![Formal Charges
Formal charge = [group number ] – [nonbonding electrons ] – ½ [shared electrons]
H3O+ NO+
• Formal charges are a way of keeping track of electrons.
• They may or may not correspond to actual charges in the molecule.](https://image.slidesharecdn.com/module1-220808135933-ae085b54/85/Module-1-pdf-18-320.jpg)
![Solved Problem 2
Draw the important resonance forms for [CH3OCH2]+. Indicate which structure is the major and minor
contributor or whether they would have the same energy.
Solution
The first (minor) structure has a carbon atom with only six electrons around it. The second (major)
structure has octets on all atoms and an additional bond.](https://image.slidesharecdn.com/module1-220808135933-ae085b54/85/Module-1-pdf-26-320.jpg)
Module 1.pdf
- 1. Chapter 1 Lecture Organic Chemistry, 9th Edition, Global Edition L. G. Wade, Jr. Structure and Bonding © 2017 Pearson Education, Inc.
- 2. Organic Chemistry • Organic chemistry is the chemistry of carbon compounds.
- 3. Electronic Structure of the Atom • An atom has a dense, positively charged nucleus surrounded by a cloud of electrons. • The electron density is highest at the nucleus and drops off exponentially with increasing distance from the nucleus in any direction.
- 4. Electronic Structure of the Atom
- 5. The 2p Orbitals • There are three 2p orbitals oriented at right angles to each other. • Each p orbital consists of two lobes. • Each is labeled according to its orientation along the x, y, or z axis.
- 6. Isotopes • Isotopes are atoms with the same number of protons but a different number of neutrons. • Mass number is the sum of the protons and neutrons in an atom. C 12 6 6C 14
- 7. Electronic Configurations of Atoms • Valence electrons are electrons on the outermost shell of the atom.
- 8. Electronic Configurations • The aufbau principle states we must fill the lowest energy orbitals first. • Hund’s rule states that when there are two or more orbitals of the same energy (degenerate), electrons will go into different orbitals rather than pairing up in the same orbital.
- 9. Ionic Bonding • To obtain a noble gas configuration (a full valence shell), atoms may transfer electrons from one atom to another. • The atoms, now bearing opposite charges, attract each other, forming an ionic bond.
- 10. Covalent Bonding • Electrons are shared between the atoms to complete the octet. • When the electrons are shared evenly, the bond is said to be nonpolar covalent, or pure covalent. • When electrons are not shared evenly between the atoms, the resulting bond will be polar covalent.
- 11. Lewis Structures CH CH4 4 NH NH3 3 H H2 2O Cl O Cl2 2 Carbon: 4 e 4 H@1 e ea: 4 e 8 e Nitrogen: 5 e 3 H@1 e ea: 3 e 8 e Oxygen: 6 e 2 H@1 e ea: 2 e 8 e 2 Cl@7 e ea: 14 e
- 12. Bonding Patterns
- 13. Nonbonding Electrons • Also called lone pairs • Nonbonding electrons are valence-shell electrons that are not shared between atoms.
- 14. Multiple Bonding • Sharing two pairs of electrons is called a double bond. • Sharing three pairs of electrons is called a triple bond.
- 15. Electronegativity and Bond Polarity • A bond with the electrons shared equally between two atoms is called a nonpolar covalent bond. An unequally shared pair of bonding electrons is called a polar covalent bond.
- 16. Dipole Moment • Dipole moment is defined to be the amount of charge separation (d) multiplied by the bond length (m). • Charge separation is shown by an electrostatic potential map (EPM), where red indicates a partially negative region and blue indicates a partially positive region.
- 17. Pauling Electronegativities • Electronegativities can be used to predict whether a bond will be polar and also predict the direction of its dipole moment. • Since the electronegativities of carbon and hydrogen are similar, C—H bonds are considered to be nonpolar.
- 18. Formal Charges Formal charge = [group number ] – [nonbonding electrons ] – ½ [shared electrons] H3O+ NO+ • Formal charges are a way of keeping track of electrons. • They may or may not correspond to actual charges in the molecule.
- 19. Solved Problem 1 Compute the formal charge (FC) on each atom in H3N—BH3. Solution
- 21. Resonance Forms • The structures of some compounds are not adequately represented by a single Lewis structure. • Resonance forms are Lewis structures that can be interconverted by moving electrons only. • The true structure will be a hybrid between the contributing resonance forms.
- 22. Resonance Forms Resonance forms can be compared using the following criteria, beginning with the most important: 1. Has as many octets as possible 2. Has as many bonds as possible 3. Has the negative charge on the most electronegative atom 4. Has as little charge separation as possible
- 23. Major and Minor Contributors • The major contributor is the one in which all the atoms have a complete octet of electrons.
- 24. Major and Minor Contributors (Continued) • When both resonance forms obey the octet rule, the major contributor is the one with the negative charge on the most electronegative atom. The oxygen is more electronegative, so it should have the negative charge.
- 25. Non-Equivalent Resonance • Opposite charges should be on adjacent atoms. The most stable one is the one with the smallest separation of oppositely charged atoms.
- 26. Solved Problem 2 Draw the important resonance forms for [CH3OCH2]+. Indicate which structure is the major and minor contributor or whether they would have the same energy. Solution The first (minor) structure has a carbon atom with only six electrons around it. The second (major) structure has octets on all atoms and an additional bond.
- 27. Solved Problem 3 Draw the resonance structures of the compound below. Indicate which structure is the major and minor contributor or whether they would have the same energy. Solution Both of these structures have octets on oxygen and both carbon atoms, and they have the same number of bonds. The first structure has the negative charge on carbon, the second on oxygen. Oxygen is the more electronegative element, so the second structure is the major contributor.
- 28. Resonance Forms for the Acetate Ion • When acetic acid loses a proton, the resulting acetate ion has a negative charge delocalized over both oxygen atoms. • Each oxygen atom bears half of the negative charge, and this delocalization stabilizes the ion. • Each of the carbon–oxygen bonds is halfway between a single bond and a double bond and is said to have a bond order of 1½.
- 29. Condensed Structural Formulas Lewis Condensed • Condensed forms are written without showing all the individual bonds. • Atoms bonded to the central atom are listed after the central atom (CH3CH3, not H3CCH3). • If there are two or more identical groups, parentheses and a subscript may be used to represent them. 1 2 1 2
- 30. Condensed Structural Formulas Lewis Condensed
- 31. Condensed Structural Formulas (Continued)
- 32. Condensed Structural Formulas (Continued)
- 33. Condensed Structural Formulas (Continued)
- 34. Line-Angle Drawings • Sometimes called skeletal structure or stick figure. • Bonds are represented by lines, and carbons are present where a line begins or ends and where two lines meet. • Hydrogens attached to carbon are not shown. • Nitrogen, oxygen, and halides must be shown.
- 35. Line–Angle Drawings • Atoms other than carbon must be shown. • Double and triple bonds must also be shown.
- 37. Effect of Resonance on pKa • The molecular formula gives the number of atoms of each element in one molecule of a compound CH3CH2CH2CH2OH butan-1-ol, molecular formula C4H10O
- 38. Calculating Empirical Formulas The following are items that need to be considered when calculating empirical formulas: • Given % composition for each element, assume 100 g. • Convert the grams of each element to moles. • Divide by the smallest number of moles to get the ratio. • The molecular formula may be a multiple of the empirical formula.
- 39. Wave Properties of Electrons • Standing wave vibrates in a fixed location. • Wave function, , is a mathematical description of size, shape, and orientation. • Amplitude may be positive or negative. • Node: Amplitude is zero.
- 40. Linear Combination of Atomic Orbitals • Combining orbitals between two different atoms is bond formation. • Combining orbitals on the same atom is hybridization. • Conservation of orbitals • Waves that are in phase add together. Amplitude increases. • Waves that are out of phase cancel out.
- 41. Sigma Bonding • Electron density lies between the nuclei. • A bond may be formed by s—s, p—p, s—p, or hybridized orbital overlaps. • The bonding molecular orbital (MO) is lower in energy than the original atomic orbitals. • The antibonding MO is higher in energy than the atomic orbitals.
- 43. s-Bonding MO Formation of a s-bonding MO: When the 1s orbitals of two hydrogen atoms overlap in phase with each other, they interact constructively to form a bonding MO.
- 44. EPM of H2
- 45. s*-Antibonding MO Formation of a s*-antibonding MO: When two 1s orbitals overlap out of phase, they interact destructively to form an antibonding MO.
- 46. H2: s—s Overlap
- 47. Cl2: p—p Overlap • When two p orbitals overlap along the line between the nuclei, a bonding orbital and an antibonding orbital result. • Most of the electron density is centered along the line between the nuclei. • This linear overlap is another type of sigma bonding MO.
- 48. s and p Orbital Overlap • Overlap of an s orbital with a p orbital gives a s-bonding MO and a s*-antibonding MO.
- 49. Pi Bonding and Antibonding The sideways overlap of two parallel p orbitals leads to a p-bonding MO and a p*- antibonding MO. A pi (p) bond is not as strong as most sigma bonds.
- 50. Multiple Bonds • A double bond (two pairs of shared electrons) consists of a sigma bond and a pi bond. • A triple bond (three pairs of shared electrons) consists of a sigma bond and two pi bonds.
- 51. Molecular Shapes • Bond angles cannot be explained with simple s and p orbitals. • Valence-shell electron-pair repulsion theory (VSEPR) is used to explain the molecular shape of molecules. • Hybridized orbitals are lower in energy because electron pairs are farther apart.
- 52. sp Hybrid Orbitals • Hybrid orbitals result when orbitals in the same atom combine. • Two orbitals (s and p) combine to form two sp orbitals. • Linear electron pair geometry • 180° bond angle
- 53. The Bonding of BeH2 • The bond angle in BeH2 is 180° and the geometry is linear.
- 54. sp2 Hybrid Orbitals • Three orbitals (one s and two p) combine to form three sp2 orbitals. • Trigonal planar geometry • 120° bond angle
- 55. Solved Problem 1 Borane (BH3) is not stable under normal conditions, but it has been detected at low pressure. (a) Draw the Lewis structure for borane. (b) Draw a diagram of the bonding in this molecule, and label the hybridization of each orbital. (c) Predict the H–B–H bond angle. Solution There are only six valence electrons in borane. Boron has a single bond to each of the three hydrogen atoms. The best bonding orbitals are those that provide the greatest electron density in the bonding region while keeping the three pairs of bonding electrons as far apart as possible. Hybridization of an s orbital with two p orbitals gives three sp2 hybrid orbitals directed 120° apart. Overlap of these orbitals with the hydrogen 1s orbitals gives a planar, trigonal molecule. (Note that the small back lobes of the hybrid orbitals have been omitted.)
- 56. sp3 Hybrid Orbitals • Four orbitals (one s and three p) combine to form four sp3 orbitals. • The atom has tetrahedral electron-pair geometry. • 109.5° bond angle.
- 57. Methane
- 58. Summary of Hybridization and Geometry
- 59. Solved Problem 2 Predict the hybridization of the nitrogen atom in ammonia, NH3. Draw a picture of the three-dimensional structure of ammonia, and predict the bond angles. Solution The hybridization depends on the number of sigma bonds plus lone pairs. A Lewis structure provides this information. In this structure, there are three sigma bonds and one pair of nonbonding electrons. Four hybrid orbitals are required, implying sp3 hybridization and tetrahedral geometry around the nitrogen atom, with bond angles slightly smaller than 109.5°.
- 60. Bonding in Ethylene • Ethylene has three sigma bonds formed by its sp2 hybrid orbitals in a trigonal planar geometry. • The unhybridized p orbital of one carbon is perpendicular to its sp2 hybrid orbitals, and it is parallel to the unhybridized p orbital of the second carbon. • Overlap of these two p orbitals will produce a pi bond (double bond) that is located above and below the sigma bond.
- 62. Rotation in Single Bonds Single bonds are allowed to rotate giving a variety of conformations.
- 63. Rotation Around Double Bonds • Double bonds cannot rotate. • Compounds that differ in how their substituents are arranged around the double bond can be isolated and separated.
- 64. Isomerism • Molecules that have the same molecular formula but differ in the arrangement of their atoms are called isomers. • Constitutional (or structural) isomers differ in their bonding sequence. • Stereoisomers differ only in the arrangement of the atoms in space.
- 65. Constitutional Isomers • Constitutional isomers have the same chemical formula, but the atoms are connected in a different order. • Constitutional isomers have different properties. • The number of isomers increases rapidly as the number of carbon atoms increases.
- 66. Geometric Isomers: Cis and Trans • Stereoisomers are compounds with the atoms bonded in the same order, but their atoms have different orientations in space. • Cis and trans are examples of geometric stereoisomers; they occur when there is a double bond in the compound. • Since there is no free rotation along the carbon–carbon double bond, the groups on these carbons can point to different places in space.