Alkoxide oxygen, protonation

Together with a shift of the proton from the a-carbon to the alkoxide oxygen, the tertiary amine is eliminated from the addition product to yield the unsaturated product 3. Early examples of the Baylis-Hillman reaction posed the problem of low conversions and slow reaction kinetics, which could not be improved with the use of simple tertiary amines. The search for catalytically active substances led to more properly adjusted, often highly specific compounds, with shorter reaction times." Suitable catalysts are, for example, the nucleophilic, sterically less hindered bases diazabicyclo[2.2.2]octane (DABCO) 6, quinuclidin-3-one 7 and quinuclidin-3-ol (3-QDL) 8. The latter compound can stabilize the zwitterionic intermediate through hydrogen bonding. ... 

Yields of i,8-anhydro derivatives, after removal of the 3,4-0 isopropyl dene blocking group, were ft-7% for the D-raannitol series, 15-20% for the -sorbitol series, and 55-60% for the L-iditol series. Tho authors asoribed this trend to the fact that formation of these 1,6-anhydrides produced two, one, and zero axial hydroxyl substituents respectively. That seven- in preference to five or aix-mentbered anhydrides were generated was attributed to the preeenoe of the 3,4-0-iflopropylidene function. The products were formulated on the assumption that no inversion had occurred either at or CtS), and that initial attack of an 0H> ion at Ct or Ct was followed by a tautomeric proton transfer to the adjacent secondary alkoxide oxygen (Eq. 597). 

The anion then adds to the carbonyl group of a second molecule of ethanal in a manner analogous to the addition of other nucleophiles to carbonyl groups (e.g., cyanide ion, Section 16-4A). The adduct so formed, 8, rapidly adds a proton to the alkoxide oxygen to form the aldol, 3-hydroxybutanal. This last step regenerates the basic catalyst, OH ... 

The enolate A or the nitronate A, respectively, initially adds to the C=0 double bond of the aldehyde or the ketone. The primary product in both cases is an alkox-ide, D, which again contains the structural motif of a fairly strong C,H-acid, namely, of an active-methylene compound or of a nitroalkane, respectively. Hence, intermediate D is protonated at the alkoxide oxygen and the C-/3 atom is deprotonated to about the same extent as in the case of the starting materials. An OH-substituted enolate C is formed (Figures 10.45 and 10.46), which then undergoes an Elcb elimination, lead-... 

A plausible reaction mechanism that accounts for a number of important observations is simple acid-catalyzed nucleophilic substitution. A hydronium ion should readily attack the basic alkoxide oxygen to form an activated complex that decomposes to an alcohol and a silanol. Although silanols are fairly acidic, at low pH they should remain protonated. Such a mechanism implies that the reaction is readily reversible from a kinetic standpoint. However, orthosilicic acid, Si(OH)4, is apparently much more thermodynamically stable than the alkoxide, and so the reaction goes to completion. 

These observations are easily explained by another simple reaction mechanism, nucleophilic substitution of an alkoxide on silicon (12). In this case, the basic alkoxide oxygens tend to repel the nucleophile, OH, and the bulkier alkyl groups tend to crowd it. Therefore, more highly hydrolyzed silicons are more prone to attack. Because this mechanism would have a pentacoordinated silicon atom in the activated complex, hydrolysis of a polymer would be more sterically hindered than hydrolysis of a monomer. Reesterification would be much more difficult in alkaline solution than in acidic solution, because silanols are more acidic than the hydroxyl protons of alcohols and would be deprotonated and negatively charged at a pH lower than the point at which the nucleophile concentration becomes significant (ii). Thus, although hydrolysis in alkaline solution is slow, it still tends to be complete and irreversible, if extensive polymerization does not occur first. 

Acid-base complexation of with the carbonyl oxygen atom first serves to make the carbonyl group a better acceptor, and nucleophilic addition of R then produces a tetrahedral magnesium alkoxide intermediate. Protonation by addition of water or dilute aqueous acid in a separate step yields the neutral alcohol (Figure 19.6). Unlike the nucleophilic additions of water and HCN, Grignard additions are irreversible because a carbanion is too poor a leaving group to be expelled in a reversal step. 

In the first step of the mechanism for imine formation, the amine attacks the carbonyl carbon. Gain of a proton by the alkoxide ion and loss of a proton by the ammonium ion forms a neutral tetrahedral intermediate. The neutral tetrahedral intermediate, called a carbinolamine, is in equilibrium with two protonated forms. Protonation can take place on either the nitrogen or the oxygen atom. Elimination of water from the oxygen-protonated intermediate forms a protonated imine that loses a proton to yield the imine. 

This can be explained by a proton transfer from water to the alkoxide oxygen and with the subsequent product being a quaternary ammonium hydroxide solution. This solution, like any containing the hydroxide ion, is strongly basic. Thus choline is prepared as follows ... 

In an extension of this chemistry the reactions of 194 with a series of aldehydes were investigated and found to direetly yield the alkoxide-alkene chelates (244—247, Scheme 14), with no evidence of iridafurans being obtained. Protonation of the alkoxide oxygen was also effected, thus affording the respective alcohol complexes (248-251, Scheme 15), the Ir-0(H)R functionalities of which are remarkably stable toward substitution by extraneous donors, a feature attributed to the chelating nature of the ligand. It is interesting to note that in this instance the trans isomers alone are obtained, with no evidence for the cis. 

In the second step the alkoxide oxygen, because it is strongly basic, removes a proton from H—Nu or some other acid. 

PROBLEM 10.20 The first step in eq. 10.26 really involves two reactions, addition of ammonia to the carbonyl carbon to form an ammonium alkox-ide followed by a proton transfer from the nitrogen to the alkoxide oxygen. Illustrate this process with equations using the arrow-pushing formalism. The second step in eq. 10.26 also involves two steps, elimination of an alkoxide (R O ) followed by deprotonation of the hydroxyl group. Write a detailed mechanism for these steps. 

The nucleophilic addition of water to an aldehyde or ketone is slow under neutral conditions but is catalyzed by both base and acid. Under basic conditions (Figure 19.4a), the nucleophile is negatively charged (OH ) and uses a pair of its electrons to form a bond to the electrophilic carbon atom of the C=0 group. At the same time, the C=0 carbon atom rehybridizes from sp to sp and two electrons from the C=0 rr bond are pushed onto the oxygen atom, giving an alkoxide ion. Protonation of the alkoxide ion by water then yields a neutral addition product plus regenerated OH . 

Protonate the resulting alkoxide oxygen to give the final hydroxyketone product ... 

IS a two step process m which the first step is rate determining In step 1 the nucleophilic hydroxide ion attacks the carbonyl group forming a bond to carbon An alkoxide ion is the product of step 1 This alkoxide ion abstracts a proton from water m step 2 yielding the gemmal diol The second step like all other proton transfers between oxygen that we have seen is fast... 

The hydrides can also be used to form primary alcohols from either terminal or internal olefins. The olefin and hydride form an alkenyl zirconium, Cp2ZrRCl, which is oxidized to the alcohol. Protonic oxidizing agents such as peroxides and peracids form the alcohol direcdy, but dry oxygen may also be used to form the alkoxide which can be hydrolyzed (234). 

The principal difference hes in the poorer ability of amide ions to act as leaving groups, compared to alkoxides. As a result, protonation at nitrogen is required for breakdown of the tetrahedral intermediate. Also, exchange between the carbonyl oxygen and water is extensive because reversal of the tetrahedral intermediate to reactants is faster than its decomposition to products. 

Once formed, and depending on the nature of the nucleophile, the tetrahedral alkoxide intermediate can undergo either of two further reactions, as shown in Figure 2. Often, the tetrahedral alkoxide intermediate is simply protonated by water or acid to form an alcohol product. Alternatively, the tetrahedral intermediate can be protonated and expel the oxygen to form a new double bond between the carbonyl carbon and the nucleophile. We ll study both processes in detail in Chapter 19. 

Formation of C—Nu The second mode of nucleophilic addition, which often occurs with amine nucleophiles, involves elimination of oxygen and formation of a C=Nu bond. For example, aldehydes and ketones react with primary amines, RNH2, to form imines, R2C=NR. These reactions proceed through exactly the same kind of tetrahedral intermediate as that formed during hydride reduction and Grignard reaction, but the initially formed alkoxide ion is not isolated. Instead, it is protonated and then loses water to form an imine, as shown in Figure 3. 

A fairly strong base is required to deprotonate an alcohol. By removing the proton, we are forming a negative charge on an oxygen atom (an alkoxide ion). Therefore, in... 

The cationic tantalum dihydride Cp2(CO)Ta(H)2]+ reacts at room temperature with acetone to generate the alcohol complex [Cp2(C0)Ta(H01Pr)]+, which was isolated and characterized [45]. The mechanism appears to involve protonation of the ketone by the dihydride, followed by hydride transfer from the neutral hydride. The OH of the coordinated alcohol in the cationic tantalum alcohol complex can be deprotonated to produce the tantalum alkoxide complex [Cp2(C0)Ta(01Pr)]. Attempts to make the reaction catalytic by carrying out the reaction under H2 at 60 °C were unsuccessful. The strong bond between oxygen and an early transition metal such as Ta appears to preclude catalytic reactivity in this example. 

As in the reductive ring-opening, titanocene—oxygen bonds have to be protonated. Here, a titanium enolate, which is generated after reductive trapping of an enol radical, has to be protonated, in addition to a simple titanocene alkoxide. As before, 2,4,6-collidine hydrochloride constitutes a suitable acid to achieve catalytic turnover, but here zinc dust turned out to be the reductant of choice [31c], The features of the stoichiometric reaction are preserved under our conditions. Acrylates and acrylonitriles are excellent radical acceptors in these reactions. Methyl vinyl ketone did not yield the desired addition product. Under the standard reaction conditions, a-substituted acceptors are readily tolerated, but (3-substitution gives the products only in low yields. 

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