Chapter 27 Pericyclic Reactions

Transcription

1 Instructor Supplemental Solutions to Problems 2010 Roberts and Company Publishers Chapter 27 Pericyclic Reactions Solutions to In-Text Problems 27.1 (b) This is a sigmatropic reaction; two electrons are involved. (d) This is a sigmatropic reaction; six electrons are involved Because there are ten p electrons, there are five occupied MOs in this alkene, each containing two electrons. The MOs alternate in symmetry; the odd-numbered ones are symmetric, and the even-numbered ones are antisymmetric. Hence, p* 6 is antisymmetric (A); it is antibonding; and it is the LUMO and is thus one of the frontier orbitals (a) The resonance structures of the allyl radical show that the unpaired electron is shared between the terminal carbons: (b) Figure 27.4, text p. 1341, shows that the unpaired electron of the allyl radical resides in the nonbonding MO, which has a node at the central carbon. The unpaired electron density therefore exists only at the carbons on either side of a node, that is, at the terminal carbons, as the resonance structures show The two different conrotatory processes are as follows: (The curved arrows indicate atomic motion, not electron flow.)

2 INSTRUCTOR SUPPLEMENTAL SOLUTIONS TO PROBLEMS CHAPTER 27 2 They are equally likely because they are enantiomeric; enantiomers have equal energies, and enantiomeric pathways have identical energies at all points and therefore identical rates (Sec. 7.8A, text p. 301) The opening of the cyclobutene is conrotatory and therefore gives (1E,3Z)-cyclodecadiene. The trans double bond does not introduce as much strain into the large ring as was present in the starting cyclobutene. Here is one way to use models to visualize this transformation. To avoid having to build two models, construct a cyclodecane (the cyclic alkane with a ten-membered ring); don t worry about the missing double bond. Now connect two carbons to make a cis-fused four-membered ring. Identify the up hydrogens on these two carbons. Using a pencil, draw a single line across the bond that is a double bond in the starting material (even though it is a single bond in your model). Now draw two lines across each of the bonds that will become double bonds as a result of the reaction. Now break the cyclobutane ring and turn the carbons bearing the up hydrogens in a conrotatory manner. Examine the stereochemistry at the bonds marked with two lines; one should have an E configuration and the other a Z configuration. Voila! Use the HOMO of the 4p-electron component and the LUMO of the 2p-electron component. The [4a + 2s] process would involve the overlap shown in (a) below. The [4s + 2a] process would involve the overlap shown in (b). In both cases an out-of-phase overlap between the p-electron systems is required at one end. Consequently, these processes are not allowed The four products are the diastereomeric 1,2,3,4-tetramethylcyclobutanes. Each of the pure alkene stereoisomers can undergo an allowed photochemical [2s + 2s] cycloaddition in two distinguishable ways. Cis-2-butene reacts to give compounds A and B; trans-2-butene reacts to give compounds B and C; and the mixture of cis- and trans-2- butene can give these three compounds plus a fourth, compound D, that results from the [2s + 2s] cycloaddition of cis-2-butene to trans-2-butene.

3 INSTRUCTOR SUPPLEMENTAL SOLUTIONS TO PROBLEMS CHAPTER (b) Two other starting materials that would give the same product as in part (a): (c) This is a [5,5] sigmatropic rearrangement If the migration were antarafacial, the opposite stereochemical result would have been observed; that is, the 3E stereoisomer of the product would have the S configuration at the asymmetric carbon stereocenter, and the 3Z stereoisomer of the product would have the R configuration at the asymmetric carbon stereocenter. This is the result that is not observed (a) Think of the five-carbon p-electron system as a 2,4-pentadienyl cation, whose MOs are shown in Fig. 27.5, text p. 1342, and think of the migrating group as a carbon anion in which the unshared electron pair resides in a 2p orbital. The orbitals involved in the rearrangement are the LUMO of the pentadienyl system and the HOMO of the carbon anion. The LUMO of the pentadienyl system, as shown by Fig. 27.5, is symmetric; that is, at each end of the system, the LUMO has the same phase on a given face. The HOMO of the carbon anion is simply the filled 2p orbital. The migration is allowed only if the carbon anion migrates suprafacially such that each end of the pentadienyl LUMO interacts with the same lobe of the carbon anion 2p orbital.

4 INSTRUCTOR SUPPLEMENTAL SOLUTIONS TO PROBLEMS CHAPTER 27 4 Alternatively, you can think of this problem in terms of the 2,4-pentadienyl anion and a methyl cation. The orbitals involved are now the HOMO of the anion and the LUMO of the cation. But these are the same orbitals used in the foregoing solution (only their occupancies are different); hence, the stereochemical result is the same (b) The hydrogen walk on 2,3-dimethyl-1,3-cyclopentadiene should proceed as shown in the following equation. Although a hydrogen that starts out on carbon-1 could migrate to four other carbons, only two of the possible products are unique. It can be argued that compounds A and B should be the predominant alkenes at equilibrium, and that compound C should be present in least amount. (Why? See Sec. 4.5B, text p. 144.) Remember that the selection rules have nothing to say about which compound is favored at equilibrium, only whether the equilibrium can be attained at a reasonable rate.) (b) Deduce the starting material by drawing the curved arrows for the reverse of a Claisen rearrangement: See Fig. 27.4, text p. 1341, for the relevant orbital diagram. The SOMOs of the two allylic radicals (p 2 ) interact because it is these molecular orbitals that contain the unpaired electrons. Since the two orbitals are the same, they have the same symmetry, and therefore interact suprafacially on both components with positive overlap (b)

5 INSTRUCTOR SUPPLEMENTAL SOLUTIONS TO PROBLEMS CHAPTER (a) The s bond that moves is indicated with an asterisk (*). It moves suprafacially on both p systems; [3,3] sigmatropic rearrangements are thermally allowed processes Heating should promote a disrotatory reaction; that is, the reaction should have a stereochemical course which is opposite to that of the photochemical reaction A [1,7] thermal sigmatropic hydrogen shift should be antarafacial. That is, a hydrogen from the methyl group migrates from a conformation in which it is up to the lower face of the ring, or a hydrogen from the methyl group migrates from a conformation in which it is down to the upper face of the ring, or both. Problem (text p. 1373) describes an effort to elucidate the stereochemistry of the previtamin D rearrangement.

6 INSTRUCTOR SUPPLEMENTAL SOLUTIONS TO PROBLEMS CHAPTER 27 6 Solutions to Additional Problems (b) Think of the Diels Alder reaction as a prototype for an allowed thermal cycloaddition. If 4n + 2 electrons are involved, a cycloaddition is allowed thermally. If 4n electrons are involved, then the thermal cycloaddition is forbidden and the photochemical cycloaddition is allowed. Therefore, a photochemical [8s + 4s] process (12 electrons) is allowed The pericyclic selection rules say absolutely nothing about the position of equilibrium in each case. These rules refer to rates of reactions, not to equilibrium constants. Other considerations must be used to decide on the position of equilibrium. The point of the problem is for you to use what you have learned about the relative energies of molecules; the molecule of lower energy is favored in each equilibrium. (b) (d) The left side of the equation is favored at equilibrium because the double bonds have more alkyl branches. (Alkyl substitution stabilizes a double bond.) The right side of the equation is favored at equilibrium because one product is aromatic and therefore particularly stable, and the other product is volatile (ethylene is a gas) (b) The stereoisomer of compound A that also gives compound C on heating is the one in which both of the terminal double bonds have the Z configuration (b) Because a thermal suprafacial [1,9] or [1,13] methyl migration must occur with retention of configuration at the migrating carbon, the methyl group that migrates, when isotopically substituted, must have the S configuration in both starting material and product (a) The structure of the ozonolysis product C shows that compound B is a cyclobutene, which must be formed in a disrotatory photochemical electrocyclic reaction. This defines the stereochemistry of B, which, in turn, defines the stereochemistry of C: (b) Compound D is the trans-fused stereoisomer of the cyclobutene derivative, which undergoes conrotatory opening upon heating to give compound A. The reason that compound B is inert under the same conditions is that conrotatory ring opening of B would give a stereoisomer of A in which one of the cyclohexene rings would contain a trans double bond within a six-membered ring. This introduces so much strain that the reaction does not occur, even though it is allowed by the selection rules.

7 INSTRUCTOR SUPPLEMENTAL SOLUTIONS TO PROBLEMS CHAPTER This cycloaddition reaction involves sixteen electrons; hence, it must be suprafacial on one component and antarafacial on the other. Thus, it must be a [14s + 2a] or a [14a + 2s] cycloaddition. The heptafulvene molecule is large enough that its p-electron system can twist without introducing too much strain or without losing too much p- electron overlap; hence, the cycloaddition is a [14a + 2s] process. In either case, the product has the following stereochemistry: (a) Toluene has added stability associated with aromaticity; compound A, although conjugated, is not aromatic. Because any equilibrium favors the more stable compound, the added stability of toluene causes it to be strongly favored in the equilibrium (b) This reaction superficially appears to involve internal rotation about both double bonds, a process that is virtually impossible. However, a more reasonable pericyclic mechanism is available: a sequence of two conrotatory electrocyclic reactions, the first one a ring closing, and the second one a ring opening. (The curved arrows refer to the clockwise rotations of groups and not to electron flow.) These transformations are very much like the last two in the solution to Problem 27.40, text p A Diels Alder reaction is followed by a reverse Diels Alder reaction. The driving force for the reverse Diels Alder is the formation of a product that is stabilized by aromaticity.

8 INSTRUCTOR SUPPLEMENTAL SOLUTIONS TO PROBLEMS CHAPTER (c) The intermediate trapped by maleic anhydride is the tetraene Z formed by a [1,9] (presumably suprafacial) sigmatropic hydrogen migration Compound A is a secondary alcohol, and the aldehyde B is formed from an enol intermediate (shown in parentheses) that results from a [3,3] sigmatropic (oxycope) rearrangement. Evidently in this case the Grignard reagent undergoes mostly 1,2-addition to the a,b-unsaturated aldehyde; none of the 1,4-addition product is formed. (See Sec A, text p ) (b) A key step in this synthesis is a Claisen rearrangement of the starting material, which is prepared in the solution to part (a) (b) Two successive photochemical disrotatory electrocyclic reactions account for the products.

9 INSTRUCTOR SUPPLEMENTAL SOLUTIONS TO PROBLEMS CHAPTER 27 9 The product, the very unstable antiaromatic cyclic diene 1,3-cyclobutadiene (see text p. 729), was generated in 1972 by this sequence of reactions carried out at a temperature of 8 K in the laboratory of Professor Orville L. Chapman ( ), then at Iowa State University, and later at the University of California, Los Angeles. This technique allowed the observation and characterization of 1,3-cyclobutadiene by infrared spectroscopy (b) Given that germacrone assumes a chairlike conformation in its transition state when it reacts to give b-elemenone, deduce the structure of germacrone by mentally reversing the Cope rearrangement. Now re-examine (or work) Problem 27.42, text p. 1370, to see why the product in that problem has E stereochemistry (b) Oxidation of the secondary alcohol and cleavage of the ether gives estrone: This is a [1,7] sigmatropic rearrangement, which, according to the selection rules, should be antarafacial. (See Problem 27.28, text p. 1367, and its solution on p. 5 of this chapter.) Migration of hydrogen gives B and migration of deuterium gives C. (Presumably, the predominance of B is due in part to a primary isotope effect, which should favor more rapid migration of hydrogen, all things being equal.) If the rearrangements are indeed antarafacial, the stereochemistry of the products should be as follows:

10 INSTRUCTOR SUPPLEMENTAL SOLUTIONS TO PROBLEMS CHAPTER This was indeed the stereochemistry observed. This work was carried out to adduce evidence for the stereochemistry of the previtamin D rearrangement to vitamin D (Eq , text p. 1366). Compound A was constructed to be a model for previtamin D. As noted in Problem 27.28, the structure of the previtamin itself does not reveal the stereochemistry of the rearrangement (a) If the formation of benzene were concerted, it would have to be a thermal disrotatory electrocyclic ringopening reaction involving four electrons. (b) The reaction that gives benzene is not allowed by the selection rules; therefore, it does not occur. In spite of the high degree of strain of Dewar benzene, and in spite of the great stability of benzene, Dewar benzene is effectively constrained to exist by the selection rules! Prismane, or Ladenburg benzene, which is discussed on text p. 718 and 1348, is another very unstable constitutional isomer of benzene that is effectively trapped into existence because its concerted conversion into benzene would violate the selection rules for pericyclic reactions.