benzyl bromide-d 7

benzyl bromide-d 7

formula | C_6D_5CD_2Br Hill formula | C_7D_7Br name | benzyl bromide-d 7 IUPAC name | 1-(bromo-dideuteriomethyl)-2, 3, 4, 5, 6-pentadeuteriobenzene alternate names | 1-(bromo-dideuterio-methyl)-2, 3, 4, 5, 6-pentadeuterio-benzene | 1-(bromo-dideuteriomethyl)-2, 3, 4, 5, 6-pentadeuteriobenzene | α-bromotoluene-d7 mass fractions | Br (bromine) 0.449% | C (carbon) 0.472% | H (hydrogen) 0.0792%

Draw the Lewis structure of benzyl bromide-d 7. Start by drawing the overall structure of the molecule, ignoring potential double and triple bonds: Count the total valence electrons of the bromine (n_Br, val = 7), carbon (n_C, val = 4), and hydrogen (n_H, val = 1) atoms: n_Br, val + 7 n_C, val + 7 n_H, val = 42 Calculate the number of electrons needed to completely fill the valence shells for bromine (n_Br, full = 8), carbon (n_C, full = 8), and hydrogen (n_H, full = 2): n_Br, full + 7 n_C, full + 7 n_H, full = 78 Subtracting these two numbers shows that 78 - 42 = 36 bonding electrons are needed. Each bond has two electrons, so in addition to the 15 bonds already present in the diagram add 3 bonds. To minimize formal charge carbon wants 4 bonds. Identify the atoms that want additional bonds and the number of electrons remaining on each atom: Fill in the 3 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 | 178.08 g/mol phase | liquid (at STP) melting point | -2 °C boiling point | 198.5 °C density | 1.497 g/cm^3

density | 1.497 g/cm^3 refractive index | 1.574

H-2 | 7

CAS number | 35656-93-0 PubChem CID number | 12217704 PubChem SID number | 24871681 SMILES identifier | C1=CC=C(C=C1)CBr InChI identifier | InChI=1/C7H7Br/c8-6-7-4-2-1-3-5-7/h1-5H, 6H2/i1D, 2D, 3D, 4D, 5D, 6D2 MDL number | MFCD01075417

flash point | 86.67 °C