ethyl(4-iodobenzoyl)acetate

ethyl(4-iodobenzoyl)acetate

formula | IC_6H_4COCH_2CO_2C_2H_5 Hill formula | C_11H_11IO_3 name | ethyl(4-iodobenzoyl)acetate IUPAC name | 3-(4-iodophenyl)-3-oxopropanoic acid ethyl ester alternate names | 3-(4-iodophenyl)-3-keto-propionic acid ethyl ester | ethyl 3-(4-iodophenyl)-3-oxo-propanoate | ethyl 3-(4-iodophenyl)-3-oxopropanoate mass fractions | C (carbon) 41.5% | H (hydrogen) 3.49% | I (iodine) 39.9% | O (oxygen) 15.1%

Draw the Lewis structure of ethyl(4-iodobenzoyl)acetate. 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 oxygen (n_O, val = 6) atoms: 11 n_C, val + 11 n_H, val + n_I, val + 3 n_O, val = 80 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 oxygen (n_O, full = 8): 11 n_C, full + 11 n_H, full + n_I, full + 3 n_O, full = 142 Subtracting these two numbers shows that 142 - 80 = 62 bonding electrons are needed. Each bond has two electrons, so in addition to the 26 bonds already present in the diagram add 5 bonds. To minimize formal charge oxygen wants 2 bonds and carbon wants 4 bonds. Identify the atoms that want additional bonds and the number of electrons remaining on each atom: Fill in the 5 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: | |

3D structure

molar mass | 318.11 g/mol density | 1.609 g/cm^3

CAS number | 63131-30-6 PubChem CID number | 3593589 PubChem SID number | 24879803 SMILES identifier | CCOC(=O)CC(=O)C1=CC=C(C=C1)I InChI identifier | InChI=1/C11H11IO3/c1-2-15-11(14)7-10(13)8-3-5-9(12)6-4-8/h3-6H, 2, 7H2, 1H3 MDL number | MFCD03701540

flash point | 110 °C