name of 1-(4-bromophenyl)-3-(4-chlorophenyl)pyrazole-4-propionic acid

1-(4-bromophenyl)-3-(4-chlorophenyl)pyrazole-4-propionic acid

molar mass | 405.7 g/mol formula | C_18H_14BrClN_2O_2 empirical formula | Br_C_18N_2O_2Cl_H_14 SMILES identifier | C1=C(C=CC(=C1)Cl)C2=NN(C=C2CCC(=O)O)C3=CC=C(C=C3)Br InChI identifier | InChI=1/C18H14BrClN2O2/c19-14-4-8-16(9-5-14)22-11-13(3-10-17(23)24)18(21-22)12-1-6-15(20)7-2-12/h1-2, 4-9, 11H, 3, 10H2, (H, 23, 24)/f/h23H InChI key | CRCKLQOTWRGUSH-UHFFFAOYSA-N

Draw the Lewis structure of 1-(4-bromophenyl)-3-(4-chlorophenyl)pyrazole-4-propionic 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 + 14 n_H, val + 2 n_N, val + 2 n_O, val = 122 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 + 14 n_H, full + 2 n_N, full + 2 n_O, full = 220 Subtracting these two numbers shows that 220 - 122 = 98 bonding electrons are needed. Each bond has two electrons, so in addition to the 40 bonds already present in the diagram add 9 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 9 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 | 428.5 °C boiling point | 757.7 °C critical temperature | 1293 K critical pressure | 2.186 MPa critical volume | 917.5 cm^3/mol molar heat of vaporization | 96.9 kJ/mol molar heat of fusion | 50.43 kJ/mol molar enthalpy | -56.4 kJ/mol molar free energy | 169.6 kJ/mol (computed using the Joback method)

longest chain length | 13 atoms longest straight chain length | 4 atoms longest aliphatic chain length | 3 atoms aromatic atom count | 17 atoms H-bond acceptor count | 3 atoms H-bond donor count | 1 atom

Find the elemental composition for 1-(4-bromophenyl)-3-(4-chlorophenyl)pyrazole-4-propionic 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_14BrClN_2O_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) | 2 O (oxygen) | 2 Cl (chlorine) | 1 H (hydrogen) | 14 N_atoms = 1 + 18 + 2 + 2 + 1 + 14 = 38 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 | 1/38 C (carbon) | 18 | 18/38 N (nitrogen) | 2 | 2/38 O (oxygen) | 2 | 2/38 Cl (chlorine) | 1 | 1/38 H (hydrogen) | 14 | 14/38 Check: 1/38 + 18/38 + 2/38 + 2/38 + 1/38 + 14/38 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent Br (bromine) | 1 | 1/38 × 100% = 2.63% C (carbon) | 18 | 18/38 × 100% = 47.4% N (nitrogen) | 2 | 2/38 × 100% = 5.26% O (oxygen) | 2 | 2/38 × 100% = 5.26% Cl (chlorine) | 1 | 1/38 × 100% = 2.63% H (hydrogen) | 14 | 14/38 × 100% = 36.8% 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 Br (bromine) | 1 | 2.63% | 79.904 C (carbon) | 18 | 47.4% | 12.011 N (nitrogen) | 2 | 5.26% | 14.007 O (oxygen) | 2 | 5.26% | 15.999 Cl (chlorine) | 1 | 2.63% | 35.45 H (hydrogen) | 14 | 36.8% | 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 Br (bromine) | 1 | 2.63% | 79.904 | 1 × 79.904 = 79.904 C (carbon) | 18 | 47.4% | 12.011 | 18 × 12.011 = 216.198 N (nitrogen) | 2 | 5.26% | 14.007 | 2 × 14.007 = 28.014 O (oxygen) | 2 | 5.26% | 15.999 | 2 × 15.999 = 31.998 Cl (chlorine) | 1 | 2.63% | 35.45 | 1 × 35.45 = 35.45 H (hydrogen) | 14 | 36.8% | 1.008 | 14 × 1.008 = 14.112 m = 79.904 u + 216.198 u + 28.014 u + 31.998 u + 35.45 u + 14.112 u = 405.676 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 | 2.63% | 79.904/405.676 C (carbon) | 18 | 47.4% | 216.198/405.676 N (nitrogen) | 2 | 5.26% | 28.014/405.676 O (oxygen) | 2 | 5.26% | 31.998/405.676 Cl (chlorine) | 1 | 2.63% | 35.45/405.676 H (hydrogen) | 14 | 36.8% | 14.112/405.676 Check: 79.904/405.676 + 216.198/405.676 + 28.014/405.676 + 31.998/405.676 + 35.45/405.676 + 14.112/405.676 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent Br (bromine) | 1 | 2.63% | 79.904/405.676 × 100% = 19.70% C (carbon) | 18 | 47.4% | 216.198/405.676 × 100% = 53.29% N (nitrogen) | 2 | 5.26% | 28.014/405.676 × 100% = 6.906% O (oxygen) | 2 | 5.26% | 31.998/405.676 × 100% = 7.888% Cl (chlorine) | 1 | 2.63% | 35.45/405.676 × 100% = 8.739% H (hydrogen) | 14 | 36.8% | 14.112/405.676 × 100% = 3.479%

The first step in finding the oxidation states (or oxidation numbers) in 1-(4-bromophenyl)-3-(4-chlorophenyl)pyrazole-4-propionic acid is to draw the structure diagram. Next set every oxidation number equal to the atom's formal charge: In 1-(4-bromophenyl)-3-(4-chlorophenyl)pyrazole-4-propionic 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, 3 carbon-nitrogen bonds, 2 carbon-oxygen bonds, 18 carbon-carbon bonds, and 1 nitrogen-nitrogen bond. 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: Next look at the nitrogen-nitrogen bond: element | electronegativity (Pauling scale) | N | 3.04 | N | 3.04 | | | 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 | C (carbon) | 2 | N (nitrogen) | 2 | O (oxygen) | 2 -1 | Br (bromine) | 1 | C (carbon) | 8 | Cl (chlorine) | 1 0 | C (carbon) | 3 +1 | C (carbon) | 3 | H (hydrogen) | 14 +2 | C (carbon) | 1 +3 | C (carbon) | 1

First draw the structure diagram for 1-(4-bromophenyl)-3-(4-chlorophenyl)pyrazole-4-propionic 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: | |