( 4-(trifluoromethoxy)benzylboronic acid pinacol ester)

( 4-(trifluoromethoxy)benzylboronic acid pinacol ester)

molar mass | 302.1 g/mol formula | C_14H_18BF_3O_3 empirical formula | F_3C_14O_3B_H_18 SMILES identifier | CC1(C)C(C)(C)OB(CC2=CC=C(C=C2)OC(F)(F)F)O1 InChI identifier | InChI=1/C14H18BF3O3/c1-12(2)13(3, 4)21-15(20-12)9-10-5-7-11(8-6-10)19-14(16, 17)18/h5-8H, 9H2, 1-4H3 InChI key | CQXXCBRJJSMFRZ-UHFFFAOYSA-N

Draw the Lewis structure of ( 4-(trifluoromethoxy)benzylboronic acid pinacol ester). 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 + 14 n_C, val + 3 n_F, val + 18 n_H, val + 3 n_O, val = 116 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 + 14 n_C, full + 3 n_F, full + 18 n_H, full + 3 n_O, full = 202 Subtracting these two numbers shows that 202 - 116 = 86 bonding electrons are needed. Each bond has two electrons, so in addition to the 40 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 | 12 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 | 0 atoms

Find the elemental composition for ( 4-(trifluoromethoxy)benzylboronic acid pinacol ester) 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_14H_18BF_3O_3 Use the chemical formula, C_14H_18BF_3O_3, to count the number of atoms, N_i, for each element and find the total number of atoms, N_atoms: | number of atoms F (fluorine) | 3 C (carbon) | 14 O (oxygen) | 3 B (boron) | 1 H (hydrogen) | 18 N_atoms = 3 + 14 + 3 + 1 + 18 = 39 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) | 3 | 3/39 C (carbon) | 14 | 14/39 O (oxygen) | 3 | 3/39 B (boron) | 1 | 1/39 H (hydrogen) | 18 | 18/39 Check: 3/39 + 14/39 + 3/39 + 1/39 + 18/39 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent F (fluorine) | 3 | 3/39 × 100% = 7.69% C (carbon) | 14 | 14/39 × 100% = 35.9% O (oxygen) | 3 | 3/39 × 100% = 7.69% B (boron) | 1 | 1/39 × 100% = 2.56% H (hydrogen) | 18 | 18/39 × 100% = 46.2% 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) | 3 | 7.69% | 18.998403163 C (carbon) | 14 | 35.9% | 12.011 O (oxygen) | 3 | 7.69% | 15.999 B (boron) | 1 | 2.56% | 10.81 H (hydrogen) | 18 | 46.2% | 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) | 3 | 7.69% | 18.998403163 | 3 × 18.998403163 = 56.995209489 C (carbon) | 14 | 35.9% | 12.011 | 14 × 12.011 = 168.154 O (oxygen) | 3 | 7.69% | 15.999 | 3 × 15.999 = 47.997 B (boron) | 1 | 2.56% | 10.81 | 1 × 10.81 = 10.81 H (hydrogen) | 18 | 46.2% | 1.008 | 18 × 1.008 = 18.144 m = 56.995209489 u + 168.154 u + 47.997 u + 10.81 u + 18.144 u = 302.100209489 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) | 3 | 7.69% | 56.995209489/302.100209489 C (carbon) | 14 | 35.9% | 168.154/302.100209489 O (oxygen) | 3 | 7.69% | 47.997/302.100209489 B (boron) | 1 | 2.56% | 10.81/302.100209489 H (hydrogen) | 18 | 46.2% | 18.144/302.100209489 Check: 56.995209489/302.100209489 + 168.154/302.100209489 + 47.997/302.100209489 + 10.81/302.100209489 + 18.144/302.100209489 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent F (fluorine) | 3 | 7.69% | 56.995209489/302.100209489 × 100% = 18.87% C (carbon) | 14 | 35.9% | 168.154/302.100209489 × 100% = 55.66% O (oxygen) | 3 | 7.69% | 47.997/302.100209489 × 100% = 15.89% B (boron) | 1 | 2.56% | 10.81/302.100209489 × 100% = 3.578% H (hydrogen) | 18 | 46.2% | 18.144/302.100209489 × 100% = 6.006%

The first step in finding the oxidation states (or oxidation numbers) in ( 4-(trifluoromethoxy)benzylboronic acid pinacol ester) is to draw the structure diagram. Next set every oxidation number equal to the atom's formal charge: In ( 4-(trifluoromethoxy)benzylboronic acid pinacol ester) 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, 3 carbon-fluorine bonds, 4 carbon-oxygen bonds, and 12 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 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 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 -3 | C (carbon) | 5 -2 | O (oxygen) | 3 -1 | C (carbon) | 4 | F (fluorine) | 3 0 | C (carbon) | 1 +1 | C (carbon) | 3 | H (hydrogen) | 18 +3 | B (boron) | 1 +4 | C (carbon) | 1

First draw the structure diagram for ( 4-(trifluoromethoxy)benzylboronic acid pinacol ester), 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 | 39 edge count | 40 Schultz index | 16109 Wiener index | 4135 Hosoya index | 1.015×10^7 Balaban index | 2.777