(1,3-dioxolan-2-ylmethyl)magnesium bromide

(1, 3-dioxolan-2-ylmethyl)magnesium bromide

molar mass | 191.3 g/mol formula | C_4H_7BrMgO_2 empirical formula | O_2C_4Mg_Br_H_7 SMILES identifier | C1COC(C[Mg]Br)O1 InChI identifier | InChI=1/C4H7O2.BrH.Mg/c1-4-5-2-3-6-4;;/h4H, 1-3H2;1H;/q;;+1/p-1/fC4H7O2.Br.Mg/h;1h;/q;-1;m InChI key | ZDNTZRAFJZERSL-UHFFFAOYSA-M

Draw the Lewis structure of (1, 3-dioxolan-2-ylmethyl)magnesium bromide. 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), magnesium (n_Mg, val = 2), and oxygen (n_O, val = 6) atoms: n_Br, val + 4 n_C, val + 7 n_H, val + n_Mg, val + 2 n_O, val = 44 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), magnesium (n_Mg, full = 4), and oxygen (n_O, full = 8): n_Br, full + 4 n_C, full + 7 n_H, full + n_Mg, full + 2 n_O, full = 74 Subtracting these two numbers shows that 74 - 44 = 30 bonding electrons are needed. Each bond has two electrons, so the above diagram has all the necessary bonds. There are 15 bonds and hence 30 bonding electrons in the diagram. Lastly, fill in the remaining unbonded electrons on each atom. In total, there remain 44 - 30 = 14 electrons left to draw: Answer: | |

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

Find the elemental composition for (1, 3-dioxolan-2-ylmethyl)magnesium bromide 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_4H_7BrMgO_2 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 O (oxygen) | 2 C (carbon) | 4 Mg (magnesium) | 1 Br (bromine) | 1 H (hydrogen) | 7 N_atoms = 2 + 4 + 1 + 1 + 7 = 15 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 O (oxygen) | 2 | 2/15 C (carbon) | 4 | 4/15 Mg (magnesium) | 1 | 1/15 Br (bromine) | 1 | 1/15 H (hydrogen) | 7 | 7/15 Check: 2/15 + 4/15 + 1/15 + 1/15 + 7/15 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent O (oxygen) | 2 | 2/15 × 100% = 13.3% C (carbon) | 4 | 4/15 × 100% = 26.7% Mg (magnesium) | 1 | 1/15 × 100% = 6.67% Br (bromine) | 1 | 1/15 × 100% = 6.67% H (hydrogen) | 7 | 7/15 × 100% = 46.7% 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 O (oxygen) | 2 | 13.3% | 15.999 C (carbon) | 4 | 26.7% | 12.011 Mg (magnesium) | 1 | 6.67% | 24.305 Br (bromine) | 1 | 6.67% | 79.904 H (hydrogen) | 7 | 46.7% | 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 O (oxygen) | 2 | 13.3% | 15.999 | 2 × 15.999 = 31.998 C (carbon) | 4 | 26.7% | 12.011 | 4 × 12.011 = 48.044 Mg (magnesium) | 1 | 6.67% | 24.305 | 1 × 24.305 = 24.305 Br (bromine) | 1 | 6.67% | 79.904 | 1 × 79.904 = 79.904 H (hydrogen) | 7 | 46.7% | 1.008 | 7 × 1.008 = 7.056 m = 31.998 u + 48.044 u + 24.305 u + 79.904 u + 7.056 u = 191.307 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 O (oxygen) | 2 | 13.3% | 31.998/191.307 C (carbon) | 4 | 26.7% | 48.044/191.307 Mg (magnesium) | 1 | 6.67% | 24.305/191.307 Br (bromine) | 1 | 6.67% | 79.904/191.307 H (hydrogen) | 7 | 46.7% | 7.056/191.307 Check: 31.998/191.307 + 48.044/191.307 + 24.305/191.307 + 79.904/191.307 + 7.056/191.307 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent O (oxygen) | 2 | 13.3% | 31.998/191.307 × 100% = 16.73% C (carbon) | 4 | 26.7% | 48.044/191.307 × 100% = 25.11% Mg (magnesium) | 1 | 6.67% | 24.305/191.307 × 100% = 12.70% Br (bromine) | 1 | 6.67% | 79.904/191.307 × 100% = 41.77% H (hydrogen) | 7 | 46.7% | 7.056/191.307 × 100% = 3.688%

The first step in finding the oxidation states (or oxidation numbers) in (1, 3-dioxolan-2-ylmethyl)magnesium bromide is to draw the structure diagram. Next set every oxidation number equal to the atom's formal charge: In (1, 3-dioxolan-2-ylmethyl)magnesium bromide 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-magnesium bond, 1 carbon-magnesium bond, 4 carbon-oxygen bonds, and 2 carbon-carbon bonds. For each of these bonds, assign the bonding electrons to the most electronegative element. First examine the bromine-magnesium bond: element | electronegativity (Pauling scale) | Br | 2.96 | Mg | 1.31 | | | Since bromine is more electronegative than magnesium, 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 magnesium accordingly: Next look at the carbon-magnesium bond: element | electronegativity (Pauling scale) | C | 2.55 | Mg | 1.31 | | | Since carbon is more electronegative than magnesium, the electrons in this bond will go to carbon: Next look at the carbon-oxygen bonds: element | electronegativity (Pauling scale) | C | 2.55 | O | 3.44 | | | Since oxygen is more electronegative than carbon, the electrons in these bonds will go to oxygen: 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 -3 | C (carbon) | 1 -2 | O (oxygen) | 2 -1 | Br (bromine) | 1 | C (carbon) | 2 +1 | C (carbon) | 1 | H (hydrogen) | 7 +2 | Mg (magnesium) | 1

First draw the structure diagram for (1, 3-dioxolan-2-ylmethyl)magnesium bromide, 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 | 15 edge count | 15 Schultz index | 1224 Wiener index | 315 Hosoya index | 454 Balaban index | 3.04