(s)-(-)-2-[(1-hydroxy-3,3-dimethylbutan-2-ylimino)methyl]-4,6-diiodophenol

(s)-(-)-2-[(1-hydroxy-3, 3-dimethylbutan-2-ylimino)methyl]-4, 6-diiodophenol

molar mass | 473.1 g/mol formula | C_13H_17I_2NO_2 empirical formula | O_2C_13N_I_2H_17 SMILES identifier | CC(C)(C)[C@@H](CO)N=CC1=CC(=CC(=C1O)I)I InChI identifier | InChI=1/C13H17I2NO2/c1-13(2, 3)11(7-17)16-6-8-4-9(14)5-10(15)12(8)18/h4-6, 11, 17-18H, 7H2, 1-3H3/t11-/m1/s1 InChI key | WVBNUUNXQNMMOB-LLVKDONJSA-N

Draw the Lewis structure of (s)-(-)-2-[(1-hydroxy-3, 3-dimethylbutan-2-ylimino)methyl]-4, 6-diiodophenol. Start by drawing the overall structure of the molecule, ignoring potential double and triple bonds: Count the total valence electrons of the carbon (n_C, val = 4), hydrogen (n_H, val = 1), iodine (n_I, val = 7), nitrogen (n_N, val = 5), and oxygen (n_O, val = 6) atoms: 13 n_C, val + 17 n_H, val + 2 n_I, val + n_N, val + 2 n_O, val = 100 Calculate the number of electrons needed to completely fill the valence shells for carbon (n_C, full = 8), hydrogen (n_H, full = 2), iodine (n_I, full = 8), nitrogen (n_N, full = 8), and oxygen (n_O, full = 8): 13 n_C, full + 17 n_H, full + 2 n_I, full + n_N, full + 2 n_O, full = 178 Subtracting these two numbers shows that 178 - 100 = 78 bonding electrons are needed. Each bond has two electrons, so in addition to the 35 bonds already present in the diagram add 4 bonds. To minimize formal charge nitrogen wants 3 bonds and 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: | |

melting point | 356.5 °C boiling point | 675.4 °C critical temperature | 1216 K critical pressure | 2.857 MPa critical volume | 823.5 cm^3/mol molar heat of vaporization | 101 kJ/mol molar heat of fusion | 34.18 kJ/mol molar enthalpy | -174 kJ/mol molar free energy | 96.97 kJ/mol (computed using the Joback method)

longest chain length | 9 atoms longest straight chain length | 5 atoms longest aliphatic chain length | 4 atoms aromatic atom count | 6 atoms H-bond acceptor count | 3 atoms H-bond donor count | 2 atoms

Find the elemental composition for (s)-(-)-2-[(1-hydroxy-3, 3-dimethylbutan-2-ylimino)methyl]-4, 6-diiodophenol 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_17I_2NO_2 Use the chemical formula, C_13H_17I_2NO_2, to count the number of atoms, N_i, for each element and find the total number of atoms, N_atoms: | number of atoms O (oxygen) | 2 C (carbon) | 13 N (nitrogen) | 1 I (iodine) | 2 H (hydrogen) | 17 N_atoms = 2 + 13 + 1 + 2 + 17 = 35 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 O (oxygen) | 2 | 2/35 C (carbon) | 13 | 13/35 N (nitrogen) | 1 | 1/35 I (iodine) | 2 | 2/35 H (hydrogen) | 17 | 17/35 Check: 2/35 + 13/35 + 1/35 + 2/35 + 17/35 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent O (oxygen) | 2 | 2/35 × 100% = 5.71% C (carbon) | 13 | 13/35 × 100% = 37.1% N (nitrogen) | 1 | 1/35 × 100% = 2.86% I (iodine) | 2 | 2/35 × 100% = 5.71% H (hydrogen) | 17 | 17/35 × 100% = 48.6% 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 O (oxygen) | 2 | 5.71% | 15.999 C (carbon) | 13 | 37.1% | 12.011 N (nitrogen) | 1 | 2.86% | 14.007 I (iodine) | 2 | 5.71% | 126.90447 H (hydrogen) | 17 | 48.6% | 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 O (oxygen) | 2 | 5.71% | 15.999 | 2 × 15.999 = 31.998 C (carbon) | 13 | 37.1% | 12.011 | 13 × 12.011 = 156.143 N (nitrogen) | 1 | 2.86% | 14.007 | 1 × 14.007 = 14.007 I (iodine) | 2 | 5.71% | 126.90447 | 2 × 126.90447 = 253.80894 H (hydrogen) | 17 | 48.6% | 1.008 | 17 × 1.008 = 17.136 m = 31.998 u + 156.143 u + 14.007 u + 253.80894 u + 17.136 u = 473.09294 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 O (oxygen) | 2 | 5.71% | 31.998/473.09294 C (carbon) | 13 | 37.1% | 156.143/473.09294 N (nitrogen) | 1 | 2.86% | 14.007/473.09294 I (iodine) | 2 | 5.71% | 253.80894/473.09294 H (hydrogen) | 17 | 48.6% | 17.136/473.09294 Check: 31.998/473.09294 + 156.143/473.09294 + 14.007/473.09294 + 253.80894/473.09294 + 17.136/473.09294 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent O (oxygen) | 2 | 5.71% | 31.998/473.09294 × 100% = 6.764% C (carbon) | 13 | 37.1% | 156.143/473.09294 × 100% = 33.00% N (nitrogen) | 1 | 2.86% | 14.007/473.09294 × 100% = 2.961% I (iodine) | 2 | 5.71% | 253.80894/473.09294 × 100% = 53.65% H (hydrogen) | 17 | 48.6% | 17.136/473.09294 × 100% = 3.622%

The first step in finding the oxidation states (or oxidation numbers) in (s)-(-)-2-[(1-hydroxy-3, 3-dimethylbutan-2-ylimino)methyl]-4, 6-diiodophenol is to draw the structure diagram. Next set every oxidation number equal to the atom's formal charge: In (s)-(-)-2-[(1-hydroxy-3, 3-dimethylbutan-2-ylimino)methyl]-4, 6-diiodophenol 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 2 carbon-iodine bonds, 2 carbon-nitrogen bonds, 2 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 carbon-iodine bonds: element | electronegativity (Pauling scale) | C | 2.55 | I | 2.66 | | | Since iodine is more electronegative than carbon, the electrons in these bonds will go to iodine. Decrease the oxidation number for iodine in every highlighted bond (by 1 for single bonds, 2 for double bonds, and 3 for triple bonds), and increase the oxidation number for carbon accordingly: Next look at the carbon-nitrogen bonds: element | electronegativity (Pauling scale) | C | 2.55 | N | 3.04 | | | Since nitrogen is more electronegative than carbon, the electrons in these bonds will go to nitrogen: 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) | 3 | N (nitrogen) | 1 -2 | O (oxygen) | 2 -1 | C (carbon) | 3 | I (iodine) | 2 0 | C (carbon) | 3 +1 | C (carbon) | 4 | H (hydrogen) | 17

First draw the structure diagram for (s)-(-)-2-[(1-hydroxy-3, 3-dimethylbutan-2-ylimino)methyl]-4, 6-diiodophenol, 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 | 35 edge count | 35 Schultz index | 11409 Wiener index | 3000 Hosoya index | 2.118×10^6 Balaban index | 3.937