2,3-methano-5,6-dichloroindene

2, 3-methano-5, 6-dichloroindene

molar mass | 197.1 g/mol formula | C_10H_6Cl_2 empirical formula | Cl_C_5H_3 SMILES identifier | C1C2=C(C2)C3=CC(=C(C=C13)Cl)Cl InChI identifier | InChI=1/C10H6Cl2/c11-9-3-6-1-5-2-7(5)8(6)4-10(9)12/h3-4H, 1-2H2 InChI key | SWDLAELXNGWDOH-UHFFFAOYSA-N

Draw the Lewis structure of 2, 3-methano-5, 6-dichloroindene. 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), chlorine (n_Cl, val = 7), and hydrogen (n_H, val = 1) atoms: 10 n_C, val + 2 n_Cl, val + 6 n_H, val = 60 Calculate the number of electrons needed to completely fill the valence shells for carbon (n_C, full = 8), chlorine (n_Cl, full = 8), and hydrogen (n_H, full = 2): 10 n_C, full + 2 n_Cl, full + 6 n_H, full = 108 Subtracting these two numbers shows that 108 - 60 = 48 bonding electrons are needed. Each bond has two electrons, so in addition to the 20 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 | 130.3 °C boiling point | 294.9 °C critical temperature | 809.6 K critical pressure | 3.526 MPa critical volume | 503.5 cm^3/mol molar heat of vaporization | 52.6 kJ/mol molar heat of fusion | 21.69 kJ/mol molar enthalpy | 154 kJ/mol molar free energy | 264.8 kJ/mol (computed using the Joback method)

longest chain length | 7 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 2, 3-methano-5, 6-dichloroindene 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_10H_6Cl_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 Cl (chlorine) | 2 C (carbon) | 10 H (hydrogen) | 6 N_atoms = 2 + 10 + 6 = 18 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 Cl (chlorine) | 2 | 2/18 C (carbon) | 10 | 10/18 H (hydrogen) | 6 | 6/18 Check: 2/18 + 10/18 + 6/18 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent Cl (chlorine) | 2 | 2/18 × 100% = 11.1% C (carbon) | 10 | 10/18 × 100% = 55.6% H (hydrogen) | 6 | 6/18 × 100% = 33.3% 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 Cl (chlorine) | 2 | 11.1% | 35.45 C (carbon) | 10 | 55.6% | 12.011 H (hydrogen) | 6 | 33.3% | 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 Cl (chlorine) | 2 | 11.1% | 35.45 | 2 × 35.45 = 70.90 C (carbon) | 10 | 55.6% | 12.011 | 10 × 12.011 = 120.110 H (hydrogen) | 6 | 33.3% | 1.008 | 6 × 1.008 = 6.048 m = 70.90 u + 120.110 u + 6.048 u = 197.058 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 Cl (chlorine) | 2 | 11.1% | 70.90/197.058 C (carbon) | 10 | 55.6% | 120.110/197.058 H (hydrogen) | 6 | 33.3% | 6.048/197.058 Check: 70.90/197.058 + 120.110/197.058 + 6.048/197.058 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent Cl (chlorine) | 2 | 11.1% | 70.90/197.058 × 100% = 35.98% C (carbon) | 10 | 55.6% | 120.110/197.058 × 100% = 60.95% H (hydrogen) | 6 | 33.3% | 6.048/197.058 × 100% = 3.069%

The first step in finding the oxidation states (or oxidation numbers) in 2, 3-methano-5, 6-dichloroindene is to draw the structure diagram. Next set every oxidation number equal to the atom's formal charge: In 2, 3-methano-5, 6-dichloroindene 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-chlorine bonds, and 12 carbon-carbon bonds. For each of these bonds, assign the bonding electrons to the most electronegative element. First examine the carbon-chlorine bonds: element | electronegativity (Pauling scale) | C | 2.55 | Cl | 3.16 | | | Since chlorine is more electronegative than carbon, the electrons in these bonds will go to chlorine. Decrease the oxidation number for chlorine 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) | 2 -1 | C (carbon) | 2 | Cl (chlorine) | 2 0 | C (carbon) | 4 +1 | C (carbon) | 2 | H (hydrogen) | 6

First draw the structure diagram for 2, 3-methano-5, 6-dichloroindene, 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 | 18 edge count | 20 Schultz index | 2133 Wiener index | 494 Hosoya index | 3495 Balaban index | 2.021