handwritten style diazene, bis[4-(pentyloxy)phenyl]-, 1-oxide

diazene, bis[4-(pentyloxy)phenyl]-, 1-oxide

formula | C_22H_30N_2O_3 name | diazene, bis[4-(pentyloxy)phenyl]-, 1-oxide IUPAC name | oxido-(4-pentoxyphenyl)-(4-pentoxyphenyl)imino-azanium alternate names | 4, 4'-bis(pentyloxy)azoxybenzene | 4, 4'-di-N-amyloxyazoxybenzene | 4, 4'-di-N-pentyloxyazoxybenzene | 4, 4'-dipentoxyazoxybenzene mass fractions | C (carbon) 71.3% | H (hydrogen) 8.16% | N (nitrogen) 7.56% | O (oxygen) 13%

Draw the Lewis structure of diazene, bis[4-(pentyloxy)phenyl]-, 1-oxide. Start by drawing the overall structure of the molecule, ignoring potential double and triple bonds: Count the total valence electrons of the carbon (n_C, val = 4), hydrogen (n_H, val = 1), nitrogen (n_N, val = 5), and oxygen (n_O, val = 6) atoms: 22 n_C, val + 30 n_H, val + 2 n_N, val + 3 n_O, val = 146 Calculate the number of electrons needed to completely fill the valence shells for carbon (n_C, full = 8), hydrogen (n_H, full = 2), nitrogen (n_N, full = 8), and oxygen (n_O, full = 8): 22 n_C, full + 30 n_H, full + 2 n_N, full + 3 n_O, full = 276 Subtracting these two numbers shows that 276 - 146 = 130 bonding electrons are needed. Each bond has two electrons, so in addition to the 58 bonds already present in the diagram add 7 bonds. To minimize formal charge oxygen wants 2 bonds, nitrogen wants 3 bonds, and carbon wants 4 bonds. Identify the atoms that want additional bonds and the number of electrons remaining on each atom: To fully fill its valence shell, nitrogen will donate one of its electrons, allowing it to form four bonds (the maximum number an element on period 2 can form). Fill in the 7 bonds by pairing electrons between adjacent highlighted atoms, noting the formal charges of the atoms. The six atom rings are aromatic, so that the single and double bonds may be rearranged: Answer: | |

3D structure

molar mass | 370.49 g/mol solubility in water | insoluble

CAS number | 19482-05-4 Beilstein number | 963337 PubChem CID number | 88090 SMILES identifier | CCCCCOC1=CC=C(C=C1)N=[N+](C2=CC=C(C=C2)OCCCCC)[O-] InChI identifier | InChI=1/C22H30N2O3/c1-3-5-7-17-26-21-13-9-19(10-14-21)23-24(25)20-11-15-22(16-12-20)27-18-8-6-4-2/h9-16H, 3-8, 17-18H2, 1-2H3/b24-23-