ethyl 3-(3, 4-dimethoxyphenyl)propionate

ethyl 3-(3, 4-dimethoxyphenyl)propionate

molar mass | 238.3 g/mol formula | C_13H_18O_4 empirical formula | C_13O_4H_18 SMILES identifier | CCOC(=O)CCC1=CC(=C(C=C1)OC)OC InChI identifier | InChI=1/C13H18O4/c1-4-17-13(14)8-6-10-5-7-11(15-2)12(9-10)16-3/h5, 7, 9H, 4, 6, 8H2, 1-3H3 InChI key | GFAGDEMAMZYNAI-UHFFFAOYSA-N

Draw the Lewis structure of ethyl 3-(3, 4-dimethoxyphenyl)propionate. 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), and oxygen (n_O, val = 6) atoms: 13 n_C, val + 18 n_H, val + 4 n_O, val = 94 Calculate the number of electrons needed to completely fill the valence shells for carbon (n_C, full = 8), hydrogen (n_H, full = 2), and oxygen (n_O, full = 8): 13 n_C, full + 18 n_H, full + 4 n_O, full = 172 Subtracting these two numbers shows that 172 - 94 = 78 bonding electrons are needed. Each bond has two electrons, so in addition to the 35 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 | 101.4 °C boiling point | 363.4 °C critical temperature | 835.5 K critical pressure | 2.077 MPa critical volume | 717.5 cm^3/mol molar heat of vaporization | 60.4 kJ/mol molar heat of fusion | 29.43 kJ/mol molar enthalpy | -679.8 kJ/mol molar free energy | -368.6 kJ/mol (computed using the Joback method)

longest chain length | 12 atoms longest straight chain length | 6 atoms longest aliphatic chain length | 3 atoms aromatic atom count | 6 atoms H-bond acceptor count | 2 atoms H-bond donor count | 0 atoms

Find the elemental composition for ethyl 3-(3, 4-dimethoxyphenyl)propionate 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_13H_18O_4 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) | 13 O (oxygen) | 4 H (hydrogen) | 18 N_atoms = 13 + 4 + 18 = 35 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) | 13 | 13/35 O (oxygen) | 4 | 4/35 H (hydrogen) | 18 | 18/35 Check: 13/35 + 4/35 + 18/35 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent C (carbon) | 13 | 13/35 × 100% = 37.1% O (oxygen) | 4 | 4/35 × 100% = 11.4% H (hydrogen) | 18 | 18/35 × 100% = 51.4% 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 C (carbon) | 13 | 37.1% | 12.011 O (oxygen) | 4 | 11.4% | 15.999 H (hydrogen) | 18 | 51.4% | 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 C (carbon) | 13 | 37.1% | 12.011 | 13 × 12.011 = 156.143 O (oxygen) | 4 | 11.4% | 15.999 | 4 × 15.999 = 63.996 H (hydrogen) | 18 | 51.4% | 1.008 | 18 × 1.008 = 18.144 m = 156.143 u + 63.996 u + 18.144 u = 238.283 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) | 13 | 37.1% | 156.143/238.283 O (oxygen) | 4 | 11.4% | 63.996/238.283 H (hydrogen) | 18 | 51.4% | 18.144/238.283 Check: 156.143/238.283 + 63.996/238.283 + 18.144/238.283 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent C (carbon) | 13 | 37.1% | 156.143/238.283 × 100% = 65.53% O (oxygen) | 4 | 11.4% | 63.996/238.283 × 100% = 26.86% H (hydrogen) | 18 | 51.4% | 18.144/238.283 × 100% = 7.614%

The first step in finding the oxidation states (or oxidation numbers) in ethyl 3-(3, 4-dimethoxyphenyl)propionate is to draw the structure diagram. Next set every oxidation number equal to the atom's formal charge: In ethyl 3-(3, 4-dimethoxyphenyl)propionate 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 7 carbon-oxygen bonds, and 10 carbon-carbon bonds. For each of these bonds, assign the bonding electrons to the most electronegative element. First examine 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. Decrease the oxidation number for oxygen 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-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 -2 | C (carbon) | 4 | O (oxygen) | 4 -1 | C (carbon) | 4 0 | C (carbon) | 1 +1 | C (carbon) | 2 | H (hydrogen) | 18 +3 | C (carbon) | 1

First draw the structure diagram for ethyl 3-(3, 4-dimethoxyphenyl)propionate, 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 | 35 edge count | 35 Schultz index | 12997 Wiener index | 3408 Hosoya index | 1.679×10^6 Balaban index | 3.476