5-bromo-2-(tert-butyldimethylsilyl)pyrimidine

5-bromo-2-(tert-butyldimethylsilyl)pyrimidine

molar mass | 273.2 g/mol formula | C_10H_17BrN_2Si empirical formula | Br_C_10N_2Si_H_17 SMILES identifier | CC(C)(C)[Si](C)(C)C1=NC=C(C=N1)Br InChI identifier | InChI=1/C10H17BrN2Si/c1-10(2, 3)14(4, 5)9-12-6-8(11)7-13-9/h6-7H, 1-5H3 InChI key | XFMYABCUJUMURC-UHFFFAOYSA-N

Draw the Lewis structure of 5-bromo-2-(tert-butyldimethylsilyl)pyrimidine. 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), hydrogen (n_H, val = 1), nitrogen (n_N, val = 5), and silicon (n_Si, val = 4) atoms: n_Br, val + 10 n_C, val + 17 n_H, val + 2 n_N, val + n_Si, val = 78 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), nitrogen (n_N, full = 8), and silicon (n_Si, full = 8): n_Br, full + 10 n_C, full + 17 n_H, full + 2 n_N, full + n_Si, full = 146 Subtracting these two numbers shows that 146 - 78 = 68 bonding electrons are needed. Each bond has two electrons, so in addition to the 31 bonds already present in the diagram add 3 bonds. To minimize formal charge carbon wants 4 bonds and nitrogen wants 3 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: | |

longest chain length | 8 atoms longest straight chain length | 4 atoms longest aliphatic chain length | 3 atoms aromatic atom count | 6 atoms H-bond acceptor count | 2 atoms H-bond donor count | 0 atoms

Find the elemental composition for 5-bromo-2-(tert-butyldimethylsilyl)pyrimidine in terms of the atom and mass percents: atom percent = N_i/N_atoms × 100% mass percent = (N_im_i)/m × 100% Plan: • Write the chemical formula and gather atomic masses from the periodic table. • Determine values for N_i, m_i, N_atoms and m using these items. • Finally, compute the percents and check the results. Write the chemical formula: C_10H_17BrN_2Si Use the chemical formula to count the number of atoms, N_i, for each element and find the total number of atoms, N_atoms, per molecule: | number of atoms Br (bromine) | 1 C (carbon) | 10 N (nitrogen) | 2 Si (silicon) | 1 H (hydrogen) | 17 N_atoms = 1 + 10 + 2 + 1 + 17 = 31 Divide each N_i by N_atoms to calculate atom fractions. Then use the property that atom fractions must sum to one to check the work: | number of atoms | atom fraction Br (bromine) | 1 | 1/31 C (carbon) | 10 | 10/31 N (nitrogen) | 2 | 2/31 Si (silicon) | 1 | 1/31 H (hydrogen) | 17 | 17/31 Check: 1/31 + 10/31 + 2/31 + 1/31 + 17/31 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent Br (bromine) | 1 | 1/31 × 100% = 3.23% C (carbon) | 10 | 10/31 × 100% = 32.3% N (nitrogen) | 2 | 2/31 × 100% = 6.45% Si (silicon) | 1 | 1/31 × 100% = 3.23% H (hydrogen) | 17 | 17/31 × 100% = 54.8% Look up the atomic mass, m_i, in unified atomic mass units, u, for each element in the periodic table: | number of atoms | atom percent | atomic mass/u Br (bromine) | 1 | 3.23% | 79.904 C (carbon) | 10 | 32.3% | 12.011 N (nitrogen) | 2 | 6.45% | 14.007 Si (silicon) | 1 | 3.23% | 28.085 H (hydrogen) | 17 | 54.8% | 1.008 Multiply N_i by m_i to compute the mass for each element. Then sum those values to compute the molecular mass, m: | number of atoms | atom percent | atomic mass/u | mass/u Br (bromine) | 1 | 3.23% | 79.904 | 1 × 79.904 = 79.904 C (carbon) | 10 | 32.3% | 12.011 | 10 × 12.011 = 120.110 N (nitrogen) | 2 | 6.45% | 14.007 | 2 × 14.007 = 28.014 Si (silicon) | 1 | 3.23% | 28.085 | 1 × 28.085 = 28.085 H (hydrogen) | 17 | 54.8% | 1.008 | 17 × 1.008 = 17.136 m = 79.904 u + 120.110 u + 28.014 u + 28.085 u + 17.136 u = 273.249 u Divide the mass for each element by m to calculate mass fractions. Then use the property that mass fractions must sum to one to check the work: | number of atoms | atom percent | mass fraction Br (bromine) | 1 | 3.23% | 79.904/273.249 C (carbon) | 10 | 32.3% | 120.110/273.249 N (nitrogen) | 2 | 6.45% | 28.014/273.249 Si (silicon) | 1 | 3.23% | 28.085/273.249 H (hydrogen) | 17 | 54.8% | 17.136/273.249 Check: 79.904/273.249 + 120.110/273.249 + 28.014/273.249 + 28.085/273.249 + 17.136/273.249 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent Br (bromine) | 1 | 3.23% | 79.904/273.249 × 100% = 29.24% C (carbon) | 10 | 32.3% | 120.110/273.249 × 100% = 43.96% N (nitrogen) | 2 | 6.45% | 28.014/273.249 × 100% = 10.25% Si (silicon) | 1 | 3.23% | 28.085/273.249 × 100% = 10.28% H (hydrogen) | 17 | 54.8% | 17.136/273.249 × 100% = 6.271%

The first step in finding the oxidation states (or oxidation numbers) in 5-bromo-2-(tert-butyldimethylsilyl)pyrimidine is to draw the structure diagram. Next set every oxidation number equal to the atom's formal charge: In 5-bromo-2-(tert-butyldimethylsilyl)pyrimidine hydrogen is not bonded to a metal with lower electronegativity, so it will have an oxidation state of +1. Any element bonded to hydrogen gains the bonding electrons, decreasing their oxidation state by 1 for every bond: With hydrogen out of the way, look at the remaining bonds. There are 1 bromine-carbon bond, 4 carbon-nitrogen bonds, 4 carbon-silicon bonds, and 5 carbon-carbon bonds. For each of these bonds, assign the bonding electrons to the most electronegative element. First examine the bromine-carbon bond: element | electronegativity (Pauling scale) | Br | 2.96 | C | 2.55 | | | Since bromine is more electronegative than carbon, the electrons in this bond will go to bromine. Decrease the oxidation number for bromine (by 1 for single bonds, 2 for double bonds, and 3 for triple bonds), and increase the oxidation number for carbon accordingly: Next look at the carbon-nitrogen bonds: element | electronegativity (Pauling scale) | C | 2.55 | N | 3.04 | | | Since nitrogen is more electronegative than carbon, the electrons in these bonds will go to nitrogen: Next look at the carbon-silicon bonds: element | electronegativity (Pauling scale) | C | 2.55 | Si | 1.90 | | | Since carbon is more electronegative than silicon, the electrons in these bonds will go to carbon: Next look at the carbon-carbon bonds: element | electronegativity (Pauling scale) | C | 2.55 | C | 2.55 | | | Since these elements are the same the bonding electrons are shared equally, and there is no change to the oxidation states: Now summarize the results: Answer: | | oxidation state | element | count -4 | C (carbon) | 2 -3 | C (carbon) | 3 | N (nitrogen) | 2 -1 | Br (bromine) | 1 | C (carbon) | 1 0 | C (carbon) | 1 +1 | C (carbon) | 2 | H (hydrogen) | 17 +2 | C (carbon) | 1 +4 | Si (silicon) | 1

First draw the structure diagram for 5-bromo-2-(tert-butyldimethylsilyl)pyrimidine, and for every non-hydrogen atom, count the σ-bonds. Note that double and triple bonds consist of one σ-bond together with one or two π-bonds: Identify those atoms with lone pairs: Find the steric number by adding the lone pair count to the number of σ-bonds: Consult the following chart to determine the hybridization from the steric number: steric number | hybridization 2 | sp 3 | sp^2 4 | sp^3 5 | dsp^3 6 | d^2sp^3 7 | d^3sp^3 Now assign the hybridization for each atom: Answer: | |

vertex count | 31 edge count | 31 Schultz index | 7233 Wiener index | 1922 Hosoya index | 258304 Balaban index | 4.37