4 trifluoromethyl benzonitrile

4 trifluoromethyl benzonitrile

molar mass | 171.1 g/mol formula | C_8H_4F_3N empirical formula | F_3C_8N_H_4 SMILES identifier | C1=C(C=CC(=C1)C(F)(F)F)C#N InChI identifier | InChI=1/C8H4F3N/c9-8(10, 11)7-3-1-6(5-12)2-4-7/h1-4H InChI key | DRNJIKRLQJRKMM-UHFFFAOYSA-N

Draw the Lewis structure of 4 trifluoromethyl benzonitrile. 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), fluorine (n_F, val = 7), hydrogen (n_H, val = 1), and nitrogen (n_N, val = 5) atoms: 8 n_C, val + 3 n_F, val + 4 n_H, val + n_N, val = 62 Calculate the number of electrons needed to completely fill the valence shells for carbon (n_C, full = 8), fluorine (n_F, full = 8), hydrogen (n_H, full = 2), and nitrogen (n_N, full = 8): 8 n_C, full + 3 n_F, full + 4 n_H, full + n_N, full = 104 Subtracting these two numbers shows that 104 - 62 = 42 bonding electrons are needed. Each bond has two electrons, so in addition to the 16 bonds already present in the diagram add 5 bonds. To minimize formal charge carbon wants 4 bonds and nitrogen wants 3 bonds. Identify the atoms that want additional bonds and the number of electrons remaining on each atom: Fill in the 5 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 | 14.89 °C boiling point | 237.6 °C critical temperature | 722.5 K critical pressure | 3.049 MPa critical volume | 444.5 cm^3/mol molar heat of vaporization | 43.1 kJ/mol molar heat of fusion | 13.45 kJ/mol molar enthalpy | -415.6 kJ/mol molar free energy | -329.2 kJ/mol (computed using the Joback method)

longest chain length | 8 atoms longest straight chain length | 3 atoms longest aliphatic chain length | 0 atoms aromatic atom count | 6 atoms H-bond acceptor count | 1 atom H-bond donor count | 0 atoms

Find the elemental composition for 4 trifluoromethyl benzonitrile 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_8H_4F_3N Use the chemical formula, C_8H_4F_3N, to count the number of atoms, N_i, for each element and find the total number of atoms, N_atoms: | number of atoms F (fluorine) | 3 C (carbon) | 8 N (nitrogen) | 1 H (hydrogen) | 4 N_atoms = 3 + 8 + 1 + 4 = 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 F (fluorine) | 3 | 3/16 C (carbon) | 8 | 8/16 N (nitrogen) | 1 | 1/16 H (hydrogen) | 4 | 4/16 Check: 3/16 + 8/16 + 1/16 + 4/16 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent F (fluorine) | 3 | 3/16 × 100% = 18.8% C (carbon) | 8 | 8/16 × 100% = 50.0% N (nitrogen) | 1 | 1/16 × 100% = 6.25% H (hydrogen) | 4 | 4/16 × 100% = 25.0% 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 F (fluorine) | 3 | 18.8% | 18.998403163 C (carbon) | 8 | 50.0% | 12.011 N (nitrogen) | 1 | 6.25% | 14.007 H (hydrogen) | 4 | 25.0% | 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 F (fluorine) | 3 | 18.8% | 18.998403163 | 3 × 18.998403163 = 56.995209489 C (carbon) | 8 | 50.0% | 12.011 | 8 × 12.011 = 96.088 N (nitrogen) | 1 | 6.25% | 14.007 | 1 × 14.007 = 14.007 H (hydrogen) | 4 | 25.0% | 1.008 | 4 × 1.008 = 4.032 m = 56.995209489 u + 96.088 u + 14.007 u + 4.032 u = 171.122209489 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 F (fluorine) | 3 | 18.8% | 56.995209489/171.122209489 C (carbon) | 8 | 50.0% | 96.088/171.122209489 N (nitrogen) | 1 | 6.25% | 14.007/171.122209489 H (hydrogen) | 4 | 25.0% | 4.032/171.122209489 Check: 56.995209489/171.122209489 + 96.088/171.122209489 + 14.007/171.122209489 + 4.032/171.122209489 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent F (fluorine) | 3 | 18.8% | 56.995209489/171.122209489 × 100% = 33.31% C (carbon) | 8 | 50.0% | 96.088/171.122209489 × 100% = 56.15% N (nitrogen) | 1 | 6.25% | 14.007/171.122209489 × 100% = 8.185% H (hydrogen) | 4 | 25.0% | 4.032/171.122209489 × 100% = 2.356%

The first step in finding the oxidation states (or oxidation numbers) in 4 trifluoromethyl benzonitrile is to draw the structure diagram. Next set every oxidation number equal to the atom's formal charge: In 4 trifluoromethyl benzonitrile 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 3 carbon-fluorine bonds, 1 carbon-nitrogen bond, and 8 carbon-carbon bonds. For each of these bonds, assign the bonding electrons to the most electronegative element. First examine the carbon-fluorine bonds: element | electronegativity (Pauling scale) | C | 2.55 | F | 3.98 | | | Since fluorine is more electronegative than carbon, the electrons in these bonds will go to fluorine. Decrease the oxidation number for fluorine 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-nitrogen bond: element | electronegativity (Pauling scale) | C | 2.55 | N | 3.04 | | | Since nitrogen is more electronegative than carbon, the electrons in this bond will go to nitrogen: 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 | N (nitrogen) | 1 -1 | C (carbon) | 4 | F (fluorine) | 3 0 | C (carbon) | 2 +1 | H (hydrogen) | 4 +3 | C (carbon) | 2

First draw the structure diagram for 4 trifluoromethyl benzonitrile, 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 | 16 edge count | 16 Schultz index | 1490 Wiener index | 385 Hosoya index | 959 Balaban index | 2.983