name of 1-(4-iodophenyl)piperazine

1-(4-iodophenyl)piperazine

molar mass | 288.1 g/mol formula | C_10H_13IN_2 empirical formula | I_C_10N_2H_13 SMILES identifier | C1=C(C=CC(=C1)N2CCNCC2)I InChI identifier | InChI=1/C10H13IN2/c11-9-1-3-10(4-2-9)13-7-5-12-6-8-13/h1-4, 12H, 5-8H2 InChI key | CFZRRGPCDUOJHX-UHFFFAOYSA-N

Draw the Lewis structure of 1-(4-iodophenyl)piperazine. 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), iodine (n_I, val = 7), and nitrogen (n_N, val = 5) atoms: 10 n_C, val + 13 n_H, val + n_I, val + 2 n_N, val = 70 Calculate the number of electrons needed to completely fill the valence shells for carbon (n_C, full = 8), hydrogen (n_H, full = 2), iodine (n_I, full = 8), and nitrogen (n_N, full = 8): 10 n_C, full + 13 n_H, full + n_I, full + 2 n_N, full = 130 Subtracting these two numbers shows that 130 - 70 = 60 bonding electrons are needed. Each bond has two electrons, so in addition to the 27 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 | 179 °C boiling point | 360.8 °C critical temperature | 908 K critical pressure | 3.682 MPa critical volume | 572.5 cm^3/mol molar heat of vaporization | 59.5 kJ/mol molar heat of fusion | 25.18 kJ/mol molar enthalpy | 238 kJ/mol molar free energy | 437 kJ/mol (computed using the Joback method)

longest chain length | 9 atoms longest straight chain length | 0 atoms longest aliphatic chain length | 0 atoms aromatic atom count | 6 atoms H-bond acceptor count | 2 atoms H-bond donor count | 1 atom

Find the elemental composition for 1-(4-iodophenyl)piperazine 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_10H_13IN_2 Use the chemical formula, C_10H_13IN_2, to count the number of atoms, N_i, for each element and find the total number of atoms, N_atoms: | number of atoms I (iodine) | 1 C (carbon) | 10 N (nitrogen) | 2 H (hydrogen) | 13 N_atoms = 1 + 10 + 2 + 13 = 26 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 I (iodine) | 1 | 1/26 C (carbon) | 10 | 10/26 N (nitrogen) | 2 | 2/26 H (hydrogen) | 13 | 13/26 Check: 1/26 + 10/26 + 2/26 + 13/26 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent I (iodine) | 1 | 1/26 × 100% = 3.85% C (carbon) | 10 | 10/26 × 100% = 38.5% N (nitrogen) | 2 | 2/26 × 100% = 7.69% H (hydrogen) | 13 | 13/26 × 100% = 50.0% 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 I (iodine) | 1 | 3.85% | 126.90447 C (carbon) | 10 | 38.5% | 12.011 N (nitrogen) | 2 | 7.69% | 14.007 H (hydrogen) | 13 | 50.0% | 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 I (iodine) | 1 | 3.85% | 126.90447 | 1 × 126.90447 = 126.90447 C (carbon) | 10 | 38.5% | 12.011 | 10 × 12.011 = 120.110 N (nitrogen) | 2 | 7.69% | 14.007 | 2 × 14.007 = 28.014 H (hydrogen) | 13 | 50.0% | 1.008 | 13 × 1.008 = 13.104 m = 126.90447 u + 120.110 u + 28.014 u + 13.104 u = 288.13247 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 I (iodine) | 1 | 3.85% | 126.90447/288.13247 C (carbon) | 10 | 38.5% | 120.110/288.13247 N (nitrogen) | 2 | 7.69% | 28.014/288.13247 H (hydrogen) | 13 | 50.0% | 13.104/288.13247 Check: 126.90447/288.13247 + 120.110/288.13247 + 28.014/288.13247 + 13.104/288.13247 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent I (iodine) | 1 | 3.85% | 126.90447/288.13247 × 100% = 44.04% C (carbon) | 10 | 38.5% | 120.110/288.13247 × 100% = 41.69% N (nitrogen) | 2 | 7.69% | 28.014/288.13247 × 100% = 9.723% H (hydrogen) | 13 | 50.0% | 13.104/288.13247 × 100% = 4.548%

The first step in finding the oxidation states (or oxidation numbers) in 1-(4-iodophenyl)piperazine is to draw the structure diagram. Next set every oxidation number equal to the atom's formal charge: In 1-(4-iodophenyl)piperazine 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-iodine bond, 5 carbon-nitrogen bonds, and 8 carbon-carbon bonds. For each of these bonds, assign the bonding electrons to the most electronegative element. First examine the carbon-iodine bond: element | electronegativity (Pauling scale) | C | 2.55 | I | 2.66 | | | Since iodine is more electronegative than carbon, the electrons in this bond will go to iodine. Decrease the oxidation number for iodine (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 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) | 2 -1 | C (carbon) | 8 | I (iodine) | 1 +1 | C (carbon) | 2 | H (hydrogen) | 13

First draw the structure diagram for 1-(4-iodophenyl)piperazine, 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 | 26 edge count | 27 Schultz index | 5180 Wiener index | 1307 Hosoya index | 71704 Balaban index | 2.675