Exam (elaborations) TEST BANK FOR The Art of Writing Reasonable Organic Reaction Mechanisms 3rd Edition By Grossman

Exam (elaborations) TEST BANK FOR The Art of Writing Reasonable Organic Reaction Mechanisms 3rd Edition By Robert B. Grossman (Solution Manual) sp2 hybridization. One of the lone pairs is in a p orbital, and the other is in an sp2 orbital. Only the lone pair in the p orbital is used in resonance. 1.5. (a) No by-products. C(1–3) and C(6–9) are the keys to numbering. (b) After numbering the major product, C6 and Br25 are left over, so make a bond between them and call it the by-product. 1.6. (a) Make C4–O12, C6–C11, C9–O12. Break C4–C6, C9–C11, C11–O12. (b) Make C8–N10, C9–C13, C12–Br24. Break O5–C6, C8–C9. 1.7. PhCºCH is much more acidic than BuCºCH because the pKb of HO– is 15, PhCºCH has a pKa ≤ 23 and BuCºCH has pKa 23. H2C O CH3 H H H H B F F F sp2 sp2 sp2 sp3 sp2 sp2 sp OH Ph O H+, H2O O Ph H 1 2 3 O 4 5 6 7 8 9 10 11 12 13 12 13 10 9 8 2 1 3 4 5 7 6 11 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 HN Br OMe O OMe H Br N O O OMe Br 19 20 21 22 23 Br Br 24 25 1 Me Br 2 3 4 5 6 7 8 10 9 19 18 17 16 15 24 14 13 20 21 11 12 25 1.8. The OH is more acidic (pKa ≈ 17) than the C a to the ketone (pKa ≈ 20). Because the by-product of the reaction is H2O, there is no need to break the O–H bond to get to product, but the C–H bond a to the ketone must be broken. Answers to Chapter 1: EndofChapter Problems 1. (a) Both N and O in amides have lone pairs that can react with electrophiles. When the O reacts with an electrophile E+, a product is obtained for which two good resonance structures can be drawn. When the N reacts, only one good resonance structure can be drawn for the product. (b) Esters are lower in energy than ketones because of resonance stabilization from the O atom. Upon addition of a nucleophile to either an ester or a ketone, a tetrahedral intermediate is obtained for which resonance is not nearly as important, and therefore the tetrahedral product from the ester is nearly the same energy as the tetrahedral product from the ketone. As a result, it costs more energy to add a nucleophile to an ester than it does to add one to a ketone. (c) Exactly the same argument as in (b) can be applied to the acidity of acyl chlorides versus the acidity of esters. Note that Cl and O have the same electronegativity, so the difference in acidity between acyl chlorides and esters cannot be due to inductive effects and must be due to resonance effects. (d) A resonance structure can be drawn for 1 in which charge is separated. Normally a charge-separated structure would be a minor contributor, but in this case the two rings are made aromatic, so it is much more important than normal. (e) The difference between 3 and 4 is that the former is cyclic. Loss of an acidic H from the g -C of 3 gives a structure for which an aromatic resonance structure can be drawn. This is not true of 4. R N R R O+ E R N R R O E reaction on O reaction on N R N R R O E (f) Both imidazole and pyridine are aromatic compounds. The lone pair of the H-bearing N in imidazole is required to maintain aromaticity, so the other N, which has its lone pair in an sp2 orbital that is perpendicular to the aromatic system, is the basic one. Protonation of this N gives a compound for which two equally good aromatic resonance structures can be drawn. By contrast, protonation of pyridine gives an aromatic compound for which only one good resonance structure can be drawn. (g) The C=C π bonds of simple hydrocarbons are usually nucleophilic, not electrophilic. However, when a nucleophile attacks the exocyclic C atom of the nonaromatic compound fulvene, the electrons from the C=C π bond go to the endocyclic C and make the ring aromatic. (h) The tautomer of 2,4-cyclohexadienone, a nonaromatic compound, is phenol, an aromatic compound. (i) Carbonyl groups C=O have an important resonance contributor C + –O – . In cyclopentadienone, this resonance contributor is antiaromatic. [Common error alert: Many cume points have been lost over the years when graduate students used cyclohexadienone or cyclopentadienone as a starting material in a synthesis problem!] (j) PhOH is considerably more acidic than EtOH (pKa= 10 vs. 17) because of resonance stabilization of the conjugate base in the former. S is larger than O, so the S(p)–C(p) overlap in PhS– is much smaller O O H3C H – H+ O O H3C O O H3C N H N H+ HN H N HN H N –Nu Nu non-aromatic aromatic than the O(p)–C(p) overlap in PhO–. The reduced overlap in PhS– leads to reduced resonance stabilization, so the presence of a Ph ring makes less of a difference for the acidity of RSH than it does for the acidity of ROH. (k) Attack of an electrophile E+ on C2 gives a carbocation for which three good resonance structures can be drawn. Attack of an electrophile E+ on C3 gives a carbocation for which only two good resonance structures can be drawn. 2. (a) The inductive electron-withdrawing effect of F is greater than that of Cl. (b) In general, AH+ is more acidic than AH. O H H H H E O H H H H E O H H H H E O H H H H 2 E+ O H H H E H O H H H E H O H H H H E+ 3 F3C O OH Cl3C O OH N H2 N H (c) Ketones are more acidic than esters. (d) Deprotonation of the 5-membered ring gives an aromatic anion, whereas deprotonation of the 7- membered ring gives an antiaromatic anion. (e) The N(sp2) lone pair derived from deprotonation of pyridinium is in a lower energy orbital than the N(sp3) lone pair derived from deprotonation of piperidinium. (f) As you move down a column in the periodic table, acidity increases due to increasing atomic size and hence worse overlap in the A–H bond. (g) The anion of phenylacetate is further stabilized by resonance with the phenyl ring. (h) Anions of 1,3-dicarbonyl compounds are stabilized by resonance into two carbonyl groups. (i) The anion of 4-nitrophenol is stabilized by resonance directly into the nitro groups. The anion of 3 -nitrophenol cannot do this. EtO CH3 O O H3C CH3 O O NH NH2 PH2 NH2 CO2Et CO2Et EtO2C CO2Et EtO2C CO2Et O2N OH O2N OH (j) More electronegative atoms are more acidic than less electronegative atoms in the same row of the periodic table. (k) C(sp) is more acidic than C(sp3), even when the anion of the latter can be delocalized into a Ph ring. (!) The anion of the latter cannot overlap with the C=O π bond, so it is not made acidic by the carbonyl group. (m) The C(sp2)–H bond on the upper atom is in the plane of the screen/paper, orthogonal to the p orbitals of the C=O π bond, so the C=O π bond provides no acidifying influence. By contrast, the C(sp3)–H bonds on the lower atom are perpendicular to the plane of the screen/paper, so there is overlap with the p orbitals of the C=O π bond. 3. (a) Free-radical. (Catalytic peroxide tips you off.) (b) Metal-mediated. (Os) (c) Polar, acidic. (Nitric acid.) (d) Polar, basic. (Fluoride ion is a good base. Clearly, it’s not acting as a nucleophile in this reaction.) N O– –O O N O –O –O H3C OH O H3C NH2 O Ph CH3 Ph H O O O this C atom this C atom (e) Free-radical. (Air.) Yes, an overall transformation can sometimes be achieved by more than one mechanism. (f) Pericyclic. (Electrons go around in circle, π bonds involved. No nucleophile or electrophile, no metal.) (g) Polar, basic. (LDA is strong base; allyl bromide is electrophile.) (h) Free-radical. (AIBN tips you off.) (i) Pericyclic. (Electrons go around in circle, π bonds involved. No nucleophile or electrophile, no acid or base, no metal.) (j) Metal-mediated. (Zr.) (k) Pericyclic. (Electrons go around in circle, π bonds involved. No nucleophile or electrophile, no acid or base, no metal.) (l) Polar, basic. (Ethoxide base. Good nucleophile, good electrophile.) (m) Pericyclic. (Electrons go around in circle, π bonds involved. No nucleophile or electrophile, no metal.) 4. (a) The mechanism is free-radical (AIBN). Sn7 and Br6 are missing from the product, so they are probably bound to one another in a by-product. Made: C5–C3, Sn7–Br6. Broken: C4–C3, C5–Br6. (b) Ag+ is a good Lewis acid, especially where halides are concerned, so polar acidic mechanism is a reasonable guess, but mechanism is actually pericyclic (bonds forming to both C10 and C13 of the furan and C3 and C7 of the enamine). Cl8 is missing from the product; it must get together with Ag to make insoluble, very stable AgCl. An extra O appears in the product; it must come from H2O during workup. One of the H’s in H2O goes with the BF4 –, while the other is attached to N1 in the by-product. Made: C3–C10, C7–C13, C2–O (water), Ag–Cl. Broken: N1–C2, C7–Cl8. 1 2 3 4 5 6 7 Br CO2Me MeO OMe CO2Me H Bu MeO OMe 3SnH cat. AIBN 1 2 4 3 5 + Bu3SnBr 7 6 (c) This mechanism is also pericyclic. Use the carbonyl, Me3SiO, and CH3 groups as anchors for numbering the atoms. Made: C2–C12, C3–C11. Broken: C2–C8. (d) Ph3P is a Lewis base. The mechanism is polar under basic conditions. Made: C1–C7, O2–C4, O3– C6. Broken: O3–C4. (e) The mechanism is polar under acidic conditions due to the strong acid RSO3H. Made: C13–C6. Broken: C13–C8. (f) The mechanism is polar under basic conditions (NaOEt). Two equivalents of cyanoacetate react with each equivalent of dibromoethane. One of the CO2Et groups from cyanoacetate is missing in the product 1 2 3 4 5 7 6 8 9 10 11 12 13 N Cl O O O + AgBF4 1 NH HBF4 AgCl 2 3 4 5 6 7 8 9 10 11 12 13 from water 1 4 3 2 5 6 7 8 9 10 11 12 H3C O OSiMe3 D H3C Me3SiO O H 1 2 3 4 5 6 7 8 9 10 11 12 1 2 4 3 5 6 7 8 Ph O CN CO2Me O O Ph NC CO2Me + cat. Ph3P OMe O OMe 1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Me Me Me OH O Me H O Me MeH Me Me O cat. RSO3H CH2Cl2, RT 1 3 2 4 5 6 7 8 9 11 10 12 13 14 and is replaced by H. The H can come from EtOH or HOH, so the CO2Et is bound to EtO or HO. The two products differ only in the location of a H atom and a π bond; their numbering is the same. Made: C2–C5, C2'–C6, C2'–C3, C1'–OEt. Broken: C1'–C2', C5–Br, C6–Br. (g) Polar under acidic conditions. The enzyme serves to guide the reaction pathway toward one particular result, but the mechanism remains fundamentally unchanged from a solution phase mechanism. The Me groups provide clues as to the numbering. Made: C1–C6, C2–C15, C9–C14. Broken: C15–O16. (h) Two types of mechanism are involved here: First polar under basic conditions, then pericyclic. At first the numbering might seem very difficult. There are two CH3 groups in the starting material, C5 and C16, and two in the product. Use these as anchors to decide the best numbering method. Made: C1–C14, C2–C12, C12–C15. Broken: C3–C12, O7–Si8. 1 2 3 5 6 EtO2C CN NaOEt, EtOH; 1/2 BrCH2CH2Br NH2 EtO2C CN + 4 1 2 3 4 5 6 2' 3' 4' EtO2C OEt 1' 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15