difluoro isopropoxyphenylboronic acid

difluoro isopropoxyphenylboronic acid

molar mass | 216 g/mol formula | C_9H_11BF_2O_3 empirical formula | F_2C_9B_O_3H_11 SMILES identifier | CC(C)OOB(C1=C(C(=CC=C1)F)F)O InChI identifier | InChI=1/C9H11BF2O3/c1-6(2)14-15-10(13)7-4-3-5-8(11)9(7)12/h3-6, 13H, 1-2H3 InChI key | GULUFKUCQNFDLC-UHFFFAOYSA-N

Draw the Lewis structure of difluoro isopropoxyphenylboronic 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), 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 + 9 n_C, val + 2 n_F, val + 11 n_H, val + 3 n_O, val = 82 Calculate the number of electrons needed to completely fill the valence shells for boron (n_B, full = 6), carbon (n_C, full = 8), fluorine (n_F, full = 8), hydrogen (n_H, full = 2), and oxygen (n_O, full = 8): n_B, full + 9 n_C, full + 2 n_F, full + 11 n_H, full + 3 n_O, full = 140 Subtracting these two numbers shows that 140 - 82 = 58 bonding electrons are needed. Each bond has two electrons, so in addition to the 26 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 | 9 atoms longest straight chain length | 6 atoms longest aliphatic chain length | 3 atoms aromatic atom count | 6 atoms H-bond acceptor count | 3 atoms H-bond donor count | 1 atom

Find the elemental composition for difluoro isopropoxyphenylboronic 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_9H_11BF_2O_3 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 F (fluorine) | 2 C (carbon) | 9 B (boron) | 1 O (oxygen) | 3 H (hydrogen) | 11 N_atoms = 2 + 9 + 1 + 3 + 11 = 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 F (fluorine) | 2 | 2/26 C (carbon) | 9 | 9/26 B (boron) | 1 | 1/26 O (oxygen) | 3 | 3/26 H (hydrogen) | 11 | 11/26 Check: 2/26 + 9/26 + 1/26 + 3/26 + 11/26 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent F (fluorine) | 2 | 2/26 × 100% = 7.69% C (carbon) | 9 | 9/26 × 100% = 34.6% B (boron) | 1 | 1/26 × 100% = 3.85% O (oxygen) | 3 | 3/26 × 100% = 11.5% H (hydrogen) | 11 | 11/26 × 100% = 42.3% 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 F (fluorine) | 2 | 7.69% | 18.998403163 C (carbon) | 9 | 34.6% | 12.011 B (boron) | 1 | 3.85% | 10.81 O (oxygen) | 3 | 11.5% | 15.999 H (hydrogen) | 11 | 42.3% | 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 F (fluorine) | 2 | 7.69% | 18.998403163 | 2 × 18.998403163 = 37.996806326 C (carbon) | 9 | 34.6% | 12.011 | 9 × 12.011 = 108.099 B (boron) | 1 | 3.85% | 10.81 | 1 × 10.81 = 10.81 O (oxygen) | 3 | 11.5% | 15.999 | 3 × 15.999 = 47.997 H (hydrogen) | 11 | 42.3% | 1.008 | 11 × 1.008 = 11.088 m = 37.996806326 u + 108.099 u + 10.81 u + 47.997 u + 11.088 u = 215.990806326 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 F (fluorine) | 2 | 7.69% | 37.996806326/215.990806326 C (carbon) | 9 | 34.6% | 108.099/215.990806326 B (boron) | 1 | 3.85% | 10.81/215.990806326 O (oxygen) | 3 | 11.5% | 47.997/215.990806326 H (hydrogen) | 11 | 42.3% | 11.088/215.990806326 Check: 37.996806326/215.990806326 + 108.099/215.990806326 + 10.81/215.990806326 + 47.997/215.990806326 + 11.088/215.990806326 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent F (fluorine) | 2 | 7.69% | 37.996806326/215.990806326 × 100% = 17.59% C (carbon) | 9 | 34.6% | 108.099/215.990806326 × 100% = 50.05% B (boron) | 1 | 3.85% | 10.81/215.990806326 × 100% = 5.005% O (oxygen) | 3 | 11.5% | 47.997/215.990806326 × 100% = 22.22% H (hydrogen) | 11 | 42.3% | 11.088/215.990806326 × 100% = 5.134%

The first step in finding the oxidation states (or oxidation numbers) in difluoro isopropoxyphenylboronic acid is to draw the structure diagram. Next set every oxidation number equal to the atom's formal charge: In difluoro isopropoxyphenylboronic 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, 2 carbon-fluorine bonds, 1 carbon-oxygen bond, 8 carbon-carbon bonds, and 1 oxygen-oxygen bond. 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 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-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: Next look at the oxygen-oxygen bond: element | electronegativity (Pauling scale) | O | 3.44 | O | 3.44 | | | 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) | 2 -2 | O (oxygen) | 1 -1 | C (carbon) | 4 | F (fluorine) | 2 | O (oxygen) | 2 0 | C (carbon) | 1 +1 | C (carbon) | 2 | H (hydrogen) | 11 +3 | B (boron) | 1

First draw the structure diagram for difluoro isopropoxyphenylboronic 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 | 26 edge count | 26 Schultz index | 6026 Wiener index | 1567 Hosoya index | 64640 Balaban index | 3.054