name of 3,4-methylenedioxyphenethylamine

3, 4-methylenedioxyphenethylamine

molar mass | 165.2 g/mol formula | C_9H_11NO_2 empirical formula | C_9O_2N_H_11 SMILES identifier | C1=CC2=C(C=C1CCN)OCO2 InChI identifier | InChI=1/C9H11NO2/c10-4-3-7-1-2-8-9(5-7)12-6-11-8/h1-2, 5H, 3-4, 6, 10H2 InChI key | RRIRDPSOCUCGBV-UHFFFAOYSA-N

Draw the Lewis structure of 3, 4-methylenedioxyphenethylamine. 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 oxygen (n_O, val = 6) atoms: 9 n_C, val + 11 n_H, val + n_N, val + 2 n_O, val = 64 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 oxygen (n_O, full = 8): 9 n_C, full + 11 n_H, full + n_N, full + 2 n_O, full = 118 Subtracting these two numbers shows that 118 - 64 = 54 bonding electrons are needed. Each bond has two electrons, so in addition to the 24 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 | 128.1 °C boiling point | 306.7 °C critical temperature | 814 K critical pressure | 4.011 MPa critical volume | 460.5 cm^3/mol molar heat of vaporization | 59.1 kJ/mol molar heat of fusion | 30.54 kJ/mol molar enthalpy | -153 kJ/mol molar free energy | 80.72 kJ/mol (computed using the Joback method)

longest chain length | 8 atoms longest straight chain length | 3 atoms longest aliphatic chain length | 2 atoms aromatic atom count | 6 atoms H-bond acceptor count | 1 atom H-bond donor count | 1 atom

Find the elemental composition for 3, 4-methylenedioxyphenethylamine 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_9H_11NO_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) | 9 O (oxygen) | 2 N (nitrogen) | 1 H (hydrogen) | 11 N_atoms = 9 + 2 + 1 + 11 = 23 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) | 9 | 9/23 O (oxygen) | 2 | 2/23 N (nitrogen) | 1 | 1/23 H (hydrogen) | 11 | 11/23 Check: 9/23 + 2/23 + 1/23 + 11/23 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent C (carbon) | 9 | 9/23 × 100% = 39.1% O (oxygen) | 2 | 2/23 × 100% = 8.70% N (nitrogen) | 1 | 1/23 × 100% = 4.35% H (hydrogen) | 11 | 11/23 × 100% = 47.8% 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) | 9 | 39.1% | 12.011 O (oxygen) | 2 | 8.70% | 15.999 N (nitrogen) | 1 | 4.35% | 14.007 H (hydrogen) | 11 | 47.8% | 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) | 9 | 39.1% | 12.011 | 9 × 12.011 = 108.099 O (oxygen) | 2 | 8.70% | 15.999 | 2 × 15.999 = 31.998 N (nitrogen) | 1 | 4.35% | 14.007 | 1 × 14.007 = 14.007 H (hydrogen) | 11 | 47.8% | 1.008 | 11 × 1.008 = 11.088 m = 108.099 u + 31.998 u + 14.007 u + 11.088 u = 165.192 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) | 9 | 39.1% | 108.099/165.192 O (oxygen) | 2 | 8.70% | 31.998/165.192 N (nitrogen) | 1 | 4.35% | 14.007/165.192 H (hydrogen) | 11 | 47.8% | 11.088/165.192 Check: 108.099/165.192 + 31.998/165.192 + 14.007/165.192 + 11.088/165.192 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent C (carbon) | 9 | 39.1% | 108.099/165.192 × 100% = 65.44% O (oxygen) | 2 | 8.70% | 31.998/165.192 × 100% = 19.37% N (nitrogen) | 1 | 4.35% | 14.007/165.192 × 100% = 8.479% H (hydrogen) | 11 | 47.8% | 11.088/165.192 × 100% = 6.712%

The first step in finding the oxidation states (or oxidation numbers) in 3, 4-methylenedioxyphenethylamine is to draw the structure diagram. Next set every oxidation number equal to the atom's formal charge: In 3, 4-methylenedioxyphenethylamine 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-nitrogen bond, 4 carbon-oxygen bonds, and 8 carbon-carbon bonds. For each of these bonds, assign the bonding electrons to the most electronegative element. First examine 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. Decrease the oxidation number for nitrogen (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-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) | 2 -1 | C (carbon) | 4 0 | C (carbon) | 2 +1 | C (carbon) | 2 | H (hydrogen) | 11

First draw the structure diagram for 3, 4-methylenedioxyphenethylamine, 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 | 23 edge count | 24 Schultz index | 4084 Wiener index | 1017 Hosoya index | 21127 Balaban index | 2.397