name of 2-chloro-4-(trifluoromethyl)benzonitrile

2-chloro-4-(trifluoromethyl)benzonitrile

molar mass | 205.6 g/mol formula | C_8H_3ClF_3N empirical formula | Cl_C_8N_F_3H_3 SMILES identifier | C1=CC(=CC(=C1C#N)Cl)C(F)(F)F InChI identifier | InChI=1/C8H3ClF3N/c9-7-3-6(8(10, 11)12)2-1-5(7)4-13/h1-3H InChI key | GEHMLBFNZKJDQM-UHFFFAOYSA-N

Draw the Lewis structure of 2-chloro-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), chlorine (n_Cl, val = 7), fluorine (n_F, val = 7), hydrogen (n_H, val = 1), and nitrogen (n_N, val = 5) atoms: 8 n_C, val + n_Cl, val + 3 n_F, val + 3 n_H, val + n_N, val = 68 Calculate the number of electrons needed to completely fill the valence shells for carbon (n_C, full = 8), chlorine (n_Cl, full = 8), fluorine (n_F, full = 8), hydrogen (n_H, full = 2), and nitrogen (n_N, full = 8): 8 n_C, full + n_Cl, full + 3 n_F, full + 3 n_H, full + n_N, full = 110 Subtracting these two numbers shows that 110 - 68 = 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 | 57.33 °C boiling point | 280 °C critical temperature | 770.8 K critical pressure | 2.881 MPa critical volume | 493.5 cm^3/mol molar heat of vaporization | 48.1 kJ/mol molar heat of fusion | 17.26 kJ/mol molar enthalpy | -442.8 kJ/mol molar free energy | -350.7 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 2-chloro-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_3ClF_3N Use the chemical formula, C_8H_3ClF_3N, to count the number of atoms, N_i, for each element and find the total number of atoms, N_atoms: | number of atoms Cl (chlorine) | 1 C (carbon) | 8 N (nitrogen) | 1 F (fluorine) | 3 H (hydrogen) | 3 N_atoms = 1 + 8 + 1 + 3 + 3 = 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 Cl (chlorine) | 1 | 1/16 C (carbon) | 8 | 8/16 N (nitrogen) | 1 | 1/16 F (fluorine) | 3 | 3/16 H (hydrogen) | 3 | 3/16 Check: 1/16 + 8/16 + 1/16 + 3/16 + 3/16 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent Cl (chlorine) | 1 | 1/16 × 100% = 6.25% C (carbon) | 8 | 8/16 × 100% = 50.0% N (nitrogen) | 1 | 1/16 × 100% = 6.25% F (fluorine) | 3 | 3/16 × 100% = 18.8% H (hydrogen) | 3 | 3/16 × 100% = 18.8% 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) | 1 | 6.25% | 35.45 C (carbon) | 8 | 50.0% | 12.011 N (nitrogen) | 1 | 6.25% | 14.007 F (fluorine) | 3 | 18.8% | 18.998403163 H (hydrogen) | 3 | 18.8% | 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) | 1 | 6.25% | 35.45 | 1 × 35.45 = 35.45 C (carbon) | 8 | 50.0% | 12.011 | 8 × 12.011 = 96.088 N (nitrogen) | 1 | 6.25% | 14.007 | 1 × 14.007 = 14.007 F (fluorine) | 3 | 18.8% | 18.998403163 | 3 × 18.998403163 = 56.995209489 H (hydrogen) | 3 | 18.8% | 1.008 | 3 × 1.008 = 3.024 m = 35.45 u + 96.088 u + 14.007 u + 56.995209489 u + 3.024 u = 205.564209489 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) | 1 | 6.25% | 35.45/205.564209489 C (carbon) | 8 | 50.0% | 96.088/205.564209489 N (nitrogen) | 1 | 6.25% | 14.007/205.564209489 F (fluorine) | 3 | 18.8% | 56.995209489/205.564209489 H (hydrogen) | 3 | 18.8% | 3.024/205.564209489 Check: 35.45/205.564209489 + 96.088/205.564209489 + 14.007/205.564209489 + 56.995209489/205.564209489 + 3.024/205.564209489 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent Cl (chlorine) | 1 | 6.25% | 35.45/205.564209489 × 100% = 17.25% C (carbon) | 8 | 50.0% | 96.088/205.564209489 × 100% = 46.74% N (nitrogen) | 1 | 6.25% | 14.007/205.564209489 × 100% = 6.814% F (fluorine) | 3 | 18.8% | 56.995209489/205.564209489 × 100% = 27.73% H (hydrogen) | 3 | 18.8% | 3.024/205.564209489 × 100% = 1.471%

The first step in finding the oxidation states (or oxidation numbers) in 2-chloro-4-(trifluoromethyl)benzonitrile is to draw the structure diagram. Next set every oxidation number equal to the atom's formal charge: In 2-chloro-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 1 carbon-chlorine bond, 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-chlorine bond: element | electronegativity (Pauling scale) | C | 2.55 | Cl | 3.16 | | | Since chlorine is more electronegative than carbon, the electrons in this bond will go to chlorine. Decrease the oxidation number for chlorine (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-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: 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) | 3 | Cl (chlorine) | 1 | F (fluorine) | 3 0 | C (carbon) | 2 +1 | C (carbon) | 1 | H (hydrogen) | 3 +3 | C (carbon) | 2

First draw the structure diagram for 2-chloro-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