α,α,α',α'-tetrakis(trifluoromethyl)-1,4-benzenedimethanol

α, α, α', α'-tetrakis(trifluoromethyl)-1, 4-benzenedimethanol

molar mass | 410.2 g/mol formula | C_12H_6F_12O_2 empirical formula | F_6C_6O_H_3 SMILES identifier | C1=C(C=CC(=C1)C(C(F)(F)F)(C(F)(F)F)O)C(C(F)(F)F)(C(F)(F)F)O InChI identifier | InChI=1/C12H6F12O2/c13-9(14, 15)7(25, 10(16, 17)18)5-1-2-6(4-3-5)8(26, 11(19, 20)21)12(22, 23)24/h1-4, 25-26H InChI key | YTJDSANDEZLYOU-UHFFFAOYSA-N

Draw the Lewis structure of α, α, α', α'-tetrakis(trifluoromethyl)-1, 4-benzenedimethanol. 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: 12 n_C, val + 12 n_F, val + 6 n_H, val + 2 n_O, val = 150 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): 12 n_C, full + 12 n_F, full + 6 n_H, full + 2 n_O, full = 220 Subtracting these two numbers shows that 220 - 150 = 70 bonding electrons are needed. Each bond has two electrons, so in addition to the 32 bonds already present in the diagram add 3 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 3 bonds by pairing electrons between adjacent highlighted atoms. Note that the six atom ring is aromatic, so that the single and double bonds may be rearranged: Answer: | |

melting point | 134 °C boiling point | 388.7 °C critical temperature | 820.2 K critical pressure | 1.897 MPa critical volume | 787.5 cm^3/mol molar heat of vaporization | 61 kJ/mol molar heat of fusion | 21.13 kJ/mol molar enthalpy | -2776 kJ/mol molar free energy | -2441 kJ/mol (computed using the Joback method)

longest chain length | 10 atoms longest straight chain length | 5 atoms longest aliphatic chain length | 3 atoms aromatic atom count | 6 atoms H-bond acceptor count | 2 atoms H-bond donor count | 2 atoms

Find the elemental composition for α, α, α', α'-tetrakis(trifluoromethyl)-1, 4-benzenedimethanol 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_12H_6F_12O_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) | 12 C (carbon) | 12 O (oxygen) | 2 H (hydrogen) | 6 N_atoms = 12 + 12 + 2 + 6 = 32 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) | 12 | 12/32 C (carbon) | 12 | 12/32 O (oxygen) | 2 | 2/32 H (hydrogen) | 6 | 6/32 Check: 12/32 + 12/32 + 2/32 + 6/32 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent F (fluorine) | 12 | 12/32 × 100% = 37.5% C (carbon) | 12 | 12/32 × 100% = 37.5% O (oxygen) | 2 | 2/32 × 100% = 6.25% H (hydrogen) | 6 | 6/32 × 100% = 18.8% 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) | 12 | 37.5% | 18.998403163 C (carbon) | 12 | 37.5% | 12.011 O (oxygen) | 2 | 6.25% | 15.999 H (hydrogen) | 6 | 18.8% | 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) | 12 | 37.5% | 18.998403163 | 12 × 18.998403163 = 227.980837956 C (carbon) | 12 | 37.5% | 12.011 | 12 × 12.011 = 144.132 O (oxygen) | 2 | 6.25% | 15.999 | 2 × 15.999 = 31.998 H (hydrogen) | 6 | 18.8% | 1.008 | 6 × 1.008 = 6.048 m = 227.980837956 u + 144.132 u + 31.998 u + 6.048 u = 410.158837956 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) | 12 | 37.5% | 227.980837956/410.158837956 C (carbon) | 12 | 37.5% | 144.132/410.158837956 O (oxygen) | 2 | 6.25% | 31.998/410.158837956 H (hydrogen) | 6 | 18.8% | 6.048/410.158837956 Check: 227.980837956/410.158837956 + 144.132/410.158837956 + 31.998/410.158837956 + 6.048/410.158837956 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent F (fluorine) | 12 | 37.5% | 227.980837956/410.158837956 × 100% = 55.58% C (carbon) | 12 | 37.5% | 144.132/410.158837956 × 100% = 35.14% O (oxygen) | 2 | 6.25% | 31.998/410.158837956 × 100% = 7.801% H (hydrogen) | 6 | 18.8% | 6.048/410.158837956 × 100% = 1.475%

The first step in finding the oxidation states (or oxidation numbers) in α, α, α', α'-tetrakis(trifluoromethyl)-1, 4-benzenedimethanol is to draw the structure diagram. Next set every oxidation number equal to the atom's formal charge: In α, α, α', α'-tetrakis(trifluoromethyl)-1, 4-benzenedimethanol 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 12 carbon-fluorine bonds, 2 carbon-oxygen bonds, and 12 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 | O (oxygen) | 2 -1 | C (carbon) | 4 | F (fluorine) | 12 0 | C (carbon) | 2 +1 | C (carbon) | 2 | H (hydrogen) | 6 +3 | C (carbon) | 4

First draw the structure diagram for α, α, α', α'-tetrakis(trifluoromethyl)-1, 4-benzenedimethanol, 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 Now assign the hybridization for each atom: Answer: | |

vertex count | 32 edge count | 32 Schultz index | 8888 Wiener index | 2356 Hosoya index | 549888 Balaban index | 3.863