[1, 2-bis(diphenylphosphino)ethane]

1, 2-bis(diphenylphosphino)ethane

formula | (C_6H_5)_2PCH_2CH_2P(C_6H_5)_2 Hill formula | C_26H_24P_2 name | 1, 2-bis(diphenylphosphino)ethane IUPAC name | 2-di(phenyl)phosphanylethyl-di(phenyl)phosphane alternate names | 2-di(phenyl)phosphanylethyl-di(phenyl)phosphane | bis(1, 2-diphenylphosphino)ethane | diphos | dppe | ethylene bis(diphenylphosphine) mass fractions | C (carbon) 78.4% | H (hydrogen) 6.07% | P (phosphorus) 15.5%

Draw the Lewis structure of 1, 2-bis(diphenylphosphino)ethane. 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), and phosphorus (n_P, val = 5) atoms: 26 n_C, val + 24 n_H, val + 2 n_P, val = 138 Calculate the number of electrons needed to completely fill the valence shells for carbon (n_C, full = 8), hydrogen (n_H, full = 2), and phosphorus (n_P, full = 8): 26 n_C, full + 24 n_H, full + 2 n_P, full = 272 Subtracting these two numbers shows that 272 - 138 = 134 bonding electrons are needed. Each bond has two electrons, so in addition to the 55 bonds already present in the diagram add 12 bonds. To minimize formal charge carbon wants 4 bonds. Identify the atoms that want additional bonds and the number of electrons remaining on each atom: Fill in the 12 bonds by pairing electrons between adjacent highlighted atoms. Note that the six atom rings are aromatic, so that the single and double bonds may be rearranged: Answer: | |

3D structure

molar mass | 398.43 g/mol phase | solid (at STP) melting point | 139.5 °C

CAS number | 1663-45-2 Beilstein number | 761261 PubChem CID number | 74267 PubChem SID number | 24846735 SMILES identifier | C1=CC=C(C=C1)P(CCP(C2=CC=CC=C2)C3=CC=CC=C3)C4=CC=CC=C4 InChI identifier | InChI=1/C26H24P2/c1-5-13-23(14-6-1)27(24-15-7-2-8-16-24)21-22-28(25-17-9-3-10-18-25)26-19-11-4-12-20-26/h1-20H, 21-22H2 MDL number | MFCD00003047