(s)-(-)-5,5-diphenyl-4-isobutyl-2-oxazolidinone

(s)-(-)-5, 5-diphenyl-4-isobutyl-2-oxazolidinone

molar mass | 295.4 g/mol formula | C_19H_21NO_2 empirical formula | C_19N_O_2H_21 SMILES identifier | CC(C)C[C@H]1C(C2=CC=CC=C2)(C3=CC=CC=C3)OC(=O)N1 InChI identifier | InChI=1/C19H21NO2/c1-14(2)13-17-19(22-18(21)20-17, 15-9-5-3-6-10-15)16-11-7-4-8-12-16/h3-12, 14, 17H, 13H2, 1-2H3, (H, 20, 21)/t17-/m0/s1/f/h20H InChI key | VZRSFLJYNPTJNK-KRWDZBQOSA-N

Draw the Lewis structure of (s)-(-)-5, 5-diphenyl-4-isobutyl-2-oxazolidinone. 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), nitrogen (n_N, val = 5), and oxygen (n_O, val = 6) atoms: 19 n_C, val + 21 n_H, val + n_N, val + 2 n_O, val = 114 Calculate the number of electrons needed to completely fill the valence shells for carbon (n_C, full = 8), hydrogen (n_H, full = 2), nitrogen (n_N, full = 8), and oxygen (n_O, full = 8): 19 n_C, full + 21 n_H, full + n_N, full + 2 n_O, full = 218 Subtracting these two numbers shows that 218 - 114 = 104 bonding electrons are needed. Each bond has two electrons, so in addition to the 45 bonds already present in the diagram add 7 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 7 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 | 253.5 °C boiling point | 523 °C critical temperature | 1043 K critical pressure | 2.066 MPa critical volume | 894.5 cm^3/mol molar heat of vaporization | 74.7 kJ/mol molar heat of fusion | 36.38 kJ/mol molar enthalpy | -179 kJ/mol molar free energy | 156.4 kJ/mol (computed using the Joback method)

longest chain length | 9 atoms longest straight chain length | 3 atoms longest aliphatic chain length | 3 atoms aromatic atom count | 12 atoms H-bond acceptor count | 3 atoms H-bond donor count | 1 atom

Find the elemental composition for (s)-(-)-5, 5-diphenyl-4-isobutyl-2-oxazolidinone 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_19H_21NO_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 C (carbon) | 19 N (nitrogen) | 1 O (oxygen) | 2 H (hydrogen) | 21 N_atoms = 19 + 1 + 2 + 21 = 43 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) | 19 | 19/43 N (nitrogen) | 1 | 1/43 O (oxygen) | 2 | 2/43 H (hydrogen) | 21 | 21/43 Check: 19/43 + 1/43 + 2/43 + 21/43 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent C (carbon) | 19 | 19/43 × 100% = 44.2% N (nitrogen) | 1 | 1/43 × 100% = 2.33% O (oxygen) | 2 | 2/43 × 100% = 4.65% H (hydrogen) | 21 | 21/43 × 100% = 48.8% 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 C (carbon) | 19 | 44.2% | 12.011 N (nitrogen) | 1 | 2.33% | 14.007 O (oxygen) | 2 | 4.65% | 15.999 H (hydrogen) | 21 | 48.8% | 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 C (carbon) | 19 | 44.2% | 12.011 | 19 × 12.011 = 228.209 N (nitrogen) | 1 | 2.33% | 14.007 | 1 × 14.007 = 14.007 O (oxygen) | 2 | 4.65% | 15.999 | 2 × 15.999 = 31.998 H (hydrogen) | 21 | 48.8% | 1.008 | 21 × 1.008 = 21.168 m = 228.209 u + 14.007 u + 31.998 u + 21.168 u = 295.382 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) | 19 | 44.2% | 228.209/295.382 N (nitrogen) | 1 | 2.33% | 14.007/295.382 O (oxygen) | 2 | 4.65% | 31.998/295.382 H (hydrogen) | 21 | 48.8% | 21.168/295.382 Check: 228.209/295.382 + 14.007/295.382 + 31.998/295.382 + 21.168/295.382 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent C (carbon) | 19 | 44.2% | 228.209/295.382 × 100% = 77.26% N (nitrogen) | 1 | 2.33% | 14.007/295.382 × 100% = 4.742% O (oxygen) | 2 | 4.65% | 31.998/295.382 × 100% = 10.83% H (hydrogen) | 21 | 48.8% | 21.168/295.382 × 100% = 7.166%

The first step in finding the oxidation states (or oxidation numbers) in (s)-(-)-5, 5-diphenyl-4-isobutyl-2-oxazolidinone is to draw the structure diagram. Next set every oxidation number equal to the atom's formal charge: In (s)-(-)-5, 5-diphenyl-4-isobutyl-2-oxazolidinone 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-nitrogen bonds, 3 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 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. Decrease the oxidation number for nitrogen 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-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) | 2 | N (nitrogen) | 1 -2 | C (carbon) | 1 | O (oxygen) | 2 -1 | C (carbon) | 11 0 | C (carbon) | 3 +1 | C (carbon) | 1 | H (hydrogen) | 21 +4 | C (carbon) | 1

First draw the structure diagram for (s)-(-)-5, 5-diphenyl-4-isobutyl-2-oxazolidinone, 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 | 43 edge count | 45 Schultz index | 18514 Wiener index | 4674 Hosoya index | 1.345×10^8 Balaban index | 2.579