Structure and Properties of Organic Molecules

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Presentation transcript:

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

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

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

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: 02-20-05un.jpg Chapter 2 4

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: 02-20-07un.jpg Chapter 2 5

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

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: 02-20-20un.jpg Chapter 2 7

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: 02-21.jpg Chapter 2 8

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

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

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

Dipole–Dipole Interaction File Name: AAAKPGJ0 Figure: 02-23.jpg Chapter 2 12

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

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

Dispersions File Name: AAAKPGM0 Figure: 02-24.jpg Chapter 2 15

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

Hydrogen Bonds File Name: AAAKPGO0 Figure: 02-25.jpg Chapter 2 17

Boiling Points and Intermolecular Forces H 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

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

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

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

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

Polar Solute in Polar Solvent File Name: AAAKPGR0 Figure: 02-26.jpg A polar solute dissolves in a polar solvent. Hydration releases energy; entropy increases. Chapter 2 23

Polar Solute in Nonpolar Solvent File Name: AAAKPGS0 Figure: 02-27.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

Nonpolar Solute in Nonpolar Solvent File Name: AAAKPGT0 Figure: 02-28.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

Nonpolar Solute with Polar Solvent File Name: AAAKPGU0 Figure: 02-29.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

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

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

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

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

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

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

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