4, 4'-dibromobenzil

4, 4'-dibromobenzil

molar mass | 368 g/mol formula | C_14H_8Br_2O_2 empirical formula | Br_C_7O_H_4 SMILES identifier | C1=C(C=CC(=C1)Br)C(=O)C(=O)C2=CC=C(C=C2)Br InChI identifier | InChI=1/C14H8Br2O2/c15-11-5-1-9(2-6-11)13(17)14(18)10-3-7-12(16)8-4-10/h1-8H InChI key | NYCBYBDDECLFPE-UHFFFAOYSA-N

Draw the Lewis structure of 4, 4'-dibromobenzil. Start by drawing the overall structure of the molecule, ignoring potential double and triple bonds: Count the total valence electrons of the bromine (n_Br, val = 7), carbon (n_C, val = 4), hydrogen (n_H, val = 1), and oxygen (n_O, val = 6) atoms: 2 n_Br, val + 14 n_C, val + 8 n_H, val + 2 n_O, val = 90 Calculate the number of electrons needed to completely fill the valence shells for bromine (n_Br, full = 8), carbon (n_C, full = 8), hydrogen (n_H, full = 2), and oxygen (n_O, full = 8): 2 n_Br, full + 14 n_C, full + 8 n_H, full + 2 n_O, full = 160 Subtracting these two numbers shows that 160 - 90 = 70 bonding electrons are needed. Each bond has two electrons, so in addition to the 27 bonds already present in the diagram add 8 bonds. To minimize formal charge carbon wants 4 bonds and oxygen wants 2 bonds. Identify the atoms that want additional bonds and the number of electrons remaining on each atom: Fill in the 8 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 | 271.7 °C boiling point | 550 °C critical temperature | 1096 K critical pressure | 3.603 MPa critical volume | 739.5 cm^3/mol molar heat of vaporization | 79 kJ/mol molar heat of fusion | 33.08 kJ/mol molar enthalpy | -54.7 kJ/mol molar free energy | 43.36 kJ/mol (computed using the Joback method)

longest chain length | 12 atoms longest straight chain length | 4 atoms longest aliphatic chain length | 2 atoms aromatic atom count | 12 atoms H-bond acceptor count | 2 atoms H-bond donor count | 0 atoms

Find the elemental composition for 4, 4'-dibromobenzil 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_14H_8Br_2O_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 Br (bromine) | 2 C (carbon) | 14 O (oxygen) | 2 H (hydrogen) | 8 N_atoms = 2 + 14 + 2 + 8 = 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 Br (bromine) | 2 | 2/26 C (carbon) | 14 | 14/26 O (oxygen) | 2 | 2/26 H (hydrogen) | 8 | 8/26 Check: 2/26 + 14/26 + 2/26 + 8/26 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent Br (bromine) | 2 | 2/26 × 100% = 7.69% C (carbon) | 14 | 14/26 × 100% = 53.8% O (oxygen) | 2 | 2/26 × 100% = 7.69% H (hydrogen) | 8 | 8/26 × 100% = 30.8% 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 Br (bromine) | 2 | 7.69% | 79.904 C (carbon) | 14 | 53.8% | 12.011 O (oxygen) | 2 | 7.69% | 15.999 H (hydrogen) | 8 | 30.8% | 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 Br (bromine) | 2 | 7.69% | 79.904 | 2 × 79.904 = 159.808 C (carbon) | 14 | 53.8% | 12.011 | 14 × 12.011 = 168.154 O (oxygen) | 2 | 7.69% | 15.999 | 2 × 15.999 = 31.998 H (hydrogen) | 8 | 30.8% | 1.008 | 8 × 1.008 = 8.064 m = 159.808 u + 168.154 u + 31.998 u + 8.064 u = 368.024 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 Br (bromine) | 2 | 7.69% | 159.808/368.024 C (carbon) | 14 | 53.8% | 168.154/368.024 O (oxygen) | 2 | 7.69% | 31.998/368.024 H (hydrogen) | 8 | 30.8% | 8.064/368.024 Check: 159.808/368.024 + 168.154/368.024 + 31.998/368.024 + 8.064/368.024 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent Br (bromine) | 2 | 7.69% | 159.808/368.024 × 100% = 43.42% C (carbon) | 14 | 53.8% | 168.154/368.024 × 100% = 45.69% O (oxygen) | 2 | 7.69% | 31.998/368.024 × 100% = 8.695% H (hydrogen) | 8 | 30.8% | 8.064/368.024 × 100% = 2.191%

The first step in finding the oxidation states (or oxidation numbers) in 4, 4'-dibromobenzil is to draw the structure diagram. Next set every oxidation number equal to the atom's formal charge: In 4, 4'-dibromobenzil 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 2 bromine-carbon bonds, 2 carbon-oxygen bonds, and 15 carbon-carbon bonds. For each of these bonds, assign the bonding electrons to the most electronegative element. First examine the bromine-carbon bonds: element | electronegativity (Pauling scale) | Br | 2.96 | C | 2.55 | | | Since bromine is more electronegative than carbon, the electrons in these bonds will go to bromine. Decrease the oxidation number for bromine 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 | Br (bromine) | 2 | C (carbon) | 8 0 | C (carbon) | 2 +1 | C (carbon) | 2 | H (hydrogen) | 8 +2 | C (carbon) | 2

First draw the structure diagram for 4, 4'-dibromobenzil, 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 | 27 Schultz index | 6100 Wiener index | 1519 Hosoya index | 122660 Balaban index | 2.258