methyl 4-amino-3-iodobenzoate

methyl 4-amino-3-iodobenzoate

formula | H_2NC_6H_3(I)CO_2CH_3 Hill formula | C_8H_8INO_2 name | methyl 4-amino-3-iodobenzoate IUPAC name | 4-amino-3-iodobenzoic acid methyl ester alternate names | 4-amino-3-iodo-benzoic acid methyl ester | methyl 4-amino-3-iodo-benzoate mass fractions | C (carbon) 34.7% | H (hydrogen) 2.91% | I (iodine) 45.8% | N (nitrogen) 5.06% | O (oxygen) 11.5%

Draw the Lewis structure of methyl 4-amino-3-iodobenzoate. 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), nitrogen (n_N, val = 5), and oxygen (n_O, val = 6) atoms: 8 n_C, val + 8 n_H, val + n_I, 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), iodine (n_I, full = 8), nitrogen (n_N, full = 8), and oxygen (n_O, full = 8): 8 n_C, full + 8 n_H, full + n_I, full + n_N, full + 2 n_O, full = 112 Subtracting these two numbers shows that 112 - 64 = 48 bonding electrons are needed. Each bond has two electrons, so in addition to the 20 bonds already present in the diagram add 4 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 4 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 | 277.06 g/mol phase | solid (at STP) melting point | 88.5 °C

CAS number | 19718-49-1 PubChem CID number | 4104649 PubChem SID number | 24873801 SMILES identifier | COC(=O)C1=CC(=C(C=C1)N)I InChI identifier | InChI=1/C8H8INO2/c1-12-8(11)5-2-3-7(10)6(9)4-5/h2-4H, 10H2, 1H3 MDL number | MFCD00090423