1,1'-(4-chlorobutylidene)bis(4-fluorobenzene)

1, 1'-(4-chlorobutylidene)bis(4-fluorobenzene)

molar mass | 280.7 g/mol formula | C_16H_15ClF_2 empirical formula | Cl_C_16F_2H_15 SMILES identifier | C(CC(C1=CC=C(C=C1)F)C2=CC=C(C=C2)F)CCl InChI identifier | InChI=1/C16H15ClF2/c17-11-1-2-16(12-3-7-14(18)8-4-12)13-5-9-15(19)10-6-13/h3-10, 16H, 1-2, 11H2 InChI key | UXXLTPGCINZEFM-UHFFFAOYSA-N

Draw the Lewis structure of 1, 1'-(4-chlorobutylidene)bis(4-fluorobenzene). 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), chlorine (n_Cl, val = 7), fluorine (n_F, val = 7), and hydrogen (n_H, val = 1) atoms: 16 n_C, val + n_Cl, val + 2 n_F, val + 15 n_H, val = 100 Calculate the number of electrons needed to completely fill the valence shells for carbon (n_C, full = 8), chlorine (n_Cl, full = 8), fluorine (n_F, full = 8), and hydrogen (n_H, full = 2): 16 n_C, full + n_Cl, full + 2 n_F, full + 15 n_H, full = 182 Subtracting these two numbers shows that 182 - 100 = 82 bonding electrons are needed. Each bond has two electrons, so in addition to the 35 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 | 90.91 °C boiling point | 391.2 °C critical temperature | 883 K critical pressure | 1.991 MPa critical volume | 794.5 cm^3/mol molar heat of vaporization | 59.4 kJ/mol molar heat of fusion | 31.32 kJ/mol molar enthalpy | -336.7 kJ/mol molar free energy | -114.6 kJ/mol (computed using the Joback method)

longest chain length | 11 atoms longest straight chain length | 5 atoms longest aliphatic chain length | 4 atoms aromatic atom count | 12 atoms H-bond acceptor count | 0 atoms H-bond donor count | 0 atoms

Find the elemental composition for 1, 1'-(4-chlorobutylidene)bis(4-fluorobenzene) 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_16H_15ClF_2 Use the chemical formula, C_16H_15ClF_2, to count the number of atoms, N_i, for each element and find the total number of atoms, N_atoms: | number of atoms Cl (chlorine) | 1 C (carbon) | 16 F (fluorine) | 2 H (hydrogen) | 15 N_atoms = 1 + 16 + 2 + 15 = 34 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 Cl (chlorine) | 1 | 1/34 C (carbon) | 16 | 16/34 F (fluorine) | 2 | 2/34 H (hydrogen) | 15 | 15/34 Check: 1/34 + 16/34 + 2/34 + 15/34 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent Cl (chlorine) | 1 | 1/34 × 100% = 2.94% C (carbon) | 16 | 16/34 × 100% = 47.1% F (fluorine) | 2 | 2/34 × 100% = 5.88% H (hydrogen) | 15 | 15/34 × 100% = 44.1% 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 Cl (chlorine) | 1 | 2.94% | 35.45 C (carbon) | 16 | 47.1% | 12.011 F (fluorine) | 2 | 5.88% | 18.998403163 H (hydrogen) | 15 | 44.1% | 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 Cl (chlorine) | 1 | 2.94% | 35.45 | 1 × 35.45 = 35.45 C (carbon) | 16 | 47.1% | 12.011 | 16 × 12.011 = 192.176 F (fluorine) | 2 | 5.88% | 18.998403163 | 2 × 18.998403163 = 37.996806326 H (hydrogen) | 15 | 44.1% | 1.008 | 15 × 1.008 = 15.120 m = 35.45 u + 192.176 u + 37.996806326 u + 15.120 u = 280.742806326 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 Cl (chlorine) | 1 | 2.94% | 35.45/280.742806326 C (carbon) | 16 | 47.1% | 192.176/280.742806326 F (fluorine) | 2 | 5.88% | 37.996806326/280.742806326 H (hydrogen) | 15 | 44.1% | 15.120/280.742806326 Check: 35.45/280.742806326 + 192.176/280.742806326 + 37.996806326/280.742806326 + 15.120/280.742806326 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent Cl (chlorine) | 1 | 2.94% | 35.45/280.742806326 × 100% = 12.63% C (carbon) | 16 | 47.1% | 192.176/280.742806326 × 100% = 68.45% F (fluorine) | 2 | 5.88% | 37.996806326/280.742806326 × 100% = 13.53% H (hydrogen) | 15 | 44.1% | 15.120/280.742806326 × 100% = 5.386%

The first step in finding the oxidation states (or oxidation numbers) in 1, 1'-(4-chlorobutylidene)bis(4-fluorobenzene) is to draw the structure diagram. Next set every oxidation number equal to the atom's formal charge: In 1, 1'-(4-chlorobutylidene)bis(4-fluorobenzene) 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 1 carbon-chlorine bond, 2 carbon-fluorine bonds, and 17 carbon-carbon bonds. For each of these bonds, assign the bonding electrons to the most electronegative element. First examine the carbon-chlorine bond: element | electronegativity (Pauling scale) | C | 2.55 | Cl | 3.16 | | | Since chlorine is more electronegative than carbon, the electrons in this bond will go to chlorine. Decrease the oxidation number for chlorine (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-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: 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 -1 | C (carbon) | 10 | Cl (chlorine) | 1 | F (fluorine) | 2 0 | C (carbon) | 2 +1 | C (carbon) | 2 | H (hydrogen) | 15

First draw the structure diagram for 1, 1'-(4-chlorobutylidene)bis(4-fluorobenzene), 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 | 34 edge count | 35 Schultz index | 10400 Wiener index | 2657 Hosoya index | 2.393×10^6 Balaban index | 2.895