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Structure and Bonding of Organic Molecules Orbital Theory Hybridization and Geometry Polarity Functional Groups.

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Presentation on theme: "Structure and Bonding of Organic Molecules Orbital Theory Hybridization and Geometry Polarity Functional Groups."— Presentation transcript:

1 Structure and Bonding of Organic Molecules Orbital Theory Hybridization and Geometry Polarity Functional Groups


3 Orbitals are Probabilities

4 Bonding in H 2 The Sigma (  Bond

5 Wave Depiction of H 2

6 Molecular Orbitals Mathematical Combination of Atomic Orbitals

7 Antibonding Molecular Orbital Destructive Overlap Creates Node

8 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. Chapter 28

9  and  * of H 2

10 Electron Configurations

11 Ground State Electron Configurations

12 p x, p y, p z

13 Electron Configuration of Carbon

14 For CARBON, In the Ground State 2 bonding sites, 1 lone pair

15 sp 3 Hybridization 4 Regions of electron Density link link

16 Hybridization of 1 s and 3 p Orbitals gives 4 sp 3 Orbitals

17 sp 3 is Tetrahedral Geometry Methane

18 Tetrahderal Geometry

19 Methane Representations

20 Ammonia Tetrahedral Geometry Pyramidal Shape

21 All Have the Same Geometry All Have 4 Regions of Electron Density All are sp 3 Hybridized

22 Orbital Depiction of Ethane, C 2 H 6, the  bond

23 © 2013 Pearson Education, Inc. Rotation of Single Bonds Ethane is composed of two methyl groups bonded by the overlap of their sp 3 hybrid orbitals. There is free rotation along single bonds. Chapter 223


25 A Model of a Saturated Hydrocarbon

26 sp 2 Hybridization 3 Regions of Electron Density

27 Hybridization of 1 s and 2 p Orbitals – sp 2

28 An sp 2 Hybridized Atom

29 The  bond Overlap of 2 parallel p Orbitals

30 Ethylene CH 2 =CH 2

31 © 2013 Pearson Education, Inc. 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. Chapter 231

32 Views of Ethylene, C 2 H 4

33 © 2013 Pearson Education, Inc. 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. Chapter 233

34 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. Chapter 234

35 © 2013 Pearson Education, Inc. 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. Chapter 235

36 © 2013 Pearson Education, Inc. 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. Chapter 236

37 Formaldehyde

38 sp Hybridization 2 Regions of Electron Density

39 The sp Orbital

40 Acetylene, C 2 H 2, 1  bond 2 perpendicular  bonds

41 © 2013 Pearson Education, Inc. 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. Chapter 241

42 Borane (BH 3 ) 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. 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 sp 2 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.) Solved Problem 2 Solution Chapter 242

43 Summary of Hybridization and Geometry Hybrid Orbitals (# of  bonds) HybridizationGeometryApproximate Bond Angle 2s + p = splinear 180 ⁰ 3s + p + p = sp 2 trigonal 120 ⁰ 4s + p + p + p = sp 3 tetrahedral 109.5 ⁰ Chapter 243

44 © 2013 Pearson Education, Inc. Molecular Dipole Moment The molecular dipole moment is the vector sum of the bond dipole moments. Depends on bond polarity and bond angles. Lone pairs of electrons contribute significantly to the dipole moment. Chapter 244

45 Lone Pairs and Dipole Moments Chapter 245

46 Intermolecular Forces Strength of attractions between molecules influences the melting point (m. p.), boiling point (b. p.), and solubility of compounds. Classification of attractive forces: – Dipole–dipole forces – London dispersions forces – Hydrogen bonding in molecules with —OH or — NH groups Chapter 246

47 Dipole–Dipole Interaction Chapter 247

48 London Dispersions Chapter 248

49 © 2013 Pearson Education, Inc. Effect of Branching on Boiling Point The long-chain isomer (n-pentane) has the greatest surface area and therefore the highest boiling point. As the amount of chain branching increases, the molecule becomes more spherical and its surface area decreases. The most highly branched isomer (neopentane) has the smallest surface area and the lowest boiling point. Chapter 249

50 Hydrogen Bonds Chapter 250

51 Rank the following compounds in order of increasing boiling points. Explain the reasons for your chosen order. Solved Problem 4 Chapter 251

52 To predict relative boiling points, we should look for differences in (1) hydrogen bonding, (2) molecular weight and surface area, and (3) dipole moments. Except for neopentane, these compounds have similar molecular weights. Neopentane is the lightest, and it is a compact spherical structure that minimizes van der Waals attractions. Neopentane is the lowest-boiling compound. Neither n-hexane nor 2,3-dimethylbutane is hydrogen bonded, so they will be next higher in boiling points. Because 2,3-dimethylbutane is more highly branched (and has a smaller surface area) than n-hexane, 2,3-dimethylbutane will have a lower boiling point than n-hexane. Solution The two remaining compounds are both hydrogen bonded, and pentan-1-ol has more area for van der Waals forces. Therefore, pentan-1-ol should be the highest-boiling compound. We predict the following order: neopentane < 2,3-dimethylbutane < n-hexane < 2-methylbutan-2-ol < pentan-1-ol The actual boiling points are given here to show that our prediction is correct. 10 °C58 °C69 °C102 °C138 °C Chapter 252

53 Polarity Effects on Solubility Like dissolves like. Polar solutes dissolve in polar solvents. Nonpolar solutes dissolve in nonpolar solvents. Molecules with similar intermolecular forces will mix freely. Chapter 253

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