1, 4-diethynylbenzene

1, 4-diethynylbenzene

molar mass | 126.2 g/mol formula | C_10H_6 empirical formula | C_5H_3 SMILES identifier | C#CC1=CC=C(C#C)C=C1 InChI identifier | InChI=1/C10H6/c1-3-9-5-7-10(4-2)8-6-9/h1-2, 5-8H InChI key | MVLGANVFCMOJHR-UHFFFAOYSA-N

Draw the Lewis structure of 1, 4-diethynylbenzene. 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: 10 n_C, val + 6 n_H, val = 46 Calculate the number of electrons needed to completely fill the valence shells for carbon (n_C, full = 8) and hydrogen (n_H, full = 2): 10 n_C, full + 6 n_H, full = 92 Subtracting these two numbers shows that 92 - 46 = 46 bonding electrons are needed. Each bond has two electrons, so in addition to the 16 bonds already present in the diagram add 7 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 7 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 | 62.19 °C boiling point | 167 °C critical temperature | 679.9 K critical pressure | 4.078 MPa critical volume | 411.5 cm^3/mol molar heat of vaporization | 40.5 kJ/mol molar heat of fusion | 21.25 kJ/mol molar enthalpy | 559 kJ/mol molar free energy | 582.2 kJ/mol (computed using the Joback method)

longest chain length | 8 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 1, 4-diethynylbenzene 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_6 Use the chemical formula, C_10H_6, to count the number of atoms, N_i, for each element and find the total number of atoms, N_atoms: | number of atoms C (carbon) | 10 H (hydrogen) | 6 N_atoms = 10 + 6 = 16 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) | 10 | 10/16 H (hydrogen) | 6 | 6/16 Check: 10/16 + 6/16 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent C (carbon) | 10 | 10/16 × 100% = 62.5% H (hydrogen) | 6 | 6/16 × 100% = 37.5% 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) | 10 | 62.5% | 12.011 H (hydrogen) | 6 | 37.5% | 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) | 10 | 62.5% | 12.011 | 10 × 12.011 = 120.110 H (hydrogen) | 6 | 37.5% | 1.008 | 6 × 1.008 = 6.048 m = 120.110 u + 6.048 u = 126.158 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) | 10 | 62.5% | 120.110/126.158 H (hydrogen) | 6 | 37.5% | 6.048/126.158 Check: 120.110/126.158 + 6.048/126.158 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent C (carbon) | 10 | 62.5% | 120.110/126.158 × 100% = 95.21% H (hydrogen) | 6 | 37.5% | 6.048/126.158 × 100% = 4.794%

The first step in finding the oxidation states (or oxidation numbers) in 1, 4-diethynylbenzene is to draw the structure diagram. Next set every oxidation number equal to the atom's formal charge: In 1, 4-diethynylbenzene 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 10 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 -1 | C (carbon) | 6 0 | C (carbon) | 4 +1 | H (hydrogen) | 6

First draw the structure diagram for 1, 4-diethynylbenzene, 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 1, 4-diethynylbenzene 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 | 16 edge count | 16 Schultz index | 1612 Wiener index | 416 Hosoya index | 1387 Balaban index | 2.767