Aldol reaction elimination product

Strategy In the aldol reaction, H2O is eliminated and a double bond is formed by removing hvo hydrogens from the acidic a position of one partner and the carbonyl oxygen from the second partner. The product is thus an a,/3-unsaturated aldehyde or ketone. 

A combination of an aldol reaction and an elimination was used by Pandolfi s group to obtain access to the natural product prelunularin [134], and a domino aldol/aldol sequence was elaborated by the group of West for the synthesis of highly... 

In the MPVO reaction, several side-reactions can occur (Scheme 20.23). For example, an aldol reaction can occur between two molecules of acetone, which then leads to the formation of diacetone alcohol. The latter acts as a good ligand for the metal of the MPVO catalyst, rendering it inactive. Moreover, the aldol product may subsequently eliminate water, which hydrolyzes the catalyst. The aldol reaction can be suppressed by adding zeolite NaA [84, 92]. 

Deprotonation of acidic C6 by DBU gives a carbanion, which undergoes a Michael reaction to CIO. The new carbanion at CIO can deprotonate C3 to give a new carbanion, and this can undergo an aldol reaction to C12. Now our two new C-C bonds have been formed. We still have to break C2-C6 and two C-0 bonds. The alkoxide at 013 can deprotonate MeOH, which can then add to C2. Fragmentation of the C2-C6 bond follows to give a C6 enolate. The C6 enolate then deprotonates 013, and intramolecular transesterification occurs to form the 013-C7 bond and to break the C7-09 bond. MeO then comes back and promotes El elimination across the C3-C12 bond to break the Cl2-013 bond and give the product. The intramolecular transesterification explains why C7 becomes an acid and C2 remains an ester in the product. 

Second step. The elements of CH4O3 are eliminated. The most likely by-products are H20 and HCOOH. Make None. Break C4-C5, C6-O8, 010-011. The base can deprotonate the OH on C5, and the lone pair on O can then push down to form a n bond with C5, causing the C4-C5 bond to break. The electrons keep getting pushed around until they end up on O again and the 0-0 bond is broken, providing the driving force for the step. A keto-aldehyde and formate anion are obtained. Now C7 (deprotonated) is nucleophilic and C6 is electrophilic, so an aldol reaction followed by dehydration gives the observed product. 

Aldoses generally undergo benzilic acid-type rearrangements to produce saccharinic acids, as well as reverse aldol (retro-aldol) reactions with j3-elimination, to afford a-dicarbonyl compounds. The products of these reactions are in considerable evidence at elevated temperatures. The conversions of ketoses and alduronic acids, however, are also of definite interest and will be emphasized as well. Furthermore, aldoses undergo anomerization and aldose-ketose isomerization (the Lobry de Bruyn-Alberda van Ekenstein transformation ) in aqueous base. However, both of these isomerizations are more appropriately studied at room temperature, and will be considered only in the context of other mechanisms. 

In similar works, a number of [AMIM] -containing ionic liquids were evaluated for proline-catalyzed aldol reactions. The catalyst in the ionic liquid with Cl as the anion was characterized by a higher reaction rate than the ionic liquids with the anions BF)T and PFg", but elimination products affected the yields in the latter ionic liquids (149). 

These investigations revealed that 2-alkyl substituents are eliminated from 47 during oxidation under standard conditions (5% aqueous alkali) as their corresponding carboxylic acids to form 48 (Scheme 10). Compounds such as 46 readily decompose when exposed to alkaline solution to give 47 (R = H) via retro aldol reaction and the final oxidation product is therefore also 48. 

Another common trapping method is an intramolecular aldol reaction of the initially formed anion, as shown in equation (91) and Schemes 53 and 54.% In the first case, an aldol-like trapping of the iminium salt produced (411 equation 91 ).96b The initial heteronucleophile in the other two cases is ultimately lost from the product by oxidation and elimination, so that the overall process is C—C bond formation at the a-center of an enone. Thus, treatment of the formyl enone (412 Scheme 53) with an aluminum thiolate afforded in 60% yield the trapped product (413) which could be oxidized and eliminated to give (414).96c Addition of the corresponding aluminate species to the ketoacrylate (415 Scheme 54) produced only one diastereomer of the aldol product (416) which was converted into the alkene (417) in excellent yield.96 1... 

The Mukaiyama reaction is a versatile crossed-aldol reaction that uses a silyl enol ether of an aldehyde, ketone, or ester as the carbon nucleophile and an aldehyde or ketone activated by a Lewis acid as the carbon electrophile. The product is a /1-hydroxy carbonyl compound typical of an aldol condensation. The advantages to this approach are that it is carried out under acidic conditions and elimination does not usually occur. 

Rate and equilibrium constants have been measured for representative intramolecular aldol condensations of dicarbonyls.60a For the four substrates studied (32 n = 2, R = Me n = 3, R = H/Me/Ph), results have been obtained for both the aldol addition to give ketol (33), and the elimination to the enone (34). A rate-equilibrium mismatch for the overall process is examined in the context of Baldwin s rales. The data are also compared with Richard and co-workers study of 2-(2-oxopropyl)benzaldehyde (35), for which the enone condensation product tautomerizes to the dienol60b (i.e. /(-naphthol). In all cases, Marcus theory can be applied to these intramolecular aldol reactions, and it predicts essentially the same intrinsic barrier as for their intermolecular counterparts. 

D is correct. Even if you don t recognize the reaction, the name aldol means that the product must be an alcohol. This eliminates C. You should recognize that die alpha hydrogen is the most reactive hydrogen on an aldehyde or ketone. For A or B to be correct, the carbonyl hydrogen must be removed while the alpha hydrogen remains intact. Not likely. Notice also that since in an aldol reaction between two aldehydes the product must be an aldehyde, A, B, and C are eliminated they are ketones. 

We need a formaldehyde equivalent that is less electrophilic than formaldehyde itself and will therefore add only once to enol(ate)s. The solution is the Mannich reaction.7 Formaldehyde is combined with a secondary amine to give an iminium salt that adds 47 to the enol of the aldehyde or ketone in slightly acidic conditions to give the amino ketone (or Mannich base ) 48. If the product of the aldol reaction 50 is wanted, alkylation on nitrogen provides a good leaving group and ElcB elimination does the trick. 

We have already met this in the formation of 16 by dehydration and in the formation of 37 by stable enolate formation. A couple more examples should make the general strategy clear. The unsymmetrical ketone 110 can form an enolate on either side and at first it seems that we shall need a specific enolate to control the aldol reaction. But one product 109 cannot eliminate water while the other 111 can. Under equilibrating conditions 112 is the only product.22... 

The conjugate addition of (K)-N-methyl-N-a-methylbenzyl amide 33 to tert-butyl cinnamate 34, followed by an asymmetric aldol reaction and subsequent N-oxidation/Cope elimination afforded the -substituted homochiral Baylis-Hillman product 39 in good yield (Scheme 7) [37]. This chemistry requires the use of stoichiometric rather than catalytic amounts of the chiral base. 

Two extreme pathways may be envisaged for this transformation. One involves a pericyclic reaction followed by desilylation and elimination of methoxide from the product silyl enol ether. The alternative proceeds via a Lewis acid promoted aldol reaction to give an intermediate which cyclizes to the product dihydropyrone (Figure Si3.6). The actual pathway in an individual case is affected to a considerable degree by the nature of the Lewis acid used and lies somewhere between these two extremes. 

If the above reaction is carried out with heating, then a different product is obtained (Fig.K). This arises from elimination of a molecule of water from the Aldol reaction product. The two reasons for occurrence of this reaction are First, the product still has an... 

The elimination is even easier in acid solution and acid-catalysed aldol reactions commonly give unsaturated products instead of aldols. In this simple example with a symmetrical cyclic ketone, the enone is formed in good yield in acid or base. We shall use the acid-catalysed reaction to illustrate the mechanism. First the ketone is enolized under acid catalysis as you saw in Chapter 21. 

None of these intermediates is detected or isolated in practice—simple treatment of the ketone with acid gives the enone in good yield. A base-catalysed reaction gives the same product via the aldol-ElcB elimination mechanism. 

ElcB eliminations often follow aldol reactions and lead to cc,fkjnsaturated products. In this case, though, DABCO is a much better leaving group than tne hydroxyl group, so enolization leads to loss of DABCO in an ElcB elimination, giving the product of the reaction. DABCO is recovered unchanged, and is a catalyst. 

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