Covalent Bonding / Molecular Structure / Introduction to Organic Chemistry and the Alkanes

1. Covalent Bonding / Molecular Structure / Introduction to Organic Chemistry and the Alkanes H Advanced Chemistry / Organic Unit 1

2. Objectives #1-4 Covalent Bonding and Lewis Structures *formation of a covalent bond: *in ionic bonding, the bonding mechanism comes about through the electrostatic attraction between valence electrons and atomic nuclei as atoms strive to achieve the stable electron arrangements of noble gases *these attractive forces result in substances usually having rigid structures, high melting points, and varying degrees of electrical conductivity *many compounds, including those within the realm of organic chemistry, lack a rigid structure, have low melting points, vaporize easily, and are generally poor conductors *these compounds complete their octets through the sharing of electrons in a covalent bond

3. Objectives #1-4 Covalent Bonding and Lewis Structures *initially when two nonmetals approach each other, they repel each other due to the negative charges of their respective electron clouds and the positive charges of their respective nuclei; due to this situation the potentialenergy of the separate atoms is too high for a stable bond to form *however at an optimum distance between the nuclei, the electron clouds of the two atoms overlap and the valence electrons are concentrated between the two nuclei involved; energy is released in this process and a stable bond can now form between the two atoms (diagram)

4. Illustration of Energy for Formation of Hydrogen Molecule vs. Atomic Distances

5. Drawing Lewis Structures / Structural Formulas / Condensed Structural Formulas / Molecular Line Drawings *examples of Lewis Structures: *examples of structural formulas: *examples of condensed structural formulas: *examples of molecular line drawings:

6. Objectives #5-7 Electronegativity / Formal Charge / Bond Length / Bond Energy *electronegativity and polarity: *when electrons are shared in a covalent bond the degree of electron sharing by the nuclei of the atoms involved is based on the electronegativity values of the atoms in the bond *electronegativity is a measure of the ability of an atom to attract electrons to itself; electronegativity generally decreases down a group on the periodic table and increases across the period *elements are assigned electronegativity values that range from .7 through 4 *electronegativity value differences can be used to predict the most likely type of bond that will form between two atoms; nonpolar covalent, polar covalent, or ionic *the approximate difference ranges for each type of bonding are as follows: 0-.3 nonpolar .4-1.9 polar 2.0 › ionic

7. Trends in Electronegativity

8. Objectives #5-7 Electronegativity / Formal Charge / Bond Length / Bond Energy *the greater the electronegativity difference between two elements, the more polar the bond is expected to be (examples) B-Cl C-Cl P-F P-Cl *since the electrons are not shared equally in a polar bond, regions of negative and positive charge, dipoles, are established *dipoles can be indicated by placing a partial positive charge (ς+) on the atom with the lower electronegativity value and a partial negative charge (ς-) on the atom with the higher electronegativity value; dipoles can also be indicated by drawing an arrow from the lower electronegative atom to the higher electronegative atom (examples) H – F H - F

9. Example of Positive and Negative Dipole Moments

10. Objectives #5-7 Electronegativity / Formal Charge / Bond Length / Bond Energy *Formal Charge *in many cases, when drawing Lewis structures more than one structure can be drawn to satisfy the octet rule *the concept of formal charge can be used to determine which structure representation is the most stable *in order to calculate the formal charge on an atom, the following procedure is used: • All unshared (nonbonding) electrons are assigned to the atom on which they are found • For any bond – single, double, or triple – half of the bonding electrons are assigned to each atom in the bond

11. Objectives #5-7 Electronegativity / Formal Charge / Bond Length / Bond Energy *the formal charge is then calculated by subtracting the number of electrons assigned to the atom from the number of valence electrons in the individual atom *whichever Lewis structure that results in a net formal charge that is closest to zero or results in having the negative charge reside on the more electronegative atom would be the most reasonable to use

12. (examples of formal charge calculations) • Determine the formal charges on the CN-1 ion. [:C=N:]-1 For carbon: 2 nonbondedelectrons assigned to C 3 bonded electrons (1/2 of 6 in the triple bond) assigned to C Total assigned: 5 Total 4 on neutral carbon atom: 4 Formal charge = 4-5 = -1

13. For nitrogen: 2 nonbonded electrons assigned to N 3 bonded electrons (1/2 of 6 in the triple bond) assigned to N Total assigned: 5 Total valence on neutral nitrogen atom: 5 Formal Charge = 5-5 = 0

14. The net overall formal charge for the ion is -1 which equals the charge on the ion

15. Formal Charge Calculation for Carbon Dioxide O=C=O O-C=O Valence electrons assigned: 6,4,6 6,4,6 Nonbonding /bonding electrons assigned to atom: 4+2, 0+4, 4+2 6+1, 0+4, 2+3 Formal Charge: 0, 0, 0 -1, 0, 1 (closest to zero)

16. Formal Charge Calculation for the Thiocyanate Ion N-C=S N=C=S N=C-S (Val) 5,4,6 5,4,6 5,4,6 (B/N) 7, 4, 5 6, 4, 6 5, 4, 7 (FC) -2, 0, 1 -1, 0, 0 0, 0, -1 --------------------------------------------- [N=C=S]-1 is most reasonable because the neg. charge is on the most electronegative element and the sum is closest to zero

17. Objectives #5-7 Electronegativity / Formal Charge / Bond Length / Bond Energy *Bond Length and Bond Energy *recall that bond breaking is usually an endothermic process while bond formation is an exothermic process *a primary source of energy in our society comes from the break down of organic molecules such as methane or propane *some relationships based on “Average Bond Energies” chart: as multiple bonds (single to double to triple) increase, bond length decreases and bond energy increases this is due to the increase in the number of electron pairs present the bond energies listed are considered average values because the type of molecule that the bond belongs to can affect its energy value *to calculate the change in energy or ∆H of a reaction, one must determine the energy required to break the bonds of the reactants and the amount of energy released to form the bonds of the products

18. Example of Bond Energy for the Combustion of Methane

19. Example I Calculate the ΔH for the reaction of hydrogen and fluorine to produce hydrogen fluoride. H2(g) + F2(g) -- 2HF(g) Consult chart for bond energies. Step I Calculate the energy needed (endothermic) to break the bonds of the reactants: H-H +432 kJ/mole F-F +154 kJ/mole Total +586 kJ

20. Step II Calculate the energy released (exothermic) to form the bonds of the products. 2H-F -565 kJ/mole X 2 = -1130 kJ Step III Add the two values together. Remember that the energy term for the reactants will have a positive value and the energy term for the products will have a negative value ∆H = +586 kJ – 1130 kJ = -544 kJ

21. Example II: Calculate the ΔH for the reaction below using the table of bond energies CH4 + 2O2 -- 2H2O + CO2 (Draw Lewis structures)

22. Reactants 4 C-H bonds X 413 kJ = 1652 kJ 2 O=O bonds X 495 kJ = 990 kJ Total Reactants = +2642 kJ Products 4 O-H bonds X 467 kJ = 1868 kJ 2 C=O bonds X 799 kJ = 1598 kJ Total Products = -3466 kJ ΔH = +2642 kJ – 3466 kJ = -824 kJ *What does the resulting ΔH value for this reaction tell you about the “ease” with which this reaction proceeds to completion?

23. Example III Calculate the ΔH for the reaction below using the table of bond energies. 3H2 + N2 --- 2NH3 (draw Lewis structures) Reactants 3 H-H bonds X 432 kJ = 1296 kJ 1 N=N bond X 941 kJ = 941 kJ Total Reactants = +2237 kJ Products 6 N-H bonds X 391 kJ = 2346 kJ Total Products = -2346 kJ ΔH = -109 kJ What does the resulting ΔH value for this reaction tell you about the “ease” with which this reaction proceeds to completion? What is happening to the level of entropy in this reaction?

24. Objectives #5-7 Electronegativity / Formal Charge / Bond Length / Bond Energy *interpreting bond enthalpy diagrams: (see diagram in lecture guide) reactants start with an elevated level of energy energy increases as bonds are broken new bonds are formed and energy decreases new products have greater stability and a lower level of energy than reactants the overall change in this process is exothermic

25. Interpreting a Bond Enthalpy Diagram

26. Objectives #8-9 The VSEPR Theory and Polarity of Molecules *the behavior and properties of organic molecules are very much dependent on their molecular shape and polarity *the VSEPR theory guides us in understanding the shape and polarity of molecules based on the following concepts: • bonding electron pairs and nonbonding electron pairs must position themselves in such a way as to minimize the repulsive forces that exist between them • nonbonding electron pairs take up more space than a bonding electron pair 3. the presence of different atoms about a central atom in a molecule and / or the presence of lone electron pairs about the central atom will usually determine the level of polarity of a molecule

27. Objectives #8-9 The VSEPR Theory and Polarity of Molecules *molecular shape can usually be found by determining the stericnumber which is the sum of the number of atoms attached to the central atom in the molecule and the number of lone electron pairs on the central atom *the steric number is then compared to the number of lone electron pairs on the central atom to predict the molecular geometry (shape)

28. Objectives #8-9 The VSEPR Theory and Polarity of Molecules CH4 has 4 atoms attached to the central carbon atom and 0 lone pairs so the steric number is 4 with 0 L.P. NH3 has 3 atoms attached to the central nitrogen atom and 1 lone pair so the steric number is also 4 but with 1 L.P. the molecular geometry of CH4 is tetrahedral and for NH3 it is trigonal pyramidal CH4 would be nonpolar in character because all of the atoms attached to the carbon are identical which results in a zero net dipole moment

29. Objectives #8-9 The VSEPR Theory and Polarity of Molecules NH3 would be polar in character because of the presence of the lone pair on the nitrogen atom which will result in a negative region of charge and a dipole moment *some atoms can violate the octet rule by expanding their valence shell to include the use of d orbitals *some atoms can violate the octet rule by utilizing fewer than 8 electrons in their outer shell (chart and examples)

30. Objectives #10-11 Orbital Hybridization / Sigma and Pi Bonds *Electron Promotion and the Formation of Hybrid Orbitals *when orbitalsoverlap during the formation of covalent bonds, a two part process occurs: Electron promotion Hybridization *electron promotion is the process of adding energy to the electrons in a particular orbital in order to move them to a nearby orbital in order to provide sufficient unpaired electrons to allow bonds to form more easily; for example, in order for the carbon atom to form 4 equivalent bonds with hydrogen in the CH4 molecule, one of the s orbital electrons in the 2s valence shell must move to occupy the empty orbital as follows: (diagram)

31. Objectives #10-11 Orbital Hybridization / Sigma and Pi Bonds *in order to now form 4 identical bonds, the single 2s orbital and the three 2p orbitals “mix” or hybridize to form four sp3orbitals (diagram) *the number of hybridorbitals formed will equal the number of orbitalsinvolved in the hybridization process *the hybridized orbitals can overlap more efficiently than the nonhybridized and therefore stronger bonds can form; energy is released as a result of this process *the type of hybridization expected in a molecule can be determined if the electron domain geometry is known as follows: (chart and examples)

32. Formation of Hybrid Orbitals

33. Geometric Arrangement and Hybrid Orbitals

34. Objectives #10-11 Orbital Hybridization / Sigma and Pi Bonds *Formation of Sigma and Pi Bonds *the first type of organic molecule we will be studying in detail will be the alkane family which involves carbon atoms exhibiting sp3 hybridization *this type of hybridization allows for the formation of 4 equivalent bonds in which each can then overlap with the s orbital of 4 non-hybridized hydrogen atoms to form the simplest type of alkane (a hydrocarbon), CH4 (methane) (diagram) *these overlapping bonds that form in the case of methane and other alkanes involve a “head to head” of the two orbital regions; this type of bond is referred to as a sigma bond (ς) (diagram)

35. Formation of Sigma Bonds

36. Objectives #10-11 Orbital Hybridization / Sigma and Pi Bonds *in order for multiple bonds to form such as in the case of alkenes (double bonded hydrocarbons) or alkynes (a triple bonded hydrocarbon), a combination of sigma bond and pi bonds (π) will be needed *when hybridization occurs in an atom, such as in carbon, unhybridizedp electrons will be left over after the formation of the sp2 hybrid orbital which is utilized to form part of the double carbon – carbon bond (diagram) *a portion of the double bond is therefore formed from the head to head overlap of the sp2hybrid orbitals which results in the formation of a sigma bond *the other portion of the double bond forms from the overlap of the two lobes of the unhybridized p orbitals that lie above and below the plane of the sigma bond between the carbon atoms (diagram)

37. Formation of Pi Bonds

38. Objectives #10-11 Orbital Hybridization / Sigma and Pi Bonds *pi bonds do not overlap as efficiently as sigma bonds and therefore they tend to be weaker than sigma bonds; however the presence of the pi bond along with the sigma bonds provides additional rigidity to the alkene molecule *therefore a double bond consists of one sigma bond and one pi bond (diagram) *in order for a triple bond to form such as in the case of alkynes (triple bonded hydrocarbons), a combination of sigma bond and pi bonds (π) will be needed as well *when hybridization occurs in an atom, such as in carbon, unhybridized p electrons will be left over after the formation of the sp hybrid orbital which is utilized to form part of the triple carbon – carbon bond (diagram)

39. Objectives #10-11 Orbital Hybridization / Sigma and Pi Bonds *a portion of the triple bond is therefore formed from the head to head overlap of the sp hybrid orbitals which results in the formation of a sigma bond *the other portion of the triple bond forms from the overlap of the four lobes of the unhybridized p orbitals that lie above and below the plane of the sigma bond between the carbon atoms *therefore a triple bond consists of one sigma bond and two pi bonds (diagram and examples)

40. Objective #12 IR and UV/Vis Spectroscopy *both IR and UV/Vis spectroscopy involve exposing samples of a compound to electromagnetic radiation *for IR, radiation with a wavelength in the 2000-15000 nm range is used while in UV/Vis a range of 200-400 nm is utilized is utilized *the radiation causes vibrations and stretching of the bonds in the molecule **each type of bond requires a specific amount of energy to cause these behaviors and therefore this analytical technique can be used to identify the type of bonding and functional groups present in organic molecules *because of the higher energy levels used by UV/Vis spectroscopy, this technique is primarily used to detect the presence of compounds containing multiple bonds (alkenes and alkynes) *IR is primarily ised to detect the presence of functional groups (such as the –OH group in alcohols) without destroying the molecule (example)

41. Example of IR Spectrogram Showing identification of Functional Groups in Methanal (formaldehyde)

43. Objective #14 Bonding Type and its Influence on Properties *some definitions: vapor pressure is the force exerted by a vapor boiling is the rapid release of vapor from throughout a liquid *the vapor pressure of a liquid can be influenced by: the strength of the intermolecular forces between the molecules of a liquid the stronger the intermolecular forces, the lower the vapor pressure the average kinetic energy or temperature of the liquid the higher the temperature the higher the vapor pressure *the boiling point of a liquid can be influenced by: the strength of the intermolecular forces between the molecules of the liquid

44. Objective #14 Bonding Type and its Influence on Properties the stronger the intermolecular forces, the higher the boiling point the magnitude of the external atmospheric pressure pressing on the liquid the higher the external atmospheric pressure, the higher the boiling point *normal boiling point vs. boiling point: boiling can occur at any temperature if the air pressure is adjusted as needed; normal boiling point occurs at 760 mm Hg or 1 atm pressure

45. Vapor Pressure Graph

46. (example graphs to complete)

47. Objective #14 Bonding Type and its Influence on Properties • Polar vs. Nonpolar Bonding *sharing of electrons to increase stability *equal sharing - nonpolar *unequal sharing – polar • Intermolecular Forces *London Dispersion – electron motion; found in all molecules; only force allowed in nonpolar molecules *Dipole-Dipole – unequal electron sharing; found in polar molecules *Hydrogen Bonding – high level of unequal electron sharing; found in extremely polar molecules with H and F, O, N

48. Objective #14 Bonding Type and its Influence on Properties • Influence on Properties *phase at room temperature *vapor pressure *boiling point

49. Objective #15 Structural Isomerism Structural isomers occur when one molecular formula can represent more than one molecule Example: C2H6O CH3OCH3 CH3CH2OH Boiling Points: -23oC 78oC (examples)

50. Objective #16 Functional Groups Organic compounds are classified by the functional groups they contain. The functional group of a molecule is the chemically active site of the molecule. (chart of functional groups and examples)