Chapter 6: Alkenes: Structure and Reactivity

Transcription

1 hapter 6: Alkenes: Structure and eactivity overage: 1. Degrees of Unsaturation 2. Nomenclature cis/trans E/Z 3. Alkene Stability 4. Electrophilic Addition eactions/markovinikov s ule 5. The ammond Postulate and Transition State Structure 6. arbocation earrangements End of hapter Problems: 6.23, 6,25, 6,26, 6.29, 6.31, 6.37, , 6.48 Goals: 1. Be able to calculate the degree of unsaturation from a formula and know what it means. 2. Be able to name both alkenes and cycloalkenes using IUPA rules. 3. Know the order of stability of alkenes and know how they are measured. 4. Know the mechanism of electrophilic addition. Be able to predict the regiochemistry of the reaction. 5. Know the ammond postulate and be able to predict the structure of a transition state based on the energetics of the reaction. 6. Know the fundamental processes for carbocation rearrangements including 1,2 ~ 3 shift and 1,2 ~ shift.

2 Nobel Prize in hemistry, 2001 William S. Knowles, 84 years, born 1917 (US citizen). PhD 1942 at olumbia University. Previously at Monsanto ompany, St Louis, USA. etired since e discovered that it was possible to use transition metals to make chiral catalysts for an important type of reaction called hydrogenation yoji Noyori, 63 years, born 1938 Kobe, Japan (Japanese citizen). PhD 1967 at Kyoto University. Since 1972 Professor of hemistry at Nagoya University and since 2000 Director of the esearch enter for Materials Science, Nagoya University, Nagoya, Japan. e led the further development of this process to today's general chiral catalysts for hydrogenation K. Barry Sharpless, 60 years, born 1941 Philadelphia, Pennsylvania, USA (US citizen). PhD 1968 at Stanford University. Since 1990 W.M. Keck Professor of hemistry at the Scripps esearch Institute, La Jolla, USA. e awarded half of the Prize for developing chiral catalysts for another important type of reaction oxidation

3 Alkenes Alkenes possess the double bond. Double Bond Energy 146 kcal/mol σ Bond Energy 83 kcal/mol π Bond Energy 63 kcal/mol onclusion: π bond is weaker than the σ bond by 20 kcal/mol The π bond is much more reactive! 1.33 Å 1.08 Å o o ompare: - single bond 1.54 Å The = double bond is shorter than the - single bond.

4 1. Find the parent hydrocarbon Alkene Nomenclature Find the longest continuous chain of carbon atoms that contains the double bond and name it accordingly using the ene ending. Named as octene (not decene) 2. Number the atoms in the main chain a. Begin at the end nearer the double bond. b. If the double bond is equidistant from the ends, begin at the end nearer 1 the first branch point. 2 The above molecule would be numbered as

5 3. Identify and number the substituents as you have done with alkanes. 4. Write the name as a single word. a. Use prefixes such as mono, di, tri, tetra, etc. to indicate multiple substituents. b. List substituents alphabetically, ignoring prefixes. c. Use dashes to separate substitutents, use commas to separate numbers. d. Indicate the position of the double bond by giving the number of the first alkene carbon and placing that number immediately before the parent names. e. If more than one double bond exists, indicate the position of both and use the suffix -diene butyl-2-octene Name these alkenes: Answer: 4,5-dimethyl-2-heptene Answer: 1,3-pentadiene

6 ycloalkene Nomenclature Name cycloalkenes in a similar manner. owever, since there is no chain end, the double \ bond must always be between carbons 1 and 2. Make sure the substitutents have the lowest numbers possible. It is not necessary to indicate the position of the double bond ( 3 ) 2 4-isopropylcyclohexene ,3-cyclopentadiene 2 3 Name this one ,7-dimethyl-1,3,5-cycloheptatriene 3 7 6

7 Stereochemistry and E/Z Nomenclature We know that some alkenes can exist as cis/trans stereoisomers (geometric isomers). trans-2-butene cis-2-butene What about something like this? Which is cis and which is trans? Answer: Who knows?????? We need a more general method for describing the geometry about the double bond.

8 E/Z Method for Double Bond Geometry E entgegen meaning opposite Z zusammen meaning together or the same side In order to use this system of nomenclature, you must first prioritize the substituents according to the wonderful ahn-ingold-prelog rules. 1. onsider the sp 2 carbons separately and identify the two substitutents attached to each carbon. l 3 Left side: and l ight side: and 3

9 2. Use the ahn-ingold-prelog rules to prioritize the two substituents on each side. a. The highest priority group has an atom with the highest atomic number directly bonded to the sp 2 carbon. l 3 Left side: AN = 35 l AN = 17 > l ight Side: AN = 1 3 AN = 6 3 > The above molecule has the two highest priority groups on the opposite side and therefore this is the E isomer (E)-1-bromo-1-chloro-1-propene

10 b. If the two directly bonded atoms have the same AN, then look at the next atoms bonded these atoms. l Left side: Directly attached atoms are both carbon. 2 nd Atoms: l,, AN 17, 1, 1,, An 1, 1, 1 ight side: Directly attached atoms are both carbon. 2 nd Atoms:,, AN= 1, 1, 1,, AN = 12, 1, 1 2 l > > 3 This is a E isomer. (E)-1-chloro-2,3-dimethyl-2-pentene

11 c. If an atom is doubly bonded to another atom, then it is treated as if it were singly bonded to two of these atoms. Which has highest priority? -= 2 vs st atoms are both (tie) -= 2 becomes nd atoms,, nd atoms,, -= 2 > 2 3 d. In the case of isotopes, use the mass number since they have the same atomic number 1 vs. 2 (D) deuterium AN = 1 for both atoms, but deuterium has greater mass number, therefore 2 > 1

12 Alkene Stability Not all alkenes are created equal! Some are more stable than others. Two considerations: 1. The degree of substitution. Unsubstituted Monosubstituted Disubstituted Trisubstituted Tetrasubstituted Least Stable 2. Stereochemistry Steric Interaction Most Stable 3 trans-2-butene cis-2-butene More Stable Less Stable

13 ow do we know about this order of stability? Answer: eats of ydrogenation 2 catalyst + eat The larger the heat of hydrogenation, the less stable the alkene Measure by alorimetry eat kcal/mol

14 alculating Degrees of Unsaturation Degrees of Unsaturation (DU) number of rings and/or multiple bonds in a molecule. 4 DU 2 DU 4 Du ow to calculate: DU = 2 # # DU = 2 (6) = 4 2

15 Simple enough! But what about when you have a heteroatom in the molecule eteroatom any atom besides and. 1. alogens (X) such as F, l,, and I. ount as a hydrogen atom, that is, add the number of halogens to the number of hydrogens. For example: Formula: Oxygen or Sulfur Ignore! O DU = 2(5) = 0 2 Formula: 6 10 O 6 10 DU = 2

16 3. Nitrogen Subtract the number of hydrogens from the number of hydrogens N Formula: 4 5 N 4 4 DU = 3

17 Electrophilic Addition of X to Alkenes

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19 What happens when an asymmetric alkene reacts with? a b + Which pathway is favored. Answer: Pathway a is favored since a more stable carbocation is formed. 1 0

20 arbocation earrangements If a carbocation can rearrange to a more stable carbocation, then it probably will. Otherwise, it will not less stable 3 0 more stable Primary consideration: There must be a viable mechanism by which rearrangement can take place. 1. 1,2 ydrogen Shift (1,2 ~) ,2 ~ Both bonding electrons go with the hydrogen, leaving a + charge on the carbon left behind.

21 2. 1,2 methyl shift (1,2 ~ 3 ) ,2 ~ What is the mechanism of this rearrangement? Answer: 1,2 ~

22 In what types of reactions do we observe carbocation rearrangements? Answer: In reactions where there are carbocation intermediates. + Write a mechanism to account for both products. Which product results from carbocation rearrangement? + +

23 Transition States and the ammond Postulate The ammond postulate states that the transition state of an exothermic reaction most closely resembles the reactants in energy and structure. onversely, the transition state of an endothermic reaction most closely resembles the products in energy and structure. This is most conveniently illustrated with a reaction energy diagram. Exothermic = Endothermic = E E eaction Progress eaction Progress

24 In the exothermic reaction, the transition state (green arrow) is closer to the the reactant (red arrow) along the reaction coordinate. Meaning: The transition state structure most closely resembles the reactant. ow can we illustrate this in a structure drawing? onsider the reaction: A-B + A + B- Exothermic The transition state structure shows that B is less than half transferred to. = = [A---B ] (not [A B---] )