2,2',3,3',5,5',6,6'-octafluoro-4,4'-dimethoxy-1,1'-biphenyl

2, 2', 3, 3', 5, 5', 6, 6'-octafluoro-4, 4'-dimethoxy-1, 1'-biphenyl

molar mass | 358.2 g/mol formula | C_14H_6F_8O_2 empirical formula | F_4C_7O_H_3 SMILES identifier | COC1=C(C(=C(C2=C(C(=C(C(=C2F)F)OC)F)F)C(=C1F)F)F)F InChI identifier | InChI=1/C14H6F8O2/c1-23-13-9(19)5(15)3(6(16)10(13)20)4-7(17)11(21)14(24-2)12(22)8(4)18/h1-2H3 InChI key | TVUPJKALGKCABC-UHFFFAOYSA-N

Draw the Lewis structure of 2, 2', 3, 3', 5, 5', 6, 6'-octafluoro-4, 4'-dimethoxy-1, 1'-biphenyl. 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), fluorine (n_F, val = 7), hydrogen (n_H, val = 1), and oxygen (n_O, val = 6) atoms: 14 n_C, val + 8 n_F, val + 6 n_H, val + 2 n_O, val = 130 Calculate the number of electrons needed to completely fill the valence shells for carbon (n_C, full = 8), fluorine (n_F, full = 8), hydrogen (n_H, full = 2), and oxygen (n_O, full = 8): 14 n_C, full + 8 n_F, full + 6 n_H, full + 2 n_O, full = 204 Subtracting these two numbers shows that 204 - 130 = 74 bonding electrons are needed. Each bond has two electrons, so in addition to the 31 bonds already present in the diagram add 6 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 6 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: | |

melting point | 201.6 °C boiling point | 388.7 °C critical temperature | 836.6 K critical pressure | 1.674 MPa critical volume | 783.5 cm^3/mol molar heat of vaporization | 56 kJ/mol molar heat of fusion | 43.21 kJ/mol molar enthalpy | -1807 kJ/mol molar free energy | -1573 kJ/mol (computed using the Joback method)

longest chain length | 12 atoms longest straight chain length | 2 atoms longest aliphatic chain length | 0 atoms aromatic atom count | 12 atoms H-bond acceptor count | 0 atoms H-bond donor count | 0 atoms

Find the elemental composition for 2, 2', 3, 3', 5, 5', 6, 6'-octafluoro-4, 4'-dimethoxy-1, 1'-biphenyl in terms of the atom and mass percents: atom percent = N_i/N_atoms × 100% mass percent = (N_im_i)/m × 100% Plan: • Write the chemical formula and gather atomic masses from the periodic table. • Determine values for N_i, m_i, N_atoms and m using these items. • Finally, compute the percents and check the results. Write the chemical formula: C_14H_6F_8O_2 Use the chemical formula to count the number of atoms, N_i, for each element and find the total number of atoms, N_atoms, per molecule: | number of atoms F (fluorine) | 8 C (carbon) | 14 O (oxygen) | 2 H (hydrogen) | 6 N_atoms = 8 + 14 + 2 + 6 = 30 Divide each N_i by N_atoms to calculate atom fractions. Then use the property that atom fractions must sum to one to check the work: | number of atoms | atom fraction F (fluorine) | 8 | 8/30 C (carbon) | 14 | 14/30 O (oxygen) | 2 | 2/30 H (hydrogen) | 6 | 6/30 Check: 8/30 + 14/30 + 2/30 + 6/30 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent F (fluorine) | 8 | 8/30 × 100% = 26.7% C (carbon) | 14 | 14/30 × 100% = 46.7% O (oxygen) | 2 | 2/30 × 100% = 6.67% H (hydrogen) | 6 | 6/30 × 100% = 20.0% Look up the atomic mass, m_i, in unified atomic mass units, u, for each element in the periodic table: | number of atoms | atom percent | atomic mass/u F (fluorine) | 8 | 26.7% | 18.998403163 C (carbon) | 14 | 46.7% | 12.011 O (oxygen) | 2 | 6.67% | 15.999 H (hydrogen) | 6 | 20.0% | 1.008 Multiply N_i by m_i to compute the mass for each element. Then sum those values to compute the molecular mass, m: | number of atoms | atom percent | atomic mass/u | mass/u F (fluorine) | 8 | 26.7% | 18.998403163 | 8 × 18.998403163 = 151.987225304 C (carbon) | 14 | 46.7% | 12.011 | 14 × 12.011 = 168.154 O (oxygen) | 2 | 6.67% | 15.999 | 2 × 15.999 = 31.998 H (hydrogen) | 6 | 20.0% | 1.008 | 6 × 1.008 = 6.048 m = 151.987225304 u + 168.154 u + 31.998 u + 6.048 u = 358.187225304 u Divide the mass for each element by m to calculate mass fractions. Then use the property that mass fractions must sum to one to check the work: | number of atoms | atom percent | mass fraction F (fluorine) | 8 | 26.7% | 151.987225304/358.187225304 C (carbon) | 14 | 46.7% | 168.154/358.187225304 O (oxygen) | 2 | 6.67% | 31.998/358.187225304 H (hydrogen) | 6 | 20.0% | 6.048/358.187225304 Check: 151.987225304/358.187225304 + 168.154/358.187225304 + 31.998/358.187225304 + 6.048/358.187225304 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent F (fluorine) | 8 | 26.7% | 151.987225304/358.187225304 × 100% = 42.43% C (carbon) | 14 | 46.7% | 168.154/358.187225304 × 100% = 46.95% O (oxygen) | 2 | 6.67% | 31.998/358.187225304 × 100% = 8.933% H (hydrogen) | 6 | 20.0% | 6.048/358.187225304 × 100% = 1.689%

The first step in finding the oxidation states (or oxidation numbers) in 2, 2', 3, 3', 5, 5', 6, 6'-octafluoro-4, 4'-dimethoxy-1, 1'-biphenyl is to draw the structure diagram. Next set every oxidation number equal to the atom's formal charge: In 2, 2', 3, 3', 5, 5', 6, 6'-octafluoro-4, 4'-dimethoxy-1, 1'-biphenyl hydrogen is not bonded to a metal with lower electronegativity, so it will have an oxidation state of +1. Any element bonded to hydrogen gains the bonding electrons, decreasing their oxidation state by 1 for every bond: With hydrogen out of the way, look at the remaining bonds. There are 8 carbon-fluorine bonds, 4 carbon-oxygen bonds, and 13 carbon-carbon bonds. For each of these bonds, assign the bonding electrons to the most electronegative element. First examine the carbon-fluorine bonds: element | electronegativity (Pauling scale) | C | 2.55 | F | 3.98 | | | Since fluorine is more electronegative than carbon, the electrons in these bonds will go to fluorine. Decrease the oxidation number for fluorine in every highlighted bond (by 1 for single bonds, 2 for double bonds, and 3 for triple bonds), and increase the oxidation number for carbon accordingly: Next look at the carbon-oxygen bonds: element | electronegativity (Pauling scale) | C | 2.55 | O | 3.44 | | | Since oxygen is more electronegative than carbon, the electrons in these bonds will go to oxygen: Next look at the carbon-carbon bonds: element | electronegativity (Pauling scale) | C | 2.55 | C | 2.55 | | | Since these elements are the same the bonding electrons are shared equally, and there is no change to the oxidation states: Now summarize the results: Answer: | | oxidation state | element | count -2 | C (carbon) | 2 | O (oxygen) | 2 -1 | F (fluorine) | 8 0 | C (carbon) | 2 +1 | C (carbon) | 10 | H (hydrogen) | 6

First draw the structure diagram for 2, 2', 3, 3', 5, 5', 6, 6'-octafluoro-4, 4'-dimethoxy-1, 1'-biphenyl, and for every non-hydrogen atom, count the σ-bonds. Note that double and triple bonds consist of one σ-bond together with one or two π-bonds: Identify those atoms with lone pairs: Find the steric number by adding the lone pair count to the number of σ-bonds: Consult the following chart to determine the hybridization from the steric number: steric number | hybridization 2 | sp 3 | sp^2 4 | sp^3 5 | dsp^3 6 | d^2sp^3 7 | d^3sp^3 Assign the provisional hybridization based on the table: Next identify any sp^3 atoms with lone pair electrons which can participate in a conjugated π-bond system. These atoms can lower their energy by placing a lone pair in a unhybridized p orbital to maximize overlap with the neighboring π-bonds. Note that halogens and elements from the third period and below do not engage in bond conjugation, except in the case of aromaticity: Adjust the provisional hybridizations to arrive at the result: Answer: | |

vertex count | 30 edge count | 31 Schultz index | 9056 Wiener index | 2289 Hosoya index | 459650 Balaban index | 2.317