9,9-didodecyl-2,7-di-1-propynyl-9 H-fluorene

9, 9-didodecyl-2, 7-di-1-propynyl-9 H-fluorene

molar mass | 579 g/mol formula | C_43H_62 SMILES identifier | CCCCCCCCCCCCC1(CCCCCCCCCCCC)C2=C(C=CC(=C2)C#CC)C3=C1C=C(C#CC)C=C3 InChI identifier | InChI=1/C43H62/c1-5-9-11-13-15-17-19-21-23-25-33-43(34-26-24-22-20-18-16-14-12-10-6-2)41-35-37(27-7-3)29-31-39(41)40-32-30-38(28-8-4)36-42(40)43/h29-32, 35-36H, 5-6, 9-26, 33-34H2, 1-4H3 InChI key | BOIQMDSZNLONSM-UHFFFAOYSA-N

Draw the Lewis structure of 9, 9-didodecyl-2, 7-di-1-propynyl-9 H-fluorene. 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) and hydrogen (n_H, val = 1) atoms: 43 n_C, val + 62 n_H, val = 234 Calculate the number of electrons needed to completely fill the valence shells for carbon (n_C, full = 8) and hydrogen (n_H, full = 2): 43 n_C, full + 62 n_H, full = 468 Subtracting these two numbers shows that 468 - 234 = 234 bonding electrons are needed. Each bond has two electrons, so in addition to the 107 bonds already present in the diagram add 10 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 10 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 | 665.2 °C boiling point | 999.8 °C critical temperature | 1595 K critical pressure | 0.5411 MPa critical volume | 2123 cm^3/mol molar heat of vaporization | 121 kJ/mol molar heat of fusion | 95.92 kJ/mol molar enthalpy | 141 kJ/mol molar free energy | 982.5 kJ/mol (computed using the Joback method)

longest chain length | 25 atoms longest straight chain length | 12 atoms longest aliphatic chain length | 12 atoms aromatic atom count | 12 atoms H-bond acceptor count | 0 atoms H-bond donor count | 0 atoms

Find the elemental composition for 9, 9-didodecyl-2, 7-di-1-propynyl-9 H-fluorene 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_43H_62 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 C (carbon) | 43 H (hydrogen) | 62 N_atoms = 43 + 62 = 105 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 C (carbon) | 43 | 43/105 H (hydrogen) | 62 | 62/105 Check: 43/105 + 62/105 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent C (carbon) | 43 | 43/105 × 100% = 41.0% H (hydrogen) | 62 | 62/105 × 100% = 59.0% 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 C (carbon) | 43 | 41.0% | 12.011 H (hydrogen) | 62 | 59.0% | 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 C (carbon) | 43 | 41.0% | 12.011 | 43 × 12.011 = 516.473 H (hydrogen) | 62 | 59.0% | 1.008 | 62 × 1.008 = 62.496 m = 516.473 u + 62.496 u = 578.969 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 C (carbon) | 43 | 41.0% | 516.473/578.969 H (hydrogen) | 62 | 59.0% | 62.496/578.969 Check: 516.473/578.969 + 62.496/578.969 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent C (carbon) | 43 | 41.0% | 516.473/578.969 × 100% = 89.21% H (hydrogen) | 62 | 59.0% | 62.496/578.969 × 100% = 10.79%

The first step in finding the oxidation states (or oxidation numbers) in 9, 9-didodecyl-2, 7-di-1-propynyl-9 H-fluorene is to draw the structure diagram. Next set every oxidation number equal to the atom's formal charge: In 9, 9-didodecyl-2, 7-di-1-propynyl-9 H-fluorene 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: There are 45 carbon-carbon bonds. Since these are bonds between the same element, 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) | 4 -2 | C (carbon) | 22 -1 | C (carbon) | 6 0 | C (carbon) | 11 +1 | H (hydrogen) | 62

First draw the structure diagram for 9, 9-didodecyl-2, 7-di-1-propynyl-9 H-fluorene, 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: For 9, 9-didodecyl-2, 7-di-1-propynyl-9 H-fluorene there are no lone pairs, so the steric number is given by the σ-bond count. 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 | 43 edge count | 45 Schultz index | 30542 Wiener index | 7573 Hosoya index | 1.013×10^9 Balaban index | 1.603