3, 4, 5, 6-tetrabromo-o-cresol

3, 4, 5, 6-tetrabromo-o-cresol

molar mass | 423.7 g/mol formula | C_7H_4Br_4O empirical formula | Br_4C_7O_H_4 SMILES identifier | CC1=C(C(=C(C(=C1O)Br)Br)Br)Br InChI identifier | InChI=1/C7H4Br4O/c1-2-3(8)4(9)5(10)6(11)7(2)12/h12H, 1H3 InChI key | GGIDUULRWQOXLR-UHFFFAOYSA-N

Draw the Lewis structure of 3, 4, 5, 6-tetrabromo-o-cresol. 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: 4 n_Br, val + 7 n_C, val + 4 n_H, val + n_O, val = 66 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): 4 n_Br, full + 7 n_C, full + 4 n_H, full + n_O, full = 104 Subtracting these two numbers shows that 104 - 66 = 38 bonding electrons are needed. Each bond has two electrons, so in addition to the 16 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 | 322.9 °C boiling point | 478.3 °C critical temperature | 1039 K critical pressure | 7.014 MPa critical volume | 533.5 cm^3/mol molar heat of vaporization | 74.8 kJ/mol molar heat of fusion | 33.29 kJ/mol molar enthalpy | -69.1 kJ/mol molar free energy | -15.39 kJ/mol (computed using the Joback method)

longest chain length | 6 atoms longest straight chain length | 0 atoms longest aliphatic chain length | 0 atoms aromatic atom count | 6 atoms H-bond acceptor count | 1 atom H-bond donor count | 1 atom

Find the elemental composition for 3, 4, 5, 6-tetrabromo-o-cresol 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_7H_4Br_4O 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) | 4 C (carbon) | 7 O (oxygen) | 1 H (hydrogen) | 4 N_atoms = 4 + 7 + 1 + 4 = 16 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) | 4 | 4/16 C (carbon) | 7 | 7/16 O (oxygen) | 1 | 1/16 H (hydrogen) | 4 | 4/16 Check: 4/16 + 7/16 + 1/16 + 4/16 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent Br (bromine) | 4 | 4/16 × 100% = 25.0% C (carbon) | 7 | 7/16 × 100% = 43.8% O (oxygen) | 1 | 1/16 × 100% = 6.25% H (hydrogen) | 4 | 4/16 × 100% = 25.0% 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 Br (bromine) | 4 | 25.0% | 79.904 C (carbon) | 7 | 43.8% | 12.011 O (oxygen) | 1 | 6.25% | 15.999 H (hydrogen) | 4 | 25.0% | 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 Br (bromine) | 4 | 25.0% | 79.904 | 4 × 79.904 = 319.616 C (carbon) | 7 | 43.8% | 12.011 | 7 × 12.011 = 84.077 O (oxygen) | 1 | 6.25% | 15.999 | 1 × 15.999 = 15.999 H (hydrogen) | 4 | 25.0% | 1.008 | 4 × 1.008 = 4.032 m = 319.616 u + 84.077 u + 15.999 u + 4.032 u = 423.724 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) | 4 | 25.0% | 319.616/423.724 C (carbon) | 7 | 43.8% | 84.077/423.724 O (oxygen) | 1 | 6.25% | 15.999/423.724 H (hydrogen) | 4 | 25.0% | 4.032/423.724 Check: 319.616/423.724 + 84.077/423.724 + 15.999/423.724 + 4.032/423.724 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent Br (bromine) | 4 | 25.0% | 319.616/423.724 × 100% = 75.43% C (carbon) | 7 | 43.8% | 84.077/423.724 × 100% = 19.84% O (oxygen) | 1 | 6.25% | 15.999/423.724 × 100% = 3.776% H (hydrogen) | 4 | 25.0% | 4.032/423.724 × 100% = 0.9516%

The first step in finding the oxidation states (or oxidation numbers) in 3, 4, 5, 6-tetrabromo-o-cresol is to draw the structure diagram. Next set every oxidation number equal to the atom's formal charge: In 3, 4, 5, 6-tetrabromo-o-cresol 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 4 bromine-carbon bonds, 1 carbon-oxygen bond, and 7 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 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) | 1 -2 | O (oxygen) | 1 -1 | Br (bromine) | 4 0 | C (carbon) | 1 +1 | C (carbon) | 5 | H (hydrogen) | 4

First draw the structure diagram for 3, 4, 5, 6-tetrabromo-o-cresol, 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 Assign the provisional hybridization based on the table: Next identify any sp^3 atoms with lone pair electrons which can participate in a conjugated π-bond system. These atoms can lower their energy by placing a lone pair in a unhybridized p orbital to maximize overlap with the neighboring π-bonds. Note that halogens and elements from the third period and below do not engage in bond conjugation, except in the case of aromaticity: Adjust the provisional hybridizations to arrive at the result: Answer: | |

vertex count | 16 edge count | 16 Schultz index | 1466 Wiener index | 379 Hosoya index | 971 Balaban index | 3.032