1. Structure and Properties of Organic Molecules
Organic Chemistry, 8th Edition L. G. Wade, Jr.
Chapter 2 Lecture
Structure and Properties of Organic Molecules

2. Lecture 4: Isomerism, Forces of Attraction and Classes Organic Compounds
Bond Dipoles & Dipole Moments
Intermolecular Forces
Dipole-Dipole, London Dispersion, H-bonding
Using FOA to Explain Physical Properties
Classes of Organic Compounds
Hydrocarbons
Oxygen Containing Compounds
Nitrogen Containing Compounds

3. 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 2 3

4. 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.
File Name: AAAKPFP0
Figure: un.jpg
Chapter 2 4

5. 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.
File Name: AAAKPFQ0
Figure: un.jpg
Chapter 2 5

6. Hint Identical groups on one of the double-bonded carbons implies
no cis-trans isomerism.
File Name: AAAKPFR0
Chapter 2 6

7. Bond Dipole Moments
Dipole moments are due to differences in electronegativity.
They depend on the amount of charge and distance of separation.
They are measured in debyes (D).
File Name: AAAKPGD0
Figure: un.jpg
Chapter 2 7

8. 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 to the dipole moment.
File Name: AAAKPGH0
Figure: jpg
Chapter 2 8

9. Lone Pairs and Dipole Moments
File Name: AACXSEK0
Figure: 02_22.jpg
Chapter 2 9

10. 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 2 10

11. Dipole–Dipole Forces
Dipole–dipole interactions result from the approach of two polar molecules.
If their positive and negative ends approach, the interaction is an attractive one.
If two negative ends or two positive ends approach, the interaction is repulsive.
In a liquid or a solid, the molecules are mostly oriented with the positive and negative ends together, and the net force is attractive.
Chapter 2 11

12. Dipole–Dipole Interaction
File Name: AAAKPGJ0
Figure: jpg
Chapter 2 12

13. London Dispersion Forces
One of the Van der Waals forces.
Major attractive force for nonpolar molecules
Temporary dipole brought on by other nearby molecules
Molecules must have slightly asymmetric dispersion of electrons
Chapter 2 13

14. London Dispersion Forces
Larger atoms are more polarizable
Polarizability is the ability of the electrons on an atom to respond to a changing electric field.
Larger surface areas result in larger London forces i.e., higher boiling points.
Branching lowers boiling points because of decreased surface contact between molecules.
Chapter 2 14

15. Dispersions
File Name: AAAKPGM0
Figure: jpg
Chapter 2 15

16. Hydrogen Bonding Strong dipole–dipole attraction.
Organic molecules must have N—H or O—H to be able to form a hydrogen bond.
The hydrogen from one molecule is strongly attracted to a lone pair of electrons on the oxygen of another molecule.
O—H is more polar than N—H, so alcohols have stronger hydrogen bonding.
Chapter 2 16

17. Hydrogen Bonds
File Name: AAAKPGO0
Figure: jpg
Chapter 2 17

18. Boiling Points and Intermolecular ForcesH 3 O
dimethyl ether, b. p. = -25 °C C H 3 2 O
ethanol, b. p. = 78 °C
Hydrogen bonding increases the b. p. of the molecule. C H 3 2 O
ethanol, b. p. = 78 °C
ethyl amine, b. p. 17 °C C H 3 2 N
O—H is more polar than N—H, so alcohols have stronger hydrogen bonding and, therefore, higher boiling points.
Chapter 2 18

19. Solved Problem 4
Rank the following compounds in order of increasing boiling points. Explain the reasons for your chosen order.
File Name: AAAKPGP0
Figure taken from “Worked Examples” in the IRC
This is Solved problem 2-10, page 65, 8th edition
Chapter 2 19

20. Solved Problem 4
Rank the following compounds in order of increasing boiling points. Explain the reasons for your chosen order.
File Name: AAAKPGP0
Figure taken from “Worked Examples” in the IRC
This is Solved problem 2-10, page 65, 8th edition
Chapter 2 20

21. Solution
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.
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.
Copyright © 2006 Pearson Prentice Hall, Inc.
10 oC
58 oC
69 oC
102 oC
138 oC
Chapter 2 21

22. 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 2 22

23. Polar Solute in Polar Solvent
File Name: AAAKPGR0
Figure: jpg
A polar solute dissolves in a polar solvent.
Hydration releases energy; entropy increases.
Chapter 2 23

24. Polar Solute in Nonpolar Solvent
File Name: AAAKPGS0
Figure: jpg
The solvent cannot break apart the intermolecular interaction of the solute, so the polar solid will not dissolve in the nonpolar solvent.
Chapter 2 24

25. Nonpolar Solute in Nonpolar Solvent
File Name: AAAKPGT0
Figure: jpg
The weak intermolecular attractions of a nonpolar substance are overcome by the weak attractions for a nonpolar solvent. The nonpolar substance dissolves.
Chapter 2 25

26. Nonpolar Solute with Polar Solvent
File Name: AAAKPGU0
Figure: jpg
If a nonpolar molecule were to dissolve in water, it would break up the hydrogen bonds between the water molecules. Therefore, nonpolar substances do not dissolve in water.
Chapter 2 26

27. Classes of Compounds Classifications are based on functional group.
Part of the molecule where reactions usually occur
Three broad classes:
Hydrocarbons: Compounds composed of only carbon and hydrogen.
Compounds containing oxygen.
Compounds containing nitrogen.
Chapter 2 27

28. Hydrocarbons Alkanes: single bonds, sp3 carbons.
Cycloalkanes: sp3 carbons form a ring.
Alkanes and cycloalkanes are not considered functional groups!
Alkenes: double bond; sp2 carbons.
Cycloalkenes: double bond in a ring.
Alkynes: triple bond; sp carbons.
Aromatic: (Arenes): contains a benzene ring.
Chapter 2 28

29. Compounds Containing Oxygen
Alcohols: R-OH
Ethers: R-O-R´
Aldehydes:
Ketones:
Chapter 2 29

30. Compounds Containing Oxygen: Carboxylic Acids & Their Derivatives
Acid Chloride:
Ester:

31. Compounds Containing Nitrogen
Amines: RNH2, RNHR´, or R3N
Amides: RCONH2, RCONHR´, RCONR2
Nitriles: RCN
Chapter 2 31

32. Skill Building: Practice Problems
Problem 2-8 thru 2-22

33. Mastering Chemistry Homework Problems
Problem Set 4: 2-27, 2-28, 2-30, 2-32 thru 2-35
Problem Set 5: 2-36, 2-38 thru 2-44