2 hydroxyhexafluoroisopropyl benzene

2 hydroxyhexafluoroisopropyl benzene

molar mass | 248.2 g/mol formula | C_9H_10F_6O empirical formula | O_C_9F_6H_10 SMILES identifier | CC(C)C1(C=CC(C(C1(F)O)(F)F)(F)F)F InChI identifier | InChI=1/C9H10F6O/c1-5(2)6(10)3-4-7(11, 12)8(13, 14)9(6, 15)16/h3-5, 16H, 1-2H3 InChI key | HLVMVGYCHCWRKX-UHFFFAOYSA-N

Draw the Lewis structure of 2 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 + n_O, val = 94 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 + n_O, full = 148 Subtracting these two numbers shows that 148 - 94 = 54 bonding electrons are needed. Each bond has two electrons, so in addition to the 26 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 | 58.42 °C boiling point | 225.2 °C critical temperature | 667 K critical pressure | 2.685 MPa critical volume | 568.5 cm^3/mol molar heat of vaporization | 42.2 kJ/mol molar heat of fusion | 9.187 kJ/mol molar enthalpy | -1451 kJ/mol molar free energy | -1274 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 | 1 atom H-bond donor count | 1 atom

Find the elemental composition for 2 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 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) | 1 C (carbon) | 9 F (fluorine) | 6 H (hydrogen) | 10 N_atoms = 1 + 9 + 6 + 10 = 26 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) | 1 | 1/26 C (carbon) | 9 | 9/26 F (fluorine) | 6 | 6/26 H (hydrogen) | 10 | 10/26 Check: 1/26 + 9/26 + 6/26 + 10/26 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent O (oxygen) | 1 | 1/26 × 100% = 3.85% C (carbon) | 9 | 9/26 × 100% = 34.6% F (fluorine) | 6 | 6/26 × 100% = 23.1% H (hydrogen) | 10 | 10/26 × 100% = 38.5% 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) | 1 | 3.85% | 15.999 C (carbon) | 9 | 34.6% | 12.011 F (fluorine) | 6 | 23.1% | 18.998403163 H (hydrogen) | 10 | 38.5% | 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) | 1 | 3.85% | 15.999 | 1 × 15.999 = 15.999 C (carbon) | 9 | 34.6% | 12.011 | 9 × 12.011 = 108.099 F (fluorine) | 6 | 23.1% | 18.998403163 | 6 × 18.998403163 = 113.990418978 H (hydrogen) | 10 | 38.5% | 1.008 | 10 × 1.008 = 10.080 m = 15.999 u + 108.099 u + 113.990418978 u + 10.080 u = 248.168418978 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) | 1 | 3.85% | 15.999/248.168418978 C (carbon) | 9 | 34.6% | 108.099/248.168418978 F (fluorine) | 6 | 23.1% | 113.990418978/248.168418978 H (hydrogen) | 10 | 38.5% | 10.080/248.168418978 Check: 15.999/248.168418978 + 108.099/248.168418978 + 113.990418978/248.168418978 + 10.080/248.168418978 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent O (oxygen) | 1 | 3.85% | 15.999/248.168418978 × 100% = 6.447% C (carbon) | 9 | 34.6% | 108.099/248.168418978 × 100% = 43.56% F (fluorine) | 6 | 23.1% | 113.990418978/248.168418978 × 100% = 45.93% H (hydrogen) | 10 | 38.5% | 10.080/248.168418978 × 100% = 4.062%

The first step in finding the oxidation states (or oxidation numbers) in 2 hydroxyhexafluoroisopropyl benzene is to draw the structure diagram. Next set every oxidation number equal to the atom's formal charge: In 2 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, 1 carbon-oxygen bond, 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 bond: element | electronegativity (Pauling scale) | C | 2.55 | O | 3.44 | | | Since oxygen is more electronegative than carbon, the electrons in this bond 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) | 1 -1 | C (carbon) | 3 | F (fluorine) | 6 +1 | C (carbon) | 1 | H (hydrogen) | 10 +2 | C (carbon) | 3

First draw the structure diagram for 2 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 Now assign the hybridization for each atom: Answer: | |

vertex count | 26 edge count | 26 Schultz index | 4586 Wiener index | 1221 Hosoya index | 39784 Balaban index | 4.071