2-amino-4-(hydroxymethyl-phosphinyl)butanoic acid

2-amino-4-(hydroxymethyl-phosphinyl)butanoic acid

molar mass | 181.1 g/mol formula | C_5H_12NO_4P empirical formula | N_C_5O_4P_H_12 SMILES identifier | C(CP(=O)CO)C(C(=O)O)N InChI identifier | InChI=1/C5H12NO4P/c6-4(5(8)9)1-2-11(10)3-7/h4, 7, 11H, 1-3, 6H2, (H, 8, 9)/f/h8H InChI key | FNJQGUKETYGBGL-UHFFFAOYSA-N

Draw the Lewis structure of 2-amino-4-(hydroxymethyl-phosphinyl)butanoic acid. 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), oxygen (n_O, val = 6), and phosphorus (n_P, val = 5) atoms: 5 n_C, val + 12 n_H, val + n_N, val + 4 n_O, val + n_P, val = 66 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), oxygen (n_O, full = 8), and phosphorus (n_P, full = 8): 5 n_C, full + 12 n_H, full + n_N, full + 4 n_O, full + n_P, full = 112 Subtracting these two numbers shows that 112 - 66 = 46 bonding electrons are needed. Each bond has two electrons, so in addition to the 22 bonds already present in the diagram we expect to add 1 bond. To minimize formal charge carbon wants 4 bonds and oxygen wants 2 bonds. Identify the atoms that want additional bonds and the number of electrons remaining on each atom: Add 1 bond by pairing electrons between adjacent highlighted atoms. Additionally, atoms with large electronegativities can minimize their formal charge by forcing atoms with smaller electronegativities on period 3 or higher to expand their valence shells. The electronegativities of the atoms are 2.19 (phosphorus), 2.20 (hydrogen), 2.55 (carbon), 3.04 (nitrogen), and 3.44 (oxygen). Because the electronegativity of phosphorus is smaller than the electronegativity of oxygen, expand the valence shell of phosphorus to 5 bonds. Therefore we add a total of 2 bonds to the diagram: Answer: | |

longest chain length | 8 atoms longest straight chain length | 8 atoms longest aliphatic chain length | 4 atoms aromatic atom count | 0 atoms H-bond acceptor count | 5 atoms H-bond donor count | 3 atoms

Find the elemental composition for 2-amino-4-(hydroxymethyl-phosphinyl)butanoic acid 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_12NO_4P 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 N (nitrogen) | 1 C (carbon) | 5 O (oxygen) | 4 P (phosphorus) | 1 H (hydrogen) | 12 N_atoms = 1 + 5 + 4 + 1 + 12 = 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 N (nitrogen) | 1 | 1/23 C (carbon) | 5 | 5/23 O (oxygen) | 4 | 4/23 P (phosphorus) | 1 | 1/23 H (hydrogen) | 12 | 12/23 Check: 1/23 + 5/23 + 4/23 + 1/23 + 12/23 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent N (nitrogen) | 1 | 1/23 × 100% = 4.35% C (carbon) | 5 | 5/23 × 100% = 21.7% O (oxygen) | 4 | 4/23 × 100% = 17.4% P (phosphorus) | 1 | 1/23 × 100% = 4.35% H (hydrogen) | 12 | 12/23 × 100% = 52.2% 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 N (nitrogen) | 1 | 4.35% | 14.007 C (carbon) | 5 | 21.7% | 12.011 O (oxygen) | 4 | 17.4% | 15.999 P (phosphorus) | 1 | 4.35% | 30.973761998 H (hydrogen) | 12 | 52.2% | 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 N (nitrogen) | 1 | 4.35% | 14.007 | 1 × 14.007 = 14.007 C (carbon) | 5 | 21.7% | 12.011 | 5 × 12.011 = 60.055 O (oxygen) | 4 | 17.4% | 15.999 | 4 × 15.999 = 63.996 P (phosphorus) | 1 | 4.35% | 30.973761998 | 1 × 30.973761998 = 30.973761998 H (hydrogen) | 12 | 52.2% | 1.008 | 12 × 1.008 = 12.096 m = 14.007 u + 60.055 u + 63.996 u + 30.973761998 u + 12.096 u = 181.127761998 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 N (nitrogen) | 1 | 4.35% | 14.007/181.127761998 C (carbon) | 5 | 21.7% | 60.055/181.127761998 O (oxygen) | 4 | 17.4% | 63.996/181.127761998 P (phosphorus) | 1 | 4.35% | 30.973761998/181.127761998 H (hydrogen) | 12 | 52.2% | 12.096/181.127761998 Check: 14.007/181.127761998 + 60.055/181.127761998 + 63.996/181.127761998 + 30.973761998/181.127761998 + 12.096/181.127761998 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent N (nitrogen) | 1 | 4.35% | 14.007/181.127761998 × 100% = 7.733% C (carbon) | 5 | 21.7% | 60.055/181.127761998 × 100% = 33.16% O (oxygen) | 4 | 17.4% | 63.996/181.127761998 × 100% = 35.33% P (phosphorus) | 1 | 4.35% | 30.973761998/181.127761998 × 100% = 17.10% H (hydrogen) | 12 | 52.2% | 12.096/181.127761998 × 100% = 6.678%

The first step in finding the oxidation states (or oxidation numbers) in 2-amino-4-(hydroxymethyl-phosphinyl)butanoic acid is to draw the structure diagram. Next set every oxidation number equal to the atom's formal charge: There are 7 carbon-hydrogen bonds, 1 carbon-nitrogen bond, 3 carbon-oxygen bonds, 2 carbon-phosphorus bonds, 2 nitrogen-hydrogen bonds, 2 oxygen-hydrogen bonds, 1 oxygen-phosphorus bond, 1 phosphorus-hydrogen bond, and 3 carbon-carbon bonds in 2-amino-4-(hydroxymethyl-phosphinyl)butanoic acid. For each of these bonds, assign the bonding electrons to the most electronegative element. First examine the carbon-hydrogen bonds: element | electronegativity (Pauling scale) | C | 2.55 | H | 2.20 | | | Since carbon is more electronegative than hydrogen, the electrons in these bonds will go to carbon. Decrease the oxidation number for carbon in every highlighted bond (by 1 for single bonds, 2 for double bonds, and 3 for triple bonds), and increase the oxidation number for hydrogen accordingly: Next look at 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: 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-phosphorus bonds: element | electronegativity (Pauling scale) | C | 2.55 | P | 2.19 | | | Since carbon is more electronegative than phosphorus, the electrons in these bonds will go to carbon: Next look at the nitrogen-hydrogen bonds: element | electronegativity (Pauling scale) | N | 3.04 | H | 2.20 | | | Since nitrogen is more electronegative than hydrogen, the electrons in these bonds will go to nitrogen: Next look at the oxygen-hydrogen bonds: element | electronegativity (Pauling scale) | O | 3.44 | H | 2.20 | | | Since oxygen is more electronegative than hydrogen, the electrons in these bonds will go to oxygen: Next look at the oxygen-phosphorus bond: element | electronegativity (Pauling scale) | O | 3.44 | P | 2.19 | | | Since oxygen is more electronegative than phosphorus, the electrons in this bond will go to oxygen: Next look at the phosphorus-hydrogen bond: element | electronegativity (Pauling scale) | P | 2.19 | H | 2.20 | | | Since hydrogen is more electronegative than phosphorus, the electrons in this bond will go to hydrogen: 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) | 1 | N (nitrogen) | 1 -2 | C (carbon) | 2 | O (oxygen) | 4 -1 | H (hydrogen) | 1 0 | C (carbon) | 1 +1 | H (hydrogen) | 11 +3 | C (carbon) | 1 +5 | P (phosphorus) | 1

First draw the structure diagram for 2-amino-4-(hydroxymethyl-phosphinyl)butanoic acid, 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 | 22 Schultz index | 3598 Wiener index | 996 Hosoya index | 11431 Balaban index | 6.342