3, 6-dibromo-2-fluorobenzaldehyde

3, 6-dibromo-2-fluorobenzaldehyde

molar mass | 281.9 g/mol formula | C_7H_3Br_2FO empirical formula | Br_2C_7O_F_H_3 SMILES identifier | C1=C(C(=C(C(=C1)Br)F)C=O)Br InChI identifier | InChI=1/C7H3Br2FO/c8-5-1-2-6(9)7(10)4(5)3-11/h1-3H InChI key | IWLKVQWSDXDHAW-UHFFFAOYSA-N

Draw the Lewis structure of 3, 6-dibromo-2-fluorobenzaldehyde. 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), fluorine (n_F, val = 7), hydrogen (n_H, val = 1), and oxygen (n_O, val = 6) atoms: 2 n_Br, val + 7 n_C, val + n_F, val + 3 n_H, val + n_O, val = 58 Calculate the number of electrons needed to completely fill the valence shells for bromine (n_Br, full = 8), carbon (n_C, full = 8), fluorine (n_F, full = 8), hydrogen (n_H, full = 2), and oxygen (n_O, full = 8): 2 n_Br, full + 7 n_C, full + n_F, full + 3 n_H, full + n_O, full = 94 Subtracting these two numbers shows that 94 - 58 = 36 bonding electrons are needed. Each bond has two electrons, so in addition to the 14 bonds already present in the diagram add 4 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 4 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 | 121.7 °C boiling point | 308.2 °C critical temperature | 823.8 K critical pressure | 4.843 MPa critical volume | 478.5 cm^3/mol molar heat of vaporization | 54.2 kJ/mol molar heat of fusion | 22.69 kJ/mol molar enthalpy | -214.7 kJ/mol molar free energy | -174.1 kJ/mol (computed using the Joback method)

longest chain length | 6 atoms longest straight chain length | 2 atoms longest aliphatic chain length | 0 atoms aromatic atom count | 6 atoms H-bond acceptor count | 1 atom H-bond donor count | 0 atoms

Find the elemental composition for 3, 6-dibromo-2-fluorobenzaldehyde 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_3Br_2FO Use the chemical formula, C_7H_3Br_2FO, to count the number of atoms, N_i, for each element and find the total number of atoms, N_atoms: | number of atoms Br (bromine) | 2 C (carbon) | 7 O (oxygen) | 1 F (fluorine) | 1 H (hydrogen) | 3 N_atoms = 2 + 7 + 1 + 1 + 3 = 14 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 | 2/14 C (carbon) | 7 | 7/14 O (oxygen) | 1 | 1/14 F (fluorine) | 1 | 1/14 H (hydrogen) | 3 | 3/14 Check: 2/14 + 7/14 + 1/14 + 1/14 + 3/14 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent Br (bromine) | 2 | 2/14 × 100% = 14.3% C (carbon) | 7 | 7/14 × 100% = 50.0% O (oxygen) | 1 | 1/14 × 100% = 7.14% F (fluorine) | 1 | 1/14 × 100% = 7.14% H (hydrogen) | 3 | 3/14 × 100% = 21.4% 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 | 14.3% | 79.904 C (carbon) | 7 | 50.0% | 12.011 O (oxygen) | 1 | 7.14% | 15.999 F (fluorine) | 1 | 7.14% | 18.998403163 H (hydrogen) | 3 | 21.4% | 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 | 14.3% | 79.904 | 2 × 79.904 = 159.808 C (carbon) | 7 | 50.0% | 12.011 | 7 × 12.011 = 84.077 O (oxygen) | 1 | 7.14% | 15.999 | 1 × 15.999 = 15.999 F (fluorine) | 1 | 7.14% | 18.998403163 | 1 × 18.998403163 = 18.998403163 H (hydrogen) | 3 | 21.4% | 1.008 | 3 × 1.008 = 3.024 m = 159.808 u + 84.077 u + 15.999 u + 18.998403163 u + 3.024 u = 281.906403163 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 | 14.3% | 159.808/281.906403163 C (carbon) | 7 | 50.0% | 84.077/281.906403163 O (oxygen) | 1 | 7.14% | 15.999/281.906403163 F (fluorine) | 1 | 7.14% | 18.998403163/281.906403163 H (hydrogen) | 3 | 21.4% | 3.024/281.906403163 Check: 159.808/281.906403163 + 84.077/281.906403163 + 15.999/281.906403163 + 18.998403163/281.906403163 + 3.024/281.906403163 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent Br (bromine) | 2 | 14.3% | 159.808/281.906403163 × 100% = 56.69% C (carbon) | 7 | 50.0% | 84.077/281.906403163 × 100% = 29.82% O (oxygen) | 1 | 7.14% | 15.999/281.906403163 × 100% = 5.675% F (fluorine) | 1 | 7.14% | 18.998403163/281.906403163 × 100% = 6.739% H (hydrogen) | 3 | 21.4% | 3.024/281.906403163 × 100% = 1.073%

The first step in finding the oxidation states (or oxidation numbers) in 3, 6-dibromo-2-fluorobenzaldehyde is to draw the structure diagram. Next set every oxidation number equal to the atom's formal charge: In 3, 6-dibromo-2-fluorobenzaldehyde 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, 1 carbon-fluorine bond, 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-fluorine bond: element | electronegativity (Pauling scale) | C | 2.55 | F | 3.98 | | | Since fluorine is more electronegative than carbon, the electrons in this bond will go to fluorine: 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 -2 | O (oxygen) | 1 -1 | Br (bromine) | 2 | C (carbon) | 2 | F (fluorine) | 1 0 | C (carbon) | 1 +1 | C (carbon) | 4 | H (hydrogen) | 3

First draw the structure diagram for 3, 6-dibromo-2-fluorobenzaldehyde, 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 | 14 edge count | 14 Schultz index | 1050 Wiener index | 268 Hosoya index | 454 Balaban index | 2.861