name of 3 iodophenylboronic acid

3 iodophenylboronic acid

molar mass | 247.8 g/mol formula | C_6H_6BIO_2 empirical formula | I_C_6B_O_2H_6 SMILES identifier | C1=CC(=CC(=C1)I)B(O)O InChI identifier | InChI=1/C6H6BIO2/c8-6-3-1-2-5(4-6)7(9)10/h1-4, 9-10H InChI key | REEUXWXIMNEIIN-UHFFFAOYSA-N

Draw the Lewis structure of 3 iodophenylboronic 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), iodine (n_I, val = 7), and oxygen (n_O, val = 6) atoms: n_B, val + 6 n_C, val + 6 n_H, val + n_I, val + 2 n_O, val = 52 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), iodine (n_I, full = 8), and oxygen (n_O, full = 8): n_B, full + 6 n_C, full + 6 n_H, full + n_I, full + 2 n_O, full = 90 Subtracting these two numbers shows that 90 - 52 = 38 bonding electrons are needed. Each bond has two electrons, so in addition to the 16 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 | 6 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 | 2 atoms

Find the elemental composition for 3 iodophenylboronic 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_6H_6BIO_2 Use the chemical formula, C_6H_6BIO_2, to count the number of atoms, N_i, for each element and find the total number of atoms, N_atoms: | number of atoms I (iodine) | 1 C (carbon) | 6 B (boron) | 1 O (oxygen) | 2 H (hydrogen) | 6 N_atoms = 1 + 6 + 1 + 2 + 6 = 16 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 I (iodine) | 1 | 1/16 C (carbon) | 6 | 6/16 B (boron) | 1 | 1/16 O (oxygen) | 2 | 2/16 H (hydrogen) | 6 | 6/16 Check: 1/16 + 6/16 + 1/16 + 2/16 + 6/16 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent I (iodine) | 1 | 1/16 × 100% = 6.25% C (carbon) | 6 | 6/16 × 100% = 37.5% B (boron) | 1 | 1/16 × 100% = 6.25% O (oxygen) | 2 | 2/16 × 100% = 12.5% H (hydrogen) | 6 | 6/16 × 100% = 37.5% 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 I (iodine) | 1 | 6.25% | 126.90447 C (carbon) | 6 | 37.5% | 12.011 B (boron) | 1 | 6.25% | 10.81 O (oxygen) | 2 | 12.5% | 15.999 H (hydrogen) | 6 | 37.5% | 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 I (iodine) | 1 | 6.25% | 126.90447 | 1 × 126.90447 = 126.90447 C (carbon) | 6 | 37.5% | 12.011 | 6 × 12.011 = 72.066 B (boron) | 1 | 6.25% | 10.81 | 1 × 10.81 = 10.81 O (oxygen) | 2 | 12.5% | 15.999 | 2 × 15.999 = 31.998 H (hydrogen) | 6 | 37.5% | 1.008 | 6 × 1.008 = 6.048 m = 126.90447 u + 72.066 u + 10.81 u + 31.998 u + 6.048 u = 247.82647 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 I (iodine) | 1 | 6.25% | 126.90447/247.82647 C (carbon) | 6 | 37.5% | 72.066/247.82647 B (boron) | 1 | 6.25% | 10.81/247.82647 O (oxygen) | 2 | 12.5% | 31.998/247.82647 H (hydrogen) | 6 | 37.5% | 6.048/247.82647 Check: 126.90447/247.82647 + 72.066/247.82647 + 10.81/247.82647 + 31.998/247.82647 + 6.048/247.82647 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent I (iodine) | 1 | 6.25% | 126.90447/247.82647 × 100% = 51.21% C (carbon) | 6 | 37.5% | 72.066/247.82647 × 100% = 29.08% B (boron) | 1 | 6.25% | 10.81/247.82647 × 100% = 4.362% O (oxygen) | 2 | 12.5% | 31.998/247.82647 × 100% = 12.91% H (hydrogen) | 6 | 37.5% | 6.048/247.82647 × 100% = 2.440%

The first step in finding the oxidation states (or oxidation numbers) in 3 iodophenylboronic acid is to draw the structure diagram. Next set every oxidation number equal to the atom's formal charge: In 3 iodophenylboronic 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 carbon-iodine bond, and 6 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-iodine bond: element | electronegativity (Pauling scale) | C | 2.55 | I | 2.66 | | | Since iodine is more electronegative than carbon, the electrons in this bond will go to iodine: 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) | 2 -1 | C (carbon) | 5 | I (iodine) | 1 +1 | C (carbon) | 1 | H (hydrogen) | 6 +3 | B (boron) | 1

First draw the structure diagram for 3 iodophenylboronic 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 | 16 edge count | 16 Schultz index | 1534 Wiener index | 396 Hosoya index | 1304 Balaban index | 2.893