3-bromo-2, 6-difluorophenylboronic acid

3-bromo-2, 6-difluorophenylboronic acid

molar mass | 236.8 g/mol formula | C_6H_4BBrF_2O_2 empirical formula | Br_C_6F_2B_O_2H_4 SMILES identifier | C1=CC(=C(C(=C1Br)F)B(O)O)F InChI identifier | InChI=1/C6H4BBrF2O2/c8-3-1-2-4(9)5(6(3)10)7(11)12/h1-2, 11-12H InChI key | IOPHTXLBAFNMMI-UHFFFAOYSA-N

Draw the Lewis structure of 3-bromo-2, 6-difluorophenylboronic acid. Start by drawing the overall structure of the molecule, ignoring potential double and triple bonds: Count the total valence electrons of the boron (n_B, val = 3), 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: n_B, val + n_Br, val + 6 n_C, val + 2 n_F, val + 4 n_H, val + 2 n_O, val = 64 Calculate the number of electrons needed to completely fill the valence shells for boron (n_B, full = 6), 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): n_B, full + n_Br, full + 6 n_C, full + 2 n_F, full + 4 n_H, full + 2 n_O, full = 102 Subtracting these two numbers shows that 102 - 64 = 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: | |

longest chain length | 6 atoms longest straight chain length | 3 atoms longest aliphatic chain length | 0 atoms aromatic atom count | 6 atoms H-bond acceptor count | 2 atoms H-bond donor count | 2 atoms

Find the elemental composition for 3-bromo-2, 6-difluorophenylboronic acid 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_6H_4BBrF_2O_2 Use the chemical formula, C_6H_4BBrF_2O_2, 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) | 1 C (carbon) | 6 F (fluorine) | 2 B (boron) | 1 O (oxygen) | 2 H (hydrogen) | 4 N_atoms = 1 + 6 + 2 + 1 + 2 + 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) | 1 | 1/16 C (carbon) | 6 | 6/16 F (fluorine) | 2 | 2/16 B (boron) | 1 | 1/16 O (oxygen) | 2 | 2/16 H (hydrogen) | 4 | 4/16 Check: 1/16 + 6/16 + 2/16 + 1/16 + 2/16 + 4/16 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent Br (bromine) | 1 | 1/16 × 100% = 6.25% C (carbon) | 6 | 6/16 × 100% = 37.5% F (fluorine) | 2 | 2/16 × 100% = 12.5% B (boron) | 1 | 1/16 × 100% = 6.25% O (oxygen) | 2 | 2/16 × 100% = 12.5% 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) | 1 | 6.25% | 79.904 C (carbon) | 6 | 37.5% | 12.011 F (fluorine) | 2 | 12.5% | 18.998403163 B (boron) | 1 | 6.25% | 10.81 O (oxygen) | 2 | 12.5% | 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) | 1 | 6.25% | 79.904 | 1 × 79.904 = 79.904 C (carbon) | 6 | 37.5% | 12.011 | 6 × 12.011 = 72.066 F (fluorine) | 2 | 12.5% | 18.998403163 | 2 × 18.998403163 = 37.996806326 B (boron) | 1 | 6.25% | 10.81 | 1 × 10.81 = 10.81 O (oxygen) | 2 | 12.5% | 15.999 | 2 × 15.999 = 31.998 H (hydrogen) | 4 | 25.0% | 1.008 | 4 × 1.008 = 4.032 m = 79.904 u + 72.066 u + 37.996806326 u + 10.81 u + 31.998 u + 4.032 u = 236.806806326 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) | 1 | 6.25% | 79.904/236.806806326 C (carbon) | 6 | 37.5% | 72.066/236.806806326 F (fluorine) | 2 | 12.5% | 37.996806326/236.806806326 B (boron) | 1 | 6.25% | 10.81/236.806806326 O (oxygen) | 2 | 12.5% | 31.998/236.806806326 H (hydrogen) | 4 | 25.0% | 4.032/236.806806326 Check: 79.904/236.806806326 + 72.066/236.806806326 + 37.996806326/236.806806326 + 10.81/236.806806326 + 31.998/236.806806326 + 4.032/236.806806326 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent Br (bromine) | 1 | 6.25% | 79.904/236.806806326 × 100% = 33.74% C (carbon) | 6 | 37.5% | 72.066/236.806806326 × 100% = 30.43% F (fluorine) | 2 | 12.5% | 37.996806326/236.806806326 × 100% = 16.05% B (boron) | 1 | 6.25% | 10.81/236.806806326 × 100% = 4.565% O (oxygen) | 2 | 12.5% | 31.998/236.806806326 × 100% = 13.51% H (hydrogen) | 4 | 25.0% | 4.032/236.806806326 × 100% = 1.703%

The first step in finding the oxidation states (or oxidation numbers) in 3-bromo-2, 6-difluorophenylboronic acid is to draw the structure diagram. Next set every oxidation number equal to the atom's formal charge: In 3-bromo-2, 6-difluorophenylboronic acid 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 boron-carbon bond, 2 boron-oxygen bonds, 1 bromine-carbon bond, 2 carbon-fluorine bonds, and 6 carbon-carbon bonds. For each of these bonds, assign the bonding electrons to the most electronegative element. First examine the boron-carbon bond: element | electronegativity (Pauling scale) | B | 2.04 | C | 2.55 | | | Since carbon is more electronegative than boron, the electrons in this bond will go to carbon. Decrease the oxidation number for carbon (by 1 for single bonds, 2 for double bonds, and 3 for triple bonds), and increase the oxidation number for boron accordingly: Next look at the boron-oxygen bonds: element | electronegativity (Pauling scale) | B | 2.04 | O | 3.44 | | | Since oxygen is more electronegative than boron, the electrons in these bonds will go to oxygen: Next look at the bromine-carbon bond: element | electronegativity (Pauling scale) | Br | 2.96 | C | 2.55 | | | Since bromine is more electronegative than carbon, the electrons in this bond will go to bromine: 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 | O (oxygen) | 2 -1 | Br (bromine) | 1 | C (carbon) | 3 | F (fluorine) | 2 +1 | C (carbon) | 3 | H (hydrogen) | 4 +3 | B (boron) | 1

First draw the structure diagram for 3-bromo-2, 6-difluorophenylboronic acid, 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 | 16 edge count | 16 Schultz index | 1534 Wiener index | 396 Hosoya index | 1304 Balaban index | 2.893