3-chloro-4-methoxybenzylamine

3-chloro-4-methoxybenzylamine

molar mass | 171.6 g/mol formula | C_8H_10ClNO empirical formula | Cl_C_8N_O_H_10 SMILES identifier | COC1=C(C=C(C=C1)CN)Cl InChI identifier | InChI=1/C8H10ClNO/c1-11-8-3-2-6(5-10)4-7(8)9/h2-4H, 5, 10H2, 1H3 InChI key | OCNMSDZALRAYEX-UHFFFAOYSA-N

Draw the Lewis structure of 3-chloro-4-methoxybenzylamine. 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), hydrogen (n_H, val = 1), nitrogen (n_N, val = 5), and oxygen (n_O, val = 6) atoms: 8 n_C, val + n_Cl, val + 10 n_H, val + n_N, val + n_O, val = 60 Calculate the number of electrons needed to completely fill the valence shells for carbon (n_C, full = 8), chlorine (n_Cl, full = 8), hydrogen (n_H, full = 2), nitrogen (n_N, full = 8), and oxygen (n_O, full = 8): 8 n_C, full + n_Cl, full + 10 n_H, full + n_N, full + n_O, full = 108 Subtracting these two numbers shows that 108 - 60 = 48 bonding electrons are needed. Each bond has two electrons, so in addition to the 21 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 | 93.64 °C boiling point | 278.3 °C critical temperature | 779.3 K critical pressure | 3.535 MPa critical volume | 471.5 cm^3/mol molar heat of vaporization | 54.4 kJ/mol molar heat of fusion | 20.32 kJ/mol molar enthalpy | -109 kJ/mol molar free energy | 59.15 kJ/mol (computed using the Joback method)

longest chain length | 8 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 3-chloro-4-methoxybenzylamine 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_10ClNO Use the chemical formula, C_8H_10ClNO, 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 O (oxygen) | 1 H (hydrogen) | 10 N_atoms = 1 + 8 + 1 + 1 + 10 = 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) | 1 | 1/21 C (carbon) | 8 | 8/21 N (nitrogen) | 1 | 1/21 O (oxygen) | 1 | 1/21 H (hydrogen) | 10 | 10/21 Check: 1/21 + 8/21 + 1/21 + 1/21 + 10/21 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent Cl (chlorine) | 1 | 1/21 × 100% = 4.76% C (carbon) | 8 | 8/21 × 100% = 38.1% N (nitrogen) | 1 | 1/21 × 100% = 4.76% O (oxygen) | 1 | 1/21 × 100% = 4.76% H (hydrogen) | 10 | 10/21 × 100% = 47.6% 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 | 4.76% | 35.45 C (carbon) | 8 | 38.1% | 12.011 N (nitrogen) | 1 | 4.76% | 14.007 O (oxygen) | 1 | 4.76% | 15.999 H (hydrogen) | 10 | 47.6% | 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 | 4.76% | 35.45 | 1 × 35.45 = 35.45 C (carbon) | 8 | 38.1% | 12.011 | 8 × 12.011 = 96.088 N (nitrogen) | 1 | 4.76% | 14.007 | 1 × 14.007 = 14.007 O (oxygen) | 1 | 4.76% | 15.999 | 1 × 15.999 = 15.999 H (hydrogen) | 10 | 47.6% | 1.008 | 10 × 1.008 = 10.080 m = 35.45 u + 96.088 u + 14.007 u + 15.999 u + 10.080 u = 171.624 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 | 4.76% | 35.45/171.624 C (carbon) | 8 | 38.1% | 96.088/171.624 N (nitrogen) | 1 | 4.76% | 14.007/171.624 O (oxygen) | 1 | 4.76% | 15.999/171.624 H (hydrogen) | 10 | 47.6% | 10.080/171.624 Check: 35.45/171.624 + 96.088/171.624 + 14.007/171.624 + 15.999/171.624 + 10.080/171.624 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent Cl (chlorine) | 1 | 4.76% | 35.45/171.624 × 100% = 20.66% C (carbon) | 8 | 38.1% | 96.088/171.624 × 100% = 55.99% N (nitrogen) | 1 | 4.76% | 14.007/171.624 × 100% = 8.161% O (oxygen) | 1 | 4.76% | 15.999/171.624 × 100% = 9.322% H (hydrogen) | 10 | 47.6% | 10.080/171.624 × 100% = 5.873%

The first step in finding the oxidation states (or oxidation numbers) in 3-chloro-4-methoxybenzylamine is to draw the structure diagram. Next set every oxidation number equal to the atom's formal charge: In 3-chloro-4-methoxybenzylamine 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, 1 carbon-nitrogen bond, 2 carbon-oxygen bonds, and 7 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-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-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 | C (carbon) | 1 | O (oxygen) | 1 -1 | C (carbon) | 4 | Cl (chlorine) | 1 0 | C (carbon) | 1 +1 | C (carbon) | 2 | H (hydrogen) | 10

First draw the structure diagram for 3-chloro-4-methoxybenzylamine, 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 | 3221 Wiener index | 842 Hosoya index | 6925 Balaban index | 3.074