4-(3-bromobenzyloxy)-3-chlorophenylboronic acid

4-(3-bromobenzyloxy)-3-chlorophenylboronic acid

molar mass | 341.4 g/mol formula | C_13H_11BBrClO_3 empirical formula | Br_C_13O_3B_Cl_H_11 SMILES identifier | C1=CC(=CC(=C1)Br)COC2=C(C=C(C=C2)B(O)O)Cl InChI identifier | InChI=1/C13H11BBrClO3/c15-11-3-1-2-9(6-11)8-19-13-5-4-10(14(17)18)7-12(13)16/h1-7, 17-18H, 8H2 InChI key | HKBRUUCYKMOURT-UHFFFAOYSA-N

Draw the Lewis structure of 4-(3-bromobenzyloxy)-3-chlorophenylboronic 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), chlorine (n_Cl, val = 7), hydrogen (n_H, val = 1), and oxygen (n_O, val = 6) atoms: n_B, val + n_Br, val + 13 n_C, val + n_Cl, val + 11 n_H, val + 3 n_O, val = 98 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), chlorine (n_Cl, full = 8), hydrogen (n_H, full = 2), and oxygen (n_O, full = 8): n_B, full + n_Br, full + 13 n_C, full + n_Cl, full + 11 n_H, full + 3 n_O, full = 172 Subtracting these two numbers shows that 172 - 98 = 74 bonding electrons are needed. Each bond has two electrons, so in addition to the 31 bonds already present in the diagram add 6 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 6 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: | |

longest chain length | 12 atoms longest straight chain length | 3 atoms longest aliphatic chain length | 0 atoms aromatic atom count | 12 atoms H-bond acceptor count | 2 atoms H-bond donor count | 2 atoms

Find the elemental composition for 4-(3-bromobenzyloxy)-3-chlorophenylboronic 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_13H_11BBrClO_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 Br (bromine) | 1 C (carbon) | 13 O (oxygen) | 3 B (boron) | 1 Cl (chlorine) | 1 H (hydrogen) | 11 N_atoms = 1 + 13 + 3 + 1 + 1 + 11 = 30 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 | 1/30 C (carbon) | 13 | 13/30 O (oxygen) | 3 | 3/30 B (boron) | 1 | 1/30 Cl (chlorine) | 1 | 1/30 H (hydrogen) | 11 | 11/30 Check: 1/30 + 13/30 + 3/30 + 1/30 + 1/30 + 11/30 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent Br (bromine) | 1 | 1/30 × 100% = 3.33% C (carbon) | 13 | 13/30 × 100% = 43.3% O (oxygen) | 3 | 3/30 × 100% = 10.00% B (boron) | 1 | 1/30 × 100% = 3.33% Cl (chlorine) | 1 | 1/30 × 100% = 3.33% H (hydrogen) | 11 | 11/30 × 100% = 36.7% 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 | 3.33% | 79.904 C (carbon) | 13 | 43.3% | 12.011 O (oxygen) | 3 | 10.00% | 15.999 B (boron) | 1 | 3.33% | 10.81 Cl (chlorine) | 1 | 3.33% | 35.45 H (hydrogen) | 11 | 36.7% | 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 | 3.33% | 79.904 | 1 × 79.904 = 79.904 C (carbon) | 13 | 43.3% | 12.011 | 13 × 12.011 = 156.143 O (oxygen) | 3 | 10.00% | 15.999 | 3 × 15.999 = 47.997 B (boron) | 1 | 3.33% | 10.81 | 1 × 10.81 = 10.81 Cl (chlorine) | 1 | 3.33% | 35.45 | 1 × 35.45 = 35.45 H (hydrogen) | 11 | 36.7% | 1.008 | 11 × 1.008 = 11.088 m = 79.904 u + 156.143 u + 47.997 u + 10.81 u + 35.45 u + 11.088 u = 341.392 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 | 3.33% | 79.904/341.392 C (carbon) | 13 | 43.3% | 156.143/341.392 O (oxygen) | 3 | 10.00% | 47.997/341.392 B (boron) | 1 | 3.33% | 10.81/341.392 Cl (chlorine) | 1 | 3.33% | 35.45/341.392 H (hydrogen) | 11 | 36.7% | 11.088/341.392 Check: 79.904/341.392 + 156.143/341.392 + 47.997/341.392 + 10.81/341.392 + 35.45/341.392 + 11.088/341.392 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent Br (bromine) | 1 | 3.33% | 79.904/341.392 × 100% = 23.41% C (carbon) | 13 | 43.3% | 156.143/341.392 × 100% = 45.74% O (oxygen) | 3 | 10.00% | 47.997/341.392 × 100% = 14.06% B (boron) | 1 | 3.33% | 10.81/341.392 × 100% = 3.166% Cl (chlorine) | 1 | 3.33% | 35.45/341.392 × 100% = 10.38% H (hydrogen) | 11 | 36.7% | 11.088/341.392 × 100% = 3.248%

The first step in finding the oxidation states (or oxidation numbers) in 4-(3-bromobenzyloxy)-3-chlorophenylboronic acid is to draw the structure diagram. Next set every oxidation number equal to the atom's formal charge: In 4-(3-bromobenzyloxy)-3-chlorophenylboronic 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, 1 carbon-carbonl bond, 2 carbon-oxygen bonds, and 13 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-carbonl bond: element | electronegativity (Pauling scale) | C | 2.55 | Cl | 3.16 | | | Since chlorine is more electronegative than carbon, the electrons in this bond will go to chlorine: 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) | 3 -1 | Br (bromine) | 1 | C (carbon) | 9 | Cl (chlorine) | 1 0 | C (carbon) | 1 +1 | C (carbon) | 3 | H (hydrogen) | 11 +3 | B (boron) | 1

First draw the structure diagram for 4-(3-bromobenzyloxy)-3-chlorophenylboronic 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 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 | 30 edge count | 31 Schultz index | 9168 Wiener index | 2304 Hosoya index | 700048 Balaban index | 2.272