5-bromo-2-fluorobenzylamine

5-bromo-2-fluorobenzylamine

molar mass | 204 g/mol formula | C_7H_7BrFN empirical formula | Br_C_7N_F_H_7 SMILES identifier | C1=C(C=C(CN)C(=C1)F)Br InChI identifier | InChI=1/C7H7BrFN/c8-6-1-2-7(9)5(3-6)4-10/h1-3H, 4, 10H2 InChI key | PDKBJZGGXHTHNC-UHFFFAOYSA-N

Draw the Lewis structure of 5-bromo-2-fluorobenzylamine. Start by drawing the overall structure of the molecule, ignoring potential double and triple bonds: Count the total valence electrons of the bromine (n_Br, val = 7), carbon (n_C, val = 4), fluorine (n_F, val = 7), hydrogen (n_H, val = 1), and nitrogen (n_N, val = 5) atoms: n_Br, val + 7 n_C, val + n_F, val + 7 n_H, val + n_N, val = 54 Calculate the number of electrons needed to completely fill the valence shells for bromine (n_Br, full = 8), carbon (n_C, full = 8), fluorine (n_F, full = 8), hydrogen (n_H, full = 2), and nitrogen (n_N, full = 8): n_Br, full + 7 n_C, full + n_F, full + 7 n_H, full + n_N, full = 94 Subtracting these two numbers shows that 94 - 54 = 40 bonding electrons are needed. Each bond has two electrons, so in addition to the 17 bonds already present in the diagram add 3 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 3 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 | 90.61 °C boiling point | 261 °C critical temperature | 767.3 K critical pressure | 4.397 MPa critical volume | 428.5 cm^3/mol molar heat of vaporization | 51 kJ/mol molar heat of fusion | 20.71 kJ/mol molar enthalpy | -110 kJ/mol molar free energy | -12.83 kJ/mol (computed using the Joback method)

longest chain length | 6 atoms longest straight chain length | 2 atoms longest aliphatic chain length | 0 atoms aromatic atom count | 6 atoms H-bond acceptor count | 1 atom H-bond donor count | 1 atom

Find the elemental composition for 5-bromo-2-fluorobenzylamine 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_7BrFN 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 Br (bromine) | 1 C (carbon) | 7 N (nitrogen) | 1 F (fluorine) | 1 H (hydrogen) | 7 N_atoms = 1 + 7 + 1 + 1 + 7 = 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 Br (bromine) | 1 | 1/17 C (carbon) | 7 | 7/17 N (nitrogen) | 1 | 1/17 F (fluorine) | 1 | 1/17 H (hydrogen) | 7 | 7/17 Check: 1/17 + 7/17 + 1/17 + 1/17 + 7/17 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent Br (bromine) | 1 | 1/17 × 100% = 5.88% C (carbon) | 7 | 7/17 × 100% = 41.2% N (nitrogen) | 1 | 1/17 × 100% = 5.88% F (fluorine) | 1 | 1/17 × 100% = 5.88% H (hydrogen) | 7 | 7/17 × 100% = 41.2% 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 Br (bromine) | 1 | 5.88% | 79.904 C (carbon) | 7 | 41.2% | 12.011 N (nitrogen) | 1 | 5.88% | 14.007 F (fluorine) | 1 | 5.88% | 18.998403163 H (hydrogen) | 7 | 41.2% | 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 Br (bromine) | 1 | 5.88% | 79.904 | 1 × 79.904 = 79.904 C (carbon) | 7 | 41.2% | 12.011 | 7 × 12.011 = 84.077 N (nitrogen) | 1 | 5.88% | 14.007 | 1 × 14.007 = 14.007 F (fluorine) | 1 | 5.88% | 18.998403163 | 1 × 18.998403163 = 18.998403163 H (hydrogen) | 7 | 41.2% | 1.008 | 7 × 1.008 = 7.056 m = 79.904 u + 84.077 u + 14.007 u + 18.998403163 u + 7.056 u = 204.042403163 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 Br (bromine) | 1 | 5.88% | 79.904/204.042403163 C (carbon) | 7 | 41.2% | 84.077/204.042403163 N (nitrogen) | 1 | 5.88% | 14.007/204.042403163 F (fluorine) | 1 | 5.88% | 18.998403163/204.042403163 H (hydrogen) | 7 | 41.2% | 7.056/204.042403163 Check: 79.904/204.042403163 + 84.077/204.042403163 + 14.007/204.042403163 + 18.998403163/204.042403163 + 7.056/204.042403163 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent Br (bromine) | 1 | 5.88% | 79.904/204.042403163 × 100% = 39.16% C (carbon) | 7 | 41.2% | 84.077/204.042403163 × 100% = 41.21% N (nitrogen) | 1 | 5.88% | 14.007/204.042403163 × 100% = 6.865% F (fluorine) | 1 | 5.88% | 18.998403163/204.042403163 × 100% = 9.311% H (hydrogen) | 7 | 41.2% | 7.056/204.042403163 × 100% = 3.458%

The first step in finding the oxidation states (or oxidation numbers) in 5-bromo-2-fluorobenzylamine is to draw the structure diagram. Next set every oxidation number equal to the atom's formal charge: In 5-bromo-2-fluorobenzylamine 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 bromine-carbon bond, 1 carbon-fluorine bond, 1 carbon-nitrogen bond, and 7 carbon-carbon bonds. For each of these bonds, assign the bonding electrons to the most electronegative element. First examine the bromine-carbon bond: element | electronegativity (Pauling scale) | Br | 2.96 | C | 2.55 | | | Since bromine is more electronegative than carbon, the electrons in this bond will go to bromine. Decrease the oxidation number for bromine (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 bond: element | electronegativity (Pauling scale) | C | 2.55 | F | 3.98 | | | Since fluorine is more electronegative than carbon, the electrons in this bond 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 | Br (bromine) | 1 | C (carbon) | 4 | F (fluorine) | 1 0 | C (carbon) | 1 +1 | C (carbon) | 2 | H (hydrogen) | 7

First draw the structure diagram for 5-bromo-2-fluorobenzylamine, 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 | 17 edge count | 17 Schultz index | 1737 Wiener index | 450 Hosoya index | 1490 Balaban index | 3.055