diethyldithiocarbamate

diethyldithiocarbamate

molar mass | 148.3 g/mol formula | (C_5H_10NS_2)^- empirical formula | C_5N_S_2H_10 SMILES identifier | CCN(CC)C(=S)[S-] InChI identifier | InChI=1/C5H11NS2/c1-3-6(4-2)5(7)8/h3-4H2, 1-2H3, (H, 7, 8)/p-1/fC5H10NS2/q-1 InChI key | LMBWSYZSUOEYSN-UHFFFAOYSA-M

Draw the Lewis structure of diethyldithiocarbamate. 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), hydrogen (n_H, val = 1), nitrogen (n_N, val = 5), and sulfur (n_S, val = 6) atoms, including the net charge: 5 n_C, val + 10 n_H, val + n_N, val + 2 n_S, val - n_charge = 48 Calculate the number of electrons needed to completely fill the valence shells for carbon (n_C, full = 8), hydrogen (n_H, full = 2), nitrogen (n_N, full = 8), and sulfur (n_S, full = 8): 5 n_C, full + 10 n_H, full + n_N, full + 2 n_S, full = 84 Subtracting these two numbers shows that 84 - 48 = 36 bonding electrons are needed. Each bond has two electrons, so in addition to the 17 bonds already present in the diagram add 1 bond. To minimize formal charge carbon wants 4 bonds and sulfur wants 2 bonds. Identify the atoms that want additional bonds and the number of electrons remaining on each atom. The net charge has been given to the most electronegative atom, sulfur: Fill in the 1 bond by pairing electrons between adjacent highlighted atoms, noting the formal charges of the atoms: Answer: | |

longest chain length | 5 atoms longest straight chain length | 5 atoms longest aliphatic chain length | 2 atoms aromatic atom count | 0 atoms H-bond acceptor count | 1 atom H-bond donor count | 0 atoms

Find the elemental composition for diethyldithiocarbamate 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_5H_10NS_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 C (carbon) | 5 N (nitrogen) | 1 S (sulfur) | 2 H (hydrogen) | 10 N_atoms = 5 + 1 + 2 + 10 = 18 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 C (carbon) | 5 | 5/18 N (nitrogen) | 1 | 1/18 S (sulfur) | 2 | 2/18 H (hydrogen) | 10 | 10/18 Check: 5/18 + 1/18 + 2/18 + 10/18 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent C (carbon) | 5 | 5/18 × 100% = 27.8% N (nitrogen) | 1 | 1/18 × 100% = 5.56% S (sulfur) | 2 | 2/18 × 100% = 11.1% H (hydrogen) | 10 | 10/18 × 100% = 55.6% 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 C (carbon) | 5 | 27.8% | 12.011 N (nitrogen) | 1 | 5.56% | 14.007 S (sulfur) | 2 | 11.1% | 32.06 H (hydrogen) | 10 | 55.6% | 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 C (carbon) | 5 | 27.8% | 12.011 | 5 × 12.011 = 60.055 N (nitrogen) | 1 | 5.56% | 14.007 | 1 × 14.007 = 14.007 S (sulfur) | 2 | 11.1% | 32.06 | 2 × 32.06 = 64.12 H (hydrogen) | 10 | 55.6% | 1.008 | 10 × 1.008 = 10.080 m = 60.055 u + 14.007 u + 64.12 u + 10.080 u = 148.262 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 C (carbon) | 5 | 27.8% | 60.055/148.262 N (nitrogen) | 1 | 5.56% | 14.007/148.262 S (sulfur) | 2 | 11.1% | 64.12/148.262 H (hydrogen) | 10 | 55.6% | 10.080/148.262 Check: 60.055/148.262 + 14.007/148.262 + 64.12/148.262 + 10.080/148.262 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent C (carbon) | 5 | 27.8% | 60.055/148.262 × 100% = 40.51% N (nitrogen) | 1 | 5.56% | 14.007/148.262 × 100% = 9.447% S (sulfur) | 2 | 11.1% | 64.12/148.262 × 100% = 43.25% H (hydrogen) | 10 | 55.6% | 10.080/148.262 × 100% = 6.799%

The first step in finding the oxidation states (or oxidation numbers) in diethyldithiocarbamate is to draw the structure diagram. Next set every oxidation number equal to the atom's formal charge: In diethyldithiocarbamate 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-nitrogen bonds, 2 carbon-sulfur bonds, and 2 carbon-carbon bonds. For each of these bonds, assign the bonding electrons to the most electronegative element. First examine 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. Decrease the oxidation number for nitrogen 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-sulfur bonds: element | electronegativity (Pauling scale) | C | 2.55 | S | 2.58 | | | Since sulfur is more electronegative than carbon, the electrons in these bonds will go to sulfur: 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 | C (carbon) | 2 | N (nitrogen) | 1 -2 | S (sulfur) | 2 -1 | C (carbon) | 2 +1 | H (hydrogen) | 10 +4 | C (carbon) | 1

First draw the structure diagram for diethyldithiocarbamate, 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 | 18 edge count | 17 Schultz index | 1848 Wiener index | 515 Hosoya index | 988 Balaban index | 5.77