3, 5-diiodoanthranilic acid

3, 5-diiodoanthranilic acid

molar mass | 388.9 g/mol formula | C_7H_5I_2NO_2 empirical formula | I_2C_7O_2N_H_5 SMILES identifier | C1=C(C=C(C(=C1C(=O)O)N)I)I InChI identifier | InChI=1/C7H5I2NO2/c8-3-1-4(7(11)12)6(10)5(9)2-3/h1-2H, 10H2, (H, 11, 12)/f/h11H InChI key | RFIBDMCPIREZKC-UHFFFAOYSA-N

Draw the Lewis structure of 3, 5-diiodoanthranilic acid. 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: 7 n_C, val + 5 n_H, val + 2 n_I, val + n_N, val + 2 n_O, val = 64 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): 7 n_C, full + 5 n_H, full + 2 n_I, full + n_N, full + 2 n_O, full = 106 Subtracting these two numbers shows that 106 - 64 = 42 bonding electrons are needed. Each bond has two electrons, so in addition to the 17 bonds already present in the diagram add 4 bonds. To minimize formal charge carbon wants 4 bonds and oxygen wants 2 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 | 319.5 °C boiling point | 532.4 °C critical temperature | 1101 K critical pressure | 4.433 MPa critical volume | 548.5 cm^3/mol molar heat of vaporization | 82 kJ/mol molar heat of fusion | 30.91 kJ/mol molar enthalpy | -148.4 kJ/mol molar free energy | -69.64 kJ/mol (computed using the Joback method)

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 | 3 atoms H-bond donor count | 2 atoms

Find the elemental composition for 3, 5-diiodoanthranilic 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_7H_5I_2NO_2 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 I (iodine) | 2 C (carbon) | 7 O (oxygen) | 2 N (nitrogen) | 1 H (hydrogen) | 5 N_atoms = 2 + 7 + 2 + 1 + 5 = 17 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) | 2 | 2/17 C (carbon) | 7 | 7/17 O (oxygen) | 2 | 2/17 N (nitrogen) | 1 | 1/17 H (hydrogen) | 5 | 5/17 Check: 2/17 + 7/17 + 2/17 + 1/17 + 5/17 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent I (iodine) | 2 | 2/17 × 100% = 11.8% C (carbon) | 7 | 7/17 × 100% = 41.2% O (oxygen) | 2 | 2/17 × 100% = 11.8% N (nitrogen) | 1 | 1/17 × 100% = 5.88% H (hydrogen) | 5 | 5/17 × 100% = 29.4% 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) | 2 | 11.8% | 126.90447 C (carbon) | 7 | 41.2% | 12.011 O (oxygen) | 2 | 11.8% | 15.999 N (nitrogen) | 1 | 5.88% | 14.007 H (hydrogen) | 5 | 29.4% | 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) | 2 | 11.8% | 126.90447 | 2 × 126.90447 = 253.80894 C (carbon) | 7 | 41.2% | 12.011 | 7 × 12.011 = 84.077 O (oxygen) | 2 | 11.8% | 15.999 | 2 × 15.999 = 31.998 N (nitrogen) | 1 | 5.88% | 14.007 | 1 × 14.007 = 14.007 H (hydrogen) | 5 | 29.4% | 1.008 | 5 × 1.008 = 5.040 m = 253.80894 u + 84.077 u + 31.998 u + 14.007 u + 5.040 u = 388.93094 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) | 2 | 11.8% | 253.80894/388.93094 C (carbon) | 7 | 41.2% | 84.077/388.93094 O (oxygen) | 2 | 11.8% | 31.998/388.93094 N (nitrogen) | 1 | 5.88% | 14.007/388.93094 H (hydrogen) | 5 | 29.4% | 5.040/388.93094 Check: 253.80894/388.93094 + 84.077/388.93094 + 31.998/388.93094 + 14.007/388.93094 + 5.040/388.93094 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent I (iodine) | 2 | 11.8% | 253.80894/388.93094 × 100% = 65.26% C (carbon) | 7 | 41.2% | 84.077/388.93094 × 100% = 21.62% O (oxygen) | 2 | 11.8% | 31.998/388.93094 × 100% = 8.227% N (nitrogen) | 1 | 5.88% | 14.007/388.93094 × 100% = 3.601% H (hydrogen) | 5 | 29.4% | 5.040/388.93094 × 100% = 1.296%

The first step in finding the oxidation states (or oxidation numbers) in 3, 5-diiodoanthranilic acid is to draw the structure diagram. Next set every oxidation number equal to the atom's formal charge: In 3, 5-diiodoanthranilic 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 2 carbon-iodine bonds, 1 carbon-nitrogen bond, 2 carbon-oxygen bonds, and 7 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 bond: element | electronegativity (Pauling scale) | C | 2.55 | N | 3.04 | | | Since nitrogen is more electronegative than carbon, the electrons in this bond 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 | N (nitrogen) | 1 -2 | O (oxygen) | 2 -1 | C (carbon) | 2 | I (iodine) | 2 0 | C (carbon) | 1 +1 | C (carbon) | 3 | H (hydrogen) | 5 +3 | C (carbon) | 1

First draw the structure diagram for 3, 5-diiodoanthranilic 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 | 17 edge count | 17 Schultz index | 1755 Wiener index | 456 Hosoya index | 1814 Balaban index | 3.025