6-bromo-4-[(dimethylamino)methyl]-5-hydroxy-1-methyl-2-[(phenylthio)methyl]-3-indolecarboxylic acid ethyl ester

6-bromo-4-[(dimethylamino)methyl]-5-hydroxy-1-methyl-2-[(phenylthio)methyl]-3-indolecarboxylic acid ethyl ester

molar mass | 477.4 g/mol formula | C_22H_25BrN_2O_3S empirical formula | C_22O_3N_2Br_S_H_25 SMILES identifier | CCOC(=O)C1=C(CSC2=CC=CC=C2)N(C)C3=C1C(=C(C(=C3)Br)O)CN(C)C InChI identifier | InChI=1/C22H25BrN2O3S/c1-5-28-22(27)20-18(13-29-14-9-7-6-8-10-14)25(4)17-11-16(23)21(26)15(19(17)20)12-24(2)3/h6-11, 26H, 5, 12-13H2, 1-4H3 InChI key | KCFYEAOKVJSACF-UHFFFAOYSA-N

Draw the Lewis structure of 6-bromo-4-[(dimethylamino)methyl]-5-hydroxy-1-methyl-2-[(phenylthio)methyl]-3-indolecarboxylic acid ethyl ester. 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), hydrogen (n_H, val = 1), nitrogen (n_N, val = 5), oxygen (n_O, val = 6), and sulfur (n_S, val = 6) atoms: n_Br, val + 22 n_C, val + 25 n_H, val + 2 n_N, val + 3 n_O, val + n_S, val = 154 Calculate the number of electrons needed to completely fill the valence shells for bromine (n_Br, full = 8), carbon (n_C, full = 8), hydrogen (n_H, full = 2), nitrogen (n_N, full = 8), oxygen (n_O, full = 8), and sulfur (n_S, full = 8): n_Br, full + 22 n_C, full + 25 n_H, full + 2 n_N, full + 3 n_O, full + n_S, full = 282 Subtracting these two numbers shows that 282 - 154 = 128 bonding electrons are needed. Each bond has two electrons, so in addition to the 56 bonds already present in the diagram add 8 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 8 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 | 519.6 °C boiling point | 812.8 °C critical temperature | 1339 K critical pressure | 1.837 MPa critical volume | 1148 cm^3/mol molar heat of vaporization | 111 kJ/mol molar heat of fusion | 64.85 kJ/mol molar enthalpy | -216 kJ/mol molar free energy | 225.6 kJ/mol (computed using the Joback method)

longest chain length | 13 atoms longest straight chain length | 5 atoms longest aliphatic chain length | 2 atoms aromatic atom count | 15 atoms H-bond acceptor count | 4 atoms H-bond donor count | 1 atom

Find the elemental composition for 6-bromo-4-[(dimethylamino)methyl]-5-hydroxy-1-methyl-2-[(phenylthio)methyl]-3-indolecarboxylic acid ethyl ester 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_22H_25BrN_2O_3S 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 C (carbon) | 22 O (oxygen) | 3 N (nitrogen) | 2 Br (bromine) | 1 S (sulfur) | 1 H (hydrogen) | 25 N_atoms = 22 + 3 + 2 + 1 + 1 + 25 = 54 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 C (carbon) | 22 | 22/54 O (oxygen) | 3 | 3/54 N (nitrogen) | 2 | 2/54 Br (bromine) | 1 | 1/54 S (sulfur) | 1 | 1/54 H (hydrogen) | 25 | 25/54 Check: 22/54 + 3/54 + 2/54 + 1/54 + 1/54 + 25/54 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent C (carbon) | 22 | 22/54 × 100% = 40.7% O (oxygen) | 3 | 3/54 × 100% = 5.56% N (nitrogen) | 2 | 2/54 × 100% = 3.70% Br (bromine) | 1 | 1/54 × 100% = 1.85% S (sulfur) | 1 | 1/54 × 100% = 1.85% H (hydrogen) | 25 | 25/54 × 100% = 46.3% 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 C (carbon) | 22 | 40.7% | 12.011 O (oxygen) | 3 | 5.56% | 15.999 N (nitrogen) | 2 | 3.70% | 14.007 Br (bromine) | 1 | 1.85% | 79.904 S (sulfur) | 1 | 1.85% | 32.06 H (hydrogen) | 25 | 46.3% | 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 C (carbon) | 22 | 40.7% | 12.011 | 22 × 12.011 = 264.242 O (oxygen) | 3 | 5.56% | 15.999 | 3 × 15.999 = 47.997 N (nitrogen) | 2 | 3.70% | 14.007 | 2 × 14.007 = 28.014 Br (bromine) | 1 | 1.85% | 79.904 | 1 × 79.904 = 79.904 S (sulfur) | 1 | 1.85% | 32.06 | 1 × 32.06 = 32.06 H (hydrogen) | 25 | 46.3% | 1.008 | 25 × 1.008 = 25.200 m = 264.242 u + 47.997 u + 28.014 u + 79.904 u + 32.06 u + 25.200 u = 477.417 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 C (carbon) | 22 | 40.7% | 264.242/477.417 O (oxygen) | 3 | 5.56% | 47.997/477.417 N (nitrogen) | 2 | 3.70% | 28.014/477.417 Br (bromine) | 1 | 1.85% | 79.904/477.417 S (sulfur) | 1 | 1.85% | 32.06/477.417 H (hydrogen) | 25 | 46.3% | 25.200/477.417 Check: 264.242/477.417 + 47.997/477.417 + 28.014/477.417 + 79.904/477.417 + 32.06/477.417 + 25.200/477.417 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent C (carbon) | 22 | 40.7% | 264.242/477.417 × 100% = 55.35% O (oxygen) | 3 | 5.56% | 47.997/477.417 × 100% = 10.05% N (nitrogen) | 2 | 3.70% | 28.014/477.417 × 100% = 5.868% Br (bromine) | 1 | 1.85% | 79.904/477.417 × 100% = 16.74% S (sulfur) | 1 | 1.85% | 32.06/477.417 × 100% = 6.715% H (hydrogen) | 25 | 46.3% | 25.200/477.417 × 100% = 5.278%

The first step in finding the oxidation states (or oxidation numbers) in 6-bromo-4-[(dimethylamino)methyl]-5-hydroxy-1-methyl-2-[(phenylthio)methyl]-3-indolecarboxylic acid ethyl ester is to draw the structure diagram. Next set every oxidation number equal to the atom's formal charge: In 6-bromo-4-[(dimethylamino)methyl]-5-hydroxy-1-methyl-2-[(phenylthio)methyl]-3-indolecarboxylic acid ethyl ester 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, 6 carbon-nitrogen bonds, 4 carbon-oxygen bonds, 2 carbon-sulfur bonds, and 18 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-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-sulfur bonds: element | electronegativity (Pauling scale) | C | 2.55 | S | 2.58 | | | Since sulfur is more electronegative than carbon, the electrons in these bonds will go to sulfur: 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) | 1 | N (nitrogen) | 2 -2 | C (carbon) | 3 | O (oxygen) | 3 | S (sulfur) | 1 -1 | Br (bromine) | 1 | C (carbon) | 9 0 | C (carbon) | 3 +1 | C (carbon) | 5 | H (hydrogen) | 25 +3 | C (carbon) | 1