name of dichloro(N, N, N', N'-tetramethylethylenediamine)

dichloro(N, N, N', N'-tetramethylethylenediamine)

molar mass | 185.1 g/mol formula | C_6H_14Cl_2N_2 empirical formula | Cl_C_3N_H_7 SMILES identifier | CN(C)C(C(Cl)N(C)C)Cl InChI identifier | InChI=1/C6H14Cl2N2/c1-9(2)5(7)6(8)10(3)4/h5-6H, 1-4H3 InChI key | XINXXEPJKIRVNV-UHFFFAOYSA-N

Draw the Lewis structure of dichloro(N, N, N', N'-tetramethylethylenediamine). Start by drawing the overall structure of the molecule: Count the total valence electrons of the carbon (n_C, val = 4), chlorine (n_Cl, val = 7), hydrogen (n_H, val = 1), and nitrogen (n_N, val = 5) atoms: 6 n_C, val + 2 n_Cl, val + 14 n_H, val + 2 n_N, val = 62 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), and nitrogen (n_N, full = 8): 6 n_C, full + 2 n_Cl, full + 14 n_H, full + 2 n_N, full = 108 Subtracting these two numbers shows that 108 - 62 = 46 bonding electrons are needed. Each bond has two electrons, so the above diagram has all the necessary bonds. There are 23 bonds and hence 46 bonding electrons in the diagram. Lastly, fill in the remaining unbonded electrons on each atom. In total, there remain 62 - 46 = 16 electrons left to draw: Answer: | |

melting point | -20.99 °C boiling point | 162.4 °C critical temperature | 620.2 K critical pressure | 2.903 MPa critical volume | 493.5 cm^3/mol molar heat of vaporization | 41 kJ/mol molar heat of fusion | 18.68 kJ/mol molar enthalpy | -74.1 kJ/mol molar free energy | 192.5 kJ/mol (computed using the Joback method)

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

Find the elemental composition for dichloro(N, N, N', N'-tetramethylethylenediamine) 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_6H_14Cl_2N_2 Use the chemical formula, C_6H_14Cl_2N_2, 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) | 2 C (carbon) | 6 N (nitrogen) | 2 H (hydrogen) | 14 N_atoms = 2 + 6 + 2 + 14 = 24 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 | 2/24 C (carbon) | 6 | 6/24 N (nitrogen) | 2 | 2/24 H (hydrogen) | 14 | 14/24 Check: 2/24 + 6/24 + 2/24 + 14/24 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent Cl (chlorine) | 2 | 2/24 × 100% = 8.33% C (carbon) | 6 | 6/24 × 100% = 25.0% N (nitrogen) | 2 | 2/24 × 100% = 8.33% H (hydrogen) | 14 | 14/24 × 100% = 58.3% 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 | 8.33% | 35.45 C (carbon) | 6 | 25.0% | 12.011 N (nitrogen) | 2 | 8.33% | 14.007 H (hydrogen) | 14 | 58.3% | 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 | 8.33% | 35.45 | 2 × 35.45 = 70.90 C (carbon) | 6 | 25.0% | 12.011 | 6 × 12.011 = 72.066 N (nitrogen) | 2 | 8.33% | 14.007 | 2 × 14.007 = 28.014 H (hydrogen) | 14 | 58.3% | 1.008 | 14 × 1.008 = 14.112 m = 70.90 u + 72.066 u + 28.014 u + 14.112 u = 185.092 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 | 8.33% | 70.90/185.092 C (carbon) | 6 | 25.0% | 72.066/185.092 N (nitrogen) | 2 | 8.33% | 28.014/185.092 H (hydrogen) | 14 | 58.3% | 14.112/185.092 Check: 70.90/185.092 + 72.066/185.092 + 28.014/185.092 + 14.112/185.092 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent Cl (chlorine) | 2 | 8.33% | 70.90/185.092 × 100% = 38.31% C (carbon) | 6 | 25.0% | 72.066/185.092 × 100% = 38.94% N (nitrogen) | 2 | 8.33% | 28.014/185.092 × 100% = 15.14% H (hydrogen) | 14 | 58.3% | 14.112/185.092 × 100% = 7.624%

The first step in finding the oxidation states (or oxidation numbers) in dichloro(N, N, N', N'-tetramethylethylenediamine) is to draw the structure diagram. Next set every oxidation number equal to the atom's formal charge: In dichloro(N, N, N', N'-tetramethylethylenediamine) 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, 6 carbon-nitrogen bonds, and 1 carbon-carbon bond. 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-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-carbon bond: 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) | 2 -2 | C (carbon) | 4 -1 | Cl (chlorine) | 2 +1 | C (carbon) | 2 | H (hydrogen) | 14

First draw the structure diagram for dichloro(N, N, N', N'-tetramethylethylenediamine), 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 | 24 edge count | 23 Schultz index | 3986 Wiener index | 1102 Hosoya index | 8320 Balaban index | 6.496