CHAPTER 8 HW SOLUTIONS: ELIMINATIONS REACTIONS

Transcription

1 APTER 8 W SLUTNS: ELMNATNS REATNS S-TRANS SMERSM 1. Use a discussion and drawing of orbitals to help explain why it is generally easy to rotate around single bonds at room temperature, while it is difficult to rotate around double bonds. A sigma bond is a cylindrically symmetrical bond made by a head-to-head overlap of orbitals. n an alkane - single bond, each atom uses an sp 3 hybrid orbital to make the sigma bon Rotation around that bond is possible without breaking the overlap of those sp 3 orbitals. σ rotate σ The pi bond of a double bond is made by side-by-side overlap of p orbitals. Rotation around a double bond would result in the p-orbitals being perpendicular, which means rotation breaks the pi bon This takes too much energy, so rotation about a double bond is difficult at room temperature. π rotate 2. Which of the following compounds can exist as cis and isomers? When this type of isomerism is possible, draw the other isomer (if is shown, draw cis). cis cis no cis/ possible cis cis no cis/ possible no cis/ possible no cis/ possible is too strained 3. raw all isomers of 2 2, including stereoisomers. Page 1

2 4. dentify the relationship between each pair of compounds. Are they identical, constitutional isomers, or stereoisomers? Pair Relationship onstitutional somers Stereoisomers Pair Relationship dentical onstitutional somers Pair 3 3 Relationship Stereoisomers dentical ALKENE STABLTY 5. Answer the questions while using the following enthalpy dat (kcal/mol) 2 Pd 2 Pd Use the data to determine the relative stability of 1-butene compared to cis-2-butene. Explain how you used the enthalpy dat See the diagram at right. Both products are the same (butane) so have the same energy. 1-butene releases more energy than cis-2-butene, so 1-butene must start at a higher energy. This means that cis-2-butene is lower energy (by 1.7 kcal/mol) Would you expect a similar reaction of -2-butene to release more or less than cis-2- butene? iefly explain. Since -2-butene is lower in energy than cis-2-butene (methyl groups are further apart, fewer electronic repulsions), it would release less energy in a similar reaction. The value would be smaller (or less negative; actual value = kcal/mol) Page 2

3 6. Answer the questions about the following alkenes. Rank the following alkenes from low to high energy. Explain the tren monosubstituted disu trisu highest energy lowest energy The more substitution around a double bond, the lower energy due to more hyperconjugation / electron delocalization of the pi bon Which alkene would you expect to release the least amount of upon combustion? Explain. The tetrasubstituted alkene is lower energy than the trisubstituted alkene. This means it starts at a lower energy and so releases less upon combustion. (Similar looking diagram to question 5a). tetrasu trisu REATNS: MEANSMS + TRANSTN STATES 7. raw the curved arrow mechanism for these reactions, including all lone pairs and charge. nclude any missing products. Na K Page 3

4 8. raw the energy diagram for this reaction, including the structure of the ition state. - TS SM PR 9. For each reaction, give the products then draw the ition state K raw all possible elimination products for the following reactions, including stereoisomers. ecide which should be the major elimination product and briefly explain why. Na( 3 ) 3 nly one alkene is possible (there is only one type of b -). K 3 There are three possible products. The major product is the lowest energy (disubstituted and ). Na ircled is the lowest energy product. t is trisubstituted and has the largest groups on each carbon of the alkene (methyl and ethyl) anti. K( 3 ) 3 nly one alkene is possible. Page 4

5 REATNS: SUMMARY F FATRS 11. n each pair of reactions, decide which will be a faster reaction. Explain. Na Na Na 2 omine is a better leaving group than chlorine. t can better stabilize the developing negative charge in the ition state due to its large size. - is charged while 2 is neutral, which makes it start at a higher energy than water. This makes it more reactive, in this case more basi Na Na Na 2 Na MS 3 RX are more reactive than 2 because the developing alkene in the ition state is more substituted, meaning lower energy. becoming disu vs. becoming trisu Aprotic solvents make 2 nd order reactions faster because the anion/base is less stabilized than in protic solvents. MS can t solvate - as well as water because the d + is crowde This makes - start at a higher energy in MS (more reactive). 12. When Na 3 is used as the base in the following reaction, the major product is M. When Na( 3 ) 3 is used as the base, the major product is N. Explain these different results. base M + N M Base N Base The ition states leading to the two products are important since is a kinetically controlled reaction. With a normal sized base like Na 3, ition state M is lower in energy since it involves a more substituted partial alkene. With a bulky base like Na( 3 ) 3, there is too much steric hindrance in the M ition state (sterics dominate over alkene stability). There is less hindrance in ition state N, where the base removes the more accessible external beta-hydrogen, so this ition state is lower in energy and the major product is N. Page 5

6 REATNS: ANT ELMNATN 13. Which should be the major product (J-M) of this reaction? Explain, including Newman projections with your answer. K J K L M The major product is not J because it s only monosubstituted (higher energy), and it definitely is not K- there s no way for the stereocenter with the to invert. The Newman projection is helpful to decide between products L and M (both are disubstituted alkenes). The beta-hydrogen removed by the base must be anti to the leaving group, which is the, not the. Thus, the deuterium remains and the major product is M. 14. nly one product is observed in reaction (1) while a mixture is formed in reaction (2). Use chair conformations to explain why only one product is formed in reaction (1). Na 2 3 (1) 100% 2 3 (2) Na n chair conformations, the mechanism occurs when the leaving group is axial and the betahydrogen is also axial (it is then anti). There is only one appropriate axial b - (see chair below). major not anti Use chair conformations to explain why two products are formed in reaction (2), then why one product is the major. n reaction (2), there are two b - that are axial (and thus anti) to the leaving group, thus two products can form. The major product is the one lower in energy, the one with the more substituted alkene. 3 anti anti Page 6

7 15. Give the major organic product for these reactions. Keep in mind that the b -hydrogen must be anti to the leaving group in an reaction. Na 3 f. Na K g. Na 3 Na 3 h. 3 K Rotation involved to get b - anti Na i. Na 3 3 β- not anti e. K j. Na 3 E1 REATNS: MEANSMS + TRANSTN STATES 16. Give the curved arrow mechanism for these E1 reactions. 3 high (major product is still S N 1, but E1 competes significantly) 3 Page 7

8 17. raw the energy diagram for this reaction, including the structure of the ition state for the rate determining step. 2 carbocation + δ+ SM PR E1 REATNS: SUMMARY F FATRS 18. n each pair of reactions, decide which should have a faster E1 reaction and explain. (Note: in some reactions E1 is not the major pathway, but still decide if an E1 were to occur, which would be faster.) S Na 2, 2 The ition state of the RS has the leaving group d -. S 2 3 stabilizes the d - better than because it spreads out the partial charge through resonance. Equal rates. E1 is first order, so the rate is dependent on the structure of the alkyl halide which is the same for both reactions. The Nu or base is not involved in the RS so doesn t affect the rate /MS 3 RX are faster in E1 reactions because the RS forms a carbocation, and 3 carbocations are lower energy than 2. 1 st order reactions are faster with protic solvents (e.g. water) because they stabilize the LG formed after the RS (or they stabilize the d - in the ition state). is not stabilized in MS (naked anion) so the intermediate E is higher, and the reaction slower. MS 2 Page 8

9 E1 VERSUS REATNS 19. For each elimination reaction, decide whether it proceeds through an E1 or mechanism. Then give the major product and draw the appropriate curved arrow mechanism. Reaction E1/? Major Product Mechanism Na E1 3 Na( 3 ) high E ELMNATN VERSUS SUBSTTUTN REATNS 20. Explain why the amount of elimination product (relative to substitution product) increases when the of the reaction is increase Elimination reactions have a favorable change in entropy because they convert 2 reactants into 3 products. Substitutions have the same number of reactants as products, so typically have a negligible S. Since G = T S, temperature is linked with the change in entropy term. As temperature increases, it makes the T S term more important to the overall G. This has little effect on substitutions ( S ~0) but makes eliminations more favorable ( S = +). Page 9

10 21. Should compound A or compound B be expected to react faster in each mechanism? iefly explain. A B A is 1 RX, B is 3 RX. S N 1 E1 1 RX faster (A) 3 RX faster (B) 3 RX faster (B) 3 RX faster (B) Least hindered at ition state. Lower energy carbocation Partial alkene in TS is more substitute Lower energy carbocation 22. Which of reactions A- should never proceed through an E1 or S N 1 mechanism (there may be >1 correct answer)? Explain why these reactions should not react through these mechanisms. A B Na will never go through an E1 or S N 1 mechanism because it involves reaction of a 1 RX, and 1 carbocations are too high energy. A also won t undergo E1 or S N 1 reactions because the base Na 3 is too strong. The base is so reactive it will go through 2 nd order reactions (will go for it!), and is the major pathway for A. Neutral nucleophiles are more common for 1 st order reactions because they are only reactive enough if paired with a carbocation. 23. Substitution and elimination reactions always compete with each other. For each reaction below, draw the probable substitution and elimination products, considering if a 1 st or 2 nd order mechanism is likely. Then identify the major product. 2 S N 1 (racemic) E1 3 K MF 3 3 NaN acetone N Page 10

11 24. Explain why reaction 1 and 2 produce different alkenes as their major products. Na 3 3 Reaction 1 Reaction 2 Reaction 1 proceeds through an mechanism, which requires the b - and leaving group to be anti ( -like ). The alkene can t form near the isopropyl group because the b - there is not anti. Reaction 2 proceeds through an E1 mechanism, which involves a carbocation, and there is no special orientation of the b - in order to eliminate. Reaction 2 therefore forms the lower energy more substituted product. 25. For each reaction, decide whether the major pathway should be S N 1,, E1, or. Then give the appropriate curved arrow mechanism that produces the major product for each. Mechanism and Major Product S N 1,, E1,? 3 3 S N Na high 2 3 E1 S 2 F 3 S NaS S Page 11

12 26. Give the major organic product for each reaction. Pay attention to stereochemistry and if a mixture of products is expected (i.e. racemic mixture), draw all products. dentify if the reaction mechanism is S N 1,, E1, or. S N racemic 3 j. 3 S N 1 mixture of diasteromers K 2 3 k. 3 2 E1 NaN 2 l. 3 S 2 F 3 K acetone 3 2 high E1 m. ( 3 ) 3 Na e. KS MS S n. 3 K 3 f. Na MF o. NaN 3 3 N N 3 g. Na p. K acetone h. 3 Na 3 q. K acetone 3 i. S KN N r. F excess 3 N - F N 3 Page 12