8-naphthalic anhydride

8-naphthalic anhydride

molar mass | 326.3 g/mol formula | C_22H_14O_3 empirical formula | C_22O_3H_14 SMILES identifier | C1=CC=C2C(=C1)C=CC=C2C(=O)OC(=O)C3=CC=CC4=CC=CC=C43 InChI identifier | InChI=1/C22H14O3/c23-21(19-13-5-9-15-7-1-3-11-17(15)19)25-22(24)20-14-6-10-16-8-2-4-12-18(16)20/h1-14H InChI key | BYVCTYDTPSKPRM-UHFFFAOYSA-N

Draw the Lewis structure of 8-naphthalic anhydride. Start by drawing the overall structure of the molecule, ignoring potential double and triple bonds: Count the total valence electrons of the carbon (n_C, val = 4), hydrogen (n_H, val = 1), and oxygen (n_O, val = 6) atoms: 22 n_C, val + 14 n_H, val + 3 n_O, val = 120 Calculate the number of electrons needed to completely fill the valence shells for carbon (n_C, full = 8), hydrogen (n_H, full = 2), and oxygen (n_O, full = 8): 22 n_C, full + 14 n_H, full + 3 n_O, full = 228 Subtracting these two numbers shows that 228 - 120 = 108 bonding electrons are needed. Each bond has two electrons, so in addition to the 42 bonds already present in the diagram add 12 bonds. To minimize formal charge carbon wants 4 bonds and oxygen wants 2 bonds. Identify the atoms that want additional bonds and the number of electrons remaining on each atom: Fill in the 12 bonds by pairing electrons between adjacent highlighted atoms. Note that the six atom rings are aromatic, so that the single and double bonds may be rearranged: Answer: | |

longest chain length | 13 atoms longest straight chain length | 5 atoms longest aliphatic chain length | 0 atoms aromatic atom count | 20 atoms H-bond acceptor count | 3 atoms H-bond donor count | 0 atoms

Find the elemental composition for 8-naphthalic anhydride 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_22H_14O_3 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 C (carbon) | 22 O (oxygen) | 3 H (hydrogen) | 14 N_atoms = 22 + 3 + 14 = 39 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 C (carbon) | 22 | 22/39 O (oxygen) | 3 | 3/39 H (hydrogen) | 14 | 14/39 Check: 22/39 + 3/39 + 14/39 = 1 Compute atom percents using the atom fractions: | number of atoms | atom percent C (carbon) | 22 | 22/39 × 100% = 56.4% O (oxygen) | 3 | 3/39 × 100% = 7.69% H (hydrogen) | 14 | 14/39 × 100% = 35.9% 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 C (carbon) | 22 | 56.4% | 12.011 O (oxygen) | 3 | 7.69% | 15.999 H (hydrogen) | 14 | 35.9% | 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 C (carbon) | 22 | 56.4% | 12.011 | 22 × 12.011 = 264.242 O (oxygen) | 3 | 7.69% | 15.999 | 3 × 15.999 = 47.997 H (hydrogen) | 14 | 35.9% | 1.008 | 14 × 1.008 = 14.112 m = 264.242 u + 47.997 u + 14.112 u = 326.351 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 C (carbon) | 22 | 56.4% | 264.242/326.351 O (oxygen) | 3 | 7.69% | 47.997/326.351 H (hydrogen) | 14 | 35.9% | 14.112/326.351 Check: 264.242/326.351 + 47.997/326.351 + 14.112/326.351 = 1 Compute mass percents using the mass fractions: Answer: | | | number of atoms | atom percent | mass percent C (carbon) | 22 | 56.4% | 264.242/326.351 × 100% = 80.97% O (oxygen) | 3 | 7.69% | 47.997/326.351 × 100% = 14.71% H (hydrogen) | 14 | 35.9% | 14.112/326.351 × 100% = 4.324%

The first step in finding the oxidation states (or oxidation numbers) in 8-naphthalic anhydride is to draw the structure diagram. Next set every oxidation number equal to the atom's formal charge: In 8-naphthalic anhydride 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 4 carbon-oxygen bonds, and 24 carbon-carbon bonds. For each of these bonds, assign the bonding electrons to the most electronegative element. First examine 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. Decrease the oxidation number for oxygen in every highlighted bond (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-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 -2 | O (oxygen) | 3 -1 | C (carbon) | 14 0 | C (carbon) | 6 +1 | H (hydrogen) | 14 +3 | C (carbon) | 2

First draw the structure diagram for 8-naphthalic anhydride, 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 Assign the provisional hybridization based on the table: Next identify any sp^3 atoms with lone pair electrons which can participate in a conjugated π-bond system. These atoms can lower their energy by placing a lone pair in a unhybridized p orbital to maximize overlap with the neighboring π-bonds. Note that halogens and elements from the third period and below do not engage in bond conjugation, except in the case of aromaticity: Adjust the provisional hybridizations to arrive at the result: Answer: | |

vertex count | 39 edge count | 42 Schultz index | 18728 Wiener index | 4492 Hosoya index | 6.576×10^7 Balaban index | 1.638