trans-2-(4-methylphenyl)vinylboronic acid

trans-2-(4-methylphenyl)vinylboronic acid

molar mass | 162 g/mol formula | C_9H_11BO_2 empirical formula | C_9B_O_2H_11 SMILES identifier | CC1=CC=C(C=C1)/C=C/B(O)O InChI identifier | InChI=1/C9H11BO2/c1-8-2-4-9(5-3-8)6-7-10(11)12/h2-7, 11-12H, 1H3/b7-6+ InChI key | JJOBVKVXRDHVRP-VOTSOKGWSA-N

Draw the Lewis structure of trans-2-(4-methylphenyl)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 + 2 n_O, val = 62 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 + 2 n_O, full = 116 Subtracting these two numbers shows that 116 - 62 = 54 bonding electrons are needed. Each bond has two electrons, so in addition to the 23 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 | 9 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-methylphenyl)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_2 Use the chemical formula, C_9H_11BO_2, to count the number of atoms, N_i, for each element and find the total number of atoms, N_atoms: | number of atoms C (carbon) | 9 B (boron) | 1 O (oxygen) | 2 H (hydrogen) | 11 N_atoms = 9 + 1 + 2 + 11 = 23 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 | 9/23 B (boron) | 1 | 1/23 O (oxygen) | 2 | 2/23 H (hydrogen) | 11 | 11/23 Check: 9/23 + 1/23 + 2/23 + 11/23 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent C (carbon) | 9 | 9/23 × 100% = 39.1% B (boron) | 1 | 1/23 × 100% = 4.35% O (oxygen) | 2 | 2/23 × 100% = 8.70% H (hydrogen) | 11 | 11/23 × 100% = 47.8% 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 | 39.1% | 12.011 B (boron) | 1 | 4.35% | 10.81 O (oxygen) | 2 | 8.70% | 15.999 H (hydrogen) | 11 | 47.8% | 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 | 39.1% | 12.011 | 9 × 12.011 = 108.099 B (boron) | 1 | 4.35% | 10.81 | 1 × 10.81 = 10.81 O (oxygen) | 2 | 8.70% | 15.999 | 2 × 15.999 = 31.998 H (hydrogen) | 11 | 47.8% | 1.008 | 11 × 1.008 = 11.088 m = 108.099 u + 10.81 u + 31.998 u + 11.088 u = 161.995 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 | 39.1% | 108.099/161.995 B (boron) | 1 | 4.35% | 10.81/161.995 O (oxygen) | 2 | 8.70% | 31.998/161.995 H (hydrogen) | 11 | 47.8% | 11.088/161.995 Check: 108.099/161.995 + 10.81/161.995 + 31.998/161.995 + 11.088/161.995 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent C (carbon) | 9 | 39.1% | 108.099/161.995 × 100% = 66.73% B (boron) | 1 | 4.35% | 10.81/161.995 × 100% = 6.673% O (oxygen) | 2 | 8.70% | 31.998/161.995 × 100% = 19.75% H (hydrogen) | 11 | 47.8% | 11.088/161.995 × 100% = 6.845%

The first step in finding the oxidation states (or oxidation numbers) in trans-2-(4-methylphenyl)vinylboronic acid is to draw the structure diagram. Next set every oxidation number equal to the atom's formal charge: In trans-2-(4-methylphenyl)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, and 9 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-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) | 1 -2 | C (carbon) | 1 | O (oxygen) | 2 -1 | C (carbon) | 5 0 | C (carbon) | 2 +1 | H (hydrogen) | 11 +3 | B (boron) | 1

First draw the structure diagram for trans-2-(4-methylphenyl)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 Now assign the hybridization for each atom: Answer: | |

vertex count | 23 edge count | 23 Schultz index | 4237 Wiener index | 1108 Hosoya index | 22196 Balaban index | 3.052