1, 4-dicyclohexyl-2, 5-diiodobenzene

1, 4-dicyclohexyl-2, 5-diiodobenzene

molar mass | 494.2 g/mol formula | C_18H_24I_2 empirical formula | C_9I_H_12 SMILES identifier | C1CCC(CC1)C2=C(C=C(C3CCCCC3)C(=C2)I)I InChI identifier | InChI=1/C18H24I2/c19-17-12-16(14-9-5-2-6-10-14)18(20)11-15(17)13-7-3-1-4-8-13/h11-14H, 1-10H2 InChI key | VASMITRTCQRQGE-UHFFFAOYSA-N

Draw the Lewis structure of 1, 4-dicyclohexyl-2, 5-diiodobenzene. 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 iodine (n_I, val = 7) atoms: 18 n_C, val + 24 n_H, val + 2 n_I, val = 110 Calculate the number of electrons needed to completely fill the valence shells for carbon (n_C, full = 8), hydrogen (n_H, full = 2), and iodine (n_I, full = 8): 18 n_C, full + 24 n_H, full + 2 n_I, full = 208 Subtracting these two numbers shows that 208 - 110 = 98 bonding electrons are needed. Each bond has two electrons, so in addition to the 46 bonds already present in the diagram add 3 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 3 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 | 214.3 °C boiling point | 605.1 °C critical temperature | 1181 K critical pressure | 1.9 MPa critical volume | 977.5 cm^3/mol molar heat of vaporization | 79.5 kJ/mol molar heat of fusion | 27.72 kJ/mol molar enthalpy | 50 kJ/mol molar free energy | 349.3 kJ/mol (computed using the Joback method)

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

Find the elemental composition for 1, 4-dicyclohexyl-2, 5-diiodobenzene 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_24I_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) | 18 I (iodine) | 2 H (hydrogen) | 24 N_atoms = 18 + 2 + 24 = 44 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) | 18 | 18/44 I (iodine) | 2 | 2/44 H (hydrogen) | 24 | 24/44 Check: 18/44 + 2/44 + 24/44 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent C (carbon) | 18 | 18/44 × 100% = 40.9% I (iodine) | 2 | 2/44 × 100% = 4.55% H (hydrogen) | 24 | 24/44 × 100% = 54.5% 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) | 18 | 40.9% | 12.011 I (iodine) | 2 | 4.55% | 126.90447 H (hydrogen) | 24 | 54.5% | 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) | 18 | 40.9% | 12.011 | 18 × 12.011 = 216.198 I (iodine) | 2 | 4.55% | 126.90447 | 2 × 126.90447 = 253.80894 H (hydrogen) | 24 | 54.5% | 1.008 | 24 × 1.008 = 24.192 m = 216.198 u + 253.80894 u + 24.192 u = 494.19894 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) | 18 | 40.9% | 216.198/494.19894 I (iodine) | 2 | 4.55% | 253.80894/494.19894 H (hydrogen) | 24 | 54.5% | 24.192/494.19894 Check: 216.198/494.19894 + 253.80894/494.19894 + 24.192/494.19894 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent C (carbon) | 18 | 40.9% | 216.198/494.19894 × 100% = 43.75% I (iodine) | 2 | 4.55% | 253.80894/494.19894 × 100% = 51.36% H (hydrogen) | 24 | 54.5% | 24.192/494.19894 × 100% = 4.895%

The first step in finding the oxidation states (or oxidation numbers) in 1, 4-dicyclohexyl-2, 5-diiodobenzene is to draw the structure diagram. Next set every oxidation number equal to the atom's formal charge: In 1, 4-dicyclohexyl-2, 5-diiodobenzene 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-iodine bonds, and 20 carbon-carbon bonds. For each of these bonds, assign the bonding electrons to the most electronegative element. First examine the carbon-iodine bonds: element | electronegativity (Pauling scale) | C | 2.55 | I | 2.66 | | | Since iodine is more electronegative than carbon, the electrons in these bonds will go to iodine. Decrease the oxidation number for iodine 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 -2 | C (carbon) | 10 -1 | C (carbon) | 4 | I (iodine) | 2 0 | C (carbon) | 2 +1 | C (carbon) | 2 | H (hydrogen) | 24

First draw the structure diagram for 1, 4-dicyclohexyl-2, 5-diiodobenzene, 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 Now assign the hybridization for each atom: Answer: | |

vertex count | 44 edge count | 46 Schultz index | 21432 Wiener index | 5378 Hosoya index | 1.067×10^8 Balaban index | 2.356