trans-2-(4-methoxyphenyl)vinylboronic acid

trans-2-(4-methoxyphenyl)vinylboronic acid

molar mass | 178 g/mol formula | C_9H_11BO_3 empirical formula | C_9O_3B_H_11 SMILES identifier | COC1=CC=C(C=C1)/C=C/B(O)O InChI identifier | InChI=1/C9H11BO3/c1-13-9-4-2-8(3-5-9)6-7-10(11)12/h2-7, 11-12H, 1H3/b7-6+ InChI key | LGSBCAPDUJYMOQ-VOTSOKGWSA-N

Draw the Lewis structure of trans-2-(4-methoxyphenyl)vinylboronic 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), hydrogen (n_H, val = 1), and oxygen (n_O, val = 6) atoms: n_B, val + 9 n_C, val + 11 n_H, val + 3 n_O, val = 68 Calculate the number of electrons needed to completely fill the valence shells for boron (n_B, full = 6), carbon (n_C, full = 8), hydrogen (n_H, full = 2), and oxygen (n_O, full = 8): n_B, full + 9 n_C, full + 11 n_H, full + 3 n_O, full = 124 Subtracting these two numbers shows that 124 - 68 = 56 bonding electrons are needed. Each bond has two electrons, so in addition to the 24 bonds already present in the diagram add 4 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 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: | |

longest chain length | 10 atoms longest straight chain length | 4 atoms longest aliphatic chain length | 2 atoms aromatic atom count | 6 atoms H-bond acceptor count | 2 atoms H-bond donor count | 2 atoms

Find the elemental composition for trans-2-(4-methoxyphenyl)vinylboronic 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_11BO_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 C (carbon) | 9 O (oxygen) | 3 B (boron) | 1 H (hydrogen) | 11 N_atoms = 9 + 3 + 1 + 11 = 24 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 C (carbon) | 9 | 9/24 O (oxygen) | 3 | 3/24 B (boron) | 1 | 1/24 H (hydrogen) | 11 | 11/24 Check: 9/24 + 3/24 + 1/24 + 11/24 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent C (carbon) | 9 | 9/24 × 100% = 37.5% O (oxygen) | 3 | 3/24 × 100% = 12.5% B (boron) | 1 | 1/24 × 100% = 4.17% H (hydrogen) | 11 | 11/24 × 100% = 45.8% 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 C (carbon) | 9 | 37.5% | 12.011 O (oxygen) | 3 | 12.5% | 15.999 B (boron) | 1 | 4.17% | 10.81 H (hydrogen) | 11 | 45.8% | 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 C (carbon) | 9 | 37.5% | 12.011 | 9 × 12.011 = 108.099 O (oxygen) | 3 | 12.5% | 15.999 | 3 × 15.999 = 47.997 B (boron) | 1 | 4.17% | 10.81 | 1 × 10.81 = 10.81 H (hydrogen) | 11 | 45.8% | 1.008 | 11 × 1.008 = 11.088 m = 108.099 u + 47.997 u + 10.81 u + 11.088 u = 177.994 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 C (carbon) | 9 | 37.5% | 108.099/177.994 O (oxygen) | 3 | 12.5% | 47.997/177.994 B (boron) | 1 | 4.17% | 10.81/177.994 H (hydrogen) | 11 | 45.8% | 11.088/177.994 Check: 108.099/177.994 + 47.997/177.994 + 10.81/177.994 + 11.088/177.994 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent C (carbon) | 9 | 37.5% | 108.099/177.994 × 100% = 60.73% O (oxygen) | 3 | 12.5% | 47.997/177.994 × 100% = 26.97% B (boron) | 1 | 4.17% | 10.81/177.994 × 100% = 6.073% H (hydrogen) | 11 | 45.8% | 11.088/177.994 × 100% = 6.229%

The first step in finding the oxidation states (or oxidation numbers) in trans-2-(4-methoxyphenyl)vinylboronic acid is to draw the structure diagram. Next set every oxidation number equal to the atom's formal charge: In trans-2-(4-methoxyphenyl)vinylboronic 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-oxygen bonds, and 8 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-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 | C (carbon) | 2 | O (oxygen) | 3 -1 | C (carbon) | 5 0 | C (carbon) | 1 +1 | C (carbon) | 1 | H (hydrogen) | 11 +3 | B (boron) | 1

First draw the structure diagram for trans-2-(4-methoxyphenyl)vinylboronic 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 | 24 edge count | 24 Schultz index | 4918 Wiener index | 1285 Hosoya index | 35060 Balaban index | 2.99