(5z,9e,14z)-(8xi,11xi,12s)-8,11,12-trihydroxyicosa-5,9,14-trienoate

(5z, 9e, 14z)-(8xi, 11xi, 12s)-8, 11, 12-trihydroxyicosa-5, 9, 14-trienoate

formula | C_20H_34O_5 name | (5z, 9e, 14z)-(8xi, 11xi, 12s)-8, 11, 12-trihydroxyicosa-5, 9, 14-trienoate mass fractions | C (carbon) 68% | H (hydrogen) 9.41% | O (oxygen) 22.6%

Draw the Lewis structure of (5z, 9e, 14z)-(8xi, 11xi, 12s)-8, 11, 12-trihydroxyicosa-5, 9, 14-trienoate. 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), and oxygen (n_O, val = 6) atoms, including the net charge: 20 n_C, val + 33 n_H, val + 5 n_O, val - n_charge = 144 Calculate the number of electrons needed to completely fill the valence shells for carbon (n_C, full = 8), hydrogen (n_H, full = 2), and oxygen (n_O, full = 8): 20 n_C, full + 33 n_H, full + 5 n_O, full = 266 Subtracting these two numbers shows that 266 - 144 = 122 bonding electrons are needed. Each bond has two electrons, so in addition to the 57 bonds already present in the diagram add 4 bonds. 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. The net charge has been given to the most electronegative atom, oxygen: Fill in the 4 bonds by pairing electrons between adjacent highlighted atoms, noting the formal charges of the atoms. Double bonding carbon to the other highlighted oxygen atom would result in an equivalent molecule: Answer: | |

molar mass | 353.48 g/mol

SMILES identifier | CCCCC/C=C\C[C@@H](C(/C=C/C(C/C=C\CCCC(=O)[O-])O)O)O