Ester formation equilibrium

Anhydrides and esters may differ in two ways. One may undergo nucleophilic addition more rapidly (kinetics), but the other may create a more favorable equilibrium constant for ester formation (thermodynamics). 

From this and the relation between the equilibrium for ester formation and the pAT of the acid, °° we estimate the free energy change for the reaction... 

Classically the base catalyst, eOEt, is introduced by adding just over one mole of sodium (as wire, or in other finely divided form) plus just a little EtOH to generate an initial small concentration of Na eOEt. Further EtOH is generated in step (1), which yields further Na eOEt with sodium, and the concentration of eOEt is thereby maintained. A whole mole is required as it is essential for the / -keto ester (114) to be converted (step 3) into its anion (115)—MeCOCH2-COzEt is more acidic than EtOH (cf. p.272)—if the overall succession of equilibria is to be displaced to the right. This is necessary because the carbanion-formation equilibrium—step (1)—lies even further over to the left than that with, for example, CH3CHO this reflects the less effective stabilisation through delocalisation in the ester carbanion (111) than in that from the aldehyde (116) ... 

One of the more difficult problems encountered in obtaining a valid assay of formic acid is that of formate ester formation. The formate is derived from the cyclic, hemi-acetal structure which is an equilibrium form of many free sugars in solution. For example, the oxidation of one of the cyclic forms of D-glucose can readily be seen to give a formate ester (as well as a C-formyl group) on the atom originally denoted as C5. It... 

Quantitative measurements of simple and enzyme-catalyzed reaction rates were under way by the 1850s. In that year Wilhelmy derived first order equations for acid-catalyzed hydrolysis of sucrose which he could follow by the inversion of rotation of plane polarized light. Berthellot (1862) derived second-order equations for the rates of ester formation and, shortly after, Harcourt observed that rates of reaction doubled for each 10 °C rise in temperature. Guldberg and Waage (1864-67) demonstrated that the equilibrium of the reaction was affected by the concentration ) of the reacting substance(s). By 1877 Arrhenius had derived the definition of the equilbrium constant for a reaction from the rate constants of the forward and backward reactions. Ostwald in 1884 showed that sucrose and ester hydrolyses were affected by H+ concentration (pH). 

In addition, the reduction of a-11 and (3-11 was s tudied in dimethyl sulfoxide as solvent, in the presence of pyridiniump-toluensulfonate. This medium makes mutarota-tion slower than the redox process.73 The two anomeric forms could reduce Cr(VI) and Cr(V) by formation of a Cr(VI) and Cr(V) ester intermediate. The equilibrium constant for this step and the rate of the redox step were different for each anomer for a-11, the equilibrium constant for ester formation is higher than for (3-11, but the redox process within this complex is faster for the latter anomer. These differences can be explained by the better chelating capacity of the 1,2-cri-diol moiety of a-11. Room-temperature CW-EPR spectra of these mixtures revealed for the a anomer several five-coordinated Cr(V)-bischelates (giso= 1.9820 [crilso 15.9xl(r4cm 1 (=47.7MHz)],... 

Since anionic ester formation proceeds with release of a proton, the reaction is also favored by the presence of proton acceptors or high pH. The observation that ester formation is favored at high pH has led some researchers to conclude that the reaction involves [B(0H)4] as reactant rather than B(0H)3 [55]. However, careful analysis of equilibrium expressions for this process leads to the conclusion that a tetrahedral reactant is not required, and other evidence supporting a tetrahedral reactant is unconvincing [56]. 

These reactions are in equilibrium at any time. When the carbonylation reaction s rate is low, side reactions, e.g. ether and ester formation, can predominate. On the other hand a high carbonylation rate results in removing the available methanol from the reaction mixture and sub-... 

Schiff s base formation occurs by condensation of the free amine base with aldehyde A in EtOAc/MeOff. The free amine base solution of glycine methyl ester in methanol is generated from the corresponding hydrochloride and triethylamine. Table 4 shows the reaction concentration profiles at 20-25°C. The Schiffs base formation is second order with respect to both the aldehyde and glycine ester. The equilibrium constant (ratio k(forward)/ k(reverse)) is calculated to be 67. 

All the steps in ester formation are reversible, but the equilibrium in the C-O bond-making and -breaking processes are not very favorable, and an excess of one reactant (usually the alcohol) or removal of one product (most often water) is required to give a good yield of ester. 

While still useful for large-scale esterification of fairly robust carboxylic acids, Fischer esterification is generally not useful in small-scale reactions because the esterification depends on an acid-catalyzed equilibrium to produce the ester. The equilibrium is usually shifted to the side of the products by adding an excess of one of the reactants—usually the alcohol—and refluxing until equilibrium is established, typically several hours. The reaction is then quenched with base to freeze the equilibrium and the ester product is separated from the excess alcohol and any unreacted acid. This separation is easily accomplished on a large scale where distillation is often used to separate the product from the by-products. For small-scale reactions where distillation is not a viable option, the separation is often difficult or tedious. Consequently Fischer esterification is not widely used for ester formation in small-scale laboratory situations. In contrast, intramolecular Fischer esterification is very effective on a small scale for the closure of hydroxy acids to lactones. Here the equilibrium is driven by tire removal of water and no other reagents are needed. Moreover the closure is favored entropically and proceeds easily. 

Various methods have been developed to shift the equilibrium between esterification and hydrolysis toward the ester formation.6 For example, as an acetyl donor, vinyl acetate and isopropenyl acetate have been used because the resulting unnecessary alcohol, vinyl or isopropenyl alcohol, spontaneously changes to the corresponding aldehyde or ketone as shown in Figure 6(c).6a... 

The kinetics of the reactions of phthalic and maleic anhydrides with Z-substituted phenols (Z = H, m-Me, p-Me, m-Cl, p-C 1, and -CN) (Scheme 8) were studied in aqueous solution at pH 8.5. Two kinetic processes well separated in time were observed. The fast process was attributed to the formation of the aryl ester in equilibrium with the anhydride and allowed the determination of the rate of nucleophilic attack of the phenol on the anhydride. From the slow kinetic process, the equilibrium constant for this reaction was determined. The Brpnsted-type plots for the nucleophilic attack of substituted phenols on the anhydrides were linear with slopes /SNuc of 0.45 and 0.56 for phthalic and maleic anhydride, respectively. The results are consistent with a mechanism involving rate-determining nucleophilic attack and also with a concerted mechanism.27... 

The combination of a carboxylic acid and an alcohol gives an ester water is eliminated. Ester formation is an equilibrium process, catalyzed by an acid catalyst. 

Ester formation is reversible howto control an equilibrium... 

Acid-catalysed ester formation and hydrolysis are the exact reverse of one another the only way we can control the reaction is by altering concentrations of reagents to drive the reaction the way we want it to go. The same principles can be used to convert to convert an ester of one alcohol into an ester of another, a process known as transesterification. It is possible, for example, to force this equilibrium to the right by distilling methanol (which has a lower boiling point than the other components of the reaction) out of the mixture,... 

In fact, acetal formation is even more difficult than ester formation while the equilibrium constant for acid-catalysed formation of ester from carboxylic acid plus alcohol is usually about 1, for... 

Co-polymerization of pentaerythritol and two other monomers—an unsaturated acid and benzene 1,3-dicarboxylic acid—gives a network of polymer chains branching out from the quaternary carbon atom at the centre of pentaerythritol. The reaction is simply ester formation by a carbonyl substitution reaction at high temperature (> 200°C). Ester formation between acids and alcohols is an equilibrium reaction but at high temperatures water is lost as steam and the equilibrium is driven over to the right. 

The same authors then introduced a second system, in which arsenate esters were used as phosphate ester mimics. Dihydroxyacetone readily reacts with arsenate and the resulting ester is accepted by FDP A as a substrate. Since the arsenate ester formation is an equilibrium reaction the desired product is released in situ and the arsenate is available again for the next catalytic cycle (Scheme 5.27) [47]. However, although only catalytic amounts of arsenate are necessary it remains a toxic metal. 

Enzymatic reactions in organic media have been a major issue in the field of biocatalysis over the last two decades. Carboxylesterases (mostly lipases) have been used in monophasic organic solution under controlled values of water activity (ajj for catalyzing ester formation the reaction equilibrium can be shifted towards ester formation by interesterification or transesterification [1]. Direct esterification is often hampered by water formation, which may increase o , thus negatively influencing the equihbrium. 

The hydrolysis of esters is accomplished by refluxing with aqueous or alcoholic alkali hydroxides. Acid-catalyzed hydrolysis is an equilibrium reaction usually favoring ester formation. High-molecular-weight esters with branching in either acid or alcohol portions are sometimes hydrolyzed with difficulty. 

Amide bond formation by direct condensation between an acid and an amine is not obvious and must overcome adverse thermodynamics (8). This dehydrative process can be achieved under forcing conditions such as high temperatures (160-180° C), which are usually incompatible with the presence of other functionalities. Contrary to the ester formation between an acid and an alcohol, which is an equilibrium, the acid and the amine undergo first an acido-basic reaction that yields the stable salt. Therefore, the acid must be activated by the attachment of a leaving group before being reacted with the amine (see Fig. 2). 

In addition to feeding of components into the reactor, if the sign on v(f) is negative, products are continuously removed, for example, reactive distillation. This is done for reactions (1) that reveal product inhibition, that is, the product slows the reaction rate, (2) that have a low equilibrium constant (removal of product does not allow equilibrium to be reached), or (3) where the product alters the reaction network that is proceeding in the reactor. A common class of reactions where product removal is necessary is ester formation where water is removed. 

For esters, all of the steps of the hydrolysis reaction are reversible, and the mechanism of ester formation is the reverse of ester hydrolysis. The course of the reaction is controlled by adjusting the reaction conditions, chiefly the choice of solvent and the concentration of water, to drive the equilibrium in the desired direction. For hydrolysis, the reaction is carried out in an excess of water for ester formation, the reaction is carried out with an excess of the alcohol component under anhydrous conditions. Frequently, an experimental set-up is designed to remove water as it is formed in order to favor ester formation. 

All of the steps in this reaction are reversible. Why, then, do the hydrolysis conditions yield the formate ester and not the starting material The key, as we saw in the preceding section with ester formation and hydrolysis, lies in the overall reaction. Water is present on the left-hand side of the equation and ethanol on the right-hand side. Thus, an excess of water would shift the equilibrium to the right, and an excess of ethanol would shift the equilibrium to the left. In fact, in order to get the reaction to go to the left, water must be removed as it is produced. Depending on the reaction conditions, the hydrolysis may proceed further to the corresponding carboxylic acid. 

Epoxides, summary of chemistry, 290 Equatorial bond, 168 Equilibrium and free energy, 36 Equivalent H s, 50 Erythro form, 82, 85 Esters, formation, 338 inorganic, 262, 272, 276 reduction, 261 Ethers, cleavage, 282 crown, 286 cyclic, 285 silyl, 286... 

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