name of 4 methylenedioxyphenethylamine hydrochloride

4 methylenedioxyphenethylamine hydrochloride

molar mass | 201.7 g/mol formula | C_9H_12ClNO_2 empirical formula | Cl_C_9O_2N_H_12 SMILES identifier | C1=CC2=C(C=C1CCN)OCO2.Cl InChI identifier | InChI=1/C9H11NO2.ClH/c10-4-3-7-1-2-8-9(5-7)12-6-11-8;/h1-2, 5H, 3-4, 6, 10H2;1H InChI key | NDYXFQODWGEGNU-UHFFFAOYSA-N

Structure diagram

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 4 methylenedioxyphenethylamine hydrochloride 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_12ClNO_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 Cl (chlorine) | 1 C (carbon) | 9 O (oxygen) | 2 N (nitrogen) | 1 H (hydrogen) | 12 N_atoms = 1 + 9 + 2 + 1 + 12 = 25 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 | 1/25 C (carbon) | 9 | 9/25 O (oxygen) | 2 | 2/25 N (nitrogen) | 1 | 1/25 H (hydrogen) | 12 | 12/25 Check: 1/25 + 9/25 + 2/25 + 1/25 + 12/25 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent Cl (chlorine) | 1 | 1/25 × 100% = 4.00% C (carbon) | 9 | 9/25 × 100% = 36.0% O (oxygen) | 2 | 2/25 × 100% = 8.00% N (nitrogen) | 1 | 1/25 × 100% = 4.00% H (hydrogen) | 12 | 12/25 × 100% = 48.0% 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 Cl (chlorine) | 1 | 4.00% | 35.45 C (carbon) | 9 | 36.0% | 12.011 O (oxygen) | 2 | 8.00% | 15.999 N (nitrogen) | 1 | 4.00% | 14.007 H (hydrogen) | 12 | 48.0% | 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 Cl (chlorine) | 1 | 4.00% | 35.45 | 1 × 35.45 = 35.45 C (carbon) | 9 | 36.0% | 12.011 | 9 × 12.011 = 108.099 O (oxygen) | 2 | 8.00% | 15.999 | 2 × 15.999 = 31.998 N (nitrogen) | 1 | 4.00% | 14.007 | 1 × 14.007 = 14.007 H (hydrogen) | 12 | 48.0% | 1.008 | 12 × 1.008 = 12.096 m = 35.45 u + 108.099 u + 31.998 u + 14.007 u + 12.096 u = 201.650 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 | 4.00% | 35.45/201.650 C (carbon) | 9 | 36.0% | 108.099/201.650 O (oxygen) | 2 | 8.00% | 31.998/201.650 N (nitrogen) | 1 | 4.00% | 14.007/201.650 H (hydrogen) | 12 | 48.0% | 12.096/201.650 Check: 35.45/201.650 + 108.099/201.650 + 31.998/201.650 + 14.007/201.650 + 12.096/201.650 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent Cl (chlorine) | 1 | 4.00% | 35.45/201.650 × 100% = 17.58% C (carbon) | 9 | 36.0% | 108.099/201.650 × 100% = 53.61% O (oxygen) | 2 | 8.00% | 31.998/201.650 × 100% = 15.87% N (nitrogen) | 1 | 4.00% | 14.007/201.650 × 100% = 6.946% H (hydrogen) | 12 | 48.0% | 12.096/201.650 × 100% = 5.999%

The first step in finding the oxidation states (or oxidation numbers) in 4 methylenedioxyphenethylamine hydrochloride is to draw the structure diagram. Next set every oxidation number equal to the atom's formal charge: In 4 methylenedioxyphenethylamine hydrochloride 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 | Cl (chlorine) | 1 0 | C (carbon) | 2 +1 | C (carbon) | 2 | H (hydrogen) | 12

hybridization | element | count sp^2 | C (carbon) | 6 | O (oxygen) | 2 sp^3 | C (carbon) | 3 | Cl (chlorine) | 1 | N (nitrogen) | 1

Orbital hybridization Structure diagram

vertex count | 25 edge count | 25 Schultz index | Wiener index | Hosoya index | Balaban index |