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Home/Compounds/Other/4,9,23,28-tetratert-butyl-13,19-diaza-1-boranonacyclo[16.12.1.12,6.119,26.07,12.014,31.020,25.013,33.030,32]tritriaconta-2,4,6(33),7(12),8,10,14(31),15,17,20(25),21,23,26(32),27,29-pentadecaene
C46H49BN2
640.7 g/mol
Other

4,9,23,28-tetratert-butyl-13,19-diaza-1-boranonacyclo[16.12.1.12,6.119,26.07,12.014,31.020,25.013,33.030,32]tritriaconta-2,4,6(33),7(12),8,10,14(31),15,17,20(25),21,23,26(32),27,29-pentadecaene

CAS2170487-30-4
Category
Other
Compound Class
n.a
Primary Use
research compound

This compound is a boron-containing polycyclic heteroaromatic molecule with multiple tert-butyl substituents. Its high logP and low polarity suggest a lipophilic, rigid structure with extensive aromatic conjugation.

Content scope: Information here supports chemical identification and industrial supply, research, and manufacturing contexts. It is not a therapeutic claim, medical advice, or drug-use instructions.

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Related Compounds

ReferenceSame Category

Tert-butyl 4-methylsulfanyl-10-oxo-8-(trifluoromethylsulfonyloxy)spiro[1,3,5,11-tetrazatricyclo[7.4.0.02,7]trideca-2,4,6,8-tetraene-13,1'-cyclohexane]-11-carboxylate

CAS2170746-98-0

research intermediate

ReferenceSame Category

4-methylsulfanylspiro[1,3,5,11-tetrazatricyclo[7.4.0.02,7]trideca-2,4,6,8-tetraene-13,1'-cyclohexane]-10-one

CAS2170747-00-7

research compound

ReferenceSame Category

2-Amino-3-fluoropyridine

CAS21717-95-3

pharmaceutical intermediate

ReferenceSame Category

Flamenol

CAS2174-64-3

Pharmaceutical intermediate

Referencesame scaffold

13,19-diaza-1-boranonacyclo[16.12.1.12,6.119,26.07,12.014,31.020,25.013,33.030,32]tritriaconta-2(33),3,5,7,9,11,14(31),15,17,20,22,24,26,28,30(32)-pentadecaene

CAS2170487-27-9

research compound

Identifiers
IUPAC Name

4,9,23,28-tetratert-butyl-13,19-diaza-1-boranonacyclo[16.12.1.12,6.119,26.07,12.014,31.020,25.013,33.030,32]tritriaconta-2,4,6(33),7(12),8,10,14(31),15,17,20(25),21,23,26(32),27,29-pentadecaene

InChIKey

QUOIONVYAPAHAM-UHFFFAOYSA-N

SMILES

B12C3=CC(=CC4=C3N(C5=C1C(=CC=C5)N6C7=C(C=C(C=C7)C(C)(C)C)C8=C6C2=CC(=C8)C(C)(C)C)C9=C4C=C(C=C9)C(C)(C)C)C(C)(C)C

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