name of 2-(trifluoromethyl)pyridine-3-carboxylic acid

2-(trifluoromethyl)pyridine-3-carboxylic acid

molar mass | 191.1 g/mol formula | C_7H_4F_3NO_2 empirical formula | F_3C_7N_O_2H_4 SMILES identifier | C1=CC(=C(C(F)(F)F)N=C1)C(=O)O InChI identifier | InChI=1/C7H4F3NO2/c8-7(9, 10)5-4(6(12)13)2-1-3-11-5/h1-3H, (H, 12, 13)/f/h12H InChI key | BFROETNLEIAWNO-UHFFFAOYSA-N

Draw the Lewis structure of 2-(trifluoromethyl)pyridine-3-carboxylic acid. 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), nitrogen (n_N, val = 5), and oxygen (n_O, val = 6) atoms: 7 n_C, val + 3 n_F, val + 4 n_H, val + n_N, val + 2 n_O, val = 70 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), nitrogen (n_N, full = 8), and oxygen (n_O, full = 8): 7 n_C, full + 3 n_F, full + 4 n_H, full + n_N, full + 2 n_O, full = 112 Subtracting these two numbers shows that 112 - 70 = 42 bonding electrons are needed. Each bond has two electrons, so in addition to the 17 bonds already present in the diagram add 4 bonds. To minimize formal charge carbon wants 4 bonds, nitrogen wants 3 bonds, and oxygen wants 2 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 | 102.4 °C boiling point | 328.9 °C critical temperature | 818.3 K critical pressure | 3.174 MPa critical volume | 401.5 cm^3/mol molar heat of vaporization | 50.5 kJ/mol molar heat of fusion | 20.99 kJ/mol molar enthalpy | -909.2 kJ/mol molar free energy | -817.5 kJ/mol (computed using the Joback method)

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

Find the elemental composition for 2-(trifluoromethyl)pyridine-3-carboxylic acid 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_7H_4F_3NO_2 Use the chemical formula, C_7H_4F_3NO_2, 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) | 7 N (nitrogen) | 1 O (oxygen) | 2 H (hydrogen) | 4 N_atoms = 3 + 7 + 1 + 2 + 4 = 17 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 | 3/17 C (carbon) | 7 | 7/17 N (nitrogen) | 1 | 1/17 O (oxygen) | 2 | 2/17 H (hydrogen) | 4 | 4/17 Check: 3/17 + 7/17 + 1/17 + 2/17 + 4/17 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent F (fluorine) | 3 | 3/17 × 100% = 17.6% C (carbon) | 7 | 7/17 × 100% = 41.2% N (nitrogen) | 1 | 1/17 × 100% = 5.88% O (oxygen) | 2 | 2/17 × 100% = 11.8% H (hydrogen) | 4 | 4/17 × 100% = 23.5% 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 F (fluorine) | 3 | 17.6% | 18.998403163 C (carbon) | 7 | 41.2% | 12.011 N (nitrogen) | 1 | 5.88% | 14.007 O (oxygen) | 2 | 11.8% | 15.999 H (hydrogen) | 4 | 23.5% | 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 F (fluorine) | 3 | 17.6% | 18.998403163 | 3 × 18.998403163 = 56.995209489 C (carbon) | 7 | 41.2% | 12.011 | 7 × 12.011 = 84.077 N (nitrogen) | 1 | 5.88% | 14.007 | 1 × 14.007 = 14.007 O (oxygen) | 2 | 11.8% | 15.999 | 2 × 15.999 = 31.998 H (hydrogen) | 4 | 23.5% | 1.008 | 4 × 1.008 = 4.032 m = 56.995209489 u + 84.077 u + 14.007 u + 31.998 u + 4.032 u = 191.109209489 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 | 17.6% | 56.995209489/191.109209489 C (carbon) | 7 | 41.2% | 84.077/191.109209489 N (nitrogen) | 1 | 5.88% | 14.007/191.109209489 O (oxygen) | 2 | 11.8% | 31.998/191.109209489 H (hydrogen) | 4 | 23.5% | 4.032/191.109209489 Check: 56.995209489/191.109209489 + 84.077/191.109209489 + 14.007/191.109209489 + 31.998/191.109209489 + 4.032/191.109209489 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent F (fluorine) | 3 | 17.6% | 56.995209489/191.109209489 × 100% = 29.82% C (carbon) | 7 | 41.2% | 84.077/191.109209489 × 100% = 43.99% N (nitrogen) | 1 | 5.88% | 14.007/191.109209489 × 100% = 7.329% O (oxygen) | 2 | 11.8% | 31.998/191.109209489 × 100% = 16.74% H (hydrogen) | 4 | 23.5% | 4.032/191.109209489 × 100% = 2.110%

The first step in finding the oxidation states (or oxidation numbers) in 2-(trifluoromethyl)pyridine-3-carboxylic acid is to draw the structure diagram. Next set every oxidation number equal to the atom's formal charge: In 2-(trifluoromethyl)pyridine-3-carboxylic acid 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, 2 carbon-nitrogen bonds, 2 carbon-oxygen bonds, and 6 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 bonds: element | electronegativity (Pauling scale) | C | 2.55 | N | 3.04 | | | Since nitrogen is more electronegative than carbon, the electrons in these bonds will go to nitrogen: Next look at the carbon-oxygen bonds: element | electronegativity (Pauling scale) | C | 2.55 | O | 3.44 | | | Since oxygen is more electronegative than carbon, the electrons in these bonds will go to oxygen: 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 -2 | O (oxygen) | 2 -1 | C (carbon) | 2 | F (fluorine) | 3 0 | C (carbon) | 2 +1 | H (hydrogen) | 4 +2 | C (carbon) | 1 +3 | C (carbon) | 2

First draw the structure diagram for 2-(trifluoromethyl)pyridine-3-carboxylic acid, 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 Assign the provisional hybridization based on the table: Next identify any sp^3 atoms with lone pair electrons which can participate in a conjugated π-bond system. These atoms can lower their energy by placing a lone pair in a unhybridized p orbital to maximize overlap with the neighboring π-bonds. Note that halogens and elements from the third period and below do not engage in bond conjugation, except in the case of aromaticity: Adjust the provisional hybridizations to arrive at the result: Answer: | |

vertex count | 17 edge count | 17 Schultz index | 1775 Wiener index | 460 Hosoya index | 1546 Balaban index | 2.993