name of (2-bromoethoxy)-tert-butyldimethylsilane

(2-bromoethoxy)-tert-butyldimethylsilane

formula | BrCH_2CH_2OSi(CH_3)_2C(CH_3)_3 Hill formula | C_8H_19BrOSi name | (2-bromoethoxy)-tert-butyldimethylsilane IUPAC name | 2-bromoethoxy-tert-butyl-dimethylsilane alternate names | 2-bromoethoxy-tert-butyl-dimethyl-silane | 2-bromoethoxy-tert-butyl-dimethylsilane mass fractions | Br (bromine) 33.4% | C (carbon) 40.2% | H (hydrogen) 8.01% | O (oxygen) 6.69% | Si (silicon) 11.7%

Draw the Lewis structure of (2-bromoethoxy)-tert-butyldimethylsilane. Start by drawing the overall structure of the molecule: Count the total valence electrons of the bromine (n_Br, val = 7), carbon (n_C, val = 4), hydrogen (n_H, val = 1), oxygen (n_O, val = 6), and silicon (n_Si, val = 4) atoms: n_Br, val + 8 n_C, val + 19 n_H, val + n_O, val + n_Si, val = 68 Calculate the number of electrons needed to completely fill the valence shells for bromine (n_Br, full = 8), carbon (n_C, full = 8), hydrogen (n_H, full = 2), oxygen (n_O, full = 8), and silicon (n_Si, full = 8): n_Br, full + 8 n_C, full + 19 n_H, full + n_O, full + n_Si, full = 126 Subtracting these two numbers shows that 126 - 68 = 58 bonding electrons are needed. Each bond has two electrons, so the above diagram has all the necessary bonds. There are 29 bonds and hence 58 bonding electrons in the diagram. Lastly, fill in the remaining unbonded electrons on each atom. In total, there remain 68 - 58 = 10 electrons left to draw: Answer: | |

3D structure

molar mass | 239.23 g/mol phase | liquid (at STP) boiling point | 72.5 °C (measured at 333.3 Pa) density | 1.115 g/cm^3

density | 1.115 g/cm^3 refractive index | 1.444

CAS number | 86864-60-0 PubChem CID number | 3608067 PubChem SID number | 24866813 SMILES identifier | CC(C)(C)[Si](C)(C)OCCBr InChI identifier | InChI=1/C8H19BrOSi/c1-8(2, 3)11(4, 5)10-7-6-9/h6-7H2, 1-5H3 MDL number | MFCD00209550

flash point | 72.78 °C