bis hydroxyhexafluoroisopropyl benzene

bis hydroxyhexafluoroisopropyl benzene

molar mass | 264.2 g/mol formula | C_9H_10F_6O_2 empirical formula | O_2C_9F_6H_10 SMILES identifier | CC(C)C1(C(=C(C(C(C1(F)F)(F)F)F)O)O)F InChI identifier | InChI=1/C9H10F6O2/c1-3(2)7(11)6(17)4(16)5(10)8(12, 13)9(7, 14)15/h3, 5, 16-17H, 1-2H3 InChI key | CXGNZHHIDVGMCD-UHFFFAOYSA-N

Draw the Lewis structure of bis hydroxyhexafluoroisopropyl benzene. 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: 9 n_C, val + 6 n_F, val + 10 n_H, val + 2 n_O, val = 100 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): 9 n_C, full + 6 n_F, full + 10 n_H, full + 2 n_O, full = 156 Subtracting these two numbers shows that 156 - 100 = 56 bonding electrons are needed. Each bond has two electrons, so in addition to the 27 bonds already present in the diagram add 1 bond. 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 1 bond by pairing electrons between adjacent highlighted atoms: Answer: | |

melting point | 120.4 °C boiling point | 327.1 °C critical temperature | 761.7 K critical pressure | 2.733 MPa critical volume | 589.5 cm^3/mol molar heat of vaporization | 61.4 kJ/mol molar heat of fusion | 18.79 kJ/mol molar enthalpy | -1642 kJ/mol molar free energy | -1424 kJ/mol (computed using the Joback method)

longest chain length | 7 atoms longest straight chain length | 3 atoms longest aliphatic chain length | 3 atoms aromatic atom count | 0 atoms H-bond acceptor count | 2 atoms H-bond donor count | 2 atoms

Find the elemental composition for bis hydroxyhexafluoroisopropyl benzene 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_9H_10F_6O_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 O (oxygen) | 2 C (carbon) | 9 F (fluorine) | 6 H (hydrogen) | 10 N_atoms = 2 + 9 + 6 + 10 = 27 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 O (oxygen) | 2 | 2/27 C (carbon) | 9 | 9/27 F (fluorine) | 6 | 6/27 H (hydrogen) | 10 | 10/27 Check: 2/27 + 9/27 + 6/27 + 10/27 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent O (oxygen) | 2 | 2/27 × 100% = 7.41% C (carbon) | 9 | 9/27 × 100% = 33.3% F (fluorine) | 6 | 6/27 × 100% = 22.2% H (hydrogen) | 10 | 10/27 × 100% = 37.0% 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 O (oxygen) | 2 | 7.41% | 15.999 C (carbon) | 9 | 33.3% | 12.011 F (fluorine) | 6 | 22.2% | 18.998403163 H (hydrogen) | 10 | 37.0% | 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 O (oxygen) | 2 | 7.41% | 15.999 | 2 × 15.999 = 31.998 C (carbon) | 9 | 33.3% | 12.011 | 9 × 12.011 = 108.099 F (fluorine) | 6 | 22.2% | 18.998403163 | 6 × 18.998403163 = 113.990418978 H (hydrogen) | 10 | 37.0% | 1.008 | 10 × 1.008 = 10.080 m = 31.998 u + 108.099 u + 113.990418978 u + 10.080 u = 264.167418978 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 O (oxygen) | 2 | 7.41% | 31.998/264.167418978 C (carbon) | 9 | 33.3% | 108.099/264.167418978 F (fluorine) | 6 | 22.2% | 113.990418978/264.167418978 H (hydrogen) | 10 | 37.0% | 10.080/264.167418978 Check: 31.998/264.167418978 + 108.099/264.167418978 + 113.990418978/264.167418978 + 10.080/264.167418978 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent O (oxygen) | 2 | 7.41% | 31.998/264.167418978 × 100% = 12.11% C (carbon) | 9 | 33.3% | 108.099/264.167418978 × 100% = 40.92% F (fluorine) | 6 | 22.2% | 113.990418978/264.167418978 × 100% = 43.15% H (hydrogen) | 10 | 37.0% | 10.080/264.167418978 × 100% = 3.816%

The first step in finding the oxidation states (or oxidation numbers) in bis hydroxyhexafluoroisopropyl benzene is to draw the structure diagram. Next set every oxidation number equal to the atom's formal charge: In bis hydroxyhexafluoroisopropyl benzene 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 6 carbon-fluorine bonds, 2 carbon-oxygen bonds, and 9 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 -3 | C (carbon) | 2 -2 | O (oxygen) | 2 -1 | C (carbon) | 1 | F (fluorine) | 6 0 | C (carbon) | 1 +1 | C (carbon) | 3 | H (hydrogen) | 10 +2 | C (carbon) | 2

First draw the structure diagram for bis hydroxyhexafluoroisopropyl benzene, 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 | 27 edge count | 27 Schultz index | 5087 Wiener index | 1356 Hosoya index | 62384 Balaban index | 4.103