2, 4, 6-trimethylstyrene

2, 4, 6-trimethylstyrene

molar mass | 146.2 g/mol formula | C_11H_14 SMILES identifier | C=CC1=C(C)C=C(C)C=C1C InChI identifier | InChI=1/C11H14/c1-5-11-9(3)6-8(2)7-10(11)4/h5-7H, 1H2, 2-4H3 InChI key | PDELBHCVXBSVPJ-UHFFFAOYSA-N

Draw the Lewis structure of 2, 4, 6-trimethylstyrene. 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) and hydrogen (n_H, val = 1) atoms: 11 n_C, val + 14 n_H, val = 58 Calculate the number of electrons needed to completely fill the valence shells for carbon (n_C, full = 8) and hydrogen (n_H, full = 2): 11 n_C, full + 14 n_H, full = 116 Subtracting these two numbers shows that 116 - 58 = 58 bonding electrons are needed. Each bond has two electrons, so in addition to the 25 bonds already present in the diagram add 4 bonds. To minimize formal charge carbon wants 4 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 | 2.8 °C boiling point | 216.2 °C critical temperature | 699.5 K critical pressure | 2.641 MPa critical volume | 524.5 cm^3/mol molar heat of vaporization | 43.7 kJ/mol molar heat of fusion | 15.84 kJ/mol molar enthalpy | 57.2 kJ/mol molar free energy | 213.1 kJ/mol (computed using the Joback method)

longest chain length | 7 atoms longest straight chain length | 2 atoms longest aliphatic chain length | 2 atoms aromatic atom count | 6 atoms H-bond acceptor count | 0 atoms H-bond donor count | 0 atoms

Find the elemental composition for 2, 4, 6-trimethylstyrene 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_11H_14 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) | 11 H (hydrogen) | 14 N_atoms = 11 + 14 = 25 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) | 11 | 11/25 H (hydrogen) | 14 | 14/25 Check: 11/25 + 14/25 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent C (carbon) | 11 | 11/25 × 100% = 44.0% H (hydrogen) | 14 | 14/25 × 100% = 56.0% 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) | 11 | 44.0% | 12.011 H (hydrogen) | 14 | 56.0% | 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) | 11 | 44.0% | 12.011 | 11 × 12.011 = 132.121 H (hydrogen) | 14 | 56.0% | 1.008 | 14 × 1.008 = 14.112 m = 132.121 u + 14.112 u = 146.233 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) | 11 | 44.0% | 132.121/146.233 H (hydrogen) | 14 | 56.0% | 14.112/146.233 Check: 132.121/146.233 + 14.112/146.233 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent C (carbon) | 11 | 44.0% | 132.121/146.233 × 100% = 90.35% H (hydrogen) | 14 | 56.0% | 14.112/146.233 × 100% = 9.650%

The first step in finding the oxidation states (or oxidation numbers) in 2, 4, 6-trimethylstyrene is to draw the structure diagram. Next set every oxidation number equal to the atom's formal charge: In 2, 4, 6-trimethylstyrene 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: There are 11 carbon-carbon bonds. Since these are bonds between the same element, 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) | 3 -2 | C (carbon) | 1 -1 | C (carbon) | 3 0 | C (carbon) | 4 +1 | H (hydrogen) | 14

First draw the structure diagram for 2, 4, 6-trimethylstyrene, 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: For 2, 4, 6-trimethylstyrene there are no lone pairs, so the steric number is given by the σ-bond count. 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 Now assign the hybridization for each atom: Answer: | |

vertex count | 25 edge count | 25 Schultz index | 4567 Wiener index | 1210 Hosoya index | 28616 Balaban index | 3.629