2-(4-bromophenyl)-6-(4-chlorophenyl)pyridine-4-carboxylic acid

2-(4-bromophenyl)-6-(4-chlorophenyl)pyridine-4-carboxylic acid

molar mass | 388.6 g/mol formula | C_18H_11BrClNO_2 empirical formula | Br_C_18N_O_2Cl_H_11 SMILES identifier | C1=C(C=CC(=C1)Br)C2=NC(=CC(=C2)C(=O)O)C3=CC=C(C=C3)Cl InChI identifier | InChI=1/C18H11BrClNO2/c19-14-5-1-11(2-6-14)16-9-13(18(22)23)10-17(21-16)12-3-7-15(20)8-4-12/h1-10H, (H, 22, 23)/f/h22H InChI key | APLDOGUHCNVFLZ-UHFFFAOYSA-N

Draw the Lewis structure of 2-(4-bromophenyl)-6-(4-chlorophenyl)pyridine-4-carboxylic acid. Start by drawing the overall structure of the molecule, ignoring potential double and triple bonds: Count the total valence electrons of the bromine (n_Br, val = 7), carbon (n_C, val = 4), chlorine (n_Cl, val = 7), hydrogen (n_H, val = 1), nitrogen (n_N, val = 5), and oxygen (n_O, val = 6) atoms: n_Br, val + 18 n_C, val + n_Cl, val + 11 n_H, val + n_N, val + 2 n_O, val = 114 Calculate the number of electrons needed to completely fill the valence shells for bromine (n_Br, full = 8), carbon (n_C, full = 8), chlorine (n_Cl, full = 8), hydrogen (n_H, full = 2), nitrogen (n_N, full = 8), and oxygen (n_O, full = 8): n_Br, full + 18 n_C, full + n_Cl, full + 11 n_H, full + n_N, full + 2 n_O, full = 206 Subtracting these two numbers shows that 206 - 114 = 92 bonding electrons are needed. Each bond has two electrons, so in addition to the 36 bonds already present in the diagram add 10 bonds. To minimize formal charge carbon wants 4 bonds, nitrogen wants 3 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 10 bonds by pairing electrons between adjacent highlighted atoms. Note that the six atom rings are aromatic, so that the single and double bonds may be rearranged: Answer: | |

melting point | 402.3 °C boiling point | 757.9 °C critical temperature | 1310 K critical pressure | 2.343 MPa critical volume | 869.5 cm^3/mol molar heat of vaporization | 96.1 kJ/mol molar heat of fusion | 44.04 kJ/mol molar enthalpy | -89.9 kJ/mol molar free energy | 54.99 kJ/mol (computed using the Joback method)

longest chain length | 13 atoms longest straight chain length | 3 atoms longest aliphatic chain length | 0 atoms aromatic atom count | 18 atoms H-bond acceptor count | 3 atoms H-bond donor count | 1 atom

Find the elemental composition for 2-(4-bromophenyl)-6-(4-chlorophenyl)pyridine-4-carboxylic 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_18H_11BrClNO_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 Br (bromine) | 1 C (carbon) | 18 N (nitrogen) | 1 O (oxygen) | 2 Cl (chlorine) | 1 H (hydrogen) | 11 N_atoms = 1 + 18 + 1 + 2 + 1 + 11 = 34 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 Br (bromine) | 1 | 1/34 C (carbon) | 18 | 18/34 N (nitrogen) | 1 | 1/34 O (oxygen) | 2 | 2/34 Cl (chlorine) | 1 | 1/34 H (hydrogen) | 11 | 11/34 Check: 1/34 + 18/34 + 1/34 + 2/34 + 1/34 + 11/34 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent Br (bromine) | 1 | 1/34 × 100% = 2.94% C (carbon) | 18 | 18/34 × 100% = 52.9% N (nitrogen) | 1 | 1/34 × 100% = 2.94% O (oxygen) | 2 | 2/34 × 100% = 5.88% Cl (chlorine) | 1 | 1/34 × 100% = 2.94% H (hydrogen) | 11 | 11/34 × 100% = 32.4% 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 Br (bromine) | 1 | 2.94% | 79.904 C (carbon) | 18 | 52.9% | 12.011 N (nitrogen) | 1 | 2.94% | 14.007 O (oxygen) | 2 | 5.88% | 15.999 Cl (chlorine) | 1 | 2.94% | 35.45 H (hydrogen) | 11 | 32.4% | 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 Br (bromine) | 1 | 2.94% | 79.904 | 1 × 79.904 = 79.904 C (carbon) | 18 | 52.9% | 12.011 | 18 × 12.011 = 216.198 N (nitrogen) | 1 | 2.94% | 14.007 | 1 × 14.007 = 14.007 O (oxygen) | 2 | 5.88% | 15.999 | 2 × 15.999 = 31.998 Cl (chlorine) | 1 | 2.94% | 35.45 | 1 × 35.45 = 35.45 H (hydrogen) | 11 | 32.4% | 1.008 | 11 × 1.008 = 11.088 m = 79.904 u + 216.198 u + 14.007 u + 31.998 u + 35.45 u + 11.088 u = 388.645 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 Br (bromine) | 1 | 2.94% | 79.904/388.645 C (carbon) | 18 | 52.9% | 216.198/388.645 N (nitrogen) | 1 | 2.94% | 14.007/388.645 O (oxygen) | 2 | 5.88% | 31.998/388.645 Cl (chlorine) | 1 | 2.94% | 35.45/388.645 H (hydrogen) | 11 | 32.4% | 11.088/388.645 Check: 79.904/388.645 + 216.198/388.645 + 14.007/388.645 + 31.998/388.645 + 35.45/388.645 + 11.088/388.645 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent Br (bromine) | 1 | 2.94% | 79.904/388.645 × 100% = 20.56% C (carbon) | 18 | 52.9% | 216.198/388.645 × 100% = 55.63% N (nitrogen) | 1 | 2.94% | 14.007/388.645 × 100% = 3.604% O (oxygen) | 2 | 5.88% | 31.998/388.645 × 100% = 8.233% Cl (chlorine) | 1 | 2.94% | 35.45/388.645 × 100% = 9.121% H (hydrogen) | 11 | 32.4% | 11.088/388.645 × 100% = 2.853%

The first step in finding the oxidation states (or oxidation numbers) in 2-(4-bromophenyl)-6-(4-chlorophenyl)pyridine-4-carboxylic acid is to draw the structure diagram. Next set every oxidation number equal to the atom's formal charge: In 2-(4-bromophenyl)-6-(4-chlorophenyl)pyridine-4-carboxylic 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 bromine-carbon bond, 1 carbon-carbonl bond, 2 carbon-nitrogen bonds, 2 carbon-oxygen bonds, and 19 carbon-carbon bonds. For each of these bonds, assign the bonding electrons to the most electronegative element. First examine the bromine-carbon bond: element | electronegativity (Pauling scale) | Br | 2.96 | C | 2.55 | | | Since bromine is more electronegative than carbon, the electrons in this bond will go to bromine. Decrease the oxidation number for bromine (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-carbonl bond: element | electronegativity (Pauling scale) | C | 2.55 | Cl | 3.16 | | | Since chlorine is more electronegative than carbon, the electrons in this bond will go to chlorine: 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 | N (nitrogen) | 1 -2 | O (oxygen) | 2 -1 | Br (bromine) | 1 | C (carbon) | 10 | Cl (chlorine) | 1 0 | C (carbon) | 3 +1 | C (carbon) | 3 | H (hydrogen) | 11 +2 | C (carbon) | 1 +3 | C (carbon) | 1

First draw the structure diagram for 2-(4-bromophenyl)-6-(4-chlorophenyl)pyridine-4-carboxylic 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 | 34 edge count | 36 Schultz index | 12008 Wiener index | 2955 Hosoya index | 6.015×10^6 Balaban index | 2.032