So far, all the reactions of carbonyl compounds that we have seen were directly at the carbonyl.
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1 Chapter 22: Carbonyl α-substitutions 326
2 So far, all the reactions of carbonyl compounds that we have seen were directly at the carbonyl. But there is another possibility if the carbonyl has α-hydrogens, α substitution 327
3 Keto-Enol Tautomerism (22-2) This substitution reaction can take place under acidic condition from the enol tautomer acting as the nucleophile. Or under basic condition. Tautomerism: Rapid conversion of two isomeric chemical species 328
4 Under weakly basic (RO - or HO - ) conditions, only a small amount of the enolate anion can be produced. In these cases the nucleophile in the substitution reaction is most likely the enol tautomer. 329
5 However, the use of stronger bases such as LDA (Lithium DiisopropylAmide) generates the enolate ion in quantitative amount, making this species the nucleophile in the substitution reaction. 330
6 Practice Questions Draw the enol tautomers for each of the following compounds. For compounds that have more than one enol tautomer, indicate which is more stable. O O O O 331
7 Alkylation of Enolate Ion (22-3) The reaction follows a S N 2 mechanism where the nucleophile is the enolate anion. Alkylation of enolate anion is normally at the carbon, but some side reaction at the oxygen is also possible. 332
8 333
9 This reaction has some serious limitations: Difficult at best with ketones, esters and nitriles Carbonyl with only one type of α-proton give good results Only a strong base such as LDA can be used Condensation is often a side reaction. Cannot be done with aldehydes. To be successful, it requires small reactive alkyl halides (R-X) reactivity order: allylic > benzylic > methyl > 1 o In practice, alkylation of enolate is rarely attempted 334
10 The main reason for all of these limitations is because similar conditions are used to carry out condensation reactions. Alkylation and condensation always compete with one another. 335
11 In order to minimize this problem, one must form the enolate ion quantitatively and rapidly. The use of LDA is recommended (yet condensation will still take place). 336
12 Formation and Alkylation of Enamines (22-4) This is a milder alternative to direct alkylation of carbonyl and is the best way to alkylate aldehydes. Aldehydes cannot be alkylated directly but using an enamine, followed by an hydrolysis, alkylated aldehydes can be obtained. Ketones can also be obtained via this method. 337
13 Enamines are very similar to enolate ions in their structures and reactivity. In the presence of an electrophile, they will act as the nucleophile and produce the iminium ion. This ion will generate the corresponding carbonyl (ketone or aldehyde) on hydrolysis. 338
14 The synthesis of the enamine follows a similar mechanism as the formation of the imine (Chapter 18). However, here the carbinolamine cannot eliminate a proton from the amine function and the elimination takes place at the α position instead. 339
15 Acylation of enamines is also easy to do by an acyl substitution mechanism with acid chlorides, β diketones are obtained. 340
16 Practice Questions How would you prepare the following compound using an enamine intermediate? 341
17 How would you prepare the following compounds from a ketone and an alkyl halide? 342
18 α-halogenation of Ketones (22-5) This reaction if possible for any halogens (Cl 2, Br 2, I 2 ) under acidic conditions (acetic acid is often used). Therefore, the mechanism will proceed via the enol intermediate. 343
19 Under basic conditions it also possible carry out this reaction. However, if more than one α-hydrogens are present, they will all be substituted. O Cl H Cl _ 2 OH, H 2 O O Cl Cl Cl O Cl Cl Cl Cl O Cl Cl 344
20 This polyhalogenation under basic conditions is the basis of the haloform reaction producing carboxylic acid. Because the CX 3 group on the carbonyl is a good leaving group, acyl substitution with the hydroxide ion takes place leading to the formation of carboxylate anions. 345
21 α Bromination of Acids: The HVZ Reaction (22-6) The HVZ (Hell-Volhard-Zelinsky) reaction is not one carried out on aldehydes/ketones but rather is a reaction of a carboxylic acid, and is very similar to the halogenation reaction we just covered. This reaction gives α-brominated carboxylic acids. 346
22 Because acids, esters and amides do not enolize enough, they cannot be brominated directly. Hence, the acid is transformed first into the acid bromide, α-brominated and hydrolyzed back to a carboxylic acid. The mechanism is shown on the next slide. The formation of the intermediate acid bromide (labelled A on the next slide) follow the same mechanism we have seen for alcohols (Chapter 11 Alcohols/phenols) 347
23 Mechanism of the HVZ Reaction A 348
24 The Aldol Condensation (22-7) In the aldol condensation, the same carbonyl can be used as the nucleophile (enolate) and as the electrophile (carbonyl). The result is the formation of a β-hydroxycarbonyl. This is the reason why alkylation was so difficult to perform, this reaction always competes with others whenever an enolate anion is produced. enolate O ketone O R CH 2 - R CH 3 nucleophile electrophile 349
25 Ketones and aldehydes will give aldol condensation. H O H O OH - H O OH O O O O OH - O OH
26 Based Catalyzed Aldol Addition 351
27 Practice Question What is the aldol addition products are formed from the following carbonyls? 352
28 The aldol condensation of aldehydes and ketones can also be catalyzed by acid. The enol tautomer is acting as the nucleophile in these cases. 353
29 Dehydration of the Aldol Product (22-8) Aldol products, β-hydroxycarbonyls are very easy to dehydrate because the resulting product is very stable (unsaturated carbonyls). Since the aldol addition is at equilibrium, this step is usually necessary to complete the reaction. These two steps together (aldol addition + dehydration) will generate α,β-unsaturated carbonyl. 354
30 The dehydration of the aldol addition product can be done under acidic or basic conditions. O H 2 SO 4 O O H NaOH O H 355
31 Base-Catalyzed Dehydration of Aldol Product 356
32 Acid-Catalyzed Dehydration of Aldol Product H + O B - + O OH OH OH OH OH H H H H + H H B - H O + + OH OH 2 OH B - OH H + H H H O H 357
33 Competition Between Condensation and α-substitution. We have seen that for both of these reactions an enolate anion must be formed. Different conditions can be used to control the reaction outcome depending on the type of reaction needed: 358
34 Crossed Aldol Condensations (22-9) The condensation reaction between two structurally different aldehydes/ketones. For example, what would be the product of this reaction? How many products can be formed? 359
35 These mixed aldol (cross aldol) reactions are very difficult to control and more than one product is usually obtained. However, it is possible to limit the possibilities by choosing the correct reaction conditions: A- One carbonyl has no α-hydrogen (cannot enolize) B- Form the enolate first, then add the second carbonyl 360
36 A- One carbonyl has no α-hydrogen (cannot enolize) 361
37 B- Form the enolate first, then add the second carbonyl O LDA O - 1) O H O OH 2) H 2 O formation of the enolate should be quantitative 362
38 Practice Question What are the products of the following crossed aldol additions? 363
39 Aldol Cyclizations (22-10) When a molecule contains two carbonyl groups within its structure, it is possible for this molecule to cyclize via an aldol addition. The stability of cyclic structures is important, hence 5 and 6-membered rings are favoured. 364
40 365
41 Practice Question What is the major product of the following reactions? 366
42 Planning Syntheses using Aldol Condensations (22-11) When planning to carry out a synthesis that will involved an aldol condensation, always look at the bond between the α and β carbons since this is the bond created during the aldol condensation. 367
43 Claisen Condensation (22-12) This reaction is identical to the aldol condensation but takes place between 2 esters. A small difference exists in the mechanism to explain the formation of the final product, a β-ketoester. You should recognize in the second step the acyl substitution mechanism that we have seen in a previous chapter. 368
44 In this reaction a second step (acidic workup) is necessary since the product of the claisen condensation is more acidic than the original esters and is deprotonated with the base that is expelled. pka~ 11 R CH 2 R'O C O O C C H R H 3 O + R CH 2 R'O C O C C - O R 369
45 The Dieckmann Condensation (22-13) This is similar to the intramolecular Aldol reaction, 5 and 6-membered rings are preferred. 370
46 Crossed Claisen Condensation (22-14) Similar to the Aldol condensation, when two different esters are used, cross condensation results. 371
47 Other carbonyl can be used to replace one of the ester. When a ketone is used, it always forms the enolate since the protons are more acidic. Reaction with an ester gives a β-diketone. 372
48 Reaction of a ketone with diethyl carbonate is another good method to get β-ketoesters. 373
49 Practice Questions Which of the following cannot give Claisen condensation? 374
50 What are the products of the following reactions? O O OCH 3 1) CH 3 O - 2) HCl 375
51 Malonic ester synthesis (22-16) In this reaction the α-hydrogen are very acidic because they are located between 2 carbonyl groups hence, we say that the carbonyl is activated. The deprotonation is quantitative and unambiguous. Carboxylic acids are obtained after an hydrolysis and decarboxylation steps. 376
52 Simple bases are used because of the increased acidity of the β-dicarbonyl compounds. 377
53 Examples of the Malonic Ester Synthesis 378
54 We have seen how to hydrolyze (undeer basic or acidic conditions) esters to acids (section on carboxylic acid derivatives). Because of the structure of this diacid, it is very easy (just requires heat) to remove CO 2 and obtain the product, a carboxylic acid. 379
55 Practice Question What are the products of the following reactions? 380
56 Acetoacetic ester synthesis (22-17) The principles seen in the malonic ester synthesis also apply here except, because there is only one possible carboxyl group to eliminate, a ketone will be obtained. This is an excellent method to get an alkylated ketone and avoid condensation. This reaction will work for any β-ketoester. 381
57 Decarboxylation follows the same mechanism, but only one carboxyl group is present, hence a ketone is produced. 382
58 Example of Acetoacetic Ester Synthesis 383
59 Practice Questions What alkyl bromide would do you need in the malonic ester synthesis of the following acids? 384
60 What alkyl bromide would you use to carry out the acetoacetic ester synthesis of the following ketones? 385
61 Michael Addition (22-18) Enolate ion are not as nucleophilic as other ion such as RLi or RMgX. This is due to the resonance stabilization. Because of this, enolate ion will add in a conjugate fashion in the presence of α,β-unsaturated carbonyl. This reaction is called the Michael Addition. 386
62 The Michael addition will work with any enolate ion, but the results are better with a stable enolate is used. Enolates from: β-ketoesters β-diketones Malonic ester 387
63 Normally the reactant forming the enolate is called: Michael Donor (because it provides the electrons for the new C-C bond) The reactant that is attacked, the α,βunsaturated compound is called the Michael Acceptor (because it accepts the electron pair). These reactants are not limited to carbonyl compounds (see next slide) 388
64 389
65 Mechanism of the Michael Addition Example of Michael Addition 390
66 Practice Question How would you prepare the following compound using a Michael addition in your synthesis? 391
67 Robinson Annulation (18.17) Annulation: Formation of a ring structure from acyclic structure. The Robinson Annulation is a two step process. (1) Michael Addition (2) Intramolecular Aldol Condensation 392
68 Mechanism of Robinson Annulation 393
69 How would you prepare the following compound using a Robinson Annulation? The key questions are, which part of this molecule came from the Michael donor, which one came from the Michael acceptor? 394
70 First identify the β and γ carbon on the target molecule count going away from the double bond. Draw a line going through the β and γ carbon and splitting the double bond in half O acceptor donor 395
71 Therefore, in order to carry out this synthesis, we need: 396
72 Practice Questions Propose a synthesis for the following compound, using a Robinson Annulation? 397
73 What are the product of the following Robinson Annulations? 398
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