4-chloro-2-methylphenylmagnesium bromide

4-chloro-2-methylphenylmagnesium bromide

molar mass | 229.8 g/mol formula | C_7H_6BrClMg empirical formula | Cl_C_7Mg_Br_H_6 InChI identifier | InChI=1/C7H6Cl.BrH.Mg/c1-6-3-2-4-7(8)5-6;;/h2, 4-5H, 1H3;1H;/q;;+1/p-1/fC7H6Cl.Br.Mg/h;1h;/q;-1;m InChI key | JTXWYWFGKJGGNN-UHFFFAOYSA-M

Draw the Lewis structure of 4-chloro-2-methylphenylmagnesium bromide. 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), chlorine (n_Cl, val = 7), hydrogen (n_H, val = 1), and magnesium (n_Mg, val = 2) atoms: n_Br, val + 7 n_C, val + n_Cl, val + 6 n_H, val + n_Mg, val = 50 Calculate the number of electrons needed to completely fill the valence shells for bromine (n_Br, full = 8), carbon (n_C, full = 8), chlorine (n_Cl, full = 8), hydrogen (n_H, full = 2), and magnesium (n_Mg, full = 4): n_Br, full + 7 n_C, full + n_Cl, full + 6 n_H, full + n_Mg, full = 88 Subtracting these two numbers shows that 88 - 50 = 38 bonding electrons are needed. Each bond has two electrons, so in addition to the 16 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: | |

longest chain length | 7 atoms longest straight chain length | 2 atoms longest aliphatic chain length | 0 atoms aromatic atom count | 6 atoms H-bond acceptor count | 0 atoms H-bond donor count | 0 atoms

Find the elemental composition for 4-chloro-2-methylphenylmagnesium 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_7H_6BrClMg 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 Cl (chlorine) | 1 C (carbon) | 7 Mg (magnesium) | 1 Br (bromine) | 1 H (hydrogen) | 6 N_atoms = 1 + 7 + 1 + 1 + 6 = 16 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 Cl (chlorine) | 1 | 1/16 C (carbon) | 7 | 7/16 Mg (magnesium) | 1 | 1/16 Br (bromine) | 1 | 1/16 H (hydrogen) | 6 | 6/16 Check: 1/16 + 7/16 + 1/16 + 1/16 + 6/16 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent Cl (chlorine) | 1 | 1/16 × 100% = 6.25% C (carbon) | 7 | 7/16 × 100% = 43.8% Mg (magnesium) | 1 | 1/16 × 100% = 6.25% Br (bromine) | 1 | 1/16 × 100% = 6.25% H (hydrogen) | 6 | 6/16 × 100% = 37.5% 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 Cl (chlorine) | 1 | 6.25% | 35.45 C (carbon) | 7 | 43.8% | 12.011 Mg (magnesium) | 1 | 6.25% | 24.305 Br (bromine) | 1 | 6.25% | 79.904 H (hydrogen) | 6 | 37.5% | 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 Cl (chlorine) | 1 | 6.25% | 35.45 | 1 × 35.45 = 35.45 C (carbon) | 7 | 43.8% | 12.011 | 7 × 12.011 = 84.077 Mg (magnesium) | 1 | 6.25% | 24.305 | 1 × 24.305 = 24.305 Br (bromine) | 1 | 6.25% | 79.904 | 1 × 79.904 = 79.904 H (hydrogen) | 6 | 37.5% | 1.008 | 6 × 1.008 = 6.048 m = 35.45 u + 84.077 u + 24.305 u + 79.904 u + 6.048 u = 229.784 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 Cl (chlorine) | 1 | 6.25% | 35.45/229.784 C (carbon) | 7 | 43.8% | 84.077/229.784 Mg (magnesium) | 1 | 6.25% | 24.305/229.784 Br (bromine) | 1 | 6.25% | 79.904/229.784 H (hydrogen) | 6 | 37.5% | 6.048/229.784 Check: 35.45/229.784 + 84.077/229.784 + 24.305/229.784 + 79.904/229.784 + 6.048/229.784 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent Cl (chlorine) | 1 | 6.25% | 35.45/229.784 × 100% = 15.43% C (carbon) | 7 | 43.8% | 84.077/229.784 × 100% = 36.59% Mg (magnesium) | 1 | 6.25% | 24.305/229.784 × 100% = 10.58% Br (bromine) | 1 | 6.25% | 79.904/229.784 × 100% = 34.77% H (hydrogen) | 6 | 37.5% | 6.048/229.784 × 100% = 2.632%

The first step in finding the oxidation states (or oxidation numbers) in 4-chloro-2-methylphenylmagnesium bromide is to draw the structure diagram. Next set every oxidation number equal to the atom's formal charge: In 4-chloro-2-methylphenylmagnesium 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-chlorine bond, 1 carbon-magnesium bond, and 7 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-chlorine bond: element | electronegativity (Pauling scale) | C | 2.55 | Cl | 3.16 | | | Since chlorine is more electronegative than carbon, the electrons in this bond will go to chlorine: 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-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 -1 | Br (bromine) | 1 | C (carbon) | 4 | Cl (chlorine) | 1 0 | C (carbon) | 1 +1 | C (carbon) | 1 | H (hydrogen) | 6 +2 | Mg (magnesium) | 1

First draw the structure diagram for 4-chloro-2-methylphenylmagnesium 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 | 16 edge count | 16 Schultz index | 1466 Wiener index | 379 Hosoya index | 971 Balaban index | 3.032