α-chloro-2-(trifluoromethyl)benzyl chloroformate

α-chloro-2-(trifluoromethyl)benzyl chloroformate

molar mass | 273 g/mol formula | C_9H_5Cl_2F_3O_2 empirical formula | Cl_2C_9O_2F_3H_5 SMILES identifier | C1=CC=C(C(=C1)C(Cl)OC(=O)Cl)C(F)(F)F InChI identifier | InChI=1/C9H5Cl2F3O2/c10-7(16-8(11)15)5-3-1-2-4-6(5)9(12, 13)14/h1-4, 7H InChI key | KQELEQGMABOYIM-UHFFFAOYSA-N

Draw the Lewis structure of α-chloro-2-(trifluoromethyl)benzyl chloroformate. 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 oxygen (n_O, val = 6) atoms: 9 n_C, val + 2 n_Cl, val + 3 n_F, val + 5 n_H, val + 2 n_O, val = 88 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 oxygen (n_O, full = 8): 9 n_C, full + 2 n_Cl, full + 3 n_F, full + 5 n_H, full + 2 n_O, full = 138 Subtracting these two numbers shows that 138 - 88 = 50 bonding electrons are needed. Each bond has two electrons, so in addition to the 21 bonds already present in the diagram add 4 bonds. To minimize formal charge carbon wants 4 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 | 48.34 °C boiling point | 291.1 °C critical temperature | 773.5 K critical pressure | 2.649 MPa critical volume | 592.5 cm^3/mol molar heat of vaporization | 50.6 kJ/mol molar heat of fusion | 23.78 kJ/mol molar enthalpy | -955.1 kJ/mol molar free energy | -790.6 kJ/mol (computed using the Joback method)

longest chain length | 8 atoms longest straight chain length | 5 atoms longest aliphatic chain length | 0 atoms aromatic atom count | 6 atoms H-bond acceptor count | 2 atoms H-bond donor count | 0 atoms

Find the elemental composition for α-chloro-2-(trifluoromethyl)benzyl chloroformate 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_9H_5Cl_2F_3O_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) | 9 O (oxygen) | 2 F (fluorine) | 3 H (hydrogen) | 5 N_atoms = 2 + 9 + 2 + 3 + 5 = 21 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 | 2/21 C (carbon) | 9 | 9/21 O (oxygen) | 2 | 2/21 F (fluorine) | 3 | 3/21 H (hydrogen) | 5 | 5/21 Check: 2/21 + 9/21 + 2/21 + 3/21 + 5/21 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent Cl (chlorine) | 2 | 2/21 × 100% = 9.52% C (carbon) | 9 | 9/21 × 100% = 42.9% O (oxygen) | 2 | 2/21 × 100% = 9.52% F (fluorine) | 3 | 3/21 × 100% = 14.3% H (hydrogen) | 5 | 5/21 × 100% = 23.8% 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) | 2 | 9.52% | 35.45 C (carbon) | 9 | 42.9% | 12.011 O (oxygen) | 2 | 9.52% | 15.999 F (fluorine) | 3 | 14.3% | 18.998403163 H (hydrogen) | 5 | 23.8% | 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) | 2 | 9.52% | 35.45 | 2 × 35.45 = 70.90 C (carbon) | 9 | 42.9% | 12.011 | 9 × 12.011 = 108.099 O (oxygen) | 2 | 9.52% | 15.999 | 2 × 15.999 = 31.998 F (fluorine) | 3 | 14.3% | 18.998403163 | 3 × 18.998403163 = 56.995209489 H (hydrogen) | 5 | 23.8% | 1.008 | 5 × 1.008 = 5.040 m = 70.90 u + 108.099 u + 31.998 u + 56.995209489 u + 5.040 u = 273.032209489 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 | 9.52% | 70.90/273.032209489 C (carbon) | 9 | 42.9% | 108.099/273.032209489 O (oxygen) | 2 | 9.52% | 31.998/273.032209489 F (fluorine) | 3 | 14.3% | 56.995209489/273.032209489 H (hydrogen) | 5 | 23.8% | 5.040/273.032209489 Check: 70.90/273.032209489 + 108.099/273.032209489 + 31.998/273.032209489 + 56.995209489/273.032209489 + 5.040/273.032209489 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent Cl (chlorine) | 2 | 9.52% | 70.90/273.032209489 × 100% = 25.97% C (carbon) | 9 | 42.9% | 108.099/273.032209489 × 100% = 39.59% O (oxygen) | 2 | 9.52% | 31.998/273.032209489 × 100% = 11.72% F (fluorine) | 3 | 14.3% | 56.995209489/273.032209489 × 100% = 20.87% H (hydrogen) | 5 | 23.8% | 5.040/273.032209489 × 100% = 1.846%

The first step in finding the oxidation states (or oxidation numbers) in α-chloro-2-(trifluoromethyl)benzyl chloroformate is to draw the structure diagram. Next set every oxidation number equal to the atom's formal charge: In α-chloro-2-(trifluoromethyl)benzyl chloroformate 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, 3 carbon-fluorine bonds, 3 carbon-oxygen bonds, and 8 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-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-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 -2 | O (oxygen) | 2 -1 | C (carbon) | 4 | Cl (chlorine) | 2 | F (fluorine) | 3 0 | C (carbon) | 2 +1 | C (carbon) | 1 | H (hydrogen) | 5 +3 | C (carbon) | 1 +4 | C (carbon) | 1

First draw the structure diagram for α-chloro-2-(trifluoromethyl)benzyl chloroformate, 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 | 21 edge count | 21 Schultz index | 3037 Wiener index | 796 Hosoya index | 6607 Balaban index | 3.266