclo[6.6.2]hexadecane (dimethyl-Bcyclam), 5,12-diethyl-1,5,8,12-tetraazabicyclo[6.6.2]hexadecane (diethyl-Bcyclam) and 5-N-octyl-12-methyl-1,5,8,12-tetraazabicyclo[6.6.2]hexadecane.