Nucleophilic Substitution & Elimination Chemistry Beauchamp 1

Transcription

1 ucleophilic ubstitution & Elimination hemistry Beauchamp 1 Problem 1 - ow can you tell whether the 2 reaction occurs with front side attack, backside attack or front and backside attack? Use the two molecules to explain you answer. Use curved arrow formalism to show electron movement for how the reaction actually works. Problem 2 - Why might 3 and 4 rings react so slowly in 2 reactions? (int-think about bond angles in the transition state versus bond angles in the starting ring structure.) Problem 3 - Why might 6 rings react slower in 2 reactions? What are the possible conformations from which a reaction is expected? Trace the path of approach for backside attack across the cyclohexane ring to see what positions block this approach. Which chair conformation would have the leaving group in a more reactive position (axial or equatorial)? Is this part of the difficulty (which conformation is preferred and present in higher concentration)? y:\files\classes\315\315 andouts\315 Fall 2013\2b 315 and E & chem catalog, answers.doc

2 ucleophilic ubstitution & Elimination hemistry Beauchamp 2 Problem 4 - In each of the following pairs of s one is a much better than its cly related partner. Propose a possible explanation. Problem 5 Write out the expected 2 product for each possible combination (4x8=32 possibilities). 1 u = A Al Li B Problem 6 Using - compounds from page 2 and reagents from page 3 to propose starting materials to make each of the following compounds. ne example is provided. TM-1 is an E2 product (see page 15), all the others are 2 products. 2-azidopropane problem solution? 3 a 3 (buy) 2 TM = target molecule TM-1 (make) K E2 TM-2 (make) a 2 TM-3 (make) a 2 y:\files\classes\315\315 andouts\315 Fall 2013\2b 315 and E & chem catalog, answers.doc

3 ucleophilic ubstitution & Elimination hemistry Beauchamp 3 TM-4 (make) TM-5 (no structure) TM-6 (make) 2 2 TM TM-8 P P (buy) 2 TM-9 (buy) a 2 TM-11 TM-10 (buy) 2 Li (make) 2 TM-12 TM-13 TM-14 a a (buy) 2 (buy) 2 a (make) 2 TM-15 TM-16 Li Li (make) 2 (make) 2 TM-17 Li Al 4 (buy) 2 Problem 7 how the acyl substitution mechanism for each functional group with hydroxide as the. l a acyl substitution l l acid/base l acyl substitution y:\files\classes\315\315 andouts\315 Fall 2013\2b 315 and E & chem catalog, answers.doc

4 ucleophilic ubstitution & Elimination hemistry Beauchamp 4 a acyl substitution acid/base a acyl substitution acid/base Problem 8 Write an arrow pushing mechanism for each of the following reactions. pk a = 17 pk a = 19 a K K eq = = acid/base K eq = = acid/base (gas) (gas) pk a Problem 9 Write an arrow pushing mechanism for the following reaction. a. Al Li + 3 l 2 3 l b. B a + 3 l 2 3 The center inverted, but the priorities changed, so still. l y:\files\classes\315\315 andouts\315 Fall 2013\2b 315 and E & chem catalog, answers.doc

5 ucleophilic ubstitution & Elimination hemistry Beauchamp 5 Problem 10 It is hard to tell where the hydride was introduced since there are usually so many other hydrogens in organic molecules. Where could have have been in the reactant molecule? There are no obvious clues. Which position(s) for would likely be more reactive with the hydride reagent? ould we tell where was if we used LiAl 4? 2 Where was ""? Al Li could be on any sp 3 carbon atom. If we used LiAl 4 there would be a '' where '' was. Problem 11 - ow many total hydrogen atoms are on carbons in the given compound? ow many different types of hydrogen atoms are on carbons (a little tricky)? ow many different products are possible? int - Be careful of the simple 2. The two hydrogen atoms appear equivalent, but E/Z (cis/trans) possibilities are often present. (ee below, problem 7, for relative expected amounts of the E2 products.) 3 = has only 1, could be or stereochemistry 2 has two, they are different, lead to E/Z stereoisomers 3 has three and they are all equivalent, no E/Z possible E and Z possible here depending on stereochemistry of and 1. 3 y:\files\classes\315\315 andouts\315 Fall 2013\2b 315 and E & chem catalog, answers.doc E and Z possible here depending on which proton is lost. stability or alkenes = tetrasub > trisub > trans-disub > cis-disub = gem disub > monosub o E and Z is possible here. Problem 12 - econsider the elimination products expected in problem 11 and identify the ones that you now expect to be the major and minor products. ow would an absolute configuration of α as, compare to α as? What about 1 as versus? (3,4)-3-bromo-3,4-dimethylheptane etc E alkene tetrasub = most stable, # E alkene trisub = middle stable, # o Z or E alkene gem disub = less stable, # Z alkene trisub = middle stable, #3

6 ucleophilic ubstitution & Elimination hemistry Beauchamp 6 Problem 13 -Provide an explanation for any unexpected deviations from our general rule for alkene stabilities above. A more negative potential energy is more stable. 7 (gr) (g) = zero reference energy f least f stable most stable teric crowding by t-butyl group changes order of stability. f Expected this alkene to be the least stable, but it's not. o f = kcal/mole o f = kcal/mole o f = kcal/mole ur rules predict that stability is trans-di > cis-di > monosubstituted alkene. owever, the data show that the cis alkene is less stable than the monosubstituted alkene. This is due to the large size of the t-butyl group crowding the medium methyl group. This would probably cause twisting of the pi bond and weaken pi overlap. Problem 14 rder the stabilities of the alkynes below (1 = most stable). Provide a possible explanation. "" = a simple alkyl group ur rules predict that more substitution at pi bonds should make them more stable by extra inductive donating effect (compared to ) at more electronegative sp hybrid orbitals, so disubstituted > monosubstituted > unsubstituted. Problem 15 - Propose an explanation for the following table of data. Write out the expected products and state by which mechanism they formed. u: -- /B: -- = (a weak base, but good ) percent substitution 100 % 100 % 11 % 0 % percent elimination 0 % 0 % 89 % 100 % our rules 2 > E2 2 > E2 2 > E2 (wrong prediction) only E2 Acetate is a pretty stable anion due to resonance stabilization on two oxygen atoms. We predict that it will favor 2 reaction at methyl, primary and secondary electrophiles and E2 at tertiary. Everything looks as expected here except example 3, which is secondary, but is tertiary so there is extra steric hindrance at both and which appears to push it to E2 > 2. We will still follow our simplistic rules in this course, but here is an example that shows our rules are a bit too simplistic. y:\files\classes\315\315 andouts\315 Fall 2013\2b 315 and E & chem catalog, answers.doc

7 ucleophilic ubstitution & Elimination hemistry Beauchamp 7 Problem 16 ne of the following reactions produces over 90% 2 product and one of them produces about 85% E2 product in contrast to our general rules (ambiguity is organic chemistry s middle name). Match these results with the correct reaction and explain why they are different. pk a = 16 l 2 > E2 l 90% 2 pk a = 19 l E2 > 2 85% E2 terically larger t-butoxide forces E2 reactions even at primary. l Problem 17 - A stronger base (as measured by a higher pk a of its conjugate acid) tends to produce more relative amounts of E2 compared to 2, relative to a second (weaker) base/. Greater substitution at and also increases the proportion of E2 product, because the greater steric hindrance slows down the competing 2 reaction. Use this information to make predictions about which set of conditions in each part would produce relatively more elimination product. iefly, explain your reasoning. Write out all expected elimination products. Are there any examples below where one reaction ( 2 or E2) would completely dominate? a. l l I...versus... I more E2 because of extra steric hindrance at position. b. l l...versus... pk a ( 2 ) = 5 pk a () = 16 c. l...versus... d. l...versus... l l more E2 because base is less stable (higher pk a of conjugate acid, more basic) more E2 because of extra steric hindrance at position. more E2 because of extra steric hindrance at position, only E2 at tertiary center. y:\files\classes\315\315 andouts\315 Fall 2013\2b 315 and E & chem catalog, answers.doc

8 ucleophilic ubstitution & Elimination hemistry Beauchamp 8 e....versus... more E2 because of extra steric hindrance at position, which is quaternary, so no 2, only E2. f. l more E2 because base is less stable (higher pk a of conjugate acid, more basic) pk a () = 25...versus... pk a () = 9 l Problem 18 - (2,3)-2-bromo-3-deuteriobutane when reacted with potassium ethoxide produces cis-2-butene having deuterium and trans-2-butene not having deuterium. The diastereomer (2,3)-2-bromo-3-deuteriobutane under the same conditions produces cis-2-butene not having deuterium and trans-2-butene having deuterium present. Explain these observations by drawing the correct 3 structures, rotating to the proper conformation for elimination and showing an arrow pushing mechanism leading to the observed products. (Protium = and deuterium = ; and are isotopes. Their chemistries are similar, but we can tell them apart.) 3 3 a cis-2-butene ( present) a cis-2-butene (no present) 3 (2,3) trans-2-butene (no present) 3 (2,3) trans-2-butene ( present) 3 (2,3) 3 (2,3) ne 3 3 ne 3 example. 3 example Z alkene (with ) E alkene (with ) Et Et Et Z alkene, with E alkene, without (2,3) (2,3) (2,3) 3 1-alkene, no E/Z y:\files\classes\315\315 andouts\315 Fall 2013\2b 315 and E & chem catalog, answers.doc

9 ucleophilic ubstitution & Elimination hemistry Beauchamp 9 Problem 19 - raw a ewman projection of the two possible conformations leading to E2 reaction. how how the orientation of the substituents about the newly formed alkene compares. B staggered anti E2 reaction 1 rotate about center bond reactant conformation 1 < 1% > 99% B ewman projections alkene stereochemistry is different diasteromers eclipsed 2 syn E2 1 reaction B reactant conformation 2 B alkene stereochemistry is different Problem 20 What are the possible products of the following reactions? What is the major product(s) and what is the minor product(s)? There are 55 possible combinations u = 3 3 alkylthiolate hydroxide alkoxide t-butoxide carboxylate hydrogenthiolate cyanide acetylide B Al 2 sulfur a Li enolates ylids imidate azide borodeuteride aluminumdeuteride dithiane anion A primary u B secondary u (E2 > 2) 3 possible (E2 > 2) 3 possible (E2 > 2) 3 possible y:\files\classes\315\315 andouts\315 Fall 2013\2b 315 and E & chem catalog, answers.doc

10 ucleophilic ubstitution & Elimination hemistry Beauchamp 10 primary u u 2 > E2 products (all except t-butoxide) no / E2 products no E/Z E secondary tertiary u u 2 > E2 products (all except hydroxide, alkoxide, t-butoxide, acetylide) E2 products (enantiomers) u only E2 products (3 ways, tri > gem di) Problem 21 - Explain the differences in stability among the following carbocations (hydride affinities = a larger number means a less stable carbocation = more energy released when combined with hydride). methyl = 315 primary = 270 secondary = 249 tertiary = 232 allyl = groups inductively donate electrons to electron deficient centers like carbocations (and free radicals). The more groups the more stable. The carbocation that releases the most energy when combined with hydride was the least stable starting energy. 2 2 Primary carbocations are 270 kcal/mole, but this primary allylic carbocation is only 256, so it is 14 kcal/mole more stable than expected because of resonance sharing by adjacent pi bond. benzyl = Primary carbocations are 270 kcal/mole, but this primary benzylic carbocation is only 239, so it is 31 kcal/mole more stable than expectedbecause of resonance sharing by adjacent pi bonds. 2 = 248 = 218 = Primary carbocations are 270 kcal/mole, but the primary carbocation next to oxygen is only 248, so it is 22 kcal/mole more stable than expected even though oxygen's inductive effect works in the opposite direction. When next to nitrogen it is more stable than expected by 52 kcal/mole because less electronegative nitrogen shares better than more electronegative oxygen. The last example, lone pair next to oxygen is 40 kcal/mole more stable than expected. For all of them it is because of resonance sharing by adjacent lone pair of electrons. In each case this helps fill the octet of the carbocation and adds an extra bond. y:\files\classes\315\315 andouts\315 Fall 2013\2b 315 and E & chem catalog, answers.doc

11 ucleophilic ubstitution & Elimination hemistry Beauchamp 11 Problem 22 - Why are vinyl carbocations so difficult to form? (int What is their hybridization?) ow does an empty sp 2 orbital (phenyl carbocation) or empty sp orbital (ethynyl carbocation) compare to a typical sp 2 carbocation carbon with an empty 2p orbital? (An empty 2p orbital is also present on the vinyl carbocation, but the carbon hybridization is sp.) sp hybridized carbon is the most electronegative carbon because it is 50% s character and makes it more difficult (higher energy) to form. The empty orbital is 2p (the least electronegative). The second example is also sp carbon, but the orbital is sp, which makes it much more difficult to take electrons from. The phenyl example cannot change its shape because of the ring and that makes the empty orbital sp2, which is also more electronegative than 2p and harder to form. The difficulty to form can be seen from the energy released when each carbocation reacts with the hydride donor. The more energy released, the less stable the starting point. 2p sp 3 sp 2 sp 2s electronegativity of orbitals least 2p orbital, but on sp carbon most hold on electrons least most sp orbital very electronegative sp 2 orbital very electronegative Problem 23 The bond energy depends on charge effects in the anions too. an you explain the differences in bond energies below? (int: Where is the charge more delocalized?) We won t emphasize these differences. 3 3 = gas phase bond energies 3 3 l I +140 The difference in energy cost is due to the different anions since the carbocation is the same in every instance. The larger and more delocalized the electrons the more stable is the anion and easier to form, just like we saw in our acid/base topic. Problem 24 raw in all of the mechanistic steps in an 1 reaction of 2-bromobutane with a. water, b. methanol and c. ethanoic acid. Add in necessary details (3 stereochemistry, curved arrows, lone pairs, formal charge). What are the final products? a c b a b and c Adds from both faces, makes and chiral center. ( 1) a ( 1) / 2E alkene 2E alkene 1 alkene from c from b y:\files\classes\315\315 andouts\315 Fall 2013\2b 315 and E & chem catalog, answers.doc

12 ucleophilic ubstitution & Elimination hemistry Beauchamp 12 3 a b a 3 3 a c Adds from both faces, makes and chiral center. ( 1) similar steps and similar products b and c (ether instead of an alcohol) 2E alkene 2E alkene from c 3 b a 3 c ( 1) / 3 ( 1) / 1 alkene from b similar steps and similar products (ester instead of an alcohol or ether), plus some minor alkene products. Problem 25 What are the likely 1 and products of the initial carbocation and the rearranged carbocations from a, b and c? a : hydride migrates and are possible here 3 3 o carbocation looks very good 3 1 (achiral) E and Z possible no E and Z possible trisubst. > gem disubst. b : methyl migrates 1 and are possible here 3 2 o carbocation looks K 3 1 and possible E and Z possible trans disubst. < trisubst. no E and Z possible y:\files\classes\315\315 andouts\315 Fall 2013\2b 315 and E & chem catalog, answers.doc

13 ucleophilic ubstitution & Elimination hemistry Beauchamp 13 c : ethyl migrates 1 and are possible here o carbocation looks K 1 and possible E and Z possible trisubst. > monosubst. no E and Z possible Problem 26 Write out your own mechanism for all reasonable products from the given - compound in water (2-halo-3-methylbutane) rearrangement 3 2 o to 3 o b 2 a c Also, from second carbocation and Problem 27 - onsider all possible rearrangements from ionization of the following reactants. Which are reasonable? What are the possible 1 and products from the reasonable carbocation possibilities? a. b subsequent + initial + subsequent + initial + 1 = 1 1 = 2 1 = 6 1 = 2 = 2 = 1 = 8 = 1 total = 6 products total = 17 products total = 17 products initial + 1 = 2 = 1 c subsequent + 1 = 6 = 8 y:\files\classes\315\315 andouts\315 Fall 2013\2b 315 and E & chem catalog, answers.doc

14 ucleophilic ubstitution & Elimination hemistry Beauchamp 14 a poor = 2 o to 1 o This rearrangement is not going to happen. b ionization 2 c 3 b b 2 a a a rearrange b good = 2 o to 3 o 2 3 b' these two are equivalent 3 K, 3 o to 3 o 3 3 add beta c good = 2 o to 3 o c' these two 3 are different 2 2 K, 3 o to 3 o 3 3 b c 3 b b 2 add Assume water is the solvent. Possible 1/ products are: b good = 2 o to 3 o ionization rearrange 3 3 beta c good = 2 o to 3 o add c' K, 3 o to 2 2 beta 3 o 3 3 Possible products add add add beta beta from initial + from subsequent + beta add beta = chiral center around 17 possible products 1 and and and 1 (,, ) 1 1 y:\files\classes\315\315 andouts\315 Fall 2013\2b 315 and E & chem catalog, answers.doc

15 ucleophilic ubstitution & Elimination hemistry Beauchamp 15 c d e 2 o carbocation e 3 o carbocation 3 3 same as b above d 3 o carbocation same as c' above 3 o carbocation same as c above and 3 3 add beta add imilar to above. = chiral center around 17 possible products beta add d. What would happen to the complexity of the above problems with a small change of an ethyl for a methyl? Use the key of b and c as a guide. This problem is a lot more messy than those above, (which is the point of asking it). There are too many possibilities to consider listing every answer initial + 1 = 2 = 1 subsequent + 1 = 12 = similar beta d. Possible products rearrange add beta add beta add beta add beta add beta 1 and and and E and Z E and Z = chiral center around 31 possible products and 1 (,, ) and 1 (,, ) E and Z 1 1 and and y:\files\classes\315\315 andouts\315 Fall 2013\2b 315 and E & chem catalog, answers.doc

16 ucleophilic ubstitution & Elimination hemistry Beauchamp 16 Problem 28 Lanosterol is the first steroid skeletal structure in your body on its way to cholesterol and other steroids in your body. It is formed in a spectacular enantiospecific cyclization of protonated squalene oxide. The initially formed 3 o carbocation rearranges 4 times before it undergoes an reaction to form lanosterol. Add in the arrows and formal charge to show the rearrangements and the final reaction. squalene acid catalysis protonated squalene 3 equires 5 arrows to show the reaction. 3 lanosterol precursor 3 3 rearrangement 1 3 rearrangement B 3 3 rearrangement rearrangement B reaction lanosterol 3 19 more steps 3 cholesterol 3 other body steroids Problem 29 - Propose a synthesis of monodeuterated cyclohexane from cyclohexanol Mg 2. 2 y:\files\classes\315\315 andouts\315 Fall 2013\2b 315 and E & chem catalog, answers.doc

17 ucleophilic ubstitution & Elimination hemistry Beauchamp 17 Problem 8 (29b) Write a detailed arrow-pushing mechanism for each of the following transformations. a. l toluenesulfonyl chloride l l l -ethyltoluenesulfonamide b. l ethanoyl chloride (acetyl chloride) l 3 l -ethylethanamide (-ethylacetamide) l 3 Problem 30 We can now make the following molecules. Propose a synthesis for each. (Tosylates formed from alcohols and tosyl chloride/pyridine via acyl substitution reaction, convert from poor leaving group into a very good leaving group, similar to iodide). 3 Ts Ts Ts Ts Ts Ts All of these tosylates can be made from alcohols + tosyl chloride/pyridine. 4 5 Ts a ( 2) no rearrangement Ts 6 Ts Tsl / pyridine Ts Tosylate is a very good leaving group ( /E chemistry is possible). Tosylates can be used when a usual '' to '' transformation would lead to rearrangement. 1. Tsl/py 2. a rearrangement major product no rearrangement major product y:\files\classes\315\315 andouts\315 Fall 2013\2b 315 and E & chem catalog, answers.doc

18 ucleophilic ubstitution & Elimination hemistry Beauchamp 18 Problem 31 Look back at the table of - structures on page 2. Include stereoisomers together. Be able to list any relevant structures under each criteria below. 1. Isomers that can react fastest in 2 reactions This would include all primary, except when a neopentyl center is at a position. 2. Isomers that give E2 reaction but not 2 with sodium methoxide This would include all tertiary and secondary neopentyl, when there is a Isomers that react fastest in 1 reactions This would include all tertiary. 4. Isomers that can react by all four mechanisms, 2, E2, 1 and (What are the necessary conditions?) This would include secondary. 1/ conditions would use weak /bases (neutral) and 2/E2 conditions would use strong /bases (usually anions). When a neopentyl center is at a position the 2 reaction will not work. 5. Isomers that might rearrange to more stable carbocation in reactions with methanol. This would include secondary structures, especially if a tertiary carbocation can form. 6. Isomers that are completely unreactive with methoxide/methanol This would include primary neopentyl centers (unreactive by all four mechanisms). 7. Isomers that are completely unreactive with methanol, alone. This would include all primary compounds. y:\files\classes\315\315 andouts\315 Fall 2013\2b 315 and E & chem catalog, answers.doc

19 ucleophilic ubstitution & Elimination hemistry Beauchamp 19 Problem 32 Predict possible products of a. water and b. ethanoate (acetate) with structures 2, 10 and 13. nly consider rearrangements to more stable carbocations where appropriate. yclohexane Examples =l I Ts Which conformation is reactive? 2. Is 2 possible? equires an open approach at and. 3. Is E2 possible? equires anti ow many possible products are there? 5. What is the relationship among the starting structures? 6. What is the relationship among the products? 7. Are any of the starting structures chiral? 8. Are any of the product structures chiral? chair 1 chair t-butyl substituent locks in chair conformation with equatorial t-butyl yclohexane structures have two chair conformations possible. K eq 1 9, severe steric repulsion Mechanism predictions with: ome examples a weak s a a K strong /bases + other anions shown above y:\files\classes\315\315 andouts\315 Fall 2013\2b 315 and E & chem catalog, answers.doc

20 ucleophilic ubstitution & Elimination hemistry Beauchamp 20 nly axial '' conformations are shown, equatorial '' is too slow reacting in 2/E2 reactions. Axial or equatorial can react via 1/. 1 = 3 o 2 = 2 o 3 = 2 o 3 4 = 2 o Q1. 1/ both conformations Q2. o 2 because 3 o Q3. E2 is K at both - Q4. 1 = 1, = 1, 2 = 0, E2 = 1 Q5. A Q6. A Q7. achiral Q8. achiral Q1. 1/ both conformations Q2. o 2 because anti 3 Q3. E2 is K at at rear - Q4. 1 = 2, = 2, rearrangement likely to 3 o, 2 = 0, E2 = 1 Q5. enantiomer with 3, diastereomer with 4 and 5 Q6. diastereomer products are possible Q7. chiral Q8. chiral Q1. 1/ both conformations Q2. o 2 because anti 3 Q3. E2 is K at at front - Q4. 1 = 2, = 2, rearrangement likely to 3 o, 2 = 0, E2 = 1 Q5. enantiomer with 2, diastereomer with 4 and 5 Q6. diastereomer products are possible Q7. chiral Q8. chiral Q1. 1/ both conformations Q2. 2 is possible Q3. E2 is K at at both - Q4. 1 = 2, = 2, rearrangement likely to 3 o, 2 = 1, E2 = 2 Q5. enantiomer with 5, diastereomer with 2 and 3 Q6. diastereomer products are possible Q7. chiral Q8. chiral 5 = 2 o 3 Q1. 1/ both conformations Q2. 2 is possible Q3. E2 is K at at both - Q4. 1 = 2, = 2, rearrangement likely to 3 o, 2 = 1, E2 = 2 Q5. enantiomer with 4, diastereomer with 2 and 3 Q6. diastereomer products are possible Q7. chiral Q8. chiral 6 = 2 o 7 = 2 o 8 = 2 o Q1. 1/ both conformations Q2. 2 is possible Q3. E2 is K at at both - Q4. 1 = 2, = 2, rearrangement possible, 2 = 1, E2 = 2 Q5. enantiomer with 7, diastereomer with 8 and 9 Q6. diastereomer products are possible Q7. chiral Q8. chiral Q1. 1/ both conformations Q2. 2 is possible Q3. E2 is K at at both - Q4. 1 = 2, = 2, rearrangement possible, 2 = 1, E2 = 2 Q5. enantiomer with 6, diastereomer with 8 and 9 Q6. diastereomer products are possible Q7. chiral Q8. chiral Q1. 1/ both conformations Q2. 2 is possible Q3. E2 is K at at both - Q4. 1 = 2, = 2, rearrangement possible, 2 = 1, E2 = 2 Q5. enantiomer with 9, diastereomer with 6 and 7 Q6. diastereomer products are possible Q7. chiral Q8. chiral 9 = 2 o 3 Q1. 1/ both conformations Q2. 2 is possible Q3. E2 is K at at both - Q4. 1 = 2, = 2, rearrangement possible, 2 = 1, E2 = 2 Q5. enantiomer with 8, diastereomer with 6 and 7 Q6. diastereomer products are possible Q7. chiral Q8. chiral 10 = 2 o 11 = 2 o 12 = 2 o Q1. 1/ both conformations Q1. 1/ both conformations Q2. 2 is possible Q2. 2 is possible Q3. E2 is K at at both - Q3. E2 is K at at both - Q4. 1 = 2, = 2, rearrangement Q4. 1 = 2, = 2, rearrangement possible, 2 = 1, E2 = 2 possible, 2 = 1, E2 = 2 Q5. achiral, diastereomer with 11 Q5. achiral, diastereomer with 10 Q6. diastereomer products are possible Q6. diastereomer products are possible Q7. achiral Q7. achiral Q8. achiral Q8. achiral Q1. 1/ above conformation only Q2. 2 is possible Q3. E2 is K at at both - Q4. 1 = 2, = 2, rearrangement possible, 2 = 1, E2 = 2 Q5. achiral, diastereomer with 13 Q6. diastereomer products are possible Q7. achiral Q8. achiral y:\files\classes\315\315 andouts\315 Fall 2013\2b 315 and E & chem catalog, answers.doc

21 ucleophilic ubstitution & Elimination hemistry Beauchamp = 2 o 14 = 2 o 15 = 2 o Q1. 1/ above conformation only Q2. 2 is possible Q3. E2 is K at at both - Q4. 1 = 2, = 2, rearrangement possible, 2 = 1, E2 = 2 Q5. achiral, diastereomer with 12 Q6. diastereomer products are possible Q7. achiral Q8. achiral Q1. 1/ above conformation only Q2. 2 is possible Q3. E2 is K at at both - Q4. 1 = 2, = 2, rearrangement possible, 2 = 1, E2 = 2 Q5. achiral, diastereomer with 15 Q6. diastereomer products are possible Q7. achiral Q8. achiral Q1. 1/ above conformation only Q2. 2 is possible Q3. E2 is K at at both - Q4. 1 = 2, = 2, rearrangement possible, 2 = 1, E2 = 2 Q5. achiral, diastereomer with 14 Q6. diastereomer products are possible Q7. achiral Q8. achiral Problem 33 What are the oxidation states of each carbon atom below? All of the atoms in these examples have a formal charge of zero. 3 oxidation state oxidation state oxidation state oxidation state of carbon =? -4 of carbon =? -2 of carbon =? 0 of carbon =? +2 oxidation state of carbon =? -3 3 oxidation state of carbon =? -1 3 oxidation state of carbon =? +1 3 oxidation state of carbon =? +3 oxidation state of carbon =? +4 2 oxidation state of carbon =? +4 Problem 34 What are the oxidation states below on the carbon atom and the chromium atom as the reaction proceeds? Which step does the oxidation/reduction occur? (P, B: = pyridine and Jones, B: = water) r oxidation state of = 0 oxidation state of r = +6 B r... = slow step oxidation state of = +2 oxidation state of r = +4 partial negative of oxygen bonds to partial positive of chromium (step 1) E2 reaction (step 3) arbon oxidation occurs in this step. r oxidation state of = 0 B oxidation state of r = +6 acid/base (step 2) B B r oxidation state of = 0 oxidation state of r = +6 y:\files\classes\315\315 andouts\315 Fall 2013\2b 315 and E & chem catalog, answers.doc

22 ucleophilic ubstitution & Elimination hemistry Beauchamp 22 Problem 35 upply all of the mechanistic details in the sequences below showing 1. the oxidation of a primary alcohol, 2. hydration of the carbonyl group and 3. oxidation of the carbonyl hydrate (Jones conditions). Under aqueous conditions, a hydrate of a carbonyl group has two groups which allow a second oxidation, if another - bond is present. This is only possible if the starting carbonyl group was an aldehyde (true when starting with methyl and primary alcohols) primary alcohols 2 addition at 3 2 r 3 2 Jones = r 3 / 2 / acid primary alcohols oxidize to carboxylic acids (water hydrates the carbonyl group, which oxidizes a second time ) chromic anhydride resonance 3 2 hydration of the aldehyde acid/base 1 addition at r 3 2 r 2 acid/base acid/base r addition at r aldehydes r E2 reaction (oxidation here) second oxidation of the carbonyl hydrate 3 r 2 r inorganic ester 1. oxidation of '' to = 2. hydration of = group 3. oxidation of '' to = r acid/base r 3 2 carboxylic acid from an aldehyde or a primary alcohol E2 reaction (oxidation here) 3 2 inorganic ester y:\files\classes\315\315 andouts\315 Fall 2013\2b 315 and E & chem catalog, answers.doc

23 ucleophilic ubstitution & Elimination hemistry Beauchamp 23 Problem 36 We can now make the following molecules. Propose a synthesis for each from our starting materials. aldehydes carboxylic acids acid chlorides 3 (not stable) 3 l 3 l l 2 3 amides esters conjugated aldehydes, carboxylic acids and esters l l ketones P 1. LA 2. workup We haven't learned how to make this epoxide yet. We will when we cover alkenes. anhydrides nitriles 3 y:\files\classes\315\315 andouts\315 Fall 2013\2b 315 and E & chem catalog, answers.doc

24 ucleophilic ubstitution & Elimination hemistry Beauchamp h alkane a 2 3 nitrile 3 bromoalkane 3 3 a LiAl 4 2. workup 2 1 o amine a alcohol 1. a 2 excess a 2 2. workup (zipper) n-buli r 3 py. 2 aldehyde epoxide 1 o amide r carboxylic acid 3 l acid chloride 3 1. a ester 3 We can't make the 1 acid chloride, but we can make the 1 anhydride, which does similar reactions more reactive = anhydride 1. a h t-butoxide (E2) 2 4 / 2 h a 2 r 3 py. r 3 2 a 2 a LiAl 4 2. workup 2 -- n-buli 3 Many of the syntheses of other target molecules are similar. A few that go farther are shown. ot all are shown. 2 h t-butoxide (E2) h l a 2 r 3 py. l 3 r h t-butoxide (E2) h a 2 r 3 py. r LA 2. workup 1. LA 2. workup r 3 py. r 3 py. r 3 2 r a a 2. 3 etc. y:\files\classes\315\315 andouts\315 Fall 2013\2b 315 and E & chem catalog, answers.doc