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A. Extrapolating Results From Model Radicals
Radical philicity first was encountered in this chapter in the reactions of simple organic compounds (Section I.B). The assumption was that information about the reactivity of these compounds could be extrapolated to more complex ones, specifically, carbohydrates. Experimental data supported the validity of this assumption by providing examples of reactions of carbohydrate radicals that qualitatively paralleled those of simpler radicals. For instance, the reactions in eq 2 showed the cyclohexyl radical (a carbohydrate model) to be nucleophilic because it added more rapidly to electron-deficient multiple bonds than to electron-rich ones.6 A similar, nucleophilic behavior was seen in the reactions of the carbohydrate radical 26 (Scheme 2). In another example, the reactions described in equations 4 and 5 for simple organic molecules showed that polarity matching increased the rate of hydrogen-atom abstraction. Extending polarity matching to carbohydrate radicals offered a rationale for the electrophilic radical 22 abstracting an electron-rich hydrogen atom more rapidly than an electron-deficient one (Scheme 1). There is, of course, a limitation to extrapolation of results obtained from simple radicals to radicals that are more complicated. Identifying a limitation, as is done in the following section, can be useful and can lead to increased understanding of the reactivity of the more complicated radical.
B. A Limitation on Extrapolating Alkyl Radical Reactivity to Carbohydrate Radicals
Extrapolating results from reactions of structurally simple intermediates sometimes provides the only information available for judging reactivity of more complex ones. Successful extrapolation builds confidence in the models selected, but since model systems by their very nature have limitations, such extrapolation is always subject to some uncertainty. The reaction shown in eq 9 illustrates the need for caution in projecting reactivity from a simple to a more complex radical. The data in Table 2 show that rate constants for hydrogen-atom abstraction from compounds with H–Sn, H–S, and H–Si bonds are essentially independent of whether the abstracting radical is primary, secondary, or tertiary. (Steric factors have little effect on rate constants for hydrogen abstracttion.41) Extending this reactivity pattern to compounds with H–Se bonds and knowing that the rate constant for hydrogen-atom abstraction from C6H5SeH by the primary radical 27 is 2.1 x 109 M-1s-1 at 25 oC14 leads to the prediction that the rate constant for reaction of the pyranos-1-yl radical 9 with C6H5SeH should have a similar value. It does not; the value is far smaller (3.6 x 106 M‑1s-1 at 78 oC).15
C. An Explanation for Unsuccessful Extrapolation: Loss of Transition-State Stabilization
In attempting to understand the smaller-than-expected rate constant for hydrogen-atom abstraction by the radical 9, it is useful to recall from Chapter 6 (Section IV.A.2.c) that the conformation of pyranos-1-yl radicals depends on the quasi-anomeric effect, that is, on the stabilizing interaction of the σ* orbital of the C2–O bond with the p-type orbitals on the ring oxygen atom and C-1. This effect is sufficiently powerful to cause 9 to adopt the otherwise unstable, boat conformation shown in Figure 12. Since quasi-anomeric stabilization can only exist in structurally complex systems, simple radicals are limited in their ability to model pyranos-1-yl radicals. The quasi-anomeric effect provides a basis for understanding the smaller-than-expected value for the rate constant for hydrogen-atom abstraction by the radical 9.
As the hydrogen-atom abstraction reaction shown in eq 9 (R· = 9) moves toward the transition state, the orbital interactions (Figure 12) that cause the electron delocalization that stabilizes the radical 9 are disappearing. (They are totally gone when the product is reached.) This loss of stabilization means that the energy of activation for hydrogen-atom abstraction by 9 will be greater than that for the radical 27, for which there is no comparable reduction in electron delocalization as the reaction progresses.15 Hydrogen-atom abstraction, in effect, forces an electron localization that causes a loss of stabilization for the delocalized radical 9 but not for the localized one 27. This decrease in stabilization at the transition state for reaction of a pyranos-1-yl radical reduces its rate constant for hydrogen-atom abstraction.
The proposal that transition-state loss of quasi-anomeric stabilization in pyranos-1-yl radicals is responsible for their reduced hydrogen-abstracting ability carries with it the prediction that carbohydrate radicals that are not so stabilized should have larger rate constants for hydrogen-atom abstraction. The quantitative information needed to evaluate this prediction does not exist, but there is qualitative information that supports the basic idea. As mentioned in Section V of Chapter 6, pyranos-1-yl radicals can be generated and observed in toluene or tetrahydrofuran but radicals centered on C-2, C-3, or C-4 (with no oxygen atom attached to the radical center and, hence, no quasi-anomeric stabilization possible) cannot be observed in these solvents because such radicals rapidly abstract hydrogen atoms from the solvent. (Only α-tetrahydrofuryl radicals are observed in reactions conducted in tetrahydrofuran and only benzyl radicals are detected in reactions in toluene.42)
The possibility that a loss of transition-state stabilization due to diminishing delocalization is responsible for a smaller-than-expected rate constant for hydrogen-atom abstraction by a pyranos-1-yl radical leads to the proposal that a similar loss of stabilization should have a similar effect on hydrogen-atom abstraction by other radicals. Rate constants for abstraction from C6H5SH have a bearing on this proposal. Simple primary, secondary, and tertiary radicals all have rate constants for hydrogen-atom abstraction from C6H5SH near 1 x 108 M-1s-1 at 25 oC (Table 2), but the rate constant for abstraction from this thiol by the benzyl radical is far smaller (3.13 x 105 M‑1s-1 25 oC).13 Because the benzyl radical loses resonance stabilization as hydrogen-atom abstraction takes place, it would be expected to parallel the pyranos-1-yl radical in having a smaller rate constant hydrogen-atom abstraction than the 1 x 108 M-1s-1 observed for unstabilized radicals. The fact that the rate constant for hydrogen-atom abstraction by the benzyl radical (3.13 x 105 M‑1s-1) is a substantially smaller than 1 x 108 M-1s-1 strengthens the diminishing-delocalization explanation for reduced reactivity of the pyranos-1-yl radical 9.
VII. Summary
Radicals often exhibit reactivity characteristic of either nucleophilic or electrophilic species. An electrophilic radical reacts more rapidly with an electron-rich center in a molecule, and a nucleophilic radical is more reactive toward an electron-deficient one.
A number of procedures exist for determining the philicity of a radical. These range from simple ones that involve assignment based on inspection of radical structure combined with a general knowledge of organic chemistry to complicated ones based on ab initio molecular orbital calculations. All procedures lead to the conclusion that nearly every carbohydrate radical is nucleophilic.
Most quantitative information about rates of reaction of carbohydrate radicals comes from extrapolation of data obtained from reaction of model radicals. Although this information is useful in understanding the philicity of carbohydrate radicals, it must be treated with caution when pyranos-1-yl radicals are under consideration. Simple radicals are unable to model the stereoelectronic effects that are critical to radical stability in pyranos-1-yl radicals and, consequently, do not always provide a good measure of their reactivity. | textbooks/chem/Organic_Chemistry/Radical_Reactions_of_Carbohydrates_(Binkley)/Radical_Reactions_of_Carbohydrates_I%3A_Structure_and_Reactivity_of_Carbohydrate_Radicals/07%3A_Radical_Philicity/VI._Rate_Constants_for_Hydrogen-atom_abstract.txt |
As discussed in Chapter 4, any reaction that involves radical intermediates actually consists of two or more elementary reactions. The rate constants for these elementary reactions are critical in determining the success of an overall reaction. Knowing these rate constants then is essential to understanding existing radical reactions and being able to predict new ones.
This chapter is different from the others in this book in that it contains a large number of tables. These tables consist of collections of rate constants. Because relatively few rate constants for reactions of carbohydrates have been determined, most of the values in these tables come from reactions of simpler organic compounds; thus, they serve as models for carbohydrate reactivity. These rate constants do not represent an exhaustive list of those that have been determined; rather, they are ones of interest in understanding the reactions of carbohydrates.
08: Radical Reactivity: Reaction Rate Constants
Two types of rate constants commonly are associated with radical reactions. One of these is the actual (sometimes call absolute) rate constant for a reaction, and the other is a relative rate constant, that is, a value determined by comparing the rate of one reaction to that of another.
The absolute rate constant ka for the reaction shown in eq 1 is defined mathematically in eq 2.1 Although the value of ka is expressed in terms of the rate of disappearance of AB or appearance of RAB· and the concentrations of R· and AB, rarely can ka be determined directly from this information. Because most radicals are highly reactive species that are present in a reaction mixture in concentrations typically too low to be measured accurately, direct determination of rate constants such as ka seldom is possible. When a direct determination cannot be made, sometimes an indirect one can.
On way for determining indirectly the rate constant for the reaction shown in eq 1 depends upon being able to measure the buildup of trace amounts of RAB· at different AB concentrations. Even though the actual concentration of RAB· is unknown, for some radicals it is possible to determine accurately their rate of appearance from the change in one of their properties (e.g., UV absorption). This information can provide a basis for indirectly determining the rate constant ka.2
Relative rate constants are far easier to determine than actual rate constants because relative ones can be obtained without knowing radical concentrations or making any measurements on radicals. When CD and EF are present in the same reaction mixture, eq 5 describes a relation between the rate constants k1 and k2 for the competing reactions shown in equations 3 and 4. The ratio k1/k2 is determined by the concentrations of CD and EF and their rates of disappearance.1 If, for example, the concentrations of CD and EF are equal and CD disappears ten times more rapidly than EF, the relative rate constants of 10 and 1 can be assigned to k1 and k2, respectively. Further, if the absolute rate constant is known for one of these two reactions, the relative rate constant for the other can be converted into an absolute one. More generally, if the absolute rate constant is known for one member of a group of reactions for which relative rate constants have been determined, the relative rate constants all can be converted into absolute ones.
Although measuring product ratios provides relative rate constants for reactions of a radical with two or more compounds, determining the relative rate constants for reaction of two different radicals with a single compound is a much more difficult task. There is no reliable way to run a competition experiment. Comparing the reactivity of two (or more) radicals with a particular compound usually requires determining absolute rate constants.3 | textbooks/chem/Organic_Chemistry/Radical_Reactions_of_Carbohydrates_(Binkley)/Radical_Reactions_of_Carbohydrates_I%3A_Structure_and_Reactivity_of_Carbohydrate_Radicals/08%3A_Radical_Reactivity%3A_Reaction_Rate_Constants/II._Absolute_and_.txt |
A. Atom-Transfer Reactions
Carbon-centered radicals often are generated by atom-transfer reactions. The transfer usually is of a halogen atom, but hydrogen-atom transfer also can take place. Absolute rate constants for producing carbon-centered radicals by reaction of halogenated compounds with Bu3Sn· are found in Table 1. Table 2 contains a similar set of rate constants that includes those for atom-transfer reactions involving (Me3Si)3Si· and Et3Si·. (Tables 1 and 2 also contain some group-transfer reactions.) To produce a radical selectively by atom transfer, one atom in the substrate must be more reactive than any other atom or group. A typical pair of propagation steps that selectively form a carbohydrate radical is shown in Scheme 1, where an iodine atom is transferred from a carbohydrate to a tin-centered radical.8
Examining the rate constants in Tables 1 and 2 offers insight into why iodides and bromides are so frequently used in carbon-centered radical generation. Reactions of compounds containing these atoms are so rapid that rarely is there competition in radical formation from replacement of other groups or atoms commonly present in a reacting molecule. Chlorides are substantially less reactive than iodides and bromides; consequently, chlorine atom abstraction is a less effective way for selectively generating carbon-centered radicals. (Fluorides are effectively unreactive.) Another factor favoring the use of iodides and bromides is a synthetic one. Sulfonate esters, which are easily prepared from carbohydrates, are converted readily into the corresponding iodides and bromides by nucleophilic displacement reaction.
Because the rate constants listed in Tables 1 and 2 are for reactions of organic compounds that are structurally simpler than carbohydrates, in using these rate constants for carbohydrate reactions the assumption is that the same reactive substituent will have a similar rate constant for reaction in a more complex compound. Although such an assumption is reasonable, often necessary, and usually valid, extrapolation of rate constants from simple compounds to carbohydrates needs to be treated with caution because some of the structural features that affect the reactivity of carbohydrates and carbohydrate radicals cannot be adequately accounted for in simpler systems. (Such a situation involving pyranos-1-yl radicals was discussed in Sections VI.B. and VI.C. of Chapter 7.)
B. Group-Transfer Reactions
Group transfer can be a more complicated process than atom transfer because atom transfer consists of a single elementary reaction, but group transfer often requires two such reactions. Since the halogen-atom-transfer reactions shown in Tables 1 and 2 are irreversible, for each of these reactions the rate constant for halogen-atom transfer is the same as that for carbon-centered radical formation. The situation is different for group-transfer reactions because the first step in group transfer often is reversible. In such a situation the absolute rate constant for reaction of a substrate with Bu3Sn·(Table 1) or (Me3Si)3Si· (Table 2) is larger than the rate constant for carbon-centered radical formation.
The effect on radical reactivity of a reversible reaction during group transfer can be seen by comparing three pairs of competing reactions.2 The common reaction in each of these three is between 1-bromooctane and (Me3Si)3Si· (Scheme 2). Since this reaction gives the octyl radical R· in a single, irreversible step, the rate constant for reaction of the bromide with (Me3Si)3Si· is the same as the rate constant for formation of R·. Also, since R· then abstracts a hydrogen atom from (Me3Si)3SiH, the amount of octane formed is directly related to the number of octyl radicals produced.
The first comparison experiment involves reaction of molar-equivalent amounts of 1-bromooctane, cyclohexyl isonitrile, and tris(trimethylsilyl)silane.2 A proposed mechanism for the reaction between the isonitrile and (Me3Si)3Si· is give in Scheme 3. If the addition of (Me3Si)3Si· to the isonitrile is irreversible, then the ratio of cyclohexane to octane in the product mixture would be the same as the ratio of the rate constants given the Table 2 for reactions of the isonitrile and the bromide, respectively. The information in Scheme 3 shows that these ratios are similar but not the same. One conclusion that can be drawn from this information is that the addition of (Me3Si)3Si· to cyclohexyl isonitrile is reversible. Whenever the reverse reaction takes place, it effectively reduces the rate of cyclohexane formation and causes the ratio of cyclohexane to octane to be smaller than that expected from the ratio of the rate constants kNC and kBr (Scheme 3).
The addition of the (Me3Si)3Si· to cyclohexyl phenyl selenide (Scheme 4) and cyclohexyl xanthate (Scheme 5) presents a picture with more dramatic differences.2 Competition experiments with 1-bromooctane show that the rate constants for group transfer from the selenide and the xanthate are substantially less than the rate constants shown in Table 2. This reduced reactivity can be explained by assuming that the addition of (Me3Si)3Si· to these compounds is a frequently reversed process.
C. Fragmentation Reactions
The basic structure of carbohydrates makes possible the formation of both carbon-centered and oxygen-centered (alkoxy) radicals. The reactions that characterize oxygen-centered radicals are hydrogen-atom abstraction and radical fragmentation. When an oxygen-centered radical fragments, the result is usually a radical centered on a carbon atom; thus, the alkoxy radical 4 fragments to give the ring-open, carbon-centered radical 5 (Scheme 6).9 The lack of an effective hydrogen donor in the reaction mixture allows fragmentation to take place without competition from hydrogen-atom abstraction.
In the reaction shown in Scheme 7 the oxygen-centered radical 6 and the carbon-centered radical 7 exist in a pseudoequilibrium. Both radicals abstract hydrogen atoms from Bu3SnH.10 Due to the differences in the rate constants for ring opening (kfra = 1.1 x 107 s–1M–1 at 80 oC) and ring closure (kcyc = 1.0 x 106 s–1M–1 at 80 oC), the ring-open radical 7 dominates the pseudoequilibrium, but because the rate constant for hydrogen-atom abstraction by 6 (kH = 4.7 x 108 s–1M–1 at 80 oC) is so much larger than that for hydrogen-atom abstraction by 7 (kH = 6.4 x 106 s–1M–1 at 80 oC), the major reaction product arises from hydrogen-atom abstraction by the oxygen-centered radical 6. A related reaction that also is controlled by the large rate constant for hydrogen-atom abstraction by an oxygen-centered radical is pictured in Scheme 8, where abstraction by the alkoxy radical 9 is responsible for the only product formed.11 There is no evidence for competing fragmentation of 9 leading to ring opening; in particular, no ring-open product is formed and no epimerization takes place at the hydroxyl-bearing carbon atom. (Epimerization would be expected if a ring opening took place that was followed by rapid ring closure.)
D. Electron-Transfer Reactions
Dissociative electron transfer takes place when a compound containing a reactive atom or group accepts an electron and undergoes fragmentation (Scheme 9). Electron capture can be extremely rapid if an electron is free in solution; thus, the rate constant for capture of a solvated electron by the nucleoside 10 is 1.6 x 1010 M–1s–1 at 22 oC.12,13
Radical formation by electron transfer also can take place by reaction between transition-metal complexes such as (NH4)2Ce(NO3)6, Mn(OAc)3, SmI2, and Cp2TiCl and carbohydrate derivatives that include iodides, bromides, and sulfones; for example, complexes involving samarium(II) iodide frequently are electron donors in reactions of carbohydrates (Scheme 10). A common reaction for SmI2 is a second electron transfer to the initially formed radical R· to produce an organosamarium compound (Scheme 10). This second electron transfer is fast enough that it can limit the ability of R· to undergo radical transforming reactions such as cyclization and group migration.
Reactions involving SmI2 typically are conducted in the presence of hexamethylphosphoramide (HMPA), a compound that complexes with SmI2 and increases its ability to donate an electron. Greater electron-donating ability not only increases the rate constant for formation of the radical R· but it also increases the rate at which this radical reacts with a second molecule of SmI2. The data in Table 3 show that when the 5-hexenyl radical reacts with SmI2, the rate constants for reaction increase from 5 x 105 M–1s–1 to 6.8 x 106 M–1 s–1 at 25 oC as the amount of added HMPA increases.14 The magnitude of these rate constants is such that if a radical is to do anything other than simple combination with a molecule of SmI2, this “other reaction” must be rapid. An example of a reaction of a radical that does take place more rapidly than combination with SmI2 is the cyclization shown in Scheme 11.15,16 (Chapter 20 in Volume II contains further information about and discussion of the reactions of carbohydrate derivatives with SmI2.)
IV. Transformatio
Although every propagation step in a radical chain reaction involves conversion of one radical into another, the potential for a synthetically useful reaction usually hinges on the transformation of a carbon-centered radical. Whether or not a transformation takes place depends on the relative rates for transforming and competing reactions. Because Bu3SnH or (Me3Si)3SiH often is present in a reaction mixture to provide a chain-carrying radical as well as to serve as a hydrogen donor after radical transformation has taken place, a common competing reaction for group migration, radical cyclization, and radical addition is hydrogen-atom abstraction before transformation occurs.
V. Chain Collapse
Chain collapse occurs when an intermediate radical undergoes a chain-stopping reaction more rapidly than the propagation reaction in which this radical is participating.61 One cause of collapse is a chain-terminating step (e.g., radical dimerization) whose rate exceeds the rate of chain propagation. A second type of chain collapse occurs when an intermediate radical in a propagation sequence becomes part of a faster step in a propagation sequence leading to a different product. In such a situation chain collapse is caused by chain shift to the new reaction sequence.
VI. Summary
Generating a carbon-centered radical represents the beginning point for most radical reactions of carbohydrates. The identity of the radical formed is determined by the rate constant for atom or group transfer from the carbohydrate derivative to a radical that usually is centered on a tin or silicon atom. Compounds that contain iodine or bromine atoms are attractive starting materials for radical reactions because the rate constants for transfer of these atoms to tin- or silicon-centered radicals are quite large. Once a carbon-centered radical has formed, most reactions of importance include a radical transforming step (e.g., addition of a radical to a multiple bond or radical cyclization). Rate constants for radical transformation must be large enough that the desired reaction can take place before a competing process, often hydrogen-atom abstraction, intervenes. If a propagation step in a reaction is slower than a chain-terminating reaction, the reaction will undergo chain collapse. | textbooks/chem/Organic_Chemistry/Radical_Reactions_of_Carbohydrates_(Binkley)/Radical_Reactions_of_Carbohydrates_I%3A_Structure_and_Reactivity_of_Carbohydrate_Radicals/08%3A_Radical_Reactivity%3A_Reaction_Rate_Constants/III._Generation_o.txt |
Chemoselectivity is a term that describes the ability of a reagent or intermediate to react with one group or atom in a molecule in preference to another group or atom present in the same molecule. Since most carbohydrate radicals trace their beginnings to reactions involving either a tin-centered [usually Bu3Sn·] or a silicon-centered [usually (Me3Si)3Si·] radical, the chemoselectivity in reactions of these radicals plays a central role in carbohydrate radical formation. An example of a chemoselective reaction is found in Scheme 1, where the tri-n-butyltin radical abstracts the O-thiocarbonyl group from the thioglycoside 1 while the potentially reactive ethylthio group remains in place.1
Chemoselective reaction also can occur when a carbohydrate radical reacts with another molecule present in the reaction mixture. Such a process is shown in Scheme 2, where the pyranos-1-yl radical 2 adds to the C–C double bond in 3 rather that reacting with the chlorine atom or phenylseleno group also present in this molecule (3).2
09: Chemoselectivity
The best method for determining the chemoselectivity of two groups in a particular reaction is to compare their reactivity in a substrate containing both. Since this type of testing rarely has taken place, another approach is needed if chemoselectivity in a particular reaction is to be established from existing data. One alternative is to construct an order of reactivity based on rates of reaction of compounds that each contains a single substituent. Such an order can be established from information that exists about rates of reaction of Bu3Sn· and (Me3Si)3Si· with compounds that have a single reactive substituent.
III. Carbon-Centered Radicals
Although the primary reason for reacting carbohydrates with (Me3Si)3Si· or Bu3Sn· is to generate carbon-centered radicals, the chemoselectivity of the reactions of carbohydrate radicals formed in this way also is a matter of importance. Nearly all carbon-centered radicals derived from carbohydrates are nucleophilic and, consequently, add more readily to carbon–carbon multiple bonds that are electron deficient than to ones that are electron rich. (The rationale behind this selectivity is discussed in Sections IV.A.1. and IV.B.1. of Chapter 7.) The cyclization reactions shown in Scheme 10 illustrate the influence of an electronegative substituent on the chemoselectivity of addition of a nucleophilic radical to a compound with two double bonds.36,37 Reaction occurs at the double bond external to the ring when an electron-withdrawing substituent (R = CO2Et) is present, but if the substituent is not electron-withdrawing (R=CH2OTBS), addition is exclusively at the endocyclic double bond. (The cis-trans isomerization that takes place during formation of 9 and 10 results from reversible addition of Bu3Sn· to the double bond external to the ring.)
A variety of groups (e.g., acetal, acetoxy, carbonyl, cyano, hydroxy, and silyloxy groups) can be present in a reactant molecule but remain unchanged during carbon-centered radical addition to a carbon–carbon multiple bond. Carbon-centered radicals also participate less readily in substitution reactions with normally reactive substituents than do tin- and silicon-centered radicals; for example, Tables 1, 2, and 5 in Chapter 8 show the absolute rate constants for reaction of tert-butyl bromide with Bu3Sn·, (Me3Si)3Si·, and CH3(CH2)6CH2· are 1.4 x 108 M-1s-1, 1.2 x 108 M-1s-1, 4.6 x 103 M-1s-1, respectively. This reluctance to become involved in substitution reactions is illustrated in Scheme 2 where the radical 2 does not react with the chlorine atom or phenylseleno group in 3 but rather adds to the C–C double bond in this molecule.2
IV. Summary
Chemoselectivity refers to ability of a reagent or intermediate (e.g., a free radical) to react with one group in a molecule in preference to a different, but potentially reactive, group present in the same molecule. Since most carbohydrate radicals trace their beginnings to reactions involving either the tri-n-butyltin [Bu3Sn·] or tris(trimethylsilyl)silyl [(Me3Si)3Si·] radical, chemoselectivity in the reactions of these radicals plays a central role in carbohydrate radical formation. In many reactions a second opportunity for chemoselectivity arises when an initially formed, carbon-centered radical reacts selectively with another molecule present in solution.
The tri-n-butyltin radical adds to carbon–carbon, carbon–oxygen, and carbon–sulfur multiple bonds in a reversible fashion; consequently, for chemoselective reaction to occur, the reverse reaction must be blocked in some manner. Preventing reversal of radical addition usually is achieved by hydrogen-atom abstraction, addition to a multiple bond, or fragmentation of the adduct radical. Silicon–carbon bonds tend to be stronger than tin–carbon bonds so addition of some silyl radicals to unsaturated compounds is not reversible at normal reaction temperatures. The tris(trimethylsilyl)silyl radical, however, does add reversibly to unsaturated compounds.
Carbon-centered radicals tend to be quite chemoselective intermediates. They add readily to electron-deficient, carbon–carbon, multiple bonds but are less reactive in group and atom replacement reactions than (Me3Si)3Si· and Bu3Sn· | textbooks/chem/Organic_Chemistry/Radical_Reactions_of_Carbohydrates_(Binkley)/Radical_Reactions_of_Carbohydrates_I%3A_Structure_and_Reactivity_of_Carbohydrate_Radicals/09%3A_Chemoselectivity/II._Formation_of_Carbon-Centered_Radicals.txt |
A. Definitions of Regiospecific and Regioselective Reactions
When the terms regioselective and regiospecific were first introduced into organic chemistry, they were defined in the following way: “If a reaction proceeds without skeletal rearrangement to give exclusively (within experimental error) one of two or more possible isomers, it is called regiospecific. If there is a significant preponderance of one isomer formed, it is said to be regioselective.1" Since a regiospecific reaction can be viewed as a special type of regioselective reaction (i.e., one that is totally selective), the term regioselective can be used to describe any reaction that produces one structural isomer in greater abundance than another.
B. Regioselectivity in Radical Reactions of Carbohydrates
Among carbohydrates there are several types of radical reaction for which regioselectivity is an important consideration. The first of these (discussed in Section II) is the addition of a radical to the multiple bond in an unsaturated compound. An example of reaction of this type is found in Scheme 1, where addition of the pyranos-1-yl radical 1 takes place only at the unsubstituted carbon atom of the double bond in acrylonitrile.2,3 A second type of regioselective reaction (described in Section III) is a variation on this addition process that occurs when the radical center and the multiple bond are part of the same molecule. In the reaction shown in Scheme 2, for example, regioselectivity arises because there is a choice between producing a five-membered or six-membered ring.4 A third type of regioselective reaction, one involving β-fragmentation, is discussed in Section IV. An example is given in Scheme 3 where the oxygen-centered radical 4 undergoes ring opening to generate the carbon-centered radical 5 rather than the oxygen-centered radical 6.5 An example of the final type of regioselective radical reaction is found in Scheme 4, where abstraction of H-5 from the pentaacetate 7 takes place even though there are other hydrogen atoms present in 7 that could have been abstracted to form isomeric products.6 This type of regioselectivity, sometimes referred to as site-selectivity, is described in Section V.
II. Intermolecular Addition Reactions
A. General Reaction Equation
A useful terminology for describing radical addition reactions is given in eq 1. According to this description, when a carbon-centered radical reacts with a carbon–carbon double bond, it adds to the β-carbon atom and creates a new radical center on the α-carbon atom. The letters X, Y, and Z in eq 1 represent substituents attached to the three carbon atoms directly involved in the reaction.
B. Reaction at the Less-Substituted Carbon Atom
A characteristic of radical addition reactions is that a carbon-centered radical adds regioselectively to the less-substituted atom in a C–C multiple bond.7–10 The reaction shown in Scheme 1 provides a typical example. Other reactions involving double bonds with different substituents (eq 2)11 and double bonds with more than one substituent (eq 3)12.13 exhibit similar regioselectivity. Explaining regioselectivity in addition reactions begins by noting that they usually are not reversible;14 therefore, information about transition-state structures is critical to understanding the selectivity in these kinetically controlled reactions.
C. Transition-State Structure
The structure for the transition state in a radical addition reaction, as determined from molecular-orbital calculations, is shown in Figure 1.8 Several aspects of this structure affect reaction regioselectivity. The first is that the structure is unsymmetrical.7,8 An unsymmetrical transition state requires that radical addition to each carbon of the multiple bond represents a distinct reaction pathway; there is no common intermediate. Also, partial σ‑bond formation between the α-carbon atom and the incoming, carbon-centered radical causes the groups attached to each of these atoms to assume a decidedly pyramidal arrangement; thus, reaction causes the groups attached to each center to move closer together.
D. Factors Controlling Regioselectivity
The unsymmetrical nature of the transition state structure shown in Figure 1 requires that addition to each carbon atom of an unsymmetrically substituted double bond has a different rate constant for reaction. Understanding regioselectivity in addition reactions then depends upon correctly analyzing the factors controlling these two rate constants. “The temperature dependence of the rate constants is well described by the Arrhenius equation k = Aexp(-Ea /RT). Thus, at a given temperature, the rate variations with radical and substrate substitution can be caused by variations in the frequency factor (A) and/or the activation energy (Ea). For polyatomic radicals, the frequency factors span a narrow range... Hence, the large variation in the rate constants is mainly because of variations in the activation energy”.8 The major factors determining activation energy [bond strengths, steric effects, stereoelectronic effects, and polar effects] are then the ones that need to be considered in determining reaction regioselectivity.8
1. Bond Strengths
A characteristic of many reactions that are similar in nature is that their energies of activation (Ea) can be determined from the Evans–Polanyi relation (eq 4).8,10 (The Evans–Polanyi relation is discussed in Section I.A. of Chapter 7.) In such situations calculating these energies depends upon determining reaction enthalpies (Hr) and establishing values for the two constants in eq 4. For the addition of carbon-centered radicals to C–C double bonds the values for the experimentally determined constants are C=50 kJmol-1 and α=0.25, when Ea and Hr are expressed in kJmol-1.8,15 The number 0.25 for the proportionality constant α means that the enthalpy change, which depends on the difference in the strengths of the bonds being broken and formed, needs to be large for it to have a significant impact on the energy of activation for the reaction. The 0.25 value for α is reasonable for a reaction with an early transition state.
2. Steric Effects
Rehybridization of the β-carbon atom from sp2 to sp3 takes place during radical addition (eq 1). The necessary repositioning of groups that this rehybridization requires forces them closer together (i.e., causes group compression) as reaction proceeds. Any resistance to group compression caused by steric hindrance raises the energy required to reach the transition state for a reaction.8–10 The transition state, therefore, becomes energetically more difficult to attain as the steric size of any of the groups attached to the β‑carbon atom increases. A similar steric compression of the groups attached to the carbon atom bearing the radical center in the adding radical also takes place, but the effect should be smaller because a typical radical center has a structure that already is at an intermediate stage between sp2 and sp3 hybridization.8–10
In addition to group compression, steric interactions at the transition state also arise between groups attached to the β-carbon atom and those bonded to the adding radical (Figure 1). Experimental support for significant interaction comes from the finding that the rate constants for radical addition to the β-carbon atom of an alkene change dramatically when sterically demanding groups are introduced on this atom.7,16 In the reactions represented in eq 5 increasing the steric size of the Y group significantly decreases the rate constant for β addition.7 While it may be difficult to decide how much rate constant reduction is attributable to group compression and how much to interaction between groups on the β-carbon atom and the incoming radical, the relative rate constants shown in eq 5 leave little doubt that steric effects play a major role in determining the rates of radical addition reactions.7
Since the separation at the transition state between the adding radical and the α-carbon atom in an addition reaction is considerable (Figure 1), it is reasonable to expect that any steric hindrance involving α-substituents should be small.7,16 The relative rate constants shown in eq 6 support this expectation because a dramatic change in the steric size of groups attached to the α‑carbon atom has only a small effect on the value of these constants; the largest and the smallest differ only by a factor of 4.2.7
Steric effects have a more important role in determining addition-reaction regioselectivity than do the strengths of the bonds being broken or formed. The reason for this situation can be traced to the nature of the addition process. In the competing reactions that determine regioselectivity [i.e., addition to either the α or β carbon atom in a multiple bond of an unsaturated compound] the same number and types of bonds are being broken and formed; consequently, there should be little difference in activation energies for these two reactions based on bond strengths alone.
Although the primary role of steric effects in determining regioselectivity in radical addition reactions is clear, these effects are not always the sole determining factor. It would be difficult, for example, to explain preferential addition to C-2 in the glycal 10 (Scheme 5) on the basis of steric effects alone because C-2 is, if anything, more hindered than C-1.17–19 Clearly, another factor also affects regioselectivity in reactions of this type.
3. Polar Effects
Polar effects are influences on reactivity caused by unequal electron distribution within a molecule or reactive intermediate. In radical addition reactions these effects can originate with substituent groups and can be transmitted to the reacting atoms either through bonds or through space. Polar effects also can arise from electron delocalization that produces unequal electron distribution.
The data shown in eq 7 illustrate the importance that polar effects have on radical addition reactions.7 These data describe the relative rate constants for addition of the nucleophilic cyclohexyl radical (C6H11·) to substituted α,β-unsaturated esters. The rate constant is large when a strongly electron-withdrawing substituent (e.g., CN, CO2Me) is attached to the α‑carbon atom in one of these unsaturated esters. Electron withdrawal from the double bond by either CN or CO2Me is due primarily to delocalization that shifts electron density to one of these the α-substituents. In the case where the α‑substituent is a methoxycarbonyl group (Z = CO2Me), the electron-density shift can be seen in the contributing resonance structures shown in Figure 2. Although the polar effects being described are those that exist in the reactants, the rate constants in eq 7 support the idea that these effects remain significant at the transition state.
Polar effects not only explain the difference in rate constants for the reactions shown in eq 7 but they also rationalize the regioselectivity of these reactions. The resonance hybrid pictured in Figure 2 indicates a reduced electron density at the β-carbon atom in the carbon–carbon double bond of the ester; consequently, this atom represents a point of attraction for a nucleophilic radical. In such a situation regioselective, β-carbon-atom addition can be expected. An example of this type of addition is shown in Scheme 6 where the nucleophilic carbohydrate radical 13 adds regioselectively to the β-carbon atom of the α,β-unsaturated ketone 12.20
Addition of a nucleophilic radical to an electron-rich double bond is too slow to compete with other radical reactions, but if the radical is electrophilic, addition takes place. The dimethylmalonyl radical 11, for example, adds to the electron-rich double bond in the D-glucal 10 (Scheme 5).17–19 As the resonance hybrid pictured in Figure 3 indicates, C-2 in 10 has greater electron density than C-1; thus, the electrophilic radical 11 not only adds to the double bond in 10 but it does so regioselectivity at C-2 (Scheme 5).
Since steric and polar effects often favor formation of the same product in a radical addition reaction (i.e., that from addition to the least substituted carbon atom in the double bond of the unsaturated reactant), it is sometimes difficult to determine the relative contribution of each effect to the regioselectivity of a reaction. A series of experiments designed to test these contributions is shown in eq 8.7 The first experiment involves addition of the cyclohexyl radical to methyl acrylate (eq 8, R = H). In this reaction both steric and polar effects favor addition of the nucleophilic cyclohexyl radical to the less substituted carbon atom in the carbon–carbon double bond, but as the R group becomes sterically larger, the regioselectivity of the reaction decreases. For the sterically largest R group the favored direction of addition actually changes. The message here is that steric effects can overwhelm polar effects in establishing reaction regioselectivity, but a sterically quite demanding group (e. g., a t-butyl group) is necessary to overcome a strong polar effect.
Another indication of the significance of polar effects in radical addition reactions can be seen by returning to the Evans-Polanyi relation (eq 4). This relation applies to radical addition reactions in which polar factors are not important. For reactions where polar factors are important, energies of activation are lower than those calculated from eq 4. In such situations a modified equation (eq 9), one including the multiplicative terms Fn and Fe, reflecting nucleophilic and electrophilic polar effects, respectively, is more accurate.8,15
4. Frontier-Orbital Interactions
Because radical addition reactions have early transition states,7 frontier-orbital interactions are able to provide an alternative approach for explaining reaction regioselectivity. The first step in this approach is identifying the frontier orbitals in the reaction of interest; for example, in the addition of the dimethylmalonyl radical 11 to the electron-rich double bond in the D-glucal 10 (Scheme 5), the primary interaction is between the SOMO of 11 and the HOMO of 10 (Figure 10 in Chapter 7). Identifying the frontier-orbital interactions in a reaction does not, by itself, explain reaction regioselectivity, but orbital identification is a critical first step for such understanding because from frontier orbitals come the atomic-orbital coefficients that form the basis for explaining regioselectivity.
Atomic orbital coefficients are valuable in determining the regioselectivity of a reaction with an early transition state because the rate constant for the bond-forming reaction between two atoms in such a reaction depends to a large extent on the magnitude of the coefficients in their interacting frontier orbitals.17,18 In the reaction pictured in Scheme 5 the most effective bonding is between the radical 11 and C-2 in the D-glucal 10 because the atomic orbital coefficient at C-2 for the HOMO in 10 is larger than that at C‑1 (Figure 4);21 consequently, regioselective addition to C-2 is favored.17,18
Frontier-orbital interactions also explain regioselectivity in the addition reaction shown in Scheme 6, where a nucleophilic radical (13) is adding to an electron-deficient double bond.20 Addition of the radical 13 to C‑4, rather than C-3, in the α,β-unsaturated ketone 12 cannot be explained by steric effects, but frontier-orbital interactions do provide a basis for understanding the observed regioselectivity. The most important interaction in this case is between the SOMO of 13 and the LUMO of 12. (A justification for this being the primary, frontier-orbital interaction is given in Section IV.B.1 of Chapter 7) For a LUMO such as that in 12 the largest atomic orbital coefficient is associated with the p orbital at C-4 (Figure 4);21 consequently, regioselective addition to C-4 is favored.22
5. Adduct-Radical Stabilization
Adduct-radical stabilization as a possibility for explaining regioselective addition of a carbon-centered radical to a multiple bond is not highly regarded because the exothermic nature of and probable early transition state for radical addition reactions argue against significant, transition-state stabilization due to the developing radical. Evidence from study of model compounds that is cited in support of this point of view is that the cyclohexyl radical adds more rapidly to acrolein and acrylonitrile than to styrene (eq 10),7,23 even though a phenyl group is more effective at stabilizing a radical center than is a carbonyl or cyano group.7,24 This information indicates that adduct-radical stabilization is less important than polar effects at the transition state for an addition reaction; thus, polar effects are primarily responsible for the differences in reactivity of the unsaturated compounds, differences such as those shown in eq 10.7 (As discussed in Section II.D.2 and seen in eq 6, steric hindrance from the α substituents used in the reactions shown in eq 10 should be inconsequential.) Since reactions between electron-deficient alkenes and nucleophilic radicals are stabilized at the transition state by polar effects, these effects could mask less important, adduct-radical stabilization. A better test of the importance of adduct-radical stabilization on regioselective addition would be one in which polar effects could not be the determining factor.
The addition reactions shown in equations 11 and 12 are ones for which polar effects should be minimal.10 The similarity in relative rates for the two reactions indicates that adduct-radical stabilization is inconsequential at the transition state. These reactions also underscore the difficulty in eliminating completely the influence of polar effects when comparing radical reactions. The slightly reduced rate for the reaction shown in eq 12, when compared to that in eq 11, could be due to the effect of the weakly electron-donating methyl group in propene reducing to a small extent the rate of addition of a nucleophilic radical to a slightly more electron-rich double bond. (As mentioned in Chapter 7, Sections III.C. and III.E., there is not complete agreement about the nucleophilicity of the methyl radical.)
The reaction shown in eq 13 supports the idea that the stability of the developing radical can be a factor in reducing transition-state energy.10 The greater rate for this reaction, when compared to those shown in equations 11 and 12, can be explained by resonance stabilization in the developing radical contributing significantly to transition-state stabilization. The limited data in equations 11-13 are consistent with the idea that adduct-radical stability is only a factor in radical addition reactions when such stabilization is considerable. Once again, however, polar effects cloud this interpretation. 1,3-Butadiene can be viewed as a molecule in which each double bond has an ethenyl substituent attached. Such a substituent should be electron-withdrawing or, at least, less electron-donating than a methyl group; consequently, the double bonds in 1,3-butadiene should be more reactive toward the methyl radical than is the double bond in propene. This difference could explain, at least in part, the difference in relative rates for the reactions shown in eq 12 and eq 13. | textbooks/chem/Organic_Chemistry/Radical_Reactions_of_Carbohydrates_(Binkley)/Radical_Reactions_of_Carbohydrates_I%3A_Structure_and_Reactivity_of_Carbohydrate_Radicals/10%3A_Regioselectivity/I._Introduction.txt |
In addition to describing cyclization reactions by the size of the ring produced, the terms exo and endo indicate the way in which the ring is formed. The meaning of these terms is illustrated in the reactions shown is Scheme 8. When the exo/endo terminology is used to describe ring formation from reaction of the 5-hexenyl radical, the five-membered ring is seen as arising from exo closure and the six-membered one from endo closure (Scheme 7).
2. Transition-State Structure
Transition-state structures for radical addition and radical cyclization are given in Figure 6 in a general form. For cyclization reactions not only ring size but also ring conformation affect transition-state energy; thus, both chair-like25,27 and boat-like27,30 structures are possible during five-membered ring formation. For the unsubstituted 5-hexenyl radical the chair-like transition state leading to a five-membered ring is calculated to be lower in energy, but only slightly so, than the boat-like transition state (Scheme 9).27 (The “flagpole” interactions that contribute to making the boat conformation of cyclohexane much less stable than the chair conformation are less severe in the boat-like transition state for radical cyclization.) Both transition states (boat-like and chair-like) leading to a five-membered ring (Scheme 9) are calculated to be lower in energy than any transition states leading to a six-membered ring. These calculations match well the experimental observation that cyclization of the 5-hexenyl radical gives a five-membered ring in a highly regioselective fashion (eq 14, R = H).25,31
3. Altering Normal Regioselectivity
a. Steric Interactions and Adduct-Radical Stability
Although ring size is the primary factor affecting regioselectivity in cyclization reactions, other factors sometimes have a modifying effect; for example, in the reaction of the 5-methyl-5-hexenyl radical the presence of the methyl group increases the amount of six-membered-ring formation (eq 14, R = CH3).25,27 In this reaction steric effects and adduct-radical stability both favor a six-membered ring. The transition state in this reaction presumably is reached late enough that either steric effects or adduct-radical stability or both have a substantial impact on regioselectivity. Predicting when the transition state in this type of reaction will be early enough to cause highly regioselective, five-membered-ring formation is not easy. In the reaction shown in Scheme 10, where steric interactions and adduct-radical stability favor six-membered-ring formation at least as much as they do in the reaction shown in eq 14 (R=CH3), only a product with a five-membered ring forms.32
b. Thermodynamic Control
Although kinetically controlled reaction is the norm in radical cyclization, thermodynamic control is observed in the reaction shown in eq 15 where the substrate is an unsaturated silyl ether and the hydrogen donor (Bu3SnH) is maintained at a low level.33 When this reaction is conducted with a high Bu3SnH concentration, kinetically controlled, five-membered ring formation is the major reaction pathway. An explanation for this dependence on hydrogen-donor concentration begins with the radical 18 cyclizing to form 19, a radical with a new five-membered ring (Scheme 11). If the concentration of Bu3SnH is high, hydrogen-atom abstraction rapidly completes the reaction, but if the donor concentration is low, rearrangement to the more stable radical 20, via the transition state 21, takes place before hydrogen-atom abstraction can occur.34 Hydrogen-atom abstraction by 20 then gives the thermodynamically favored product. An alternative mechanism for this reaction, also shown in Scheme 11, is that ring opening of 19 produces a silicon-centered radical that undergoes ring closure to give the intermediate radical 20.35
c. Reversal Due to Stereochemistry
One situation where six-membered ring formation is favored consistently over reaction producing a five-membered ring is when cyclization would produce a pair of trans-fused, five-membered rings. Reactions of iodides 22 and 23 illustrate the effect that stereochemistry can have on radical cyclization. The acyclic iodide 22 undergoes an expected cyclization to give a five-membered ring (Scheme 12),36 but reaction of the iodide 23 forms a six-membered ring (Scheme 13).37 Since a trans fusion between two five-membered rings would produce the highly strained radical 25, the stereochemistry of the radical 24 dictates the regioselectivity of the cyclization reaction. Six-membered-ring formation also occurs in the reaction shown in Scheme 1438, again, because the other option would force the formation of trans-fused, five-membered rings.
IV. β-Fragmentation Reactions
β-Fragmentation is an elementary reaction that exhibits regioselectivity in ring opening and in radical expulsion. Regioselective ring opening occurs when a radical centered on an atom attached to a ring preferentially fragments one of the ring bonds. Regioselective radical expulsion takes place when one of the bonds to an atom β-related to a radical center preferentially cleaves to generate two fragments, one a new radical and the other an unsaturated compound.
V. Site-Selective Reactions
The regioselectivity discussed thus far has involved reactions in which a compound with a single functional group generates products that include two or more structural isomers. The term regioselectivity also can be used to describe the preference for reaction of a particular atom or group in a molecule that contains at least one other atom or group of the same type. Regioselectivity of this sort is sometimes referred to as site selectivity. An example of a site-selective reaction is shown in Scheme 4, where H‑5 is abstracted even though there are other hydrogen atoms present in the molecule that potentially could have been abstracted.6
Although there are a large number of radical reactions in carbohydrate chemistry that involve group and atom replacement, only for hydrogen-atom abstraction is regioselectivity a common consideration. Nearly all carbohydrates have the hydrogen atoms necessary to make site-selective abstraction conceivable, but few carbohydrates have the two or more other groups or atoms required to make selective reaction of one of these groups (or atoms) a possibility. Hydrogen-atom abstraction, therefore, provides the pool from which most site-selective reactions are drawn.
VI. Summary
Intramolecular hydrogen-atom abstraction often involves reaction of an oxygen-centered radical. Since the most stable transition state for internal hydrogen-atom transfer nearly always has a six-membered ring, the 1,5-hydrogen migration (1,5-radical translocation) that occurs is a highly regioselective reaction. Only the most reactive carbon-centered radicals (vinylic and primary) consistently are able to abstract hydrogen atoms from carbon–hydrogen bonds. Such abstraction usually is an intramolecular reaction.
Site selectivity occurs in reactions of compounds with groups and atoms other than hydrogen, but such reactions are rare because few carbohydrates contain two or more of the same reactive groups (or atoms). Compounds with two of the same O-thiocarbonyl groups or two isocyano groups are known to react selectively with the tri-n-butyltin radical. In such reactions significant selectivity exists if there is a choice between forming a primary or a secondary radical. | textbooks/chem/Organic_Chemistry/Radical_Reactions_of_Carbohydrates_(Binkley)/Radical_Reactions_of_Carbohydrates_I%3A_Structure_and_Reactivity_of_Carbohydrate_Radicals/10%3A_Regioselectivity/III._Intramolecular_Addition_%28Cyclization%29.txt |
A. Definitions
Stereoselectivity is “the preferential formation of one stereoisomer over another in a chemical reaction”.1 This selectivity can be divided into diastereoselectivity and enantioselectivity. “Enantioselectivity in a reaction is either the preferential formation of one enantiomer of the product over the other or the preferential reaction of one enantiomer of the (usually racemic) starting material over the other...Diastereoselectivity is the preferential formation in a reaction of one diastereoisomer of the product over others.”2 Although diastereoselectivity almost always refers to product formation, it also can apply to preferential consumption of one diastereomer.3
B. Factors Affecting Stereoselectivity
Stereoselectivity in radical reactions is determined by a combination of factors that includes steric, stereoelectronic, conformational, torsional, and configurational effects as well as reaction temperature.4 Each of these effects can be linked to a particular aspect of structure. Steric effects are the repulsive interactions that develop between closely approaching species (e.g., a neutral molecule and a free radical) or between two groups within the same structure. Stereoelectronic effects are geometry-dependent, orbital interactions that favor formation or consumption of one stereoisomer over another. Conformational effects are differences in stereoselectivity due to differences in the population of various conformers. Torsional effects are the destabilizing interactions that develop as electrons in bonds on adjacent atoms move closer to each other. Finally, configurational effects in radical reactions are differences in stereoselectivity due to pyramidal radicals that undergo reaction faster than inversion of configuration.
Although stereoselectivity in a reaction often results from a combination of the effects just described, it is possible to identify three important situations where a particular effect appears to be dominant. First, in addition and abstraction reactions, where the radical center is not adjacent to a ring oxygen atom, the greater role of steric effects causes reaction to occur along the least-hindered pathway. When a radical is centered on an atom adjacent to a ring oxygen atom (as occurs in pyranos-1-yl and furanos-1-yl radicals) orbital interactions become the factor most frequently determining stereoselectivity. Finally, in reactions that form new five- and six-membered rings, stereoselectivity usually is determined by maximizing stability of a chair-like or boat-like transition state.
II. Minimizing Steric Interactions: The Lea
As mentioned in the previous section, stereoselectivity in reactions of carbohydrate radicals depends on the location of the radical center. When a radical is centered on a carbon atom adjacent to a ring oxygen atom a combination of stereoelectronic, conformational, and steric effects determines stereoselectivity, but reactions of radicals centered on other carbon atoms are controlled primarily by steric effects. For the latter group the major stereoisomer in a bimolecular reaction is the one produced by a molecule and a radical approaching each other along the least-hindered pathway.
III. Maximizing Transition-State Stabilizatio
A. Radical Formation
The reactions of the glycosyl chlorides 23 and 24 provide examples of stereoselectivity in pyranos-1-yl radical formation that depend on orbital interactions (Schemes 9 and 10).25,26 The difference in their reaction rates is linked to the orientation of the C2–O bond in each of these stereoisomers. The D-mannopyranosyl chloride 23 reacts with tri-n-butyltin hydride 7.8 times more rapidly than does the epimeric D-glucopyranosyl chloride 24. For 23 the C-2 acetoxy group has an orientation that allows transition-state stabilization by the orbital interactions shown in Scheme 9. These are the same interactions associated with the quasi-anomeric effect. (The quasi-anomeric effect, discussed in Section V of Chapter 6, provides an explanation for the conformations adopted by pyranos-1-yl radicals.27,28) Since these orbital interactions are developing at the transition state, they should be responsible, at least in part, for the greater reactivity of 23 when compared to 24. Such stabilization is minimal for reaction of 24 because the σ* orbital associated with the C2–O bond does not have the proper, transition-state orientation to assist significantly in stabilization (Scheme 10).
A critical assumption about the reactions shown in Schemes 9 and 10 is that the first step (chlorine-atom abstraction by the tri-n-butyltin radical), rather than the second step (hydrogen-atom abstraction from tri-n-butyltin hydride by the carbohydrate radical) is rate-determining. Such an assumption is reasonable because chlorine-atom abstraction in free-radical dehalogenation of alkyl chlorides is known to be rate-determining,29 but it is not a certainty because for reaction of iodides, bromides, and some very reactive chlorides, hydrogen-atom abstraction from tri-n-butyltin hydride is the rate-determining step.
B. Radical Reaction
1. The Role of Radical Conformation
a. The Curtin-Hammett Principle
Consider conformations X and Y of a radical that is undergoing an addition reaction to give the products Px and Py (Scheme 11). One possibility is that interconversion of X and Y is rapid compared to their reactions. In this situation the Curtin–Hammett principle applies; that is, the ratio of the products depends only on the relative energies of the transition states leading to their formation. (A more detailed statement of the Curtin-Hammet principle is “the relative amounts of products formed from two pertinent conformers are completely independent of the relative populations of the conformers and depend only on the difference in free energy of the transition states, provided the rates of reaction are slower than the rates of conformational interconversion”.30) An energy diagram showing a situation in which the Curtin-Hammet principle applies is found in Figure 2. If in this reaction Px and Py are stereoisomers, the overall process will form Py stereoselectively, even though conformer X is present in greater amount.
A different situation exists for the reaction shown in Figure 3. In this case the interconversion of radicals X and Y is slow compared to their reactions. Under these conditions (the Curtin–Hammett principle not in effect) the conformation present in greatest amount determines the stereoselectivity of the reaction; thus, if X is the major conformer, Px will be formed preferentially.
b. An Initial Proposal
The explanation for the stereoselectivity of the reactions of pyranos-1-yl radicals has changed over time as new information about radical structure has become available. The first proposal for the stereoselective formation of the reduction products 30 and 31 from reaction of the D‑glucopyranosyl bromide 27 with tri-n-butyltin deuteride was that product yields reflected the relative amounts of 28 and 29 present in the reaction mixture (Scheme 12).31 (Inherent in this proposal was the assumption that 28 and 29 were the most stable structures for this pyranos-1-yl radical and that they reacted more rapidly than they equilibrated.) The major component in this proposed pseudoequilibrium was thought to be 28 because this radical would be stabilized more effectively than 29 by interaction of the orbital centered on C-1 with the p‑type orbital on the ring oxygen atom (Scheme 12). This interpretation subsequently had to be revised because ESR spectral analysis showed that neither 28 nor 29 was as stable as the distorted B2,5 boat conformation 26 (Scheme 13).27,28
c. A Revised Explanation
(1). Steric Effects
One possible explanation for stereoselectivity in reactions involving the D-glucopyranos-1-yl radical 26 is that steric effects are determining this selectivity just as they do in the reactions of many other carbohydrate radicals. For this explanation to be correct, approach to C-1 by Bu3SnD from the α face of 26 would have to be less hindered than a similar approach to its β face. With the neighboring 2-0-acetyl group shielding α-face reaction at C-1 (Scheme 13), it is difficult to see how the least hindered pathway to C-1 would involve approach to the α face of the radical.
Another view of the difficulty with steric interactions being the controlling factor in stereoselectivity of the reactions of 26 comes from comparing these reactions with those of the pyranos-4-yl radical 7 and the pyranos-3-yl radical 8 (Table 3). For neither 7 nor 8 is the stereoselectivity of reaction with Bu3SnD large, but for each the major stereoisomer comes from reaction on the face of the radical opposite to that containing the shielding groups on the adjacent carbon atoms. This result stands in contrast to the reaction of the pyranos-1-yl radical 26, where the stereoselectivity is not only much larger, but Bu3SnD approaches this radical from the face containing the only shielding group on a neighboring carbon atom. The α‑face stereoselectivity of 26 is not limited to reaction with Bu3SnD. Addition of this radical to acrylonitrile also is stereoselectively from the more hindered α face (eq 4).
The D-mannopyranos-1-yl radical 25 exhibits even greater α-face selectivity in deuterium abstraction (Scheme 14) and addition to acrylonitrile (eq 5) than does its D-gluco epimer 26 (Scheme 13, eq 4). While steric effects could explain the stereoselectivity in the reactions of 25, they are unable to rationalize the selectivity in the reactions of both 25 and 26; however, for each of these radicals stereoselectivity does have a link to radical conformation.
(2). Interconversion of Conformations
For the D‑mannopyranos-1-yl (25) and D‑lyxopyranos-1-yl (35) radicals only the 4C1 chair conformation can be detected in the ESR spectrum of each (Scheme 15).27,28,32 Structurally these radicals differ only at C-5 where the equatorial CH2OAc substituent in 25 is replaced by a hydrogen atom in 35. Even though the structures 25 and 35 are quite similar in the vicinity of the radical center, the stereoselectivity of their reactions is quite different. The radical 25 adds stereoselectively to acrylonitrile to give only the product arising from reaction at its α face, but the radical 35 reacts from both its α and β faces (Scheme 15).32
Since an equatorial substituent at C-5 is remote from the reacting center at C-1, steric effects alone cannot explain the difference in reactivity between these two (25 and 35). Findings concerning the reactivity of 25 and 35 (Scheme 15) are echoed in the reactions of the D-glucopyranos-1-yl radical 26 and the D‑xylopyranos-1-yl radical 36 (Scheme 16), further; observable conformations and reactivity 26 and 36 point to a possible explanation for the difference in stereoselectivity in the reactions of this pair (26 and 36) as well as the difference in stereoselectivity in the reactions of 25 and 35.
The ESR spectrum of 26 shows it to exist in a distorted B2,5 boat conformation, but for 36 B2,5 boat, 1,4B boat, and 1C4 chair conformations also can be detected.6 Conformational population, therefore, changes considerable when the CH2OAc substituent at C-5 in 26 is replaced by a hydrogen atom. This finding raises the possibility that the difference in stereoselectivity in the reactions of these radicals many be related to population and reactivity of conformational isomers (Scheme 16).
(3). Conformational Mobility
If B2,5 conformations react from their α face, and 1,4B or 4C1 conformers (or both) from their β face (Scheme 16), conformational population and mobility could explain the observed stereoselectivity in the reactions of radicals 26 and 36.6 According to this explanation, interconversion among conformers of the radical 36 would need to be more rapid than reactions of these radicals (Curtin-Hammett principle in effect). Since the assumption is that 1,4B boat or 1C4 conformers give β-face reaction products,6 the conclusion is that one or both of these conformers is much more reactive than the B2,5 conformation and that this very reactive conformation does so in a highly stereoselective fashion. A further part of this argument is that the CH2OAc substituent attached to C-5 in the radical 26 stabilizes the B2,5 conformation and, in so doing, increases its population to the point that little reaction occurs from other conformers (Scheme 16).
A similar explanation exists for the difference in stereoselectivity between radicals 25 and 35 (Scheme 15). If interconversion among the conformers of the radical 35 is faster than reaction with acrylonitrile, it is possible to form a β-C-glycoside by reaction with a minor conformer. For the radical 25 the equatorial CH2OAc substituent at C-5 must stabilize the 4C1 conformation sufficiently that the concentration of minor conformers is too small for them to account for significant product formation.
(4). Transition-State Stabilization: The Kinetic Anomeric Effect
To understand how radical conformation can have such a pronounced effect on stereoselectivity in the reactions of pyranos-1-yl radicals, it is instructive to examine transition-state stabilization. The major stereoisomer formed in these reactions results from a radical adopting a conformation that allows the stabilizing interaction between the p-type orbitals on C-1 and the ring oxygen atom to be maintained to the greatest extent in the transition-state. This conformation-dependent, stereoelectronic, transition-state stabilization is referred to as the kinetic33,34 or radical35,36 anomeric effect. (We will use “kinetic anomeric effect” in describing this phenomenon.)
The kinetic anomeric effect offers a rationale for the D‑mannopyranos-1‑yl radical 25 adding to acrylonitrile exclusively from the α face of the pyranoid ring (Scheme 15). Only for α-face reaction will stabilizing interaction between the p-type orbitals on the ring oxygen atom and the singly occupied orbital on C-1 be maintained in the transition state (Scheme 17). Predicting stereoselectivity in the reaction of a conformationally mobile radical can be a challenging task because more than one conformation is accessible and determining which conformation is the most reactive may be difficult. This means that minor conformers, even ones that cannot be detected by ESR spectroscopy, can play a major role in determining reaction stereoselectivity. For the D-glucopyranos-1-yl radical 26 the only conformation detectable by ESR spectroscopy is a distorted B2,5 boat,27,28 but 1,4B and 1C4 conformations may be only modestly higher in energy and, therefore, easily accessible. For the B2,5 conformation, reaction from the α-face of the pyranoid ring maintains stabilizing orbital interaction more effectively than reaction from its β face (Scheme 18).
Reaction does occur, however, to a small extent (5%) from the β face of the pyranoid ring in the radical 26. The β anomer formed as a minor product could come from an accessible but undetected conformation, such as the 1C4. Maintaining orbital interactions in a 1C4 chair conformation would favor the β-face addition that leads to the minor product (Scheme 19). If this is the way in which β-face addition takes place, then stereoselectivity depends not only on which conformations are accessible and highly reactive but also on which face of a given conformer maintains the reaction-promoting orbital interaction that stabilizes the transition state.
Another explanation for the formation of the minor product in the reaction of the radical 26 is that this product results from a small amount of β‑face addition to the radical in its B2,5 conformation. Such an explanation, however, fails to explain greater β-face addition for the radical 36, when compared to 26 (Scheme 16), and is incompatible with the information, discussed in the next section, on reactions of radicals with restricted conformations.
Although the case is strong for radical conformation playing a critical role in stereoselectivity of reactions of pyranos-1-yl radicals, uncertainty remains about how to identify the reactive conformation when several may be present and undergoing reaction. This uncertainty is effecttively eliminated where radicals with restricted conformations are concerned. Study of such radicals has provided considerable insight into the power of the kinetic anomeric effect in determining reaction stereoselectivity.
(5). Restricted Conformations
A restricted conformation for a radical is one that is highly favored over others due to some structural feature, such as an additional ring system. The radicals 37 (Scheme 20) and 47 (Scheme 21) have pyranoid rings restricted to 4C1 and 1C4 chair conformations, respectively. This restriction is caused by the presence of appropriately placed, additional rings.33 A second ring restricts conformational change in the radical 37 by creating a trans-decalin-type structure. For the radical 47 a bridge produces a bicyclic structure that creates a rigid, conformationally restricted system.
(a). Explanation for Reaction Stereoselectivity
For the radical 47 maintaining stabilizing orbital interaction in the transition state requires approach of a reacting molecule, such as tri-n-butyltin deuteride, to the β-face of the pyranoid ring (Scheme 21). This means that reaction proceeding according to the kinetic anomeric effect will give a product (48) with an axial deuterium atom at C-1. The high stereoselectivity of this reaction (48/49 = 99/1) supports the idea that conformational and stereoelectronic effects together have a powerful influence on the reactions of pyranos-1-yl radicals. Such an idea is reinforced in a dramatic fashion by the reaction of the radical 37 (Scheme 20). In this case in order to benefit from kinetic anomeric stabilization, the deuterium donor must approach the radical from the α face of the pyranoid ring. α‑Face selectivity (39/40 = 97/3) in this reaction (the pyranoid ring restricted to a 4C1 conformation) is nearly as great as the β‑face selectivity (48/49= 99/1) in the reaction of 47 (pyranoid ring restricted to a 1C4 conformation). Changing the radical conformation then completely changes reaction stereoselectivity;33 furthermore, the selectivity observed in each case is that predicted by the kinetic anomeric effect.
When Bu3SnCH2CH=CH2 replaces Bu3SnD as the molecule reacting with the radical 37, the stereoselectivity decreases somewhat, although it remains high.33 [The ratio of α-face to β-face reaction decreases from 97/3 (39/40) to 91/9 (41/42) (Scheme 20).] This modest decrease in stereoselectivity disappears when the hydroxy group at C-2 is replaced by a hydrogen atom: that is, the stereoselectivity for the reaction of the 2-deoxy radical 38 with Bu3SnD is essentially the same as that for reaction with Bu3SnCH2CH=CH2 (Scheme 20). Steric hindrance associated with the C-2 hydroxy group, therefore, may be responsible for the small difference in stereoselectivity observed for reactions of the radical 37 with Bu3SnD and Bu3SnCH2CH=CH2. The overall picture, however, remains one in which steric effects have, at most, a minor role in determining stereoselectivity in the reactions of these two radicals (37 and 38).
(b). σ*-Orbital Interaction
The σ* orbital of the C2–O bond in a pyranos-1-yl radical contributes to the stability of these radicals by interacting with the p-type orbitals on C-1 and the ring oxygen atom. Radical stabilization of this type plays a critical role in determining preferred radical conformation (see Section IV.A.2.d. of Chapter 6). A question raised by these orbital interactions concerns whether this type of stabilization also affects stereoselectivity in reactions of pyranos-1-yl radicals. The more rapid reaction of the D-mannopyranosyl chloride 23 (Scheme 9) when compared to the epimeric D‑glucopyranosyl chloride 24 (Scheme 10) indicates that it may. Even for 24, however, σ*-orbital interaction is not essential because reaction still takes place (although less rapidly) even with minimal participation from this orbital (Scheme 10). Study of radicals with restricted conformations helps to clarify the role of the C2–O σ* orbital in stereoselectivity of reactions of pyranos-1-yl radicals.
Because the pyranoid rings in radicals 37 and 38 (Scheme 20) are restricted to 4C1 conformations by their trans-decalin-like ring systems, the reactions of these radicals provide insight into the effect of a C2–O bond (in particular, its σ*-orbital) on the stereoselectivity of the reactions of pyranos-1-yl radicals. The radical 37, for example, reacts in a highly stereoselective fashion with tri-n-butyltin deuteride even though the σ* orbital of the C2–O bond is not aligned to assist in transition-state stabilization (Figure 4). The radical 38 completely lacks a C-2 substituent; yet, it also undergoes highly stereoselective reaction (Scheme 20). σ*-Orbital interaction involving the C2–O bond, therefore, is not essential to the stereoselectivity of the reactions of the radicals 37 and 38. The reactions of these radicals, however, point to the critical factor in determining reaction stereoselectivity, namely, maintaining effective interaction in the transition state between the singly occupied, p-type orbital on C-1 and the p-type orbital on the ring oxygen atom.
Study of the reactions of radicals with restricted conformations generates a powerful argument in favor of the control that conformational and stereoelectronic effects have on the reactivity of pyranos-1-yl radicals. Among the radicals of this type discussed thus far, there is little indication that steric effects have other than a minor role in determining stereoselectivity. Study of compounds that have particularly well shielded radical centers, however, shows that steric effects can be significant in the stereoselectivity of some pyranos-1-yl radical reactions.
(6). Steric Effects Revisited
(a). Pyranos-1-yl Radicals
Even though conformational and stereoelectronic effects usually have the dominant role in determining stereoselectivity in the reactions of pyranos-1-yl radicals, comparing the reactions shown in equations 6 and 7 shows that steric effects can assert themselves when a sufficiently effective shielding group is present in a reacting molecule. Reaction of the glycosyl chloride 52 with allyltri-n-butyltin follows the now familiar pattern of α-face reaction that maintains transition-state interaction between orbitals on C-1 and the ring oxygen atom (eq 6).37 This reaction stands in contrast to that shown in eq 7 where the sterically demanding phthalimido group at C-2 causes a reversal of stereoselectivity in C-glycoside formation. This second reaction (eq 7) is a striking example of steric effects overwhelming the conformational and stereoelectronic effects that normally control stereoselectivity in reactions of pyranos-1-yl radicals.37,38
Reaction of the bromide 53 with tert-butyl isocyanide gives additional insight into the competition between steric and stereoelectronic effects in determining the stereoselectivity of reaction of pyranos-1-yl radicals (Scheme 22).35 The C-1 configuration in products 55 and 56 is determined by the stereoselectivity of the reaction between the radical 54 and tert-butyl isocyanide. In this reaction steric effects, which favor β-face reaction of 54, are more powerful than the stereoelectronic effects, which favor forming the product with an α configuration. In addition to shielding due to the methylene carbon atom attached to C-2, the shape of the radical in the vicinity of the ethoxy group has major impact on the steric effects directing the approach of tert-butyl isocyanide (Scheme 22).
(b). Furanos-1-yl Radicals
Study of pyranos-1‑yl radicals has identified conformational mobility as a “key” factor in the stereoselectivity of their reactions. This selectivity depends not only on which conformation is the major one but also on the accessibility and reactivity of other conformations. For furanos-1-yl radicals the same factors are involved, but their relative importance may differ. Conformational mobility is greater for furanos-1-yl radicals than for their pyranos-1-yl counterparts. Also, due to the shape of the furanoid ring, maintaining effective, transition-state interaction between p-type orbitals on C-1 and the ring oxygen atom (i.e., the interaction necessary for kinetic-anomeric stabilization) in many conformations is possible during reaction from both α and β faces of the ring. These observations lead to the proposal that conformational and stereoelectronic effects may be less important in determining stereoselectivity in furanos-1-yl radical reaction than in reactions of pyranos-1-yl radicals. The available information on stereoselectivity of the reactions of furanos-1-yl radicals is not sufficient to draw a firm conclusion about the relative importance of the kinetic anomeric effect when compared to steric effects. Exclusive hydrogen-atom transfer to the less-hindered β face of the radical 57 is consistent with steric effects controlling stereoselectivity; in addition, the fact that the radical 58, for which the β face is more hindered, is less stereoselective in its reaction also is consistent with steric effects primarily determining reaction stereoselectivity (Scheme 23).39
(c). Supersteric Radicals
Although bonding between two radicals normally is a rare event, such a reaction can become significant when other processes (e.g., hydrogen-atom abstraction or addition to an unsaturated compound) take place too slowly. Bonding between two radicals begins at a sufficiently long distance that differences in transition-state energies leading to different products sometimes are too small to be of consequence; as a result, little or no stereoselectivity is observed (eq 8).40
If one of the radicals involved in a coupling process has steric requirements that are great enough, even radical coupling becomes stereoselective. The "supersteric" Co(dmgH)2py radical, for example, has such a large steric requirement that its reaction with the pyranos-1-yl radical 59 occurs much more rapidly from the β face of the pyranoid ring in 59 than from its α face (Scheme 24).41 Having the Co(dmgH)2py substituent in an equatorial position (β anomer) is so much more stable than having it in the more crowded axial orientation (α anomer) that this difference significantly affects the transition-state energies leading to these two anomers. The steric size of the Co(dmgH)2py substituent is so great that it overcomes the usually more important kinetic anomeric effect in determining reaction stereoselectivity.
2. Effect of Temperature on Stereoselectivity
A final point with respect to stereoselectivity in bimolecular radical reactions concerns reaction temperature. In the process shown in Scheme 25 stereoselectivity is determined by the approach of Bu3SnD to the intermediate pyranos-1-yl radical 26.31 According to the kinetic anomeric effect [Section III.B.1.c.(4).] stereoselective deuterium transfer to the α face of 26 (to give 30) should have a lower transition-state energy than transfer to its β face (to give 31). At low temperature the stereoselectivity is high, but as the temperature rises, stereoselectivity decreases because a progressively larger percentage of radicals are able to react with Bu3SnD and transcend the barriers leading to both products (30 and 31). The results shown in Scheme 25 can be viewed as a general response of reaction stereoselectivity to changes in temperature. | textbooks/chem/Organic_Chemistry/Radical_Reactions_of_Carbohydrates_(Binkley)/Radical_Reactions_of_Carbohydrates_I%3A_Structure_and_Reactivity_of_Carbohydrate_Radicals/11%3A_Stereoselectivity/I._Introduction.txt |
Whenever a radical center and the group with which it is reacting are part of the same molecule, new factors become important in determining both regio- and stereoselectivity.42–44 Since the reaction taking place is an internal radical addition to a π system, the size of the ring being formed and the constraints placed on reactivity by existing structural features (e.g., other ring systems) both contribute to determining selectivity. Regioselectivity in ring formation is discussed in Chapter 10 (Section III). The discussion here is concerned with the stereoselectivity of reactions that create new ring systems.
V. Stereoselectivity in Synthesis
The primary thrust of the discussion in this chapter is to understand the factors controlling stereoselectivity in radical reactions. Many of the reactions examined provide both basic understanding of radical-reaction stereoselectivity and examples of how stereoselectivity can achieve a particular synthetic goal. The emphasis in discussion in this part of the chapter shifts to reactions where the primary goal is to solve a particular synthetic problem by taking advantage of knowledge about stereoselectivity in radical reactions. Even though the purpose in conducting these reactions is synthetic, their investigation has increased basic understanding of radical-reaction stereoselectivity.
VI. Summary
Stereoselectivity is the preferential formation or consumption of one stereoisomer rather than another in a chemical reaction. In radical reactions stereoselectivity is controlled by a combination of conformational, steric, stereoelectronic, and torsional effects. The stereoselectivity caused by these effects is generally increased by conducting reactions at lower temperature. For radicals not centered on C-1, steric effects direct reaction to occur along the least-hindered pathway. A primary factor in determining this pathway is the way in which various groups shield a radical center.
As the steric size of a molecule reacting with a carbohydrate radical increases, the extent to which the least-hindered pathway is followed also increases. As this size of reacting molecules becomes smaller, stereoselectivity decreases but does not completely disappear; rather, a low level of selectivity remains due to torsional interactions.
Stereoelectronic effects operate in conjunction with conformational effects to determine stereoselectivity in reactions of pyranos-1-yl radicals. The critical factor in forming a particular stereoisomer in a reaction is the ability of the reactants to maintain in the transition state a stabilizing interaction between orbitals on C-1 and the ring oxygen atom. Maintaining this interaction causes different conformations of a radical to yield stereoisomerically different products. This stereoelectronic, conformation-dependent, transition-state stabilization gives rise to a phenomenon known as the kinetic anomeric effect. This effect provides a basis for predicting and rationalizing stereoselectivity of pyranos-1-yl radical reactions.
Radical cyclization places additional requirements on reaction stereoselectivity. Prominent among these is that in most situations a reaction proceeds through a chair-like transition state that has substituents located in pseudoequatorial positions. In some instances a boat-like transition state is lower in energy than a chair-like one. This is often the case when structural features such as allylic strain or pseudo-1,3-diaxial interactions destabilize a chair-like transition state. The stereoselectivity of radical reactions provides the basis for several synthetic processes. These include the synthesis of β-glycosides, the use carbohydrates as chiral auxiliaries, and the incorporation of carbohydrates into enantioselective syntheses. | textbooks/chem/Organic_Chemistry/Radical_Reactions_of_Carbohydrates_(Binkley)/Radical_Reactions_of_Carbohydrates_I%3A_Structure_and_Reactivity_of_Carbohydrate_Radicals/11%3A_Stereoselectivity/IV._Maximizing_Transition-State_Stability_Dur.txt |
When deciding whether or not to conduct a radical reaction, certain information is crucial. It is important to know as completely as possible how intermediate radicals will form and, once formed, how these radicals will react with various reagents and solvents present in the reaction mixture. This information not only points to the expected product but also answers questions such as: What side reactions could take place? How might these reactions be avoided or minimized? What is the outcome of reactions that have been reported for similar compounds? A fitting way to begin framing the answers to these questions is by looking at the advantages and disadvantages of radical reactions.
01: Advantages and Disadvantages of Radical Reactions
Some advantages of radical reactions can be traced to the mild conditions under which they are conducted; others to the stability of common protecting groups during reaction. Ease in performing group and atom replacement is still another benefit, but the greatest value radical reactions bring to carbohydrate synthesis is in forming new, carbon–carbon bonds; in other words, even though radical reactions are capable of causing a variety of structural changes, their most valuable role is their most basic one, namely, building more complicated structures from less complicated ones.
III. Disadvantages of Radic
In addition to the advantages associated with radical reactions, there also are disadvantages. Deciding whether to use a radical reaction depends, in part, on knowing what these disadvantages are. Problems with radical reactions are described in the next four sections.
IV. Looking Ahead
The advantages of radical reactions discussed in this chapter provide an introduction to the possibilities these reactions bring to carbohydrate chemistry. The chapters immediately ahead are devoted to a systematic presentation of the various types of carbohydrate derivatives that are able to produce radicals and the reactions that these radicals undergo. Later chapters contain information about reactions that take place when radicals are generated by electron transfer from transition-metal complexes to various carbohydrate derivatives.
I. Introduction
Halogen-atom abstraction to give a carbon-centered radical is the first step in many radical reactions. Among the more common of these are simple reduction, radical addition, and radical cyclization.
02: Halogenated Compounds
Halogen-atom abstraction to give a carbon-centered radical is the first step in many radical reactions. Among the more common of these are simple reduction, radical addition, and radical cyclization. A typical example is give in Scheme 1, where chlorine-atom abstraction produces a pyranos-1-yl radical that then adds to the carbon–carbon double bond in allyltri-n-butyltin.1 Group migration, ring-opening, elimination, hydrogen-atom abstraction, and radical combination are less common reactions that also often begin with halogen-atom abstraction that forms a carbohydrate radical.
Another way of viewing reactions that begin with halogen-atom abstraction is one taken from the perspective of the involvement of such reactions in carbohydrate synthesis. This view describes these reactions in terms of the changes in carbohydrate structure that they make possible; thus, halogen-atom abstraction is involved in synthesizing deoxygenated sugars and nucleosides, establishing glycosidic linkages, extending carbon-atom chains, forming new ring systems, introducing unsaturation, and modifying protecting group reactivity. This broad range of applications combines with generally good product yields to make halogen-atom abstraction the first step in many, useful synthetic reactions.
In addition to the variety of reactions that begin with loss of a halogen atom, there are others that reverse this process; that is, there are reactions that cause a halogen atom to be incorporated into a carbohydrate. The range of synthetically useful compounds prepared in this way is limited. Radical halogenation is not part of a general synthesis of carbohydrate halides, but it sometimes provides a route to compounds that are difficult to prepare in other ways. The specialized, but useful, nature of radical halogenation is illustrated by the conversion of a 4,6-O-benzylidene acetal into the corresponding bromodeoxy benzoate (eq 1).2
II. Radical Formation by Dehalogenation
Most mechanistic understanding of radical formation by dehalogenation comes from study of alkyl halides. It is reasonable to assume that the findings from these investigations also are applicable to reactions of halogenated carbohydrates but, at the same time, to recognize that such application does involve extrapolation of the data to more complex structures.
III. Radical Reactions
A. Simple Reduction
1. Typical Reaction Conditions
Simple reduction of a halogenated carbohydrate is a reaction in which the only change that occurs in the substrate is replacement of a halogen atom with a hydrogen atom. This change typically is brought about by heating a solution of a halogenated carbohydrate, tri-n-butyltin hydride, and a catalytic amount of 2,2'-azobis(isobutyronitrile) (AIBN) in benzene or toluene at 80-110 oC for a few hours.
2. Thermodynamic Driving Force for Reaction
There is a powerful thermodynamic driving force for simple reduction of halogenated compounds with tri-n-butyltin hydride.14 This driving force is apparent in the highly exothermic reaction of bromomethane with tri-n-butyltin hydride (eq 6).14 The exothermic nature of this reaction derives from a carbon–bromine bond being replaced by a stronger carbon–hydrogen bond and a tin–hydrogen bond being exchanged for a stronger tin–bromine bond. The data in eq 7 show that simple reduction of chloromethane is also quite exothermic.14
3. Synthesis of Deoxygenated Sugars and Nucleosides
Simple reduction is a common reaction for halogenated sugars and nucleosides. A typical example, one in which halogen replacement takes place at C-4 in a pyranoid ring, is shown in eq 8.23 Other reactions, ones where the halogen atom is located at C-1,24,25 C‑2,26,27 C-3,28 C‑4,29,30 C-5,31 or C-632,33 in a pyranoid ring, are common. There are fewer reports of simple reduction when the halogen atom is attached to a furanoid ring (C‑1,34,35 C‑5,36,37 or C-638 ), but many nucleoside reactions are known where the halogen atom is bonded to C‑2',39,40 C-3',41,42 or C-5' 43,44 in the substrate. (Each reference listed above refers to a simple-reduction reaction representative of those taking place at the indicated carbon atom.)
When a halogen atom is replaced by a deuterium atom, simple reduction can provide information about reaction stereoselectivity; thus, in the reaction shown in Scheme 7 the approach of Bu3SnD to the nucleoside radical 3 is from the less hindered face of the furanoid ring.45 Reaction at the 2'‑position in other halogenated nucleosides also is stereoselective, but this selectivity, which is heavily influenced by the structure and positioning of ring substituents, sometimes is modest.46–48
4. Protecting Group Modification
Simple reduction sometimes is used to modify protecting group reactivity.49–57 In the reaction shown in eq 9, for example, replacement of the three chlorine atoms in the trichloroethylidene group with three hydrogen atoms makes hydrolytic removal of this group much easier.49 In a related example, the trichloroacetamido group is converted to the more easily hydrolyzed acetamido group by three, successive simple reduction reactions (eq 10).56 Reaction with tri-n-butyltin hydride as a means of replacing all the chlorine atoms in a protecting group with hydrogen atoms is not always successful because complete dechlorination sometimes is elusive (eq 1158).58–60
5. Acyloxy Group Migration
When an anomeric halide with a neighboring acyloxy group reacts with tri-n-butyltin hydride, acyloxy group migration leading to a 2-deoxy sugar competes with simple reduction (Scheme 8).61 If 2-deoxy sugar synthesis (and not simple reduction) is the goal of the reaction,61–64 conditions that maximize group migration need to be selected. Generally, this means maintaining the tri-n-butyltin hydride concentration at a low enough level to allow time for migration to occur.
B. Addition
Halogenated compounds are common radical precursors for addition reactions involving carbohydrates, but other carbohydrate derivatives also function in this capacity; consequently, radical addition reactions are mentioned in most of the chapters in this book. They are discussed extensively not only in this chapter but also in Chapters 10, 12, 13, and 18. In Chapter 18 radical addition is considered from a different point of view, that is, with a focus on the compound to which addition is occurring rather than on the radical precursor.
Addition reactions can be divided into three groups based on what happens after formation of the adduct radical. The first group (addition-abstraction reactions) contains those reactions where the adduct radical abstracts a group or atom from a donor present in solution. Atom abstraction is more common that group abstraction and nearly always involves a hydrogen atom. The second type of reaction (addition-combination) is one in which the adduct radical combines with another radical or an organometallic reagent present in the reaction mixture. In the third reaction type (addition-elimination or addition-fragmentation) the adduct radical forms an unsaturated compound by a β-fragmentation process that eliminates a radical. Examples in which halogenated carbohydrates serve as radical precursors for each of these reaction types are given in the following three sections.
1. Addition-Abstraction Reactions
Formation of a carbon-centered radical by reaction of a glycosyl halide with a tin- or silicon-centered radical is the first step in many addition reactions. An addition-abstraction reaction takes place when a radical produced by halogen-atom abstraction adds to a multiple bond to generate an adduct radical that then abstracts a hydrogen atom. The hydrogen-atom transfer is nearly always a tin- or silicon hydride. An example of this type of reaction is given in eq 12, where a pyranos-1-yl radical generated from the D-mannopyranosyl bromide 4 adds to the α,β-unsaturated ketone 5.65 Pyranos-1-yl radicals generated from halogenated carbohydrates are known to add to α,β-unsaturated nitriles,66,67 esters,67,68 aldehydes,67 ketones,69,70 lactones,71,72 and phosphonates.73,74 In all of these reactions formation of the adduct radical is followed by hydrogen-atom abstraction. Pyranos-1-yl and other carbohydrate radicals formed from halogenated compounds also add to oximes75–77 and electron-deficient enol ethers.78,79 The reactions shown in equations 13 and 14 underscore a critical feature of radical addition reactions, namely, that unless a multiple bond is electron-deficient (eq 13), addition of a nucleophilic radical (which nearly all carbohydrate radicals are) will be too slow to compete effectively with simple reduction (eq 14).80
Addition-abstraction reactions that do not involve pyranos-1-yl radicals are much less common than those that do. An example of a non-pyranos-1-yl radical reaction is shown in eq 15.81 The pyranos-5-yl radical formed in this reaction has reactivity similar to that of a pyranos-1-yl radical because both are nucleophilic species that are stabilized by a ring oxygen atom. This similarity in reactivity can be seen when comparing the reaction shown in eq 1581 with that in eq 16.82 Several addition-abstraction reactions of radicals centered on C-683–86 are known, but reaction of a radical located on an atom in or attached to a pyranoid ring (other than C-1or C-6) is rare.87 There are reports of addition-abstraction reactions where the radical is centered on a carbon atom that is part of or attached to a furanoid ring.88–90
2. Addition-Combination Reactions
Although radical formation from a glycosyl halide normally involves halogen-atom abstraction by a tin or silicon centered radical, a radical also can be generated by electron transfer to the glycosyl halide from a metal ion, such as the samarium ion in SmI2 or the titanium ion in Cp2TiCl. An example of this type of reaction is shown in Scheme 9, where the pyranos-1-yl radical (R·) is formed by electron transfer from titanium to the glycosyl bromide (RBr).91 In the reaction shown in Scheme 9, radical addition to an electron-deficient multiple bond is faster than combination of the radical with a second molecule of Cp2TiCl. Once addition occurs, however, the situation changes. The reactivity of the adduct radical is so different from the far more nucleophilic pyranos-1-yl radical that the fastest reaction for the adduct radical is combination with a molecule of Cp2TiCl. (Chapters 20-24 contain further discussion of addition-combination reactions brought about by electron-transfer from organometallic complexes to carbohydrate halides.)
3. Addition-Elimination Reactions
When a carbon-centered radical adds to allyltri-n-butyltin, it begins a sequence of reactions that replaces a halogen atom with an allyl group (Scheme 1).1 This type of transformation (an addition-elimination reaction) often occurs when the halogen atom is attached to C‑11,92–94 but also takes place when such an atom is bonded elsewhere in a carbohydrate framework.95,96 For example, the reaction between the tri-n-butyltin acrylate 6 and the deoxyiodo sugar 7 involves a radical centered at C-6 (eq 17).97
A reaction mechanistically similar to that shown in eq 17, but one with a quite different outcome, occurs when a halogen atom is abstracted by Bu3Sn· in the presence of t-butyl isocyanide. When this happens, an addition-elimination reaction produces a nitrile (Scheme 10).98
Most addition-elimination reactions transfer an allyl or substituted allyl group from a tin-containing compound to a carbon-centered radical; however, this transfer can be tin-free.99,100 In the reaction shown in eq 18 the allyltin reactant is replaced by allyl ethyl sulfone.99
C. Cyclization
New ring formation is pervasive in the radical reactions of carbohydrates; consequently, radical cyclization, like radical addition, is mentioned in many of the chapters in this book. Significant discussion exists in this chapter because halogenated carbohydrates often are the precursors for radical-based formation of new ring systems. A large portion of Chapter 12 also is devoted to radical cyclization because O-thiocarbonyl compounds frequently are precursors for radicals involved in new ring formation. Chapter 19 is devoted entirely to cyclization reactions and is concerned less with radical formation and more with the internal radical addition that produces new rings.
1. Substrate Reactivity
As with simple reduction reactions, the abstracting radical that begins radical cyclization nearly always is derived from a tin or silicon hydride. The well-established order of reactivity for halogenated compounds with silicon and tin hydrides (RI > RBr > RCl >> RF), mentioned in Section II.B, accounts for the usual selection of a carbohydrate iodide or bromide as the substrate in a cyclization reaction.
Halogen identity is especially critical to dehalogenation with either SmI2 or Cp2TiCl. In the cyclization reaction shown in Scheme 11 the iodide gives a good yield of the substituted cyclopentane, but the bromide is enough less reactive that it produces only bromine-containing dimers arising from reduction of the double bond by SmI2.101 Reaction of a similar compound with Cp2TiCl results in an 82% yield of substituted cyclopentanes from the iodide but only an 18% yield from the corresponding bromide (Scheme 12).102 More forcing reaction conditions might have improved product yield from the bromide because its attempted cyclization returned primarily unreacted starting material.
Successful cyclization of halogenated carbohydrates by reaction with SmI2 or Cp2TiCl depends upon reaction conditions and additives. The presence of hexamethylphosphoramide (HMPA) is so important to the reactivity of SmI2 that the cyclic product shown in Scheme 11 does not form unless HMPA is present in the reaction mixture.101 In a similar fashion, UV radiation is critical to the reaction shown in Scheme 12. If it is omitted, little reaction takes place.102
2. Competition between Cyclization and Hydrogen-Atom Abstraction
Whenever a cyclization reaction is conducted in the presence of a tin or silicon hydride, hydrogen-atom abstraction by the initially formed radical, leading to simple reduction, competes with cyclization (Scheme 13).38 Other reagents that are not hydrogen-atom transfers but are capable of generating radicals from halogenated carbohydrates (e.g., Bu3SnSnBu3,103 SmI2,101,104,105 and Cp2TiCl102) have the advantage that simple reduction is suppressed. Even though simple reduction is not a problem when using SmI2 and Cp2TiCl, each of these compounds can prevent cyclization by combining with a carbon-centered radical before ring formation takes place; therefore, rapid ring closure remains an important criterion for successful reaction.
3. Organization of Cyclization Reactions
Successful radical cyclization requires a multiple bond and radical center that are suitably positioned with respect to each other. The radical center in such a reaction can be either on an atom that is part of the carbohydrate framework (Figure 2) or on a substituent group. The same possibilities exist for the multiple bond. For purposes of discussion it is useful to divide cyclization reactions into the four basic types shown in Figure 3. This Figure also contains a short-hand terminology that has been proposed to identify each reaction type.105 Chapter 19 contains extensive tables of cyclization reactions in which radicals are formed from halogenated and nonhalogenated carbohydrates.
a. Addition of a Framework Radical to a Framework Multiple Bond
The possibilities for a framework radical adding internally to a framework multiple bond are limited by the size of the rings that are easily produced; thus, only five- and six-membered rings generally form rapidly enough to compete with other radical reactions. Since radical cyclization produces five-membered rings more quickly than six-membered ones, the typical cyclization reaction produces a pair of heavily substituted, cyclopentane derivatives (eq 19106).106–111 When five-membered ring formation is not possible but producing a six-membered ring is, cyclization gives a substituted, cyclohexane derivative (eq 20).112 Addition of a framework radical to a framework multiple bond also can produce a bicyclic compound (Scheme 14).113
Although forming either a five- or six-membered ring is the typical result of radical cyclization, larger rings are possible if the carbohydrate framework is held so that the radical center easily can approach the multiple bond. Striking examples of such a situation are found in the reactions shown in equations 21 and 22, where the O-isopropylidene groups cause the iodides 8 and 9 to adopt conformations that favor formation of seven- and eight-membered rings, respectively.114
One way to increase the reactivity of a multiple bond in a radical cyclization reaction is to replace one of the carbon atoms with an electronegative heteroatom.115–121 The remaining carbon atom then will have electron density drawn from it and, as a result, have enhanced reactivity toward nucleophilic radicals. The oxime 10 contains a double bond activated in this way (eq 23).115 Having reactive centers able easily to come within bonding distance translates into cyclic products being formed in reactions involving other carbon–nitrogen116–119 and even carbon–oxygen120,121 multiple bonds. In the reaction shown in Scheme 15, for example, ring formation occurs because the radical center easily comes into contact with the cyano group.118
b. Addition of a Framework Radical to a Substituent Multiple Bond
An example of a reaction in which a framework radical adds to a substituent multiple bond to form a five-membered ring is shown in eq 24.122 If five-membered-ring formation introduces too much strain into a system, reaction to give a larger ring becomes a possibility; thus, the radical centered at C-5 in 11 adds to the substituent double bond to form a six-membered ring (Scheme 16) and, in so doing, avoids producing highly strained, trans-fused, five-membered rings.123
Although bimolecular addition of a carbohydrate radical to a multiple bond is fast enough to be observed only when the multiple bond is electron-deficient, radical cyclization is not limited in this way. Having a radical center, such as the one at C-5 in 11 (Scheme 16), in close proximity to a double bond makes cyclization competitive with other radical reactions (e.g., simple reduction) even though the multiple bond is not electron-deficient.
Radical cyclization followed by ring opening that breaks the newly formed bond is a degenerate addition-elimination reaction that is rarely useful or even detectable. Sometimes, however, ring opening breaks a different bond from that produced during cyclization.124,125 The result of such a reaction is an addition-elimination process, such as that shown in Scheme17, where the effect of the reaction is migration of a part of a substituent group.124
A unique type of cyclization between a framework radical and a substituent multiple bond takes place in nucleosides that have properly placed dibromovinyl groups.103,126–131 Bromine-atom abstraction by Bu3Sn· produces a vinyl radical that begins a sequential reaction leading to two spiro compounds (Scheme 18).103 Using Bu3SnSnBu3 in this reaction (rather than Bu3SnH) improves the yield of the cyclization product because competing simple reduction involving a tin hydride is not an option.103
c. Addition of a Substituent Radical to a Framework Multiple Bond
Cyclization involving halogenated carbohydrates often occurs when the halogen atom is part of a substituent group and the multiple bond is located in the carbohydrate framework. The substrate in many of these reactions is a silyl ether with a bromine atom incorporated into the silicon-containing group.132–139 An example is shown in Scheme 19, which describes a reaction that stereoselectively creates a new C–C bond at C-3 in the product.132 Nonradical reaction of the resulting product transforms it into a diol (Scheme 19) that can be readily converted into other compounds. An extension of this type of reaction to a pair of saccharide units connected by a silaketal tether leads to formation of a protected C-disaccharide (12) from which the tether is easily removed (Scheme 20).140
Cyclization of an unsaturated carbohydrate in which an acetal substituent contains a halogen atom is similar to cyclization of brominated silyl ethers such as that pictured in Scheme 19. High stereoselectivity is the norm in these reactions.141–147 In the process shown in Scheme 21, for example, the stereochemistry at C-3 is controlled by the kinetically and thermodynamically favored formation of a cis-fused ring system.141 [There is a second chiral center (C-4) created during this reaction. The stereochemistry at this center is determined by the least-hindered approach of tri-n-butyltin hydride to the reacting radical.] The reaction shown in eq 25 provides an example of new ring formation when a radical formed from a halogen-containing, acetal substituent adds internally to a triple bond.142
d. Addition of a Substituent Radical to a Substituent Multiple Bond
Sometimes in a radical cyclization reaction neither the carbon atom bearing the radical center nor the carbon atoms of the multiple bond are part of the carbohydrate framework.148,149 When this occurs, the carbohydrate portion of the molecule has only an indirect influence on the reaction; that is, it can affect reaction stereoselectivity by acting as a chiral auxiliary, or its cyclic structure can bring reactive centers into bonding distance. It is the latter role that assists eleven-membered ring formation in the reaction shown in eq 26.148
D. Elimination
Free-radical elimination takes place when a vicinal dihalide reacts with tri-n-butyltin hydride (eq 27).150 A similar reaction occurs if one of the vicinal substituents is a halogen atom and the other contains an O-thiocarbonyl group.151,152 In the reaction shown in eq 28151 there is little doubt that the initial interaction between Bu3Sn· and compound 13 in most instances is with the substituent at C-3' because the rate constant for reaction of Bu3Sn· with a compound containing an O-thiocarbonyl group is much larger than that for reaction with a chlorinated compound.
Electrochemical reduction of glycosyl halides (and their sulfur-containing analogs154a) begins with electron transfer to the halide to produce a radical anion that reacts further to give the corresponding glycal (Scheme 22153). Reaction of glycosyl bromides with zinc dust also leads to glycal formation by electron-transfer to a glycosyl halide (eq 29).154b The mechanism for this reaction may parallel that shown in Scheme 22, but the possibility also exists that an organozinc intermediate forms following the initial electron transfer.
E. Ring-Opening
If halogen-atom abstraction produces a cyclopropylcarbinyl radical, cyclopropane ring opening takes place. The radical remaining after ring opening then can undergo hydrogen-atom abstraction,155 addition,156 or, as shown in eq 30, cyclization.157
F. Internal Hydrogen-Atom Abstraction
Reaction of a halogenated carbohydrate to form a carbon-centered radical rarely results in this radical abstracting a hydrogen atom from a carbon–hydrogen bond in another molecule because such abstraction is, in most instances, too slow to compete with other radical reactions. If, however, the abstraction is internal and the radical is quite reactive (primary, vinyl or aryl), hydrogen-atom abstraction can take place.158–160 Dehalogenation of the bromide 14 begins such a reaction by producing a primary radical that abstracts a hydrogen atom from C‑1, a process that leads to epimerization at this carbon atom (Scheme 23).158 Internal hydrogen-atom abstraction by a vinyl radical takes place in the reaction pictured in Scheme 18.103
G. Radical Combination
1. Dimerization
Radical combination is not a common synthetic reaction for carbohydrates, but it does take place when conditions are adjusted so that reactions such as hydrogen-atom abstraction are to slow to compete.161–163 Having the source of the chain-carrying radical be Me3SnSnMe3 (as opposed to a tin hydride) reduces the rate of hydrogen-atom abstraction by the carbohydrate radical to the point that the three dimers shown in eq 31 are formed. {Similar dimer formation takes place during reaction of glycosyl phenyl sulfones [Chapter 3, Section VII.B.1.c] and glycosyl phenyl selenides [Chapter 4, Section II.B.6]} Electrochemical reactions of glycosyl halides also produce radical dimers.163
2. Reaction with Molecular Oxygen
Reaction of halogenated carbohydrates with tri-n-butyltin hydride in the presence of oxygen replaces the halogen atom with a hydroxyl group.164–168 Since combination of a carbon-centered radical with molecular oxygen is either diffusion-controlled or nearly so (k \(\cong\) 2 x 109 M‑1s-1),169 any other reaction of the radical taking place in the presence of oxygen must be rapid enough to happen before O2 reaches the radical center. An example of a cyclization reaction that is fast enough to meet this criterion is shown in eq 32.164 Since the oxygen concentration in the reaction mixture is approximately 1 x 10-3 M, the rate constant for cyclization needs to be at least 1 x 107 s-1 in order for an acceptable yield of a cyclic product to be obtained.166b A proposed mechanism for replacement of a halogen atom with a hydroxyl group is given in Scheme 24.170
Other reagents and reaction conditions also cause replacement of halogen atoms with hydroxyl groups. These include Et3B–O2 initiated reaction of a deoxyiodo sugar with molecular oxygen171,172 and adding O2 to a reaction mixture in which AIBN is both initiator and reactant.166 Another set of conditions leading to replacement of a halogen atom with a hydroxyl group consists of reacting a deoxyiodo sugar with NaBH4 and O2 in the presence of a catalytic amount of Co(salen) (eq 33173).162,173–175 A comparative study found Co(salen)-catalyzed oxidation reactions to be experimentally more convenient than those with a tin hydride.162
H. Water-Soluble Halides
Most halogenated carbohydrates used in synthesis are rendered soluble in organic solvents by the introduction of various hydroxyl protecting groups, but sometimes it is useful to be able to conduct dehalogenation reactions in aqueous solution on unprotected or partially protected carbohydrates. Such a situation requires a water-soluble replacement for the tin and silicon hydrides typically used. Water-soluble hydrogen-atom transfers can be formed by replacing the alkyl substituents normally attached to tin or silicon with more polar ones. This replacement produces hydrides that are sufficiently soluble in water to allow simple reduction to take place in aqueous solution (eq 34176 and eq 35177).
There are potential advantages to conducting reactions in water, advantages that extend beyond substrate solubility.176,177 One of these is that reactions conducted in aqueous solution may exhibit new reactivity because, rather than taking place in an essentially nonpolar liquid, these reactions occur in a polar, heavily hydrogen-bonded solvent. Since water generally does not participate in radical reactions, it is effectively an inert solvent. Also, using water as the reaction medium can reduce or even eliminate the need for recycling or disposal of organic solvents.
I. Hypohalites
Hypohalites are intermediates in the formation of alkoxy radicals. Reactions of hypohalites and the alkoxy radicals they produce are discussed in Chapter 6.
J. Organotin Hydrides
The majority of reactions of halogenated carbohydrates use an organotin hydride (nearly always Bu3SnH) as a hydrogen-atom transfer and as a source of a chain-carrying-radical; however, there are serious problems associated with the toxicity of tin-containing compounds and the difficulty in removing their residues from reaction products. A variety of solutions to these problems have been proposed. Since most of these solutions apply not only to reactions of halogenated carbohydrates but also to those of a broad range of carbohydrate derivatives, the solutions will not be discussed here (and then repeated in later chapters); rather, they are gathered together in Appendix 1, which is found at the end of this book. | textbooks/chem/Organic_Chemistry/Radical_Reactions_of_Carbohydrates_(Binkley)/Radical_Reactions_of_Carbohydrates_II%3A_Radical_Reactions_of_Carbohydrates/01%3A_Advantages_and_Disadvantages_of_Radical_Reactions/II._Advantages_of_Radical_R.txt |
Regioselectively replacing a particular hydrogen atom in a carbohydrate with a halogen atom depends upon a radical abstracting one hydrogen atom from among the many present in a typical molecule. When this type of selectivity occurs, it sometimes is linked to radical philicity; that is, a hydrogen atom that is more electron rich than others in a molecule can be abstracted preferentially by a highly electrophilic radical. This regioselectivity caused by radical philicity only occurs when the transition state is early in a reaction. When the transition state is late, selective reaction also can take place, but in this case selectivity exists because abstraction of a particular hydrogen atom produces a carbon-centered radical that is much more stable than radicals formed by abstraction of other hydrogen atoms. Since there are relatively few carbohydrates where highly selective hydrogen-atom abstraction occurs and since hydrogen-atom abstraction typically is the first step in halogenation, regioselective, free-radical halogenation reactions are limited in number.
V. Summary
Halogen-atom abstraction by a tin-centered radical is a common reaction in carbohydrate chemistry. Ample evidence supporting radical intermediates in this type of reaction comes from chemical reactivity and from direct radical observation by ESR spectroscopy. Iodides are the most reactive of the halogenated carbohydrates. Bromides are slightly less so, but the reactivity of chlorides is considerably reduced. Fluorides are essentially unreactive. In dehalogenation reactions the transition-state structure is thought to involve partial tin–halogen and carbon–halogen bonds.
Simple reduction (replacement of a halogen atom with a hydrogen atom) occurs under mild reaction conditions. The halogen atom being abstracted can be attached to any carbon atom in the carbohydrate framework. Although the primary role of simple reduction is in the synthesis of deoxy sugars and deoxy nucleosides, this reaction also can be used to modify the reactivity of halogenated protecting groups. Replacement of halogen atoms with hydrogen atoms can convert a group that is difficult to hydrolyze into one that does so more easily.
Simple reduction of anomeric halides must compete with group migration when there is an acyloxy group at C-2. This migration, which is useful in the synthesis of 2-deoxy sugars, is most likely to occur when the concentration of the hydrogen-atom transfer (e.g., tri-n-butyltin hydride) is held at a very low level.
Halogen-atom abstraction often is the first step in the addition of a carbohydrate radical to a compound containing a multiple bond. Such addition will occur in an intramolecular fashion if the multiple bond is electron-deficient because under the proper conditions a nucleophilic carbohydrate radical adds to an electron-deficient multiple bond more rapidly than it abstracts a hydrogen atom. When the radical center and the multiple bond are in the same molecule and easily come within bonding distance, cyclization takes place so readily that it will occur even if the multiple bond is not electron-deficient. A characteristic of cyclization reactions is that formation of the new carbon–carbon bond often occurs in a highly stereoselective fashion.
Halogenated carbohydrates participate in a variety of less common reactions. These include double bond formation, internal hydrogen-atom abstraction, addition to molecular oxygen, cyclopropane ring opening, and radical anion formation.
Free-radical bromination produces several types of brominated carbohydrates. Bromination of benzylidene acetals leading to formation of bromodeoxy benzoates is a standard reaction in carbohydrate synthesis. Modifications of this reaction also are known; thus, the cation produced following benzylidene acetal bromination can be intercepted by water to give a hydroxy benzoate. Carbohydrates protected as benzyl ethers react with bromine to produce unstable bromides that, in turn, react with water to give benzaldehyde and the deprotected carbohydrate. Reaction with bromine of carbohydrates that do not contain benzyl or benzylidene protection regioselectively replaces a hydrogen atom on one of the carbon atoms attached to the ring oxygen atom. | textbooks/chem/Organic_Chemistry/Radical_Reactions_of_Carbohydrates_(Binkley)/Radical_Reactions_of_Carbohydrates_II%3A_Radical_Reactions_of_Carbohydrates/02%3A_Halogenated_Compounds/IV._Halogenation.txt |
A carbohydrate derivative that contains a sulfur atom bonded to two carbon atoms is capable of forming carbon-centered radicals. A common pathway for radical formation in compounds of this type is homolytic cleavage of a carbon–sulfur bond brought about by group abstraction.
03: Compounds with Carbon–Sulfur Single Bonds
A carbohydrate derivative that contains a sulfur atom bonded to two carbon atoms is capable of forming carbon-centered radicals. A common pathway for radical formation in compounds of this type is homolytic cleavage of a carbon–sulfur bond brought about by group abstraction (eq 1). When the sulfur atom in a C–S bond is part of an electronegative group, as is the case in a glycosyl phenyl sulfone, electron transfer of the type shown in Scheme 1 represents another pathway to carbon-centered radical formation. A beginning point for discussing these reactions is examining their possible mechanisms.
2. Reaction Mechanisms
A. Group Abstraction
Two mechanisms are considered to be reasonable possibilities for the carbon–sulfur bond cleavage described by the reaction shown in eq 1.1 The first is the bimolecular, homolytic, substitution (SH2) reaction pictured in Scheme 2, and the second is a stepwise process that involves formation of an intermediate (1) with a hypervalent sulfur atom (Scheme 3). The choice between these two hinges on the existence of 1.
There is little experimental evidence upon which to base a decision about formation of a compound with a hypervalent sulfur atom during carbon–sulfur bond cleavage, but reaction of the thioacetal 2 with the tri-n-butyltin radical provides some suggestive information (Scheme 4).2 Although this reaction produces BuOCH2· (3), the effect of temperature on the ESR signal for this radical is unexpected because the intensity of the signal increases as the temperature in the ESR cavity rises. (Signals due to radicals arising from reaction of bromides with Bu3Sn· decrease with rising temperature due to leveling of the Boltzmann distribution of spin states.2) A possible explanation for this behavior is that a slow, temperature-dependent reaction between the thioacetal 2 and Bu3Sn· produces the hypervalent, sulfur-centered radical 4 (not observable by ESR), an intermediate that then fragments rapidly to give the ESR observable radical 3 (Scheme 4).2
Molecular orbital calculations also have been used to study the possibility of formation of intermediates with hypervalent sulfur atoms. When these calculations focus on the reactions of sulfides, they do no support the existence of such intermediates.1,3–6
B. Electron-Transfer
Electron transfer to a sulfur-containing carbohydrate naturally depends upon such a compound having a group that readily accepts electrons. Sulfones meet this requirement and, thus, are prime candidates for electron-transfer reaction.7 Two proposed mechanisms showing how such transfer could lead to cleavage of a carbon–sulfur bond are shown in Scheme 5. In one of these a sulfone reacts with an electron donor (e.g., SmI2) to produce a radical anion (5) that then fragments to give an anion and a carbon-centered radical. In the other, dissociative electron transfer forms an anion and a carbon-centered radical directly from reaction of a sulfone with SmI2.
3. Alkylthio and Arylthio S
The identity of the carbon–sulfur single bond broken during reaction of a carbohydrate that has two such bonds depends upon the stability of the carbon-centered radical being formed. If the sulfur atom is part of a methylthio,8–11 ethylthio,12–15 or arylthio15–24 group, radical stability favors producing a carbohydrate radical rather than a methyl, ethyl, phenyl, or p-tolyl radical (Scheme 6).
4. Dithioacetals
Dithioacetals react with tri-n-butyltin hydride to replace first one, and then the second, alkylthio group with a hydrogen atom (Scheme 10).30 Because the first group is replaced more rapidly than the second, good yields of compounds with a single sulfur atom are obtained under the proper reaction conditions.31–35 The greater reactivity of the first ethylthio group in these compounds is due to formation of an intermediate, carbon-centered radical that is stabilized by the sulfur atom in the remaining ethylthio group.
An unsaturated dithioacetal in which the double bond is properly positioned undergoes intramolecular radical addition.31–34 Reaction typically involves capture of the first-formed, carbon-centered radical by the multiple bond; thus, in the reaction shown in Scheme 11, the major product has a new ring system with an ethylthio substituent.31 Here again the greater reactivity of the first ethylthio group allows reaction to occur with no detectable loss of the second.
5. Thiocarbonates and Dithi
Thiocarbonates and dithiocarbonates are compounds in which at least one sulfur atom is bonded to the carbon atom of a carbonyl group. The reactivity of these compounds is similar to that of the sulfur-containing compounds already discussed in that reaction begins with carbon–sulfur bond cleavage producing the more stable of the possible carbon-centered radicals; thus, in the reaction shown in eq 5, product identity is consistent with forming an intermediate allylic radical from reaction of a thiocarbonate.36
Addition of Bu3Sn· to the dithiocarbonate 9 is the first step in an addition-elimination reaction that produces the tin-containing compound 11 (Scheme 12).37 The stability of CH3SC(=O)S· is critical to this type of reaction because it, rather than Bu3Sn·, is expelled when a radical such as 10 forms a tin-containing product.37,38 Since Bu3Sn· addition to a double bond often is reversible, 10 sometimes may break a carbon–tin bond causing an undetectable regeneration of Bu3Sn· and the substrate 9. | textbooks/chem/Organic_Chemistry/Radical_Reactions_of_Carbohydrates_(Binkley)/Radical_Reactions_of_Carbohydrates_II%3A_Radical_Reactions_of_Carbohydrates/03%3A_Compounds_with_Carbon%E2%80%93Sulfur_Single_Bonds/1._Introduction.txt |
Compounds with carbon–sulfur single bonds are substantially less reactive with tin- and silicon-centered radicals than are compounds with C–S double bonds. Among carbohydrates these double bonds are almost always part of O-thiocarbonyl groups. (The reactions of O-thiocarbonyl carbohydrate derivatives are discussed in Chapter 12.) The reaction shown in eq 6 illustrates the greater reactivity of a C–S double bond when compared to a C–S single bond because only the O-thiocarbonyl group in the 1‑thioglycoside 12 reacts even though an ethylthio group is present in the molecule.39 Greater reactivity of a carbon–sulfur double bond also can be seen in the reaction shown in Scheme 13, where Bu3Sn· reacts only with the O‑thiocarbonyl group.26 A quantitative measure of the reactivity of C–S single and double bonds comes from comparing absolute rate constants for their reactions; thus, rate constants for reaction of (Me3Si)3Si· with C10H21SC6H5 and C6H11OC(=S)SMe are less than 5 x 106 and 1.1 x 109 M-1s-1, respectively, at 21 oC.40 The reactions in Schemes 12 and 13 also illustrate the ease of fragmentation of a carbon–sulfur single bond when a radical is centered on an adjacent carbon atom.
7. Sulfones
Radicals are involved in both the synthesis and the reactions of carbohydrate sulfones. Sulfones produce carbon-centered radicals by group abstraction, dissociative electron-transfer, and photochemical bond homolysis. Sulfone synthesis takes place when a sulfonyl radical adds to an unsaturated carbohydrate.
8. Thiols and Thiyl Radical
In compounds with an H–S bond, hydrogen-atom abstraction to produce a sulfur-centered radical (eq 10) is a significant (sometimes the exclusive) reaction pathway. Such reactivity exists because thiols are among the most effective hydrogen-atom transfers in organic chemistry. Rate constants for hydrogen-atom abstraction by primary, secondary, and tertiary, carbon-centered radicals from thiophenol range from 0.8 x 108 to 1.5 x 108 M-1s-1 at 25 oC.70
A characteristic reaction of a thiyl radical is addition to a carbon–carbon multiple bond.71–85 In the reaction shown in Scheme 21, for example, addition of the thiyl radical 23 to the unsaturated carbohydrate 21 leads to formation of the S-disaccharide 22.77 This reaction is not only regiospecific but hydrogen-atom abstraction from 20 is so much faster than reaction with the molecular oxygen dissolved in the reaction mixture that an inert atmosphere is not required for successful S-disaccharide formation. Similar radical addition takes place between the thiol 20 and various D-glycals, including the D-glucal 24 (eq 11).78
Even though the most common radical reaction of a compound with an H–S bond is hydrogen-atom abstraction, under some conditions the HS group is replaced by a hydrogen atom (eq 12).86
Although a carbohydrate containing a sulfur-centered radical typically is generated by hydrogen-atom abstraction from a thiol, the reaction shown in eq 13 forms a thiyl radical by the addition-elimination sequence pictured in Scheme 22.87 Critical to chain propagation in this reaction is the removal of the sulfur atom from 25 by reaction with triphenylphosphine.
9. Summary
Tin-centered radicals react with carbohydrates that contain methylthio, ethylthio, or phenylthio groups to produce carbon-centered radicals. Two mechanisms have been proposed for such a reaction. The first is a concerted SH2 process, and the second is a stepwise reaction that forms an intermediate with a hypervalent sulfur atom. Molecular orbital calculations favor the concerted process.
Compounds with alkylthio or arylthio groups break the C–S bond that produces the more stable, carbon-centered radical. This means that when fragmentation takes place in a carbohydrate containing a methylthio, ethylthio, or phenylthio substituent, a carbohydrate radical forms rather than an alkyl or aryl radical. Reactions that begin with carbon–sulfur bond cleavage often lead to either simple reduction or radical cyclization. Similar reactions occur when the sulfur atom is part of a dithioacetal, thiocarbonate, or dithiocarbonate.
When the carbon–sulfur bonds in a carbohydrate are part of a sulfone and when an electron-donor (usually SmI2) is present, bond cleavage occurs via an electron-transfer reaction. The resulting radical combines rapidly with a second molecule of SmI2 to produce an organosamarium intermediate that undergoes reactions typical of an organometallic compound (e.g., proton abstraction, β elimination, or addition to an aldehyde or ketone). Radical cyclization is one of the few reactions fast enough to occur before this combination takes place.
If a compound has a hydrogen–sulfur bond, the major reaction pathway usually is hydrogen-atom abstraction to form a sulfur-centered radical. This radical adds readily to a carbon–carbon double bond. | textbooks/chem/Organic_Chemistry/Radical_Reactions_of_Carbohydrates_(Binkley)/Radical_Reactions_of_Carbohydrates_II%3A_Radical_Reactions_of_Carbohydrates/03%3A_Compounds_with_Carbon%E2%80%93Sulfur_Single_Bonds/6._O-Thiocarbonyl_Compounds.txt |
Carbohydrates containing selenium–carbon bonds react with tin and silicon hydrides to generate carbon-centered radicals. Phenyl selenides are the most common type of selenium-containing carbohydrate used in radical formation. As radical precursors, phenyl selenides rival the reactivity of bromides and iodides. (Absolute rate constants for reaction of simple organic iodides,1 phenyl selenides,2 and bromides2,3 with (CH3Si)3Si· are 4.0 x 109, 9.6 x 107, and 2.0 x 107 M-1s-1, respectively.)
Selenophenyl glycosides have a distinct advantage over the corresponding iodides and bromides when it comes to radical formation because anomeric phenyl selenides are thermally more stable than the corresponding anomeric halides. Anomeric iodides are, in fact, too unstable to have a significant role in generating pyranos-1-yl or furanos-1-yl radicals. Anomeric bromides are acceptable radical precursors in many instances, but when they are too unstable, phenyl selenides become attractive alternatives. Phenyl selenide advantage is apparent in the generation of furanos-1-yl radicals where glycosyl bromides typically are unable to survive the heating at reflux in benzene or toluene that normally is used in such reactions. Selenophenyl glycosides are stable enough under these conditions to avoid nonradical, thermal decomposition and, therefore, they are able to form the desired radicals.4,5
Organotellurium compounds represent another source of carbohydrate radicals. Although selenides are used much more frequently as starting materials for radical formation, tellurides undergo many of the same types of reaction. A problem with many tellurium-containing compounds is that they decompose so readily that they can be difficult to purify and store.
04: Selenides and Tellurides
A. Reaction Mechanism
The SH2 mechanism pictured in Scheme 1 and the stepwise process shown in Scheme 2 both are considered possibilities for explaining the reaction between a phenyl selenide and a tin- or silicon-centered radical.6 The tris(trimethylsilyl)silyl radical (1) is used in illustrating these two mechanisms because it plays a significant role in the choice between them.2 A way for making this choice begins with the observation that the absolute rate constant for reaction of 1 with cyclohexyl phenyl selenide to give cyclohexane (kSe) is 9.6 x 107 M-1s-1 and the rate constant for reaction of 1-bromooctane to give octane (kBr) is 2.0 x 107 M-1s-1. If every cyclohexyl and 1-octyl radical is formed irreversibly and abstracts a hydrogen atom from (CH3Si)3SiH (Scheme 3), then the ratio kSe/kBr should be equal to the ratio of cyclohexane to octane formed when an equal-molar mixture of the selenide and bromide react with a limited amount of 1. When an experiment to test this possibility is conducted, the ratio of cyclohexane to octane in the product mixture is 0.08, a value far less than the 4.8 ratio predicted from the absolute rate constants (Scheme 3).2 This result is inconsistent with a process in which both the bromide and selenide react according to the SH2 mechanism shown in Scheme 3. The 0.08 ratio is consistent with the bromide reacting as pictured in Scheme 3 but the selenide producing an intermediate that can return to the starting materials (Scheme 4). A likely intermediate in such a reaction is one with a hypervalent selenium atom.2 The results from this comparative experiment, therefore, favor the stepwise mechanism for selenide reaction shown in Scheme 2 over the concerted process pictured in Scheme 1.
B. Reactions
1. Reduction
Carbohydrates that have a selenophenyl group attached to a pyranoid ring react with tri-n-butyltin hydride, triphenyltin hydride, or tris(trimethylsilyl)silane to replace the selenium-containing group with a hydrogen atom.7–19 Such a reaction is the final step in the disaccharide synthesis shown in Scheme 5.7 Although reduction involving selenophenyl group replacement is usually at C-2 in monosaccharides or at C-2' in disaccharides and nucleosides, reaction in monosaccharides also has been observed at C-117,18 and at C‑6.20
The polymer 3,21,22 with selenium attached to the aromatic rings in polystyrene, reacts with the glycal 2 in the presence of the partially protected sugar 4 to produce the carbohydrate-containing polymers 5 and 6 (Scheme 6).21 (The polymer-bound reagent 3 has the advantage that it is odorless, safer, and more convenient to handle than C6H5SeCl, which is toxic and foul smelling.21) Reaction of 5 and 6 with tri-n-butyltin hydride releases the carbohydrates from the polymers and, at the same time, completes the reduction process.
Replacing a selenophenyl group in a five-membered ring by a hydrogen atom is a common reaction for nucleosides and nucleoside analogs.23–32 This replacement can be conducted either at 80-110 oC with AIBN initiation (eq 1),23 or at room temperature with Et3B–O2 as the initiator (eq 2).26 [Selenophenyl group replacement, when initiated by Et3B–O2, can occur at temperatures as low as ‑75 oC (eq 3).31] Tri-n-butyltin hydride is the normal hydrogen-atom transfer in such reactions, but tris(trimethylsilyl)silane (eq 4)28 and 1,4-cyclohexadiene (eq 5)30 also are effective in this role. Yields from reaction of Bu3SnH with carbohydrates containing selenophenyl groups remain high when the oxygen atom in a furanoid ring is replaced by a sulfur atom.33–35
2. Addition
Carbohydrate radicals generated from phenyl selenides undergo characteristic addition reactions with compounds containing multiple bonds.19,36–39 These radicals add not only to decidedly electron-deficient double bonds, such as that found in t-butyl acrylate, but also to less electron-deficient double bonds, such as that present in styrene.19,37,38 Product yields from addition to styrene are lower, however, because hydrogen-atom abstraction from tri-n-butyltin hydride to give the reduction product competes effectively with addition to a less electron-deficient multiple bond (eq 6). Conditions are critical to the success of these addition reactions because only hydrogen-atom abstraction is observed unless Et3B–O2 is the initiator and the reaction is run at room temperature.38 As is typical for reactions of this type (i.e., ones that form intermediate pyranos-1-yl radicals), the stereoselectivity of addition is controlled by the kinetic anomeric effect [Section III.B of Chapter 11 in Volume I].
3. Cyclization
The reaction shown in Scheme 740 illustrates the established preference of unsaturated radicals for forming five-membered rings even when six-membered ones are possible.40–43 This reaction (Scheme 7) also reveals a complication in radical cyclization caused by internal hydrogen-atom abstraction, a process that leads in this instance to epimerization at C-5. Only carbon-centered radicals that are very reactive, such as the primary radical 7, are able to abstract a hydrogen atom from a carbon–hydrogen bond fast enough to be of consequence. Epimerization at C-5 in this reaction can be reduced or even eliminated by increasing the tri-n-butyltin hydride concentration to the point that internal hydrogen-atom abstraction by 7 no longer competes successfully with abstraction from Bu3SnH (Scheme 7).40
Cyclization of unsaturated carbohydrates in which a selenophenyl group is attached to a pyranoid ring is marked by a surprising variety of new ring systems that can be produced. In addition to the expected five-40–44 and six-membered45 rings, formation of seven-membered,46 eight-membered,47,48 and even nine-membered49–54 rings also takes place. Larger rings usually are generated when a radical center and a multiple bond are linked through a silicon–oxygen connector.46–52 Reactions of this type often produce carbohydrates in which two saccharide units are linked by a methylene bridge (Scheme 8.)48 Although bridges containing silicon and oxygen atoms are common, reactions also occur between monosaccharides connected by other combinations of atoms.53,54
In cyclization reactions a selenophenyl group attached to a furanoid ring behaves in a manner similar to one attached to a pyranoid ring; that is, reaction produces a radical that adds to a connected multiple bond. The connecting group sometimes contains a nitrogen atom (eq 7)55 or an oxygen atom56–58 (eq 856) or the collection of atoms that make up an ester linkage,59,60 but as is the case for compounds with pyranoid rings, a radical centered in a furanoid ring frequently has the unsaturated group tethered to the five-membered ring through a silicon–oxygen bridge61–69 (eq 961). Reported radical cyclization of this type, such as that shown in eq 10,62 often involves reaction of a nucleoside.
Although in most carbohydrates a selenophenyl group undergoing reaction is bonded to a ring carbon atom, cyclization70,71 (and addition72) reactions also can start with a ring-open structure. An example is given in eq 11.70 Cyclization of the ring-open selenide 8 begins with electron transfer from samarium(II) iodide. The intermediate samarium ketyl formed during this reaction displaces a benzyl group from selenium to give a ring system that contains a selenium atom (Scheme 9).73 This is an unusual method for ring formation because it takes place by group displacement rather than addition to a multiple bond.
4. Group Migration
Group migration is a characteristic reaction of a pyranos-1-yl radical that has an acyloxy group attached to C-2. Since phenyl selenides are one type of precursor for these radicals, it is reasonable to expect selenides to be substrates for such a migration.5,74,75 The reaction shown in eq 12 justifies this expectation.5 (Acyloxy-group-migration reactions are discussed in Section V. of Chapter 8.)
5. Radical-Cation Formation
In the reaction shown in Scheme 10 abstraction of the selenophenyl group from 9 by Bu3Sn· gives the pyranos-1‑yl radical 10, which then fragments to produce the radical cation 11.76 This radical cation then undergoes a combination of cyclization, proton loss, and hydrogen-atom abstraction to give the final product. Investigating radical-cation formation from nucleotides containing selenophenyl groups is used to study the mechanism of DNA strand scission.77,78
6. Radical Combination
Replacement of a selenophenyl group with a hydrogen atom typically depends on the ability of a reagent such as tri-n-butyltin hydride both to provide a chain-carrying radical (Bu3Sn·) and to serve as a hydrogen-atom transfer. If this reagent is replaced by one that lacks hydrogen-donating ability but retains the capacity to generate a chain-carrying radical, selenophenyl group loss still will occur, but hydrogen-atom abstraction cannot be depended upon to complete the reaction. If unsaturated reactants are present, radical addition is possible, but if such compounds are absent, radical combination can take place (eq 13).4 {[Combination of the type shown in eq 13 also happens when pyranos-1-yl radicals are formed from glycosyl bromides [Chapter 2, Section III.G.1] and glycosyl phenyl sulfones [Chapter 3, Section VII.B.1.c.]}
III. Tellurides
The organotellurium compounds that are used as radical precursors in carbohydrate chemistry usually are synthesized by a nucleophilic displacement reaction such as that pictured in eq 14.79
The majority of compounds prepared in this way are anomeric tellurides. Furanosyl tellurides are relatively unstable and tend to decompose within a few days,80–82 but although their pyranosyl counterparts can exist unchanged in the solid state for months,79 heating or exposing pyranosyl tellurides to UV light causes epimerization (eq 15).83
Two procedures, both of which involve photolysis, cause radical reaction of carbohydrate tellurides. The first of these is photochemical homolysis of a carbon–tellurium bond, a reaction that generates the more stable of the two possible, carbon-centered radicals (Scheme 11).
An example of reaction brought about in this way is found eq 16.83 The second procedure for radical formation from a carbohydrate telluride calls for photochemical decomposition of N‑acetoxy-2-thiopyridone to produce a methyl radical that then reacts with the telluride (Scheme 12).
Equation 17 describes a cyclization reaction initiated in this way.84 Reactions of carbohydrate radicals formed from tellurides include cyclization,84,85 addition,86–89 reduction,83 and group migration.83 It is reasonable to assume that reaction of a carbohydrate telluride with a methyl radical involves, as molecular orbital calculations indicate, formation of an intermediate with a hypervalent tellurium atom (Scheme 13).90
IV. Summary
Reactions of carbohydrate selenides with tin-centered and silicon-centered radicals produce carbon-centered, carbohydrate radicals. These radicals can undergo hydrogen-atom abstraction, intra- and intermolecular addition, group migration, radical-cation formation, and radical combination. Reduction and radical cyclization are the two most common of these reactions. Reduction is involved in the synthesis of 2-deoxy sugars and 2'-deoxy disaccharides and nucleosides. Cyclization, which is characterized by the formation of compounds with a variety of ring sizes, is the central reaction in a general procedure for converting monosaccharides into C-disaccharides. In this procedure glycosyl phenyl selenides are chosen to generate pyranos-1-yl and furanos-1-yl radicals because a carbohydrate with a selenophenyl group at C‑1 is thermally more stable than other substituents usually used for radical generation at this position. Carbohydrate tellurides are relatively unstable compounds that form carbohydrate radicals upon absorption of light or reaction with methyl radicals. | textbooks/chem/Organic_Chemistry/Radical_Reactions_of_Carbohydrates_(Binkley)/Radical_Reactions_of_Carbohydrates_II%3A_Radical_Reactions_of_Carbohydrates/04%3A_Selenides_and_Tellurides/II._Selenides.txt |
I. Introduction
Acetals are pervasive in carbohydrate chemistry. They link together saccharide units in oligo- and polysaccharides, provide the bonding in glycosides that joins carbohydrate and aglycon portions of a molecule, and furnish protection for hydroxyl groups during synthetic transformations. Because acetals have these vital protective and connective roles, their stability in the presence of free radicals is critical in enabling radical reactions selectively to modify other parts of a carbohydrate structure. Even though most acetals are stable in the presence of the carbon-centered radicals typically encountered in carbohydrate chemistry, there are reactions between heteroatom-centered radicals and acetals that are useful in carbohydrate transformation.
Ethers also serve as hydroxyl protecting groups during carbohydrate synthesis because, like acetals, they are unreactive in the presence of most carbon-centered radicals. When reaction of an ether or acetal does occur, it typically is hydrogen-atom abstraction from a carbon atom that has an attached oxygen atom.
05: Acetals and Ethers
Free-radical bromination of acetals and ethers is discussed in Section IV of Chapter 2.
III. Thiol-Catalyzed Reactions of Acetals: Polarity-Reve
Thiols act as catalysts for hydrogen-atom abstraction from acetals.1–10 The initiation phase in these reactions generates a thiyl radical that then abstracts a hydrogen atom from the acetal in the first propagation step (Scheme 1). This first step is reversible and a pseudo equilibrium is established that favors the reactants (RH and XS·). (The position of this equilibrium is based on the enthalpy calculated from the bond-dissociation energies given in eq 1.11) The overall process is driven by the second propagation step, a reaction that irreversibly converts one carbon-centered radical (R·) into another (R’·). The final step is rapid hydrogen-atom abstraction from the thiol by the newly formed, carbon-centered radical R’.
Hydrogen-atom abstraction from a molecule of substrate should have a lower transition-state energy when the abstracting radical is sulfur-centered (Scheme 1) rather than carbon-centered (Scheme 2). In reactions of this type the transition state can be described as a hybrid of valence-bond structures 1-4 (Figure 1).10 If the abstracting radical is carbon-centered, structures 1 and 2 are the major contributors to the hybrid; the charge-separated structures 3 and 4 are of little consequence. If, however, abstraction is done by a thiyl radical, not only are structures 1 and 2 important, but contribution from the charge-separated structure 3 also is significant.10 (In the case where a thiyl radical abstracts a hydrogen atom from an acetal, the valence-bond structures 1-4 can be represented in the more descriptive manner shown in Figure 2.) Significant contribution from 3 means that the energy required to reach the transition state for abstraction of a hydrogen atom is less than that needed when a carbon-centered radical abstracts the same hydrogen atom. Faster hydrogen-atom abstraction in propagation step 1 (Scheme 1) means that as R· is converted into R’· in step 2, the R· needed to continue the propagation sequence will be replenished more rapidly.
A structure such as 3 is a transition-state-stabilizing contributor in any hydrogen-abstraction reaction where a change in radical philicity takes place. Such a change occurs in propagation steps 1 and 3 in the thiol-catalyzed mechanism pictured in Scheme 1, but it does not take place at all in the uncatalyzed mechanism shown in Scheme 2. When a change in radical philicity occurs during a reaction, either due to abstraction of an electron-rich hydrogen atom by an electrophilic radical or abstraction of an electron-deficient hydrogen atom by a nucleophilic radical (propagation steps 1 and 3, respectively, in Scheme 1), the reaction is described as polarity-matched.4,10 If one radical must be converted into another by hydrogen-atom abstraction without benefit from a change in radical philicity (step 2 in Scheme 2), the reaction is described as being polarity-mismatched. The transition state for a polarity-matched reaction will be stabilized by contribution from the charge-separated, valence-bond structure 3 (Figures 1 and 2), but a polarity-mismatched reaction will not experience similar, transition-state stabilization. A polarity-matched reaction, therefore, will have transition-state stabilization that is denied to a polarity-mismatched reaction.
Although the combination of steps 1 and 3 in Scheme 1 achieves the same result as step 2 in Scheme 2 (see Figure 3), the polarity-matched steps in Scheme 1 can be fast enough that in combination they are more rapid, sometimes much more rapid, than the single, polarity-mismatched step in Scheme 2. When this occurs, the added thiol is said to catalyze the entire reaction by polarity-reversal catalysis.10 The next three sections describe reactions that either are made possible by or have improved yields due to polarity-reversal catalysis.
IV. Ring Opening of Specially Designed Acetals
Search for compounds with more versatile reactivity than that provided by a 4,6-O-benzylidene group has stimulated development of some specially designed structures.14–18 The acetal 23, which fits into this “specially designed” category, reacts with Bu3Sn· to form the aryl radical 24. The iodine-atom abstraction that generates 24 is the first step in a sequence of radical reactions that culminates in producing the protected glycoside 25 (Scheme 9).14–16 An example of the synthetic usefulness of this reaction is found in the conversion of a tetrasaccharide containing four such protecting groups into one in which each group is transformed into an O-benzoyl group.15 The glycoside 26 is another cyclic benzylidene acetal with an aromatic iodo substituent that undergoes a sequential radical reaction that leads to the corresponding deoxy benzoate 27 (eq 10).17 The reactions pictured in Scheme 9 and eq 10 are two more examples (in addition to those shown in equations 2 and 3) where trans-fused rings open to produce primary rather than a secondary radicals. Ring opening of the 4,6-O-benzylidene acetal 28 to give a secondary radical (eq 11) further supports the proposal made for the acetal 11 (eq 4) that for a more flexible, cis-fused ring system the direction of ring opening is controlled by radical stability rather than ring strain at the transition state. | textbooks/chem/Organic_Chemistry/Radical_Reactions_of_Carbohydrates_(Binkley)/Radical_Reactions_of_Carbohydrates_II%3A_Radical_Reactions_of_Carbohydrates/05%3A_Acetals_and_Ethers/II._Bromination_of_Acetals_and_Ethers.txt |
A. Abstraction by Alkoxy Radicals
Hydrogen-atom abstraction by alkoxy radicals from acetals and ethers is described in the next several sections. More information about the formation and reactions of alkoxy radicals is found in Chapter 6.
1. Abstraction From an Acetal
Intramolecular hydrogen-atom abstraction by an oxygen-centered radical from the central carbon atom in an acetal linkage is the “key” step in the orthoester formation pictured in Scheme 10.19 The radical phase of this reaction begins with photochemically initiated fragmentation of the hypoiodite 29. Internal hydrogen-atom abstraction followed by carbon–iodine bond formation completes the radical phase of the reaction. Formation of the orthoester 31 from the iodide 30 then occurs by an ionic process.
2. Abstraction From an Ether
Internal hydrogen-atom abstraction from a benzyloxy group produces a highly stabilized radical (32) that can be an intermediate in the formation of a benzylidene acetal (Scheme 11). This type of reaction takes place in good yield when the substrate contains adjacent O-benzyl and hydroxyl groups (Scheme 11).20 The reaction in Scheme 12 illustrates the type of transformation possible. In this reaction the hypoiodite 33 is not just assumed to exist but is actually observed by 13C NMR spectroscopy. Such direct observation of a hypoiodite is rare.
It is not essential to have aromatic stabilization in the developing radical for internal hydrogen-atom abstraction to take place.21–23 In the alkoxy radical 35 abstraction from a nearby methoxy group begins a process that ultimately unites the interacting groups as an acetal (Scheme 13).21 This reaction constitutes a regioselective transformation of a methoxy group that is in close proximity to an oxygen-centered radical.
3. Abstraction From an α-Aminoether
Internal hydrogen-atom abstraction by an alkoxy radical from an α‑aminoether linkage can lead to the same type of ring formation observed in reactions of acetals and other ethers. For example, 1,6-hydrogen-atom abstraction converts the alkoxy radical 36 into the α-amino radical 37. Combination of 37 with an iodine atom or reaction of 37 with I2 then produces a reactive iodide that cyclizes to give the spiro nucleoside 38 (Scheme 14).24,25
B. Abstraction by Carbon-Centered Radicals
Although internal hydrogen-atom abstraction usually involves an alkoxy radical, some carbon-centered radicals are capable of such reaction. One element associated with successful hydrogen-atom abstraction is that ring strain in the transition state be minimal. (Ring strain usually is minimized when hydrogen-atom abstraction involves a six-membered-ring transition state.26 Such a reaction can be described as a 1,5-hydrogen-atom transfer or 1,5-HAT.) A second characteristic of successful abstraction is that stabilization of the developing radical contribute to lowering the transition-state barrier.26 The need for radical stabilization means that primary27 and vinylic28,29 radicals are prime candidates for hydrogen-atom abstraction because their reactions typically lead to much more stable radicals; however, even a secondary radical will abstract a hydrogen atom internally if the developing radical is sufficiently stabilized.30 In the reaction shown in Scheme 15, the vinylic radical 39 abstracts a hydrogen atom from the adjacent O-benzyl group in route to the major products 41 and 42 (80% combined yield). The product 40, formed when 39 abstracts a hydrogen atom from (C6H5)3SnH, is produced in only 8% yield, demonstrating that intermolecular reaction from this tin hydride has difficulty competing with internal hydrogen-atom abstraction.29
It is often difficult to predict the extent of internal hydrogen-atom abstraction when a reactive, carbon-centered radical is formed in the presence of an effective hydrogen-atom transfer. For example, generating the radical 39 with (C6H5)3SnH present in solution still results primarily in internal reaction (Scheme 15);29 in contrast, in the reaction shown in eq 12 deuterium incorporation demonstrates that even though a primary radical is formed, abstraction from Bu3SnH is more rapid than internal 1,4- or 1,5-HAT.31 | textbooks/chem/Organic_Chemistry/Radical_Reactions_of_Carbohydrates_(Binkley)/Radical_Reactions_of_Carbohydrates_II%3A_Radical_Reactions_of_Carbohydrates/05%3A_Acetals_and_Ethers/V._Internal_Hydrogen-atom_abstraction_in_Acetals_and_Ether.txt |
Radical cyclization depends upon having a radical center and multiple bond held in close enough proximity for internal addition to take place. In carbohydrates an ether linkage often is the means for connecting these two reactive centers. In the reaction shown in Scheme 16, for example, a carbohydrate (43) with a radical precursor at C-1 is connected by a silyl ether linkage at C-2 to a substituent containing a double bond.32 Radical cyclization to give the silyl ether 44 creates a new carbon–carbon bond. Nonradical ring opening of 44 produces a silicon-free carbohydrate with an extended, carbon-atom chain.
Acetals and nonsilyl ethers also act as tethers that connect reactive centers during radical cyclization. In the reaction shown in Scheme 17, for example, the acetal linkage holds the double bond and the radical center in 45 close enough for ring formation to occur.33 In a similar manner an ether linkage connects the reactive centers during the cyclization reaction shown in eq 13.34 Unlike silyl ethers, the rings formed when acetals and nonsilyl ethers act as tethers usually are not destined for immediate ring opening. (Section IV.C of Chapter 19 contains additional examples of ethers and acetals serving as tethers in radical cyclization reactions.)
VII. Silyl Ether Rearrangement
Rearrangement takes place during radical cyclization involving some silyl ethers. The primary evidence for this rearrangement is the dependence of product ring size on the concentration of Bu3SnH, the hydrogen-atom transfer in these reactions. When the reaction shown in eq 14 is conducted in dilute Bu3SnH solution, the major product contains a six-membered ring,35 but at high Bu3SnH concentration reaction regioselectivity changes to give a product with a five‑membered ring.36,37 This concentration dependence can be explained by the more rapidly formed, but less stable, radical 46 having sufficient time and energy, when the concentration of Bu3SnH is low, to be converted into the more stable radical 47, either by a rearrangement that involves a cyclic transition state or by a fragmentation-addition sequence (Scheme 18).37 At high Bu3SnH concentration hydrogen-atom abstraction occurs before ring expansion can take place.
VIII. Summary
The free-radical bromination of a benzylidene acetal is a standard procedure in carbohydrate chemistry for ring opening that results in the formation of bromodeoxy sugars. Ring opening in the absence of bromine occurs when 4,6-O-benzylidene acetals react with peroxides in the presence of a thiol catalyst. Hydrogen-atom abstraction by an electrophilic, thiyl radical is the first step in this reaction. This is also the first step in reactions of other acetals leading to epimerization and deoxygenation.
Ethers, like acetals, serve as protecting groups during carbohydrate synthesis, but this protection is not total because both ethers and acetals undergo hydrogen-atom abstraction in the presence of reactive, electrophilic radicals. These radicals can be sulfur-, oxygen-, or bromine-centered. When hydrogen-atom abstraction by an alkoxy radical is intramolecular, it typically is highly regioselective and can lead to formation of a new ring system.
Acetals and ethers, including silyl ethers, have a connective role in radical cyclization reactions. The radical center and the multiple bond involved in a cyclization reaction are often joined together by an acetal or ether linkage.
II. Alkoxy Radicals
Radical formation by abstraction of a hydrogen atom potentially can convert a partially protected carbohydrate into either an oxygen-centered or carbon-centered radical (Scheme 1). In practice, however, such abstraction produces only carbon-centered radicals because their greater stability is reflected in the transition state leading to their formation. Because hydroxyl groups in unprotected or partially protected carbohydrates do not react with radicals typically present during transformation of carbohydrates (e.g., Bu3Sn·, (Me3Si)3Si·, or various carbon-centered radicals), protecting these groups is not necessary to prevent them from becoming involved in radical reactions.
06: Alkoxy Radicals
Although oxygen-centered radicals are not formed in significant number by direct hydrogen-atom abstraction from carbohydrates, these radicals can be produced indirectly in high yield from carbohydrate derivatives. Since such radicals are capable of internal reaction, intramolecular abstraction provides a pathway for oxygen-centered radicals to produce specific, carbon-centered radicals and, in so doing, to participate in synthetically useful reactions. (Alkoxy and alkoxyl are both terms used in describing oxygen-centered radicals derived from alcohols. The alkoxy designation will be used here.)
III. Summary
Alcohols are converted indirectly into alkoxy radicals through intermediate hypoiodites, nitrates, and phthalimides. A common reaction of alkoxy radicals is hydrogen-atom abstraction. This reaction becomes synthetically useful when the abstraction is internal because regioselective formation of a carbon-centered radical takes place. This selectivity depends on a combination of factors that include transition-state ring size, stability of the developing radical, and polarity matching between reacting atoms. Alkoxy radicals also can undergo carbon–carbon bond fragmentation that produces a carbonyl group and a carbon-centered radical. No matter which pathway is taken by the alkoxy radical (hydrogen-atom abstraction or β fragmentation), the resulting carbon-centered radical can undergo new ring formation, epimerization at a chiral center, ring opening, and other reactions characteristic of this radical intermediate. | textbooks/chem/Organic_Chemistry/Radical_Reactions_of_Carbohydrates_(Binkley)/Radical_Reactions_of_Carbohydrates_II%3A_Radical_Reactions_of_Carbohydrates/05%3A_Acetals_and_Ethers/VI._Radical_Cyclization%3A_The_Role_of_Ethers_and_Acetals.txt |
Radical reactions of unprotected carbohydrates begin with hydrogen-atom abstraction from a carbon–hydrogen bond in the carbohydrate structure. Such reaction requires a radical more reactive than the tin- or silicon-centered ones that are common in reactions of carbohydrate derivatives. A hydrogen atom usually is abstracted from an unprotected carbohydrate by a hydroxyl radical (HO·), but sometimes the sulfate radical anion (SO4·-) is the abstracting agent. A beginning point for discussing radical reactions of unprotected carbohydrates is to examine how the abstracting radicals HO· and SO4·- are formed.
07: Unprotected Carbohydrates
A. Hydroxyl Radicals
1. Ionizing Radiation and Ultraviolet Light
γ-Radiolysis of water produces hydroxyl radicals along with the compounds, ions, and other radicals shown in eq 1.1–4 The yield of hydroxyl radicals in this reaction can be increased by adding N2O to the reaction mixture because hydrated electrons react with N2O to form hydroxyl radicals (eq 2).2–4 In N2O-containing solutions 85% of the radicals are HO· and 15% are H·.4 Both HO· and H· abstract hydrogen atoms from carbon–hydrogen bonds.4 Hydroxyl radicals (and hydrogen atoms) also can be produced by photolysis of water with ultraviolet light of wavelength less than 185 nm (eq 3).1
2. Reaction of H2O2 with Fe2+ and Ti3+
Reaction of H2O2 with Fe2+ (eq 4) or Ti3+ (eq 5) produces hydroxyl radicals.5–8 (These reagent combinations are sometimes described as radiomimetic, that is, imitating radiation.) Each hydroxyl radical produced is capable of abstracting a hydrogen atom from a carbon–hydrogen bond present in a molecule of substrate (eq 6). The Ti4+ generated by the reaction shown in eq 5 does not react further with the carbon-centered radical produced, but the Fe3+ formed in the reaction shown in eq 4 does;8 Fe3+ oxidizes an α-hydroxy radical to a carbonyl group while itself being reduced to Fe2+. Regeneration of Fe2+ from Fe3+ by the reaction shown in eq 7 means that when this reaction occurs, only a catalytic amount of Fe2+ may be necessary for complete decomposition of H2O2 (eq 4). (The combination of H2O2 and Fe2+ is known as Fenton’s reagent,9 and that of H2O2 and Te3+ as a Fenton-type reagent.8)
B. Sulfate Radical Anions
The sulfate radical anion (SO4·-) forms when the peroxydisulfate dianion (S2O82-) reacts with Ti3+ (eq 8).10 A low concentration of Cu2+ present in the reaction mixture enhances the rate of generation of SO4·- via the reactions shown in equations 9 and 10. The sulfate radical anion also forms from direct photolysis of S2O82- (eq 11).10
III. First Formed Radicals: Radicals Produced by
A hydroxyl radical is sufficiently reactive to abstract a hydrogen atom from any of the carbon atoms in an unprotected carbohydrate.3,6 The radicals produced by such a reaction often are referred to as “first-formed” radicals, a terminology that correctly implies further transformation is likely.6 The ESR spectrum produced by the mixture of radicals generated from reaction of even a simple sugar with hydroxyl radicals is understandably complex; nevertheless, in the reaction of D-glucose (the most heavily studied of the simple sugars) signals for all six of the first-formed radicals can be detected.
IV. Reactions of First-Formed Radicals
A. Reactions in Neutral Solution
In discussing the various products arising from reaction in neutral solution of first-formed radicals (typically reactions in which the hydroxyl radical is generated by γ-radiolysis), it is convenient to distinguish between products with a molecular weight less than or equal to that of the substrate and those with a higher molecular weight. Because first-formed radicals can undergo dimerization, disproportionation, elimination, and rearrangement, the number of possible reaction products is staggering; nevertheless, many of them have been identified.1,2,24,25
Product yields for reactions begun by γ-radiolysis of water can be expressed in terms of G-values; that is, the molecules or radicals formed per 100 eV of energy absorbed. A G‑value also can be used as a measure of substrate reacted. In the reaction of D-glucose shown in Scheme 2 the G-value for consumption of starting material is 5.6.2 The values for formation of products 11-13 are 0.95, 0.15, and <0.08, respectively. The G-values cited for D‑glucose and compounds 11-13 were determined in the presence of N2O to maximize the formation of the hydroxyl radical.
1. Low-Molecular-Weight Products
Although many low-molecular-weight products are formed in detectable amounts from reactions of simple sugars, the yields of most are quite low. Many of these compounds are produced by reactions of the first-formed radicals; for example, the major, low-molecular-weight product (11) from reaction of D-glucose is believed to arise by loss of the elements of water from the first-formed radical 10 (Scheme 2, path a).2 Another reaction of 10 that forms a low-molecular-weight product is loss of a hydrogen atom to give D-glucono-1,5-lactone (12) (Scheme 2, path b), and a third reaction is opening of the pyranoid ring in 10 by fragmentation of the bond between C-5 and the ring oxygen atom to give, after hydrogen-atom abstraction, a carboxylic acid (13) (Scheme 2, path c).2 All of the reactions of the radical 10 shown in Scheme 2 are driven, at least in part, by the stability gained from forming a C–O double bond.
2. High-Molecular-Weight Products
The products 11-13 (Scheme 2) and the other low molecular-weight products (more than twenty identified) account for less than half of the D‑glucose consumed during γ-radiolysis because most products formed have high molecular weights. Little is known about either the structure of the high-molecular-weight materials or the mechanism of their formation. One proposal is that dimerization of radicals that have lost the elements of water may be the first step in formation of some high-molecular-weight products (eq 12).2
B. Acid-Catalyzed Reactions
Under strongly acidic conditions (pH = 1) four of the first-formed radicals generated from D-glucose eliminate the elements of water to give in each case a carbonyl-conjugated radical.6,26 A proposed mechanism for this reaction, shown in Scheme 3, involves protonation of the hydroxyl group adjacent to a radical center in the first-formed radical 1 to produce an intermediate (14) with an excellent leaving group that departs to form a radical cation (15). This radical cation then deprotonates to give the carbonyl-conjugated radical 16. Another mechanistic possibility for forming 16 is a concerted reaction beginning with the protonated radical 14 (Scheme 3). Forming carbonyl-conjugated radicals by acid-catalyzed reaction also has been studied in noncarbohydrate systems.5,27,28
The acid-catalyzed reactions of three first-formed radicals produced from D-glucose deserve further comment. Two of these radicals, 17 (Scheme 4) and 20 (Scheme 5), do not undergo the carbonyl-group-forming reaction characteristic of the other first-formed radicals (Scheme 3).6 Although 17 could start along this pathway by producing the radical cation 18, deprotonation of 18 to give a carbon-centered radical with an adjacent carbonyl group cannot take place. Formation of a carbonyl-conjugated radical from 18 would require opening of the pyranoid ring (Scheme 4). Evidence against such reaction is that 17 is less reactive than other first-formed radicals, and when it does react, no carbonyl-conjugated radical can be detected. The first-formed radical 20 also must undergo ring opening if a carbonyl-conjugated radical is to be produced; in fact, ring opening in this case is necessary to form the radical-cation 21 (Scheme 5). The radical 20, which is the least reactive of the first-formed radicals derived from D-glucose, also gives no indication of forming a carbonyl-conjugated radical.6
The radical 22 is the third, first-formed radical produced from D-glucose that deserves further comment. This radical is noteworthy because it is the most reactive of the first-formed radicals. Protonation of 22 gives the intermediate 23 in which the leaving group has an axial orientation and, therefore, the p-type orbitals on C‑2 and the ring oxygen atom in 23 begin stabilizing the radical cation 24 as it starts to develop (Scheme 6).6
C. Base -Catalyzed Reaction
Proton abstraction from O-2 in the first-formed radical 25 begins a process that generates the ring-open radical anion 26 (Scheme 7). (Since an α-hydroxy radical is far more acidic than its parent alcohol,2 the proton attached to O-2 should be removed much more readily than any other proton in 25.) The radical anion 26 rapidly undergoes a proton transfer to produce the semidone 27, one of two semidiones formed from base-catalyzed reaction of a first-formed, D-glucopyranosyl radical.11,29 The second of these two (30) is proposed to arise from the radical 28 according to the mechanism outlined in Scheme 8.11 (Semidiones 27 and 30 are readily detected because for each of them the negative charge slows the rates of dimerization and reduction and thus leads to more prominent ESR spectra.11) | textbooks/chem/Organic_Chemistry/Radical_Reactions_of_Carbohydrates_(Binkley)/Radical_Reactions_of_Carbohydrates_II%3A_Radical_Reactions_of_Carbohydrates/07%3A_Unprotected_Carbohydrates/II._Radicals_That_Abstract_Hydrogen_Atoms_from_Unpr.txt |
In addition to dimerization (eq 12) carbonyl-conjugated radicals also can be reduced to anions in a pH-dependent reduction by Ti3+ (eq 13).30 At pH 1 the reaction shown in eq 13 is negligible, but at pH 7 this reaction becomes an important pathway for removing carbonyl-conjugated radicals from a reaction mixture.30
VI. Oxidative Degradation of Carbohydrates
The oxidative degradation of carbohydrates in the presence of base is another reaction that involves radical intermediates. Such reaction of D‑glucose begins with ring opening and deprotonation to give the enediolate anion 31 (Scheme 9).31,32 Oxidation of this anion with O2 produces the resonance stabilized radical 32, which then is converted to the peroxy radical 33 by addition of O2. Subsequent reduction of 33 gives an anion that ultimately fragments the C1–C2 bond to give a five-carbon aldonic acid (Scheme 9). Fragmentation of other carbon–carbon bonds also takes place because base-catalyzed isomerization of the 1,2-enediolate anion 31 produces the 2,3-enediolate anion 34 (Scheme 10). Reaction of 34 with O2 and fragmentation analogous to that shown in Scheme 9 cleaves the D-glucose structure into two-carbon-atom and four-carbon-atom carboxylic acids. Continued isomerization of this type (31 to 34) produces other enediolates that undergo similar fragmentation reactions.32
VII. Reactions of Polysaccharides
Study of radical reactions of polymeric carbohydrates is a challenging undertaking. The reactions that occur are documented by changes in physical properties that come from polymer degradation (e.g., reduced solution viscosity, differences in solubility, and gel formation). Changes in physical properties have been recorded in the reactions of the hydroxyl radical with cellulose,8,32 hemicellulose,32 starch,16 and various dextrans.33
Studies of dextrans [polymers of 1$\rightarrow$6 linked α-D-glucose with α-(1$\rightarrow$3) linked side chains], for example, show essentially indiscriminate attack by hydroxyl radicals produced from reaction of Ti3+ with H2O2.33 (The polymers studied had molecular weights ranging from 10,000 to 500,000 Da) These reactions cause depolymerizations (as evidenced by reduced solution viscosity) and an increase in the number of carbonyl and carboxyl groups in the polymer fragments. Upon lowering or raising the solution pH, the first-formed radicals appear to rearrange in a manner similar to the rearrangement of first-formed radicals derived from D-glucose.
VIII. Summary
The hydroxyl radical and the sulfate radical anion both abstract hydrogen atoms from unprotected carbohydrates. In each case first-formed, carbon-centered radicals are produced. The hydroxyl radical is so reactive that it shows little regioselectivity when reacting with D-glucose; that is, spectroscopic evidence indicates that hydrogen-atom abstraction occurs from each of the six carbon atoms in this molecule. The hydroxyl radical remains unselective in reaction with other simple sugars that contain only pyranoid rings, but it does regioselectively abstract H5' from the furanoid ring in sucrose. The sulfate radical anion is a more selective abstracting agent. Hydrogen-atom abstraction occurs primarily from C-2, C-5, and C-6 in α-D-glucopyranose and at C-1, C-5, and C-6 in β-D-glucopyranose.
First-formed radicals derived from D-glucose undergo acid-catalyzed rearrangement under strongly acidic conditions to produce carbonyl-conjugated radicals. Under basic conditions first-formed radicals produce radical anions that form semidiones. When oxygen is present in the reaction mixture, first-formed radicals react to give peroxy radicals. | textbooks/chem/Organic_Chemistry/Radical_Reactions_of_Carbohydrates_(Binkley)/Radical_Reactions_of_Carbohydrates_II%3A_Radical_Reactions_of_Carbohydrates/07%3A_Unprotected_Carbohydrates/V._Reactions_of_Carbonyl-Conjugated_Radicals.txt |
Carbohydrates containing typical O-acyl groups are unreactive under the reduction conditions (AIBN initiation, Bu3SnH, 80-110 oC) normally used for radical reactions. This lack of reactivity changes when O‑acyl groups become part of the more complex structures found in α-acyloxy ketones, methyl oxalyl esters, and p‑cyanobenzoates. For such compounds radical reaction with Bu3SnH under normal reaction conditions replaces the acyloxy group with a hydrogen atom.
There are conditions under which a less complex O‑acyl group (e.g., an O‑acetyl or O-benzoyl group) is replaced with a hydrogen atom. One set of conditions includes raising the reaction temperature dramatically, a change with potentially destructive consequences for the compounds involved. A more attractive approach depends upon photochemically promoted electron transfer to an esterified carbohydrate. Electron transfer (both photochemical and nonphotochemical) permeates the radical reactions of carboxylic acid esters; that is, many of these reactions either involve (or may involve) electron transfer.
Another way in which O-acyl groups participate in radical reactions is by group migration. When a radical centered at C-1 in a pyranoid or furanoid ring has an O-acyl group attached to C-2, this group will migrate to C-1 when the conditions are properly selected. Such migration provides an effective method for producing 2-deoxy sugars.
Although esters of carboxylic acids are rich sources for substrates in radical-forming reactions, the acids themselves also can produce radicals. Under the proper conditions carboxylic acids generate carboxyl radicals, intermediates that lose carbon dioxide to form carbon-centered radicals. Carboxyl radicals are generated by electrolysis of carboxylate anions and by the reaction of carboxylic acids with hypervalent iodine compounds.
08: Carboxylic Acids and Esters
A. α-Acyloxy Ketones
α‑Acyloxy ketones react with tri-n-butyltin hydride by replacing the acyloxy group with a hydrogen atom (eq 1).1 The importance of the carbonyl group to this replacement process is evident in the two reactions shown in eq 2. In the first of these a benzoate (1) containing a keto group forms a deoxy sugar in good yield, but in the second a benzoate (2) lacking such a group is unreactive.1 Even though reactions of α-acyloxy ketones lead to formation of deoxy sugars, the usefulness of such reactions is limited by the relatively small number of carbohydrates that either have the necessary substituents or easily can be converted into compounds that do.1,2
A proposed mechanism for group replacement in α-acyloxy ketones is pictured in Scheme 1. Both addition/elimination and electron-transfer/elimination sequences are presented as possibilities for acyloxy group loss. The addition-elimination possibility was proposed at the time of the discovery of this reaction,1 but the electron-transfer option was recognized as a viable alternative later when loss of the benzoyloxy group from α-(benzoyloxy)acetophenone was shown to involve electron transfer from Bu3Sn· to this α-acyloxy ketone.3 There is no decisive evidence favoring either mechanism.
B. Methyl Oxalyl Esters
Methyl oxalyl esters can be prepared easily by esterification of partially protected carbohydrates with methyl oxalyl chloride (Scheme 2).4 These esters react with tri-n-butyltin hydride to replace the methyl oxalyloxy group with a hydrogen atom.4–19 Studies of noncarbohydrate esters show that those derived from secondary and tertiary alcohols are suitable starting materials in this deoxygenation process, but esters of primary alcohols are not because they regenerate the alcohols from which they were synthesized.20 Most of the reactions of methyl oxalyl esters of carbohydrates are of compounds in which a tertiary hydroxyl group has been esterified. Many of these compounds are nucleosides.4,7–16 One reason that most methyl oxalyl esters are formed from tertiary alcohols is that the O-thiocarbonyl compounds commonly used for deoxygenation in the Barton-McCombie reaction (Section II in Chapter 12) sometimes have difficulty forming when an alcohol is tertiary.6 Methyl oxalyl chloride typically esterifies tertiary alcohols without difficulty.4,6–19Another reason for selecting methyl oxalyl esters is that they are less likely to experience the thermal elimination (Chugaev reaction) that is common for tertiary O-thiocarbonyl compounds. In molecules with the proper structure cyclization can precede hydrogen-atom abstraction.13
A proposed mechanism for reaction of methyl oxalyl esters with tri-n-butyltin hydride is shown in Scheme 3. According to this mechanism the tri-n-butyltin radical transfers an electron to the π system of the ester to produce a highly stabilized radical anion (a semidione).20 (Supporting the idea that such a transfer takes place is the observation that Bu3Sn· reacts with oxalate esters to produce intermediates with ESR spectra characteristic of radical anions.21) Fragmentation of such a radical anion then generates a carbon-centered radical that abstracts a hydrogen atom from Bu3SnH (Scheme 3).
There are two significant problems associated with the synthesis and reaction of methyl oxalyl esters. One of these is the difficulty in starting- material purification that arises because these esters hydrolyze readily, in particular, during chromatography on silica gel.4,22 A second problem has to do with alcohol regeneration, a significant side reaction from treatment of some methyl oxalyl esters with tri-n-butyltin hydride.5,20
C. Acetates and Trifluoroacetates
Acetylated carbohydrates do not react with tri-n-butyltin hydride under normal conditions (80-110 oC, 2 h, AIBN initiation), but under different, more vigorous conditions (triphenylsilane, 140 oC, 12 h, two equivalents of benzoyl peroxide) these compounds produce the corresponding deoxy sugars (eq 3).23 These more vigorous conditions cause similar reaction in O-trifluoroacetyl substituted carbohydrates.24 The need for two equivalents of benzoyl peroxide in the reaction shown in eq 3 indicates that a nonchain process is taking place.
D. p-Cyanobenzoates
Replacement of the benzoyl group in compound 2 with a p-cyanobenzoyl group converts an unreactive compound (2) into a reactive one (3) (eq 4).25 One explanation for this difference in reactivity is that because a cyano group is quite effective at stabilizing a radical anion, electron transfer to compound 3 is taking place where analogous transfer to the unsubstituted benzoate 2 does not occur. Since radical anions can form by electron transfer from the tri-n-butyltin radical to easily reduced organic compounds,21,26 the electron-transfer mechanism pictured in Scheme 4 represents a possible pathway for replacement of a p-cyanobenzoyloxy group with a hydrogen atom. | textbooks/chem/Organic_Chemistry/Radical_Reactions_of_Carbohydrates_(Binkley)/Radical_Reactions_of_Carbohydrates_II%3A_Radical_Reactions_of_Carbohydrates/08%3A_Carboxylic_Acids_and_Esters/II._Replacement_of_an_Acyloxy_Group_with_a_Hydrog.txt |
A. Acetates and Pivalates
Photochemical electron transfer from excited hexamethylphosphoramide (HMPA) to an O-acyl group in a carbohydrate begins a series of events that result in replacing each O-acyl group with a hydrogen atom. An example of a typical reaction is shown in eq 5.27 The temperature at which this reaction can be conducted (~25 oC) is synthetically far more attractive than the 140 oC needed for the corresponding thermal reaction of an acetylated carbohydrate (eq 3).23 Photochemical electron transfer to acetates has been used for the synthesis of a number of deoxy sugars.27–34 Esters of pivalic acid, which also can serve as substrates in this type of reaction,28,35–41 sometimes give better yields than the corresponding acetates.28,35,36
1. Reaction Mechanism
Photochemical electron transfer begins with absorption of light by HMPA to produce a highly reactive, electronically excited molecule that ejects an electron into the solution (Scheme 5).42.43 The ejected electron is captured by the acylated carbohydrate to produce a radical anion that cleaves to give a carboxylate anion and a carbohydrate radical (R·).43 Hydrogen-atom abstraction by the carbohydrate radical then completes the replacement process (Scheme 5). The water present in the reaction mixture extends the lifetime of the solvated electron and, in so doing, increases the probability that this electron will be captured by a molecule of ester.43 (The importance of water to the success of this process is demonstrated by the yields of the reactions shown in eq 6.44,45)
2. Alcohol Regeneration
Ester photolysis in aqueous HMPA sometimes regenerates the alcohol from which the ester was synthesized (eq 7).34 In some instances, alcohol formation may be due to nonphotochemical ester hydrolysis. Nonphotochemical reaction provides a reasonable explanation for the easily hydrolyzed, anomeric acetate shown in eq 8 undergoing only hydrolysis (no deoxygenation) when photolyzed in aqueous HMPA.34 Even though simple hydrolysis may be significant for some compounds, as described below, alcohol regeneration during photolysis of other, probably most, esters must occur in a different way.
Alcohol formation during ester photolysis cannot be explained, in general, by simple hydrolysis because, as is shown by the reaction pictured in Scheme 6, the yield of the alcohol can depend on the concentration of the starting ester.43 One explanation for this dependence begins with the ester 5 capturing a solvated electron to form the radical anion 6. This radical anion then abstracts a hydrogen atom from a second molecule of 5 to produce the anion 7, which then forms an alkoxide ion that protonates to give the observed alcohol.43 Since, according to this explanation, raising ester concentration should increase the rate of hydrogen-atom abstraction to give 7 but not the rate of the competing β-cleavage that forms R·, greater ester concentration should increase the amount of alcohol (ROH) produced at the expense of the deoxygenated product (RH).
A critical question about the mechanism for alcohol formation presented in Scheme 6 concerns whether the radical anion 6 can abstract a hydrogen atom from the ester 5. The evidence found in eq 9 supports the idea that 5 can function as a hydrogen-atom donor. Irradiation of 5 in HMPA‑d18/D2O gives a 27% yield of 8, a reduction product that contains no deuterium. Since the only source for the second hydrogen atom at C-3 in 8 is one of the carbohydrates in the reaction mixture and since the ester 5 is the only carbohydrate present at the beginning of the reaction, abstraction from 5, at least in the early stages of reaction, seems unavoidable. If 5 can act as a hydrogen-atom donor in the formation of 8, it becomes a strong candidate for the same role in the conversion of the radical anion 6 into the alkoxide ion 7 (Scheme 6).
3. Competition From Light-Absorbing Chromophores
If an ester contains a strongly absorbing chromophore, HMPA excitation will be effectively precluded because most of the incident light will be absorbed by the ester. Failure to excite HMPA will forestall replacement of the acyloxy group with a hydrogen atom by preventing electron transfer. The light absorbing properties of the benzoyloxy group, for example, render benzoates much less desirable participants in these electron-transfer reactions because far less incident light reaches the HMPA. This means that an important factor in reaction of simple acetates and pivalates is that these compounds contain no strongly absorbing chromophore.42 An example of an ester that fails to undergo replacement of the acyloxy group due to the presence of a light-absorbing substituent (i.e., the 4,6-O-benzylidene group) is shown in eq 10.41 Even though the benzylidene group is removed during photolysis, the aromatic chromophore remains in the solution and continues to absorb the incident light. Changing 4,6-O-benzylidene to 4,6-O-isopropylidene protection allows reaction to proceed in the normal fashion (eq 11).41
B. m-(Trifluoromethyl)benzoates
A m-(trifluoromethyl)benzoate will accept an electron from excited N‑methylcarbazole (11) in a reaction that leads to replacement of the acyloxy group with a hydrogen atom; for example, photolysis of 2',3',5'-tri-O-[m-(trifluoromethyl)benzoyl]-adenosine (9) produces the 2',3'-dideoxyadenosine derivative 10 (eq 12).46 (Most,46–52 but not all,48 compounds reported to undergo this type of reaction are nucleosides.) Reactions, such as the one shown in eq 12, are regioselective because the radical anion generated from a m-(trifluoromethyl)benzoyl group does not fragment to give a primary radical.
Photochemical electron transfer involving m-(trifluoromethyl)benzoates and N-methylcarbazole (11) has several advantages over electron transfer between HMPA and acetates or pivalates. One of these is that N‑methylcarbazole has greater molar absorptivity than HMPA, a fact that renders the carbohydrate reactant less likely to stop the reaction by absorbing the incident light.52 From a safety point of view, eliminating HMPA from the reaction mixture avoids handling a highly toxic, cancer-suspect agent. Because the m‑(trifluoromethyl)benzoyl group is an effective electron acceptor (better than an acetyl or pivaloyl group) few substituents in the carbohydrate will compete with this group for an electron donated by excited N-methylcarbazole (11); consequently, reactions of m-(trifluoromethyl)benzoates usually are highly chemoselective. An example of this selectivity is shown in eq 13, where the benzoyl group remains bonded to C-3' while the m‑(trifluoromethyl)benzoyl group at C-2' is replaced by a hydrogen atom.49
When reaction is conducted in the presence of Mg(ClO4)2, it is possible to replace even an unsubstituted benzoyloxy group with a hydrogen atom (eq 1450).50–52 Magnesium perchlorate affects this reaction by hindering back electron transfer, a process that competes with the fragmentation of the radical anion 13 (Scheme 7). Another factor that affects the reaction of a benzoyloxy group is the choice of the electron donor; thus, replacing N-methylcarbazole (11) with 3,6-dimethyl-9-ethylcarbazole (12) causes deoxygenation to take place more rapidly.50 Compound 12 is superior to 11 because it forms a more stable radical cation upon electron transfer.50,53,54 | textbooks/chem/Organic_Chemistry/Radical_Reactions_of_Carbohydrates_(Binkley)/Radical_Reactions_of_Carbohydrates_II%3A_Radical_Reactions_of_Carbohydrates/08%3A_Carboxylic_Acids_and_Esters/III._Photochemical_Electron_Transfer_to_Carboxyli.txt |
Electron transfer to carboxylic acid esters also can occur via nonphotochemical reaction. Transfer of an electron from SmI2 to a carbohydrate p‑methylbenzoate produces a radical anion that fragments to give a carbon-centered, carbohydrate radical and a p-methylbenzoate anion (Scheme 8).54.55 The carbohydrate radical abstracts a hydrogen atom from a donor present in the solution to form a deoxy sugar. An example of such a reaction is shown in eq 15.
V. Acyloxy Group Migration
Group migration in radical reactions follows one of two basic pathways. For aldehydo, cyano, and aryl groups, migration takes place by a sequence of elementary reactions consisting of cyclization and β-fragmentation steps. (An example of this type of reaction is shown in Scheme 8 of Chapter 10). Group migration in esters is governed by a different mechanism, one that has been the subject of considerable investigation. To follow the progress in understanding this reaction, it is useful to view the advances in mechanistic discovery in a chronological order.
VI. Reactions of Carboxylic Acids
Carboxylic acids cannot be converted directly into carboxyl radicals, but they can form these radicals indirectly. One method for indirect formation calls for converting the acid into its anion, which then is subjected to electrolysis (Scheme 22).95 Other indirect methods require formation of carboxylic acid derivatives, such as esters of N-hydroxypyridine-2-thione, compounds that produce carboxyl radicals by photochemically initiated reaction (Scheme 23).96 Carboxyl radicals expel carbon dioxide to produce carbon-centered radicals (eq 22).
VII. Summary
Carbohydrates with simple acyloxy groups are unreactive under conditions normally used in reduction reactions. Reduction does occur, however, if the reaction temperature is raised to 140 oC and the reaction time is greatly extended. Esters with special structural features undergo reduction at lower temperatures; thus, both α-acyloxy ketones and methyl oxalyl esters react with tri-n-butyltin hydride at or below 110 oC.
Photochemical electron transfer provides a way for acyloxy groups to be replaced by hydrogen atoms under mild reaction conditions (at room temperature in neutral solution). Electron transfer occurs when either excited HMPA or N-methylcarbazole donates an electron to an ester to form a radical anion. Fragmentation of the radical anion generates a carbon-centered radical that then abstracts a hydrogen atom to produce a deoxygenated compound. Regeneration of the partially protected carbohydrate from which the ester was synthesized sometimes competes with deoxygenation.
Acyloxy group migration to a radical center on an adjacent carbon atom is a reaction that is useful in the synthesis of 2-deoxy sugars. Early proposals for the mechanism of this reaction turned out not to be correct. The considerable investigation that has taken place since then has shown that this reaction is likely to involve the formation of an intimate ion pair consisting of a carboxylate anion and a radical cation. Recombination of this pair produces a new radical, one that has undergone group migration.
Carboxylic acids produce carboxyl radicals by reaction with hypervalent iodine reagents or electrolysis of carboxylate anions. These radicals expel carbon dioxide to form carbon-centered radicals. Electrochemical reaction results in radical coupling, or if the radical is further oxidized, carbocation formation. Reaction of carboxylic acids with hypervalent iodine reagents often is conducted in the presence of heteroaromatic compounds, where radical addition to the aromatic ring takes place. | textbooks/chem/Organic_Chemistry/Radical_Reactions_of_Carbohydrates_(Binkley)/Radical_Reactions_of_Carbohydrates_II%3A_Radical_Reactions_of_Carbohydrates/08%3A_Carboxylic_Acids_and_Esters/IV._Nonphotochemcal_Electron_Transfer_to_Carboxyl.txt |
Two types of radical reaction of phosphoric acid esters are important in carbohydrate chemistry. One of these is migration of a phosphatoxy group from C-2 to C-1 in a pyranoid or furanoid ring (eq 1),1 and the other is elimination of this group from C-3' in nucleotide derivatives (eq 2).2 Even though these reactions are different in their outcome and very specific in terms of the type of structure undergoing reaction, they are mechanistically similar. An indication of this similarity is that each reaction begins by forming a radical in which a phosphatoxy group is β-related to the radical center (eq 3).
09: Phosphoric Acid Esters
A. Reaction Mechanism
Initially two mechanisms were considered as possibilities for phosphatoxy group migration of the type shown in eq 1. (This same mechanistic choice exists for acyloxy group migration and is discussed in Section V.A of Chapter 8.) The first of these mechanisms consisted of a pair of competing, concerted reactions, each of which passed through a cyclic transition state (Scheme 1).3–7 A basic difference between this pair was that in one reaction the same oxygen atom was bonded to the carbon-atom framework both before and after migration, but in the other the framework had a different oxygen atom attached after migration. Proposing migration via a combination of these two reactions made it possible to explain experiments with oxygen-labeled substrates in which only a portion of the labeled oxygen was attached to the carbon-atom framework after migration. The results from early studies favored this two-reaction explanation,3–7 but those from later investigations required it to be changed because the later studies showed that ionic intermediates were involved in the migration process.
A mechanism that satisfies the ionic-intermediate requirement is shown in Scheme 2, where the β-phosphatoxy radical 1 fragments heterolytically to give the contact ion pair (CIP) 2.8–13 This ion pair recombines in low polarity solvents to form the group-migrated radical 3, but in more polar solvents the CIP also can separate to become a solvent-separated ion pair (SSIP, 4) and then free ions 5.10
Critical support for the ion-pair mechanism for phosphatoxy group migration comes from laser-flash-photolysis (LFP) experiments. Both the SSIP 9 and the diffusively free radical cation 10 can be detected in studies where LFP generates the radical 6 (Scheme 3).12 Evidence for the CIP in this reaction is indirect presumably because its lifetime is too short to permit direct detection. Study of reaction rates in solvents of different polarity supports the idea that the radical 6 is passing through a common intermediate in forming either the migrated radical 8 or the SSIP 9. A reasonable conclusion is that the common intermediate is the contact ion pair 7 (Scheme 3).12 Entropies of activation, which are the same for ion-pair formation in high polarity solvents and group migration in solvents of low polarity, also favor a common intermediate for which 7 is the prime candidate.10,12 Generalizing these results leads to the reaction mechanism proposed in Scheme 2. (The wording in this paragraph also is found in Section V.A.5 of Chapter 8 because the information contained is pertinent to the mechanism of acyloxy group migration.)
Phosphatoxy group migrations are not wide-spread in carbohydrate chemistry; in fact, all reported reactions involve migration from C-2 to C-1 in a pyranoid or furanoid ring. This situation exists because the stabilization afforded by the ring oxygen atom is critical at the transition state leading to radical-cation formation. Examining the reactivity of the noncarbohydrate radicals shown in equations 4 and 5 is instructive. An oxygen atom must be fully able to participate in radical-cation stabilization for heterolytic bond breaking to occur (eq 4).14 Replacing the methoxy group in the substrate in the reaction shown in eq 4 with an acetyl group, as is done in the reaction shown in eq 5, prevents radical-cation formation because an oxygen atom with an electron-withdrawing group attached is unable to stabilize sufficiently the transition state leading to the radical-cation intermediate.14
B. Relative Reaction Rates
Relative rate constants for phosphatoxy group migration in the reactions of the five hexopyranosyl bromides 11-15 are given in Table 1.15 The rate constant for reaction of the 6-deoxy bromide 14 is substantially larger than those for the other bromides. Replacing the electron-withdrawing acyloxy group at C-6 with a hydrogen atom makes the radical cation 16 more stable and, in so doing, stabilizes the transition state leading to it (eq 6).
The rate constant for reaction of the D-mannopyranosyl bromide 15 is decidedly smaller than those for reactions of the bromides 11-14. One factor that contributes to this reduced reactivity is the enhanced stability of the radical 17 when compared to the corresponding radicals derived from the other bromides (11-14). Only 17 remains in a relatively strain-free, 4C1 conformation while taking advantage of the stabilizing interaction of parallel po, pc, and σ* orbitals (Table 2).15 To benefit from parallel-orbital stabilization, the radicals derived from bromides 11-14 must assume less stable conformations; for example, the radical derived from 11 adopts the B2,5 boat conformation 18.15 As migration takes place in each of the radicals 17-19, po, pc, σ* orbital stabilization is lost, but for radicals 18 and 19 this loss is compensated for, at least in part, by movement toward a more stable, 4C1 conformation. Such compensation means that the transition states for radical-cation formation from 18 and 19 are not as high in energy as that for reaction of 17; consequently, group migration for the radical 17 is slower than for 18 and 19.
Pertinent information about formation and reactivity of radical cations comes from the study of noncarbohydrate systems.16,17 Nucleophilic trapping of the radical cation 21 by methanol (k< 1 x 103 M-1s-1) is slow compared to hydrogen-atom abstraction from 1,4-cyclohexadiene (k = 6 x 105 M-1s-1) (Scheme 4).17 To the extent that this observation is a general one, radical cations can be expected to have greater radical reactivity than cationic reactivity.
C. Stereoselectivity
The reactions pictured in Schemes 5 and 6 show that phosphatoxy group migration is a stereospecific process; thus, the epimeric radicals 22 and 24 give the product radicals 23 and 25, respectively.1,15 Once migration has taken place, stereoselective deuterium abstraction completes the reaction. For the radical 23 abstraction is highly stereoselective, but it is much less so for the radical 25. Shielding of the α face of 23 by the axial phosphatoxy group causes deuterium to be abstracted from the β face of this radical (Scheme 5). The equatorial phosphatoxy group in 25 is not nearly as effective at forcing Bu3SnD to the opposite face of the pyranoid ring (Scheme 6).
Differential shielding of the faces of a pyranoid ring affects the ability of the phosphates 26 and 27 to undergo new ring formation (eq 7). The phosphate 26 gives a decidedly higher yield of the glycal 28 than does its epimer 27.18 The substantially lower product yield from reaction of 27 is attributed to steric hindrance by the nearby phosphate counter ion during cyclization of the radical cation 29 (Scheme 7). When 29 is generated from 26, however, the counter ion is on the opposite face of the ring and does not impede cyclization (Scheme 8). | textbooks/chem/Organic_Chemistry/Radical_Reactions_of_Carbohydrates_(Binkley)/Radical_Reactions_of_Carbohydrates_II%3A_Radical_Reactions_of_Carbohydrates/09%3A_Phosphoric_Acid_Esters/II._Phosphatoxy_Group_Migration.txt |
Addition of the phenylthiyl radical to the unsaturated nucleotide 30 produces the carbon-centered radical 31 (Scheme 9).2 This radical (31) either abstracts a hydrogen atom to give an epimeric mixture of reduced nucleotides or fragments the C–O bond at C-3' heterolytically to give a phosphate anion and a radical cation.19 Fragmentation is assisted by polar solvents.20 Since heterolytic fragmentation of 31 also depends on the ability of the substituent at C-3' to stabilize a negative charge as the C–O bond breaks, forming a highly stabilized anion is critical; thus, fragmentation is competitive with hydrogen-atom abstraction when the anion produced is a phosphate (Scheme 9) but not a benzoate (eq 8).2
Additional details concerning ion-pair formation from the unsaturated nucleotide 30 are given in Scheme 10.21–24 A contact ion pair (CIP), a solvent-separated ion pair (SSIP), and diffusively free ions all are included in this Scheme. Labeling experiments show how the various ion pairs participate in the reaction. Since no scrambling of the oxygen label in the phosphate group in the substrate 30 occurs after partial reaction, the CIP either cannot return to the radical 31 or if it does, no reorganization occurs within this ion pair.23 Since oxygen scrambling can take place in the SSIP, the labeling experiments show that once this intermediate is reached, there is no return to the radical 31.
Mechanistic studies using both alkyl- and arylthiols show that the equilibrium between the nucleotide 30 and the adduct radical 31 depends on the identity of the R group in the thiyl radical (Scheme 10). In this equilibrium alkylthiyl radicals favor adduct formation to a greater extent than do arylthiyl radicals. When the method of formation produces a low radical concentration, conditions can exist in which alkylthiyl radicals will form a sufficient concentration of adduct radicals (31) for reaction to proceed at an observable rate, but arylthiyl radicals do not produce the necessary radical concentration.22
Study of the rates of radical reaction of thymidine, cytidine, adenosine, and guanosine derivatives show that guanosines are by far the most reactive.23 (The rate of reaction of guanosine derivatives is too fast to be measured.) This enhanced reactivity is attributed to rapid, internal electron transfer from the guanine moiety to the radical-cation portion of the molecule (Scheme 11).23,25–27 Electron transfer of this type may be fast enough (k > 1 x 109 s-1) to be taking place within the CIP.23 One estimate of the rate constant for this type of electron transfer is 1.4 x 108 s-1.27
IV. Migration Reactions in Other β-Ester Radicals
Other groups (sulfonyloxy,28,29 bromo,29 nitroxy,28 and protonated hydroxyl30) that are β-related to a carbon-centered radical can react to give radical cations. The migration of acyloxy groups, discussed in Section V of Chapter 8, also is likely to involve radical cations.
V. Summary
Under the proper conditions a phosphatoxy group that is β-related to a radical center will fragment to produce an ion pair consisting of a phosphate anion and a radical cation. This heterolytic fragmentation is favored by polar solvents and formation of radical cations at least as stable as those arising from enol ethers. When the radical center is at C-1 in a pyranoid or furanoid ring, the ion pair recombines to give a new radical in which phosphatoxy group migration to C-1 is accompanied by radical translocation to C-2. If the radical center in a nucleotide is at C-4' and a phosphatoxy group is located at C-3', heterolytic cleavage of the C3'-O bond does not lead to group migration; rather, products arising from hydrogen-atom abstraction, solvent capture, or proton loss are observed. | textbooks/chem/Organic_Chemistry/Radical_Reactions_of_Carbohydrates_(Binkley)/Radical_Reactions_of_Carbohydrates_II%3A_Radical_Reactions_of_Carbohydrates/09%3A_Phosphoric_Acid_Esters/III._Radical_Cation_Formation_from_Nucleotides.txt |
Aldehydes and, to a lesser extent, ketones participate in radical reactions of carbohydrates by generating intermediate, oxygen-centered and carbon-centered radicals. The radical addition pictured below provides an example of conversion of a carbonyl compound into an oxygen-centered radical.
10: Aldehydes and Ketones
Aldehydes and, to a lesser extent, ketones participate in radical reactions of carbohydrates by generating intermediate, oxygen-centered and carbon-centered radicals. The radical addition pictured in eq 1 provides an example of conversion of a carbonyl compound into an oxygen-centered radical, while that in eq 2 involves transforming an aldehyde or ketone into a carbon-centered radical. [The radicals produced in the latter reaction (eq 2) are described as samarium ketyls in recognition of their partial radical-anion character.] Other reactions that generate radicals from aldehydes and ketones are photochemical bond homolysis (eq 3) and fragmentation of α‑acyloxy ketones (eq 4). The discussion in this chapter centers on the types of compounds that can be produced by these reactions and the mechanisms for their formation.
02. Intramolecular Addition of Carbon-Centered Radicals
The possibility of isolating a product from intermolecular addition of a carbon-centered radical to an aldehyde or ketone is small due to the ready reversibility of this reaction (eq 1), but the possibility of product isolation increases considerably if the reaction becomes an intramolecular addition of a carbon-centered radical to an aldehydo or keto group to give a radical centered on an oxygen atom that is attached to a five- or six-membered ring.
An example of such a reaction is shown in Scheme 1, where the carbon-centered radical 2, generated from 6-bromohexanal (1), is converted reversibly into the cyclic alkoxy radical 3.1 Hydrogen-atom abstraction by 3 from tri-n-butyltin hydride has a substantially larger rate constant than that for abstraction by 2; consequently, even though ring opening is more rapid than ring closure, reaction produces cyclohexanol as the major product and hexanal as a minor one.
Intramolecular hydrogen-atom abstraction from the aldehydo group in 2 is a very minor process. The inability of this abstraction to compete with ring formation in a noncarbohydrate system is echoed in the reactions of carbohydrate radicals containing aldehydo groups. The reaction shown in Scheme 2 is one of several discussed in this chapter where hydrogen-atom abstraction from an aldehydo group is possible but does not take place.2
Even though ring opening always is a possibility for cyclic alkoxy radicals, this transformation sometimes does not take place; for example, the reaction producing the alkoxy radical 5 from the ring-open radical 4 is not reversible (Scheme 2).2,3 Failure of the cyclohexane ring in 5 to open is demonstrated by reaction of the nitrate ester 9 (Scheme 3).3 Treatment of 9 with Bu3SnH produces 5 (and ultimately the product 7) but ring opening to give 4 does not happen. If the ring-open radical 4 were formed, the product 8 also would be produced in this reaction, but since no 8 could be detected, the conclusion is that the alkoxy radical 5 does not undergo ring opening.3
In contrast to cyclization of the aldehydo radical 4 (Scheme 2) the closely related keto radical 11 (Scheme 4) does not form a new ring system.4 Either the greater steric hindrance inherent in producing a tertiary alkoxy radical or rapid ring opening of such a strained intermediate or both are sufficient to prevent 11 from forming a new ring system. These reasons for failure to form a new ring draw support from the reactions of noncarbohydrate radicals 13 and 14 (Scheme 5).5 In the reaction shown in Scheme 5 where R is a methyl group, hydrogen-atom abstraction from tri-n-butyltin hydride is done exclusively by the open-chain radical 13. When R is a hydrogen atom, abstraction from Bu3SnH occurs only after conversion of the open-chain radical 12 into the cyclic alkoxy radical 14.
The reactivates of the aldehydo radical 4 (Scheme 2) and the keto radical 11 (Scheme 4) raise a number of questions (listed below) about participation of keto and aldehydo groups in radical cyclization reactions. Many of these questions have been answered by study of related compounds. Their answers provide insight into the factors that control the cyclization process. These questions and their answers are:
1. Roger W. Binkley (Cleveland State University) and Edith R. Binkley (Cleveland Heights-University Heights school system)
03. Migration of Aldehydo Groups
A possible fate for an alkoxy radical formed by cyclization is ring opening to produce a radical different from the one that initially formed the ring.4,13–17 A new direction in ring opening is likely if it produces a more stable radical. In the reaction shown in Scheme 8 such a situation exists.14,15 Ring opening of the alkoxy radical 20 gives the resonance-stabilized, benzylic radical 21 rather than the unstabilized radical 19 that reacted to produce the ring system. This alternative ring opening (20$\rightarrow$21) completes an addition-fragmentation sequence that causes migration of the aldehydo group.
04. Addition of Tin- and Silicon-Centered Radicals to A
Although the reactions discussed thus far have involved addition of carbon-centered radicals to carbonyl groups, other types of radicals, including tin- and silicon-centered ones, also add to aldehydes and ketones. Reaction of the tri-n-butyltin radical with a carbonyl group generates a tin ketyl, a radical with considerable negative charge on the oxygen atom. As the reaction in Scheme 918 shows, tin ketyls undergo internal radical addition to electron-deficient, C–C multiple bonds.18–22 These ketyls also react with C–N double bonds,23 and they produce pinacols upon addition to carbonyl groups (eq 8).24 Internal addition also can occur when a silicon-centered radical adds to an aldehydo group, as happens in the reaction shown in eq 9.25
05. Reaction of Samarium(II) Iodide with Aldehydes
Reaction of an aldehyde or ketone with samarium(II) iodide produces a samarium ketyl.26–39 These ketyls add intramolecularly to appropriately positioned carbon–carbon25–33 (Scheme 10)26 and carbon–nitrogen34–37 (eq 10)34 double bonds. Such reactions are reminiscent of the addition of typical carbon-centered radicals to multiple bonds.
When samarium(II) iodide reacts with compounds containing two aldehydo groups, the first is converted into a samarium ketyl that then adds to the second. This addition depends upon proper separation between the reacting groups;40–53 accordingly, pinacols with five-membered40,49‑53 (eq 11)40 and six-membered41–48 (eq 12)41 rings form easily. It is not necessary for both interacting groups in a molecule to be aldehydo groups; pinacols also arise when one49–52 (eq 13, R = H)49 or both (eq 13, R = CH2SiMe2C6H5)53 are keto groups. Complexation of the ketyl and carbonyl oxygen atoms with SmI2 forces a cis relation between the hydroxyl groups in the products (Scheme 11).42 Pinacol formation and other reactions of aldehydes and ketones with samarium(II) iodide is revisited in Chapter 20, where a broader discussion of the interaction of SmI2 with carbohydrate derivatives takes place. | textbooks/chem/Organic_Chemistry/Radical_Reactions_of_Carbohydrates_(Binkley)/Radical_Reactions_of_Carbohydrates_II%3A_Radical_Reactions_of_Carbohydrates/10%3A_Aldehydes_and_Ketones/01._Introduction.txt |
A. α-Cleavage Reactions
When photolysis of ketones causes homolytic cleavage of a bond between the carbonyl group and one of the α carbon atoms either a pair of radicals or a diradical forms. (Such a reaction is known as an α-cleavage or Norrish Type I reaction.) α-Cleavage of simple ketones does not take place in solution, although it does occur in the gas phase. Cleavage in solution happens only when stabilized radicals are produced. This means that the reaction shown in eq 3 is successful for nonvolatile compounds such as carbohydrates only when the radical center in R· is stabilized in some way (e.g., by having an oxygen or nitrogen atom attached). Most carbohydrates that contain a keto group will have at least one pathway for forming an oxygen-stabilized radical by α-cleavage.54
When the keto group in a carbohydrate is not part of a ring system, α‑cleavage produces a radical pair. Most reactions of this type involve derivatives of nucleosides, nucleotides, or oligonucleotides.55–67 Scheme 12 describes such a reaction, one in which the nucleoside member (22) of the radical pair produced by α-cleavage undergoes two characteristic radical reactions, namely, hydrogen-atom abstraction (when an effective donor, such as a thiol, is present) and addition of O2 (when molecular oxygen is one of the reactants).61
Although α-cleavage in nucleotides produces radicals that undergo typical radical reactions, such as those shown in Scheme 12, many of these radicals also undergo a heterolytic cleavage to form radical cations and phosphate anions. An example of such a reaction is shown in Scheme 13, where the radical 23 cleaves its C-3'–O bond to generate the radical cation 24 and a phosphate anion.56 (Radical-cation formation of the type shown in Scheme 13 is discussed in Section III of Chapter 9.)
α-Cleavage in a cyclic ketone, a reaction that occurs in many carbohydrates, is an internal process that produces a diradical.68 Diradicals of this type usually reform a ring system, but often after the loss of carbon monoxide.68 Scheme 14 describes formation of the diradical 25, a reaction that is followed by loss of carbon monoxide to give a second diradical, one that produces a new ring system by radical combination.69 The α-cleavage shown in Scheme 14 is driven, at least in part, by transition-state stabilization due to the developing radical center at C-6 being stabilized by an attached oxygen atom.
B. Hydrogen-Atom Abstraction Reactions
Ketones that do not undergo α-cleavage have another option for diradical formation, namely, internal abstraction that occurs when a hydrogen atom comes with bonding distance of an excited carbonyl group; thus, in the reaction is shown in Scheme 15, 1,6-hydrogen-atom abstraction produces a diradical that then forms a spiro compound by radical combination.70 If a 1,5-hydrogen-atom transfer takes place, the resulting 1,4-diradical fragments as shown in Scheme 16.71 (Many carbohydrates undergo this type of reaction, which is known as a Norrish Type II process.72)
If an excited ketone does not undergo internal abstraction or α-cleavage, hydrogen-atom abstraction from another molecule sometimes takes place.73 Such abstraction requires a transition state in which there is considerable radical stabilization. Hydrogen-atom abstraction by excited benzophenone from the benzylidene acetal 26 meets this requirement by producing the highly stabilized radical 27 (Scheme 17).74 Radical combination completes this reaction.
07. Cyclization of Acylsilanes
Acylsilanes undergo radical cyclization that involves addition of a carbon-centered radical to the carbonyl carbon atom in the acylsilyl group (Scheme 18).75 This reaction is unusual in that migration of the silyl group to the radical center on oxygen stops reaction that would reverse ring formation.
08. Reactions of α-Acyloxyketones
α-Acyloxyketones are compounds that undergo replacement of the acyloxy group with a hydrogen atom upon reaction with tri-n-butyltin hydride. Reactions of these compounds are discussed in Section II.A of Chapter 8 along with other reactions of carboxylic acid esters.
09. Summary
Aldehydo and keto groups in carbohydrates react internally with carbon-centered radicals to form cyclic alkoxy radicals. Most of these reactions involve aldehydes; both five- and six-membered rings can be formed. When ring opening of the newly formed alkoxy radical takes place, it does so to produce the more stable of the two possible, carbon-centered radicals. Such ring opening can be part of a process that causes aldehydo group migration.
Tin-centered radicals add to aldehydes to generate tin ketyls, intermediates that can add to multiple bonds. Samarium ketyls, more common than their tin counterparts, undergo similar reaction. Reaction of samarium(II) iodide with carbohydrates containing two appropriately placed aldehydo groups converts one group to a ketyl that then adds to the second in route to formation of a pinacol.
Ketone photolysis forms carbon-centered radicals by breaking the bond between the carbonyl carbon atom and one of the α carbon atoms (an α-cleavage). Typical radical reactions then take place; in addition, α-cleavage in some nucleotides and oligonucleotides is followed by radical cation formation. α-Cleavage and hydrogen-atom abstraction take place in cyclic ketones to produce diradicals. | textbooks/chem/Organic_Chemistry/Radical_Reactions_of_Carbohydrates_(Binkley)/Radical_Reactions_of_Carbohydrates_II%3A_Radical_Reactions_of_Carbohydrates/10%3A_Aldehydes_and_Ketones/06._Ketone_Photolysis.txt |
The first step in conducting most radical reactions is the preparation of a radical precursor. For many of such compounds (e.g., halides, esters, and acetals) this preparation needs little, if any, discussion, as the reactions involved are among the most common in organic chemistry. O‑Thiocarbonyl compounds [xanthates, (thiocarbonyl)imidazolides, aryl thionocarbonates, cyclic thionocarbonates, and thionoesters] are different because their preparation is less common, and the potential difficulties in their formation less well known. Because these compounds are rich sources of carbon-centered radicals and because being able to prepare them efficiently is vital to their use, understanding the synthesis of O-thiocarbonyl compounds is integral to using them in radical formation. This chapter, targeted at the synthesis of these compounds, is a companion to the one that follows, where radical reactions of O-thiocarbonyl compounds are discussed.
11: Synthesis of O-Thiocarbonyl Compounds
A. The Carbon Disulfide–Methyl Iodide Procedure (The Standard Procedure for Xanthate Synthesis)
The most common method for synthesizing O-[(alkylthio)thiocarbonyl] esters of carbohydrates (carbohydrate xanthates) begins by forming an alkoxide ion from reaction of sodium hydride with a compound containing an unprotected hydroxyl group.1 (Imidazole usually is present in the reaction mixture to promote alkoxide ion formation.) Once formed, the alkoxide ion adds to carbon disulfide, and the resulting anion is alkylated by methyl iodide. This procedure, which is the standard one for xanthate synthesis, is summarized in Scheme 1.
B. Modifications of the Standard Procedure
1. Reagents and Reaction Conditions
Modification of the procedure outlined in Scheme 1 sometimes is necessary to improve reactant solubility and reactivity. Minor changes take the form of replacing the normal reaction solvent (THF) with N,N-dimethylformamide2,3 or methyl sulfoxide.4–8 When methyl sulfoxide is the reaction solvent, sodium hydroxide usually replaces sodium hydride as the deprotonating base.4,6,7
2. Phase-Transfer Reaction
Phase-transfer reaction provides a way for synthesizing xanthates that are difficult or impossible to prepare by the standard procedure. Anomeric xanthates, compounds that provide a synthetic challenge due to their instability, can be produced by phase-transfer reaction (eq 1).9 This reaction also demonstrates the potential of the phase-transfer procedure in preparing xanthates containing base-labile groups.
C. The Phenyl Chlorodithioformate Procedure
Esterification of an alcohol with an acid chloride provides another, but rarely used procedure for xanthate synthesis. Heating the partially protected disaccharide 1 with dibutyltin oxide produces a stannylene complex that then reacts regioselectively with phenyl chlorodithioformate to give the xanthate 2 (Scheme 2).10 This reaction provides a route to carbohydrate xanthates that contain an O-[(arylthio)thiocarbonyl] group. These arylthio derivatives cannot be prepared by the iodide displacement that is part of the standard xanthate synthesis.
III. (Thiocarbonyl)imidazolides
Formation of a (thiocarbonyl)imidazolide (3) generally involves heating a partially protected carbohydrate with N,N-thiocarbonyldiimidazole (4, TCDI) under reflux in tetrahydrofuran (or 1,2-dichloroethane) and isolating the reaction product by chromatography (eq 2).1,11 Nearly every synthesis of a (thiocarbonyl)imidazolide follows this procedure, although acetonitrile,12–14 toluene,15–17 and N,N-dimethylformamide18,19 occasionally are used as reaction solvents.
There are scattered reports of (thiocarbonyl)imidazolides forming more slowly than might be expected under typical reaction conditions. One such report concerns the methyl glycoside 5, a compound that reacts so slowly that prior activation with bis(tributyltin)oxide is necessary to increase the nucleophilicity of 5 to the point that (thiocarbonyl)imidazolide formation proceeds at an acceptable rate (Scheme 3).20
Reduced reactivity in nucleosides sometimes is brought about by N‑benzoylation. The N-benzoylguanosine and adenosine derivatives 6 and 8 require treatment with TCDI (4) for 70 and 85 hours, respectively, for complete reaction to take place; in contrast, derivatives lacking the N-benzoyl group (7 and 9), need only four hours for reaction to reach completion (eq 3).21
Even though (thiocarbonyl)imidazolides (3) can be prepared readily by the reaction shown in eq 2, this procedure has several minor drawbacks. One of these is that N,N-thiocarbonyldiimidazole (4, TCDI) needs to be kept in a dry atmosphere because it is unstable in the presence of atmospheric moisture.22 Another is that imidazole, produced as a byproduct in this reaction (eq 2), may catalyze unwanted transformation of some compounds.22 Finally, the cost of TCDI (4) is high enough to be a factor in deciding upon its use, particularly in large-scale reactions.
In an effort to overcome possible disadvantages associated with use of (thiocarbonyl)imidazolides, some researchers have proposed switching to related compounds. Thionocarbamates formed from 1,1'-thiocarbonyldi-2,2'-pyridone (10), a reagent stable to atmospheric moisture, are effective replacements for (thiocarbonyl)imidazolides (eq 4),22 but detracting from the use of this new reagent (10) is its even greater cost that TCDI.
Some thionocarbamates synthesized from the inexpensive phenyl isothiocyanate (11) (eq 5) are capable of radical formation.23,24 Although producing a thionocarbamate by reacting a partially protected carbohydrate with phenyl isothiocyanate (11) solves the “cost problem”, it has the disadvantage that this reaction requires the presence of a strong base because hydroxyl group deprotonation is needed for this reaction to occur at an acceptable rate (eq 5). Also, not all thionocarbamates prepared from 11 form radicals under typical reaction conditions.25 None of the alternatives to (thiocarbonyl)imidazolides have been widely adopted. | textbooks/chem/Organic_Chemistry/Radical_Reactions_of_Carbohydrates_(Binkley)/Radical_Reactions_of_Carbohydrates_II%3A_Radical_Reactions_of_Carbohydrates/11%3A_Synthesis_of_O-Thiocarbonyl_Compounds/II._Xanthates.txt |
A. Reaction with Phenoxythiocarbonyl Chloride
1. DMAP-Catalyzed Reactions
The standard procedure for synthesis of phenyl thionocarbonates is illustrated by the reaction shown in eq 6.26,27 Phenoxythiocarbonyl chloride (12) in the presence of the powerful, acylation catalyst 4-dimethylaminopyridine (13, DMAP) esterifies most carbohydrates with ease. Even though pyridine itself can be used in some situations, the superior catalytic effect of DMAP makes it the reagent of choice. In those rare instances when thionocarbonate formation by the standard procedure is too slow, switching the reaction solvent from acetonitrile to N,N-dimethylformamide or methyl sulfoxide often is sufficient to increase the rate of reaction to a synthetically acceptable level.28
DMAP (13) causes acylation rates to increase by factors as large as 10,000 when compared to reactions catalyzed by pyridine.29 One possibility for the greater catalytic effect of DMAP is that it is a stronger base than pyridine. (The pKb for pyridine is 8.71 and that for DMAP is 4.30.29) This explanation for the difference in reactivity, however, is not sufficient to explain DMAP’s superior catalytic ability because triethylamine (pKb = 3.35), an even stronger base than DMAP, has a catalytic effect similar to that of pyridine.29
A better explanation for DMAP being such an effective catalyst is that it reacts with acid chlorides, such as 12, to form high concentrations of N‑acylpyridinium salts (eq 7).29 These salts are better able to transfer an acyl group to a nucleophile than is the acid chloride itself. Resonance stabilization (two of the principal resonance contributors are shown in eq 7) increases the equilibrium concentration of an N-acylpyridinium salt, and charge delocalization increases the reactivity of this powerful acylating agent by creating loosely bound ion pairs.
The mildly basic conditions for thionocarbonate synthesis stand in contrast to the strongly basic ones used for xanthate preparation.26 Avoiding strongly basic conditions often is necessary in nucleoside synthesis; for example, thionocarbonates such as 14 can be prepared without difficulty (by the procedure outlined in eq 6), but attempted synthesis of the corresponding xanthates results in starting material decomposition.27 The specific reason xanthate synthesis fails in this case is that it requires conditions too basic for the stability of nucleosides protected by the 1,1,3,3-tetraisopropyl-1,3-disiloxanediyl group, a common protecting group for nucleosides.
2. N-Hydroxysuccinimide-Catalyzed Reactions
Although DMAP (13) is the catalyst of choice in most syntheses of phenyl thionocarbonates, sometimes, in an effort to avoid an undesired, competing reaction or to improve product yield, DMAP is replaced by another reagent. The most common replacement is N-hydroxysuccinimide (NHS, 15).30–35 In the reaction shown eq 8, the methyl pyranoside 16 gives a better yield of the corresponding phenoxythionocarbonate when NHS (15) is the catalyst rather than DMAP.16
There is a similarity in the mode of action of DMAP and NHS in that each of them typically reacts with an acid chloride to produce a better acylating agent. For DMAP (13) this agent is the N-acylpyridinium salt shown in eq 7, and for NHS (15) the new acylating agent is the ester 17 (eq 9). Because esters of NHS react unusually rapidly with nucleophiles, they are sometimes referred to as "activated esters".36
Extensive mechanistic study of esters of N-hydroxysuccinimide with amines has shown their reaction kinetics to be consistent with a process in which reversible formation of the zwitterionic intermediate 18 is followed by a rate-determining breakdown of this intermediate by either an uncatalyzed or base-catalyzed process (Scheme 4).37,38 Since the hydroxyl group in NHS (15) is quite acidic (pKa = 6.039), its conjugate base (19) is more stable than most alkoxide ions. To the extent that the stability of the departing anion contributes to transition state stabilization (Scheme 4), esters derived from NHS should be particularly reactive.
The mechanism shown in Scheme 5 is based on the assumption that the findings from amine acylation (Scheme 4) can be extended to thioacylation of carbohydrates. Base-catalyzed reaction seems most reasonable, but the mechanism shown in Scheme 5 also includes an uncatalyzed process in which the initially formed, tetrahedral intermediate 20 undergoes proton transfer to give the zwitterion 21. This intermediate eliminates separation of charge by expelling a tautomer of NHS to form the desired phenyl thionocarbonate 22.
B. Reaction with Thiophosgene and a Phenol
An alternative synthesis for an aryl thionocarbonate consists of treating a partially protected carbohydrate with thiophosgene and then reacting the product with a phenol (Scheme 6).40–47 Since phenoxythiocarbonyl chloride (12) is commercially available, the thiophosgene procedure normally is reserved for preparing aryl thionocarbonates in which the aromatic ring contains one or more electron-withdrawing substituents. In direct reactivity comparisons, substituted aryl thionocarbonates usually give better product yields.40 In some cases these substituents are necessary for reaction to take place.41
C. Reaction of (Thiocarbonyl)imidazolides with Phenols
Thionocarbonates are sometimes synthesized by reacting (thiocarbonyl)imidazolides with a substituted phenol. Such a reaction converts a less reactive O-thiocarbonyl derivative into a more reactive one (Scheme 7).48 It also provides another method for synthesizing aryl thionocarbonates in which the aromatic ring contains one or more electron-withdrawing substituents. Affecting the change shown in Scheme 7 causes the deoxygenated product yield to rise from 38% (starting with 23) to 70% (starting with 24).
D. Reaction With Phenoxythiocarbonyltetrazole
The thioacylating agent 25 can be used to synthesize phenyl thionocarbonates under conditions that avoid the base-catalyzed side reactions that sometimes occur in the presence of DMAP (eq 10).49 | textbooks/chem/Organic_Chemistry/Radical_Reactions_of_Carbohydrates_(Binkley)/Radical_Reactions_of_Carbohydrates_II%3A_Radical_Reactions_of_Carbohydrates/11%3A_Synthesis_of_O-Thiocarbonyl_Compounds/IV._Aryl_Thionocarbonates.txt |
Two basic procedures for the synthesis of cyclic thionocarbonates are in common use. The first involves reacting a compound containing adjacent hydroxyl groups with N,N-thiocarbonyldiimidazole (eq 11).50 Most cyclic thionocarbonates are synthesized by this procedure. The second approach involves initial formation of a stannylene complex and then treatment of this complex with thiophosgene (Scheme 8)51 or phenoxythiocarbonyl chloride52–54 (Scheme 9)54.
A third, but seldom used, reaction for cyclic thionocarbonate formation is one conducted under phase-transfer conditions. This synthesis is capable of producing either a bisxanthate55 or a cyclic thionocarbonate (Scheme 10).56 The critical factors in determining which type of product will be produced are the timing of reagent addition and the relative amounts of the reagents used. To maximize the cyclic-thionocarbonate yield, methyl iodide needs to be added to the reaction mixture after the other reagents; also, the phase-transfer catalyst, and the remaining reagents, need to be limited to molar amounts equal to that of the substrate (Scheme 10).56
Synthesis of a cyclic thionocarbonate by initial stannylene complex formation can be complicated if more than one complex is possible because a dynamic equilibrium will exist between the possible structures.57–59 The equilibrium population of the various complexes is determined by their stability, which is a function of factors such as ring strain, steric hindrance, and inductive effects. The relative amounts of the various complexes do not by themselves determine final product distribution because "the steric inaccessibility of the activated oxygen atoms may retard or prevent a major complex from reacting, thus allowing a minor complex to determine the product".57 An illustration of how these factors can cause quite different cyclic thionocarbonates to form from structurally similar compounds is provided by the reactions shown in equations 12 and 13.57
Although there can be uncertainty about which cyclic thionocarbonate will form from compounds where more than one stannylene complex is possible, this uncertainty disappears for molecules with cis-related, vicinal hydroxyl groups. For such compounds the major (sometimes exclusive) product will come from a complex involving these cis-related groups (eq 1256).51,52,56
In some situations a competition exists between formation of cyclic and noncyclic thionocarbonates. In the reaction shown in Scheme 11 there is such a competition between the cyclic thionocarbonate 27 and the noncyclic thionocarbonate 26.57 Complete cyclic thionocarbonate formation is only temporarily delayed if reaction is allowed to continue because compound 26 is converted into 27 under the reaction conditions.
VI. Thionoesters
The standard synthesis of thionoesters is shown in eq 14. Scheme 12, which contains a more detailed picture of this sequence, includes a proposed mechanism for this reaction.1 Although this method of thionoester preparation is effective, it requires handling the toxic gases phosgene and hydrogen sulfide.27 This added difficulty in preparation is a factor in thionoesters being used less frequently than other, O-thiocarbonyl carbohydrate derivatives.
Thionobenzoates are used for radical formation more often than other thionoesters. Although conditions for preparation of thionobenzoates make them less attractive starting materials that other O-thiocarbonyl compounds, these esters become more desirable reactants if the O-thiobenzoyl group has an additional role in the reaction. In the transformation shown in Scheme 13 the 2-O-thiobenzoyl group anchimerically assists glycoside formation prior to participating in radical reaction.60,61. | textbooks/chem/Organic_Chemistry/Radical_Reactions_of_Carbohydrates_(Binkley)/Radical_Reactions_of_Carbohydrates_II%3A_Radical_Reactions_of_Carbohydrates/11%3A_Synthesis_of_O-Thiocarbonyl_Compounds/V._Cyclic_Thionocarbonates.txt |
A. The Strength of the Participating Nucleophile
Some compounds do not form every type of O-thiocarbonyl derivative. The tetrasaccharide 28, for example, does not produce a (thiocarbonyl)imidazolide (31) but does form a xanthate (30) (Scheme 14).62 A possible explanation for this difference in behavior is based upon the reactivity of the nucleophiles involved in preparation of each derivative. The first step in xanthate formation is conversion of 28 into the powerful nucleophile 29 by deprotonation of the C-2' hydroxyl group with sodium hydride. These reaction conditions stand in contrast to those for (thiocarbonyl)imidazolide synthesis, which depends upon the less effective nucleophile 28. (The small equilibrium concentration of the alkoxide ion 29, produced by DMAP deprotonation of 28, is insufficient to cause the (thiocarbonyl)imidazolide 31 to form in detectable amounts).
Another example illustrating the role of nucleophilicity in producing O‑thiocarbonyl compounds concerns the phenoxythionocarbonate 34, which cannot be prepared from the diol 32, even though the xanthate 35 easily forms from this compound (32) by way of the alkoxide ion 33 (Scheme 15).63 Once again, greater ease in xanthate formation can be linked to greater nucleophilicity of an alkoxide ion when compared to its corresponding alcohol.
A method for increasing the nucleophilicity of a partially protected carbohydrate without converting it into a fully ionic compound consists of forming a derivative containing a tin–oxygen bond. This is the approach adopted in several of the reactions (Schemes 2, 3, 8, and 11 and equations 12 and 13) discussed thus far. In the derivatization shown in Scheme 3, for example, combining the methyl glycoside 5 with bis(tributyltin)oxide forms a nucleophile able to produce a (thiocarbonyl)imidazolide, but reaction of 5 without increasing its nucleophilicity is unsuccessful.20
Another advantage of the nucleophilicity of an alkoxide ion when participating in xanthate synthesis is that reaction can take place at low temperatures.64,65 Reaction occurring under these conditions is particularly important for forming tertiary xanthates (eq 1564) because these compounds readily undergo thermal rearrangement and elimination reactions.66
B. Protecting-Group Migration and Loss
Although increasing the nucleophilicity of a hydroxyl group by deprotonation is sometimes helpful in forming an O-thiocarbonyl compound, deprotonation also promotes protecting group migration. Compound 36, for example, forms a (thiocarbonyl)imidazolide with the silyl group remaining in place, but attempted synthesis of the corresponding xanthate causes complete O-2' to O-3' silyl-group migration (Scheme 16).67 In another example, compound 37 forms a xanthate in only 31% yield, but (thiocarbonyl)imidazolide formation is quantitative (Scheme 17).68 Group migration (Scheme 16) and reduced product yield (Scheme 17) (possibly through benzoyl group loss or migration or both) are linked to the nucleophilicity of the alkoxide ions formed during xanthate synthesis.
Even though the absence of a strong base during (thiocarbonyl)imidazolide formation reduces the likelihood of group migration, it does not eliminate this possibility entirely. Whenever a carbon atom bearing a hydroxyl group has an acyloxy or silyloxy group on a neighboring (or nearby) atom, group migration is a possibility15,69,70 because the organic base (and catalyst) imidazole is generated as the reaction proceeds (eq 2). An example of a migration reaction that takes place during (thiocarbonyl)imidazolide synthesis is shown in eq 16, where the benzoyl group at O-3 in the starting material migrates to O-4 in forming the minor product.15
The possibility that imidazole causes group migration during (thiocarbonyl)imidazolide formation garners support from the observation that DMAP causes such reaction during thionocarbonate synthesis. Phenyl thionocarbonates 40 and 41 both form when either nucleoside 38 or 39 reacts with phenoxythiocarbonyl chloride (12) (Scheme 18).71 The formation of this mixture of products (40 and 41) is the result of DMAP-catalyzed, silyl-group migration in compounds 38 and 39 prior to esterification (Scheme 18).
Group migration sometimes can be avoided by modification in the reaction conditions. The xanthate 42, for instance, cannot be synthesized by the standard procedure, but it forms in excellent yield when carbon disulfide is the reaction solvent (Scheme 19).72 When carbon disulfide is present in large excess, the increased rate of xanthate formation suppresses competing, unimolecular reactions such as group migration.
C. Displacement Reactions
O-Thiocarbonyl groups can function as nucleofuges in displacement reactions. They are not particularly effective in this role; consequently, their participation is limited to internal reaction in which the nucleophile is created by deprotonation and is held in an advantageous position for reaction. An example of internal displacement of this type is shown in Scheme 20 where the thionocarbonate 44 forms in good yield from reaction of the nucleoside 43 with phenoxythiocarbonyl chloride (12) in the presence of pyridine, but when the stronger base DMAP (13) is used, internal SN2 displacement produces the anhydro nucleoside 45.73 Support for the idea that 44 is an intermediate in this reaction comes from its quantitative conversion into 45 by reaction with DMAP.
Cyclic thionocarbonates also can be substrates in nucleophilic substitution reactions.74,75 In the reaction shown in Scheme 21, for example, formation of the 2',3'-O-thiocarbonyl derivative 46 places nucleofuges at C-2' and C-3'. The C-2' substituent then is displaced by an oxygen atom in the nitrogenous base portion of the molecule.74
Another example of nucleophilic substitution involving an O-thiocarbonyl compound is found in Scheme 22, where attempted acetylation of the disaccharide 47 causes replacement of the O-imidazol-1-ylthiocarbonyl group with an acetyl group.10 A reasonable assumption is that the desired acetate 48 actually forms, but the O‑imidazol-1-ylthiocarbonyl group is a sufficiently good nucleofuge that it is displaced by the neighboring O-acetyl group in a reaction that leads to the pentaacetate 49. Acetylation of the closely related xanthate 50 (eq 17) without internal displacement indicates that the O-[(phenylthio)thiocarbonyl] group is a less effective nucleofuge.10
D. Regioselective Reactions
In a carbohydrate with more than one unprotected hydroxyl group, it is sometimes possible to predict which group will react preferentially with a limited amount of a thioacylating agent. For example, reaction of the less hindered of two hydroxyl groups will occur if there is a substantial difference in their steric shielding; thus, in the reaction shown in eq 18, regioselective thioacylation takes place at the primary, rather than the secondary, hydroxyl group.76
Even if there is little difference in steric shielding of two hydroxyl groups, site selectivity sometimes can be predicted if one of the groups is attached to C-2 and deprotonation is the first step in the reaction. Under these conditions the typically greater acidity of the C-2 hydroxyl group determines which of the two possible alkoxide ions will form to a greater extent. This preferential formation leads to regioselective reaction at C-2 (Scheme 1563 and eq 1930). Comparing the reaction shown in eq 1930 with that in eq 141 demonstrates that predicting regioselective reaction at C-2 must be done cautiously. In these two reactions the same compound exhibits different selectivity when the reagents and the reaction conditions change.
Regioselectivity extends to reactions where esterification is preceded by formation of a stannylene complex (equations 12 and 13). Since this selectivity is dependent upon the stability and reactivity of the various stannylene complexes that are in equilibrium in the reaction mixture, predicting or even rationalizing the formation of reaction products is complicated by esterification being a two-step process with selectivity involved in each step. Although regioselective reaction of stannylene complexes is often high, it is far from assured, as is illustrated by the nearly unselective reaction of the methyl glycoside 51 (eq 20).52 The difficulty in predicting site selectivity is underscored when comparing the reaction shown in eq 20 with that in eq 21, where an essentially unselective reaction becomes highly selective upon changing the configuration of the methoxy group at C-1.52 Under carefully selected conditions reaction of organotin complexes of a variety of unprotected methyl glycosides with phenoxythiocarbonyl chloride leads to highly regioselective thionocarbonate formation.53
VIII. Summary
Synthesis of O-thiocarbonyl compounds [(xanthates, (thiocarbonyl)imidazolides, aryl thionocarbonates, cyclic thionocarbonates, thionoesters)] is the first step in using them to generate carbon-centered radicals.
Xanthates usually are prepared by deprotonating a partially protected carbohydrate and then reacting the resulting alkoxide ion with carbon disulfide and methyl iodide. The primary limitation of this approach is that it involves conditions in which base-sensitive compounds are unstable. Xanthate synthesis by phase-transfer reaction or by reaction with phenyl chlorodithioformate avoids this difficulty.
(Thiocarbonyl)imidazolides are formed by reacting a partially protected carbohydrate with N,N-thiocarbonyldiimidazole. These conditions for synthesis are much less basic that those used for preparing xanthates.
Aryl thionocarbonates typically come from reaction of a partially protected carbohydrate with phenoxythiocarbonyl chloride in the presence of DMAP (4-dimethylaminopyridine). Side reactions are rare and tend to arise when DMAP promotes base-catalyzed reactions that compete with thionocarbonate formation. Phenyl thionocarbonates also can be prepared in reactions catalyzed by N‑hydroxysuccinimide (NHS). This alternative procedure normally is implemented to improve product yields or avoid side reactions caused by DMAP. An additional option for phenyl thionocarbonate preparation consists of reacting a partially protected sugar with thiophosgene and treating the product with a phenol. This procedure is useful in preparing phenyl thionocarbonates with groups, usually electron-withdrawing ones, in the aromatic ring.
If a partially protected carbohydrate has vicinal, cis-related hydroxyl groups, reaction with N,N-thiocarbonyldiimidazole will form a cyclic thionocarbonate. A second procedure for synthesis of these compounds consists of formation of a stannylene complex of a carbohydrate, and then reaction of this complex with thiophosgene or phenoxythiocarbonyl chloride.
Thionoesters are less frequently used in deoxygenation reactions than other O-thiocarbonyl compounds, in part, due to the difficulty in their preparation. The only thionoesters used to a significant extent in deoxygenation are thionobenzoates.
When O-thiocarbonyl compounds are unable to form under the standard reaction conditions, sometimes they can be synthesized by converting the partially protected carbohydrate reactant into its corresponding alkoxide ion. Forming an alkoxide ion also increases the possibility that group migration will compete with formation of an O-thiocarbonyl, carbohydrate derivative. | textbooks/chem/Organic_Chemistry/Radical_Reactions_of_Carbohydrates_(Binkley)/Radical_Reactions_of_Carbohydrates_II%3A_Radical_Reactions_of_Carbohydrates/11%3A_Synthesis_of_O-Thiocarbonyl_Compounds/VII._Factors_Affecting_O-Thiocarbonyl_C.txt |
Reaction of an O-thiocarbonyl derivative of a carbohydrate with a tin- or silicon-centered radical generates a carbon-centered radical that undergoes reactions typical of such an intermediate. These reactions include abstracting a hydrogen atom from a donor molecule (almost always a tin or silicon hydride), adding to a compound containing a multiple bond, or forming a new ring system by adding internally to a multiple bond within the radical. These reactions place O‑thiocarbonyl compounds among the most useful substrates for radical formation from carbohydrates. The current chapter, where the reactions of these compounds are discussed, is a close companion to the preceding one (Chapter 11), where synthesis of O-thiocarbonyl carbohydrate derivatives is described.
12: Reactions of O-Thiocarbonyl Compounds
A. A Two-Step Sequence
In 1975 Barton and McCombie reported a two-step sequence for hydroxyl-group replacement by a hydrogen atom.1 The first step in this process was the conversion of the hydroxyl group into an O-thiocarbonyl group, and the second step (the Barton-McCombie reaction) was a free-radical chain reaction that replaced the O-thiocarbonyl group with a hydrogen atom. A typical example of this widely used, reaction sequence is shown in Scheme 1.2
Various types of O-thiocarbonyl compounds undergo the Barton-McCombie reaction. Initially this group consisted of xanthates (1), thionobenzoates (2), thiocarbonylimidazolides (3), and thionoformates (4) (Figure 1).1 Subsequently, this list was expanded to contain phenyl thionocarbonates (5),3,4 including those with electron-withdrawing substituents in the aromatic ring (6-8),5,6 and cyclic thionocarbonates (9).7,8
B. Proposed Reaction Mechanisms
A proposed mechanism for the Barton-McCombie reaction is shown in Scheme 2.1,9 In the initiation phase of this reaction thermal decomposition of 2,2'-azobis(isobutyronitrile) (AIBN) (eq 1), the most common initiator for the Barton-McCombie reaction, produces a radical that abstracts a hydrogen atom from tri-n-butyltin hydride (eq 2). In the first propagation step the tri-n-butyltin radical adds to a carbon–sulfur double bond to create the adduct radical 10 (eq 3). Reaction reaches a critical stage at this point because its success requires 10 to fragment to give the radical 11 (eq 4) before competing reactions can intervene. Once fragmentation takes place, hydrogen-atom abstraction by 11 from tri-n-butyltin hydride completes the overall reaction and generates a new, chain- carrying, tri-n-butyltin radical (eq 5). (Equations 1-5 are found in Scheme 2.)
The propagation steps for a revised mechanistic proposal for the Barton-McCombie reaction are shown in Scheme 3.10 (The initiation and termination steps for this mechanism are the same as those pictured in Scheme 2.) The primary change introduced in the revised mechanism (Scheme 3) is that Bu3Sn· does not add to the thiocarbonyl group but rather abstracts the SCH3 group. Identification of the radical 15 in the ESR spectrum of the reaction mixture supports the revised mechanism; however, an argument against mechanistic significance of 15 is that this intermediate is observed under conditions quite different from those of the Barton-McCombie reaction (e.g., no effective hydrogen-atom donor (Bu3SnH) was present in the reaction mixture because the tri-n-butyltin radicals were generated from photolysis of Bu3SnSnBu3.10)
Subsequent competition experiments returned support to the original mechanism (Scheme 2).11–13 In addition to these experiments, 119Sn NMR identification of the intermediate 12 (R = SCH3) in a Barton-McCombie reaction mixture provided evidence for the addition of Bu3Sn· to the thiocarbonyl group, as proposed in Scheme 2, rather than abstraction of SCH3, as proposed in Scheme 3.13 Further mechanistic analysis led to the conclusion that the tri-n-butyltin radical must be adding reversibly to the thiocarbonyl group to give the radical 10, which then fragments as shown in eq 4 (Scheme 2). Additional support for this mechanism is based on a ring-forming reaction that is described at the end of this chapter, after cyclization reactions have been discussed.
C. Hydrogen-Atom Donors/Chain-Transfer Agents
Critical to the success of the Barton-McCombie reaction is the compound that donates a hydrogen atom to the carbohydrate radical (eq 5) to complete the propagation sequence. Tri-n- butyltin hydride is particularly well suited for this role because it rapidly donates a hydrogen atom to a carbon-centered radical, and in the same reaction generates the chain-carrying radical Bu3Sn·, an intermediate needed to begin a new sequence of propagation steps (eq 3). (A critical feature of the reactivity of Bu3Sn· is that it does not cause side reactions by abstracting hydrogen atoms from carbon-hydrogen bonds.)
Although use of tri-n-butyltin hydride has significant advantages, it also suffers from substantial drawbacks. There are serious problems associated with the toxicity of tin-containing compounds and the difficulty in removing residues of these compounds from reaction products. A variety of solutions to these problems have been proposed. Because these solutions apply not just to O-thiocarbonyl compounds but also to a broad range of carbohydrate derivatives, they will not be discussed here; rather, they have been gathered together and are found in Appendix I.
D. The Scope and Reactivity of O-Thiocarbonyl Compounds
The number of O-thiocarbonyl carbohydrate derivatives that undergo the Barton-McCombie reaction is large and continues to grow. Although all of these compounds react basically in the same way, some are better suited than others for particular situations. The following several sections focus on the scope and special reactivity of various O-thiocarbonyl compounds. To emphasize their broad range of reactivity, references are provided to Barton-McCombie reaction taking place at various positions in substituted carbohydrates. These references are not meant to represent the total number that exists, but rather to provide examples of the reactions possible in cyclic and open-chain carbohydrates.
1. Xanthates
A striking feature of xanthate reactivity is the number and variety of carbohydrates that can be deoxygenated by xanthate formation followed by Barton-McCombie reaction. This sequence replaces hydroxyl groups at the 2-,14,15 3-,16,17 4-,2,18 and 6-19,20 positions with hydrogen atoms in compounds containing pyranoid rings, as well as the 1-,21 2-,22,23 3-,24,25 5-,26 and 6-27positions in those with furanoid rings. Equations 615 and 728 provide typical examples of reactions of carbohydrates containing pyranoid and furanoid rings, respectively. (Hypophosphorous acid, one of the substitutes for tri-n-butyltin hydride mentioned in Appendix I, is the hydrogen-atom donor in the reaction shown in eq 6.) Xanthates also are used in Barton-McCombie reaction of alditols,29,30 cyclitols,31,32 and nucleosides.33,34
2. Thionocarbamates
Among thionocarbamates the (thiocarbonyl)imidazolides (3, Figure 1) are easily the most frequently used substrates for the Barton-McCombie reaction. The range of types of compounds involved is broad and includes (thiocarbonyl)imidazolides formed from carbohydrates with hydroxyl groups at the 2-,35 3‑,36–38 4-,39–41 and 6-42 positions in compounds with pyranoid rings, and at the 2‑,43,44 3-,45,46 and 5‑47,48 positions in compounds with furanoid rings. There also are numerous reports of Barton-McCombie reactions of nucleosides with O‑imidazol-1-ylthiocarbonyl groups at C-2'‑49,50 and C-3'.51,52
3. Thionocarbonates
a. Phenyl Thionocarbonates
Phenyl thionocarbonates are yet another O-thiocarbonyl, carbohydrate derivative that undergoes the Barton-McCombie reaction. These derivatives participate in reaction at C-2,53,54 C-3,55,56 C-4,57,58 and C‑659,60 in compounds with pyranoid rings, and C-1,61,62 C-2,63,64 C‑3,65,66 and C‑567,68 in compounds with furanoid rings. Further, phenyl thionocarbonates are the preferred intermediates for nucleoside deoxygenation. They are involved in reaction at the 2'-position not only for a large number of 1,1,3,3-tetraisopropyl-1,3-disiloxanediyl-protected nucleosides,69,70 but also for compounds protected by benzyl,71 benzoyl,72,73t-butyldimethylsilyl,74 and pivaloyl groups.75 Similar reactions take place at the 3'- and 5'-positions.65,76
b. Substituted Phenyl Thionocarbonates
Phenyl thionocarbonates typically undergo a Barton-McCombie reaction that produces deoxygenated products in good yield; however, when product yields are low, introducing one or more electron-withdrawing substituents into the aromatic ring often raises these yields.77–80 Two examples illustrate the effect of these substituents. First, attempted Barton-McCombie reaction of the phenyl thionocarbonate 17 produces a complex mixture of products, but the 2,4-dichlorophenyl analog 18 forms the desired dideoxy nucleoside (Scheme 4).77 In the second example, the phenyl thionocarbonate 19 is unreactive under typical Barton-McCombie conditions, but its pentafluorophenyl analog 20 reacts with tri-n-butyltin hydride to give the corresponding deoxy nucleoside 21 (eq 8).78
In light of the better yields produced by the substituted phenyl thionocarbonates 18 and 20, it is surprising to find that the p-fluoro-, pentafluoro-, p-chloro-, 2,4,6-trichloro-, and pentachlorophenoxythiocarbonyl derivatives of cyclododecanol all react more slowly than the unsubstituted compound.6 In attempting to understand such a finding it is useful to consider the various ways in which ring substituents might influence reactivity. Evidence from the study of O-thiocarbonyl compounds suggests that the formation of the radical 23 (eq 9) is a reversible process and that the rate-determining step in this reaction sequence (equations 9 and 10) is the fragmentation of 23 shown in eq 10.9 An aromatic ring substituent is likely to impact this process in several ways. These include altering the radicophilicity of 22, the stability of 23, and the strength of the carbon–oxygen bond being broken to produce R· and 24 (eq 10); therefore, since an aromatic-ring substituent can exert influence on reactivity in several ways, understanding and predicting the overall effect that such a substituent will have on a reaction can be difficult.
c. Cyclic Thionocarbonates
Cyclic thionocarbonates undergoing Barton-McCombie reaction include compounds in which 2,3-O-thiocarbonyl,81–83 3,4-O-thiocarbonyl,53,84 and 4,6-O-thiocarbonyl85 groups are attached to pyranoid rings. This reaction also takes place with 2,3-O-thiocarbonyl7,8 and 5,6-O- thiocarbonyl67,86 groups in compounds with furanoid rings. Cyclic thionocarbonates, formed from acyclic structures, also undergo the Barton-McCombie reaction.87
Reaction of a cyclic thionocarbonate is more complex than reaction of other O-thiocarbonyl compounds because it also involves ring opening. Since ring opening potentially can place a radical center on either of two carbon atoms, reaction often produces a mixture of products (eq 11).7,8 Formation of this mixture not only can reduce the amount of the desired product but also can complicate its isolation.
A proposed mechanism for reaction of a cyclic thionocarbonate with tri-n-butyltin hydride is given in Scheme 5.8 A "key" intermediate in this reaction is the radical 25, formed by addition of the tri-n-butyltin radical to the thiocarbonyl group. Radical stability usually controls the direction of ring opening; thus, even though 25 can produce either a primary or a secondary radical by ring opening, the pathway followed leads exclusively to the secondary radical (Scheme 5).7,8,88,89
Although radical stability normally controls the direction of ring opening in a cyclic thionocarbonate, relief of angle strain in the transition state sometimes is the major factor.90 Ring opening of compound 26, for example, leads to the product derived from a secondary, rather than a tertiary, radical (eq 12).91 (A mechanism for this reaction is shown in Scheme 6.) Molecular mechanics calculations on noncarbohydrates indicate that fragmentation to give a less stable radical will occur if relief of ring strain in the transition state is great enough (eq 13).90 This relief of strain provides an explanation for the unexpected conversion of 26 into 27 rather than 28 (eq 12).
There is a concentration effect associated with the reaction of the cyclic thionocarbonate 26. In concentrated Bu3SnH solution only the kinetically favored product 27 is formed, but in dilute solution some of the thermodynamically favored isomer 28 is produced (Scheme 6). One explanation for this behavior is based on reversible formation of the secondary radical 30 from the cyclic radical 29. In concentrated solution 30 abstracts a hydrogen atom rapidly enough from Bu3SnH to prevent significant return to 29. Under these conditions only the product 27 is formed. In dilute solution hydrogen-atom abstraction by 30 is slowed to the point that reversible formation of 29 becomes significant and creates greater opportunity for 29 to be converted (irreversibly) into the thermodynamically favored tertiary radical 31. Since in dilute Bu3SnH solution both kinetically and thermodynamically favored pathways are followed, a mixture of the products 27 and 28 is produced.
4. Thionoesters
Primarily because their synthesis is more challenging, thionoesters are selected less frequently as starting materials for the Barton-McCombie reaction than are other O-thiocarbonyl derivatives. Thionoesters with O-thiocarbonyl groups at C-292 and C-31 in pyranoid rings and C-228 in furanoid rings are known substrates for deoxy sugar formation. A typical example is shown in eq 14.1 The thionoesters that undergo reaction are, with rare exception,93 thionobenzoates.28,92,94–101
E. Competing Reactions
Frequent application of the Barton-McCombie reaction in carbohydrate chemistry occurs because this reaction has a variety of attractive features. These include the broad range of compounds that undergo this reaction, the generally good product yields, and the freedom, in most cases, from significant, competing reactions. Even though side reactions usually do not represent a major concern, Barton-McCombie reaction has been conducted on so many compounds that quite a number of these reactions have been identified. They range in importance from alcohol regeneration, a common but usually minor side reaction, to phenyl-group migration, a rare event.
1. Alcohol Regeneration
Regenerating the alcohol from which an O-thiocarbonyl compound originally was synthesized is a side reaction sometimes accompanying deoxygenation. An example of such a reaction is shown in eq 15.102 In rare instances alcohol regeneration is the major reaction pathway (eq 16).42
Not all O-thiocarbonyl derivatives of carbohydrates are equally prone to alcohol regeneration. One of the advantages initially associated with phenyl thionocarbonates was that they did not undergo this reaction.3,4 Continued study of these compounds, however, showed that they are not immune to the alcohol-reforming process (eq 17).103
a. Proposed Reaction Mechanisms
Although alcohol regeneration is the most common competing process during Barton-McCombie reaction, there is not general agreement on how the regeneration process proceeds. The two most frequently cited possibilities are both multi-step processes with many intermediates in common. The difference between these two rests with thiocarbonyl group reduction, one mechanism (Scheme 7) involves radical intermediates and the other (Scheme 8) does not.
(1). A Radical-Based Process
If one assumes that the radical 10 is an intermediate in the Barton-McCombie reaction (Scheme 2, X = SCH3), the possibility exists that before this radical fragments, it could abstract a hydrogen atom. Such a reaction would produce the intermediate 32 (Scheme 7).1,9 Formation of 32 not only would reduce the deoxygenated product yield, but it also could provide an explanation for the formation of the regenerated alcohol 36 as a side-product in the reaction.1,9 When viewed in this way, 32 is the beginning point in a series of events that leads first to the thionoformate 33, which reacts with tri-n-butyltin hydride to give the tin-containing intermediate 34, a substance that hydrolyzes during workup to the hemithioacetal 35. Compound 35 then decomposes spontaneously to give the alcohol 36 and thioformaldehyde (Scheme 7).9
(2). A Hydride-Transfer-Based Reaction Sequence
The conditions originally used for the Barton-McCombie reaction did not include an added initiator; rather, reaction depended upon adventitious initiation. Within a few years, however, adding 2,2'-azobis(isobutyronitrile) (AIBN) became standard procedure because dependable initiation was recognized as a significant factor in maximizing deoxygenation.4,13 The reaction shown in eq 18 is one that provides an illustration of the improvement in product yield brought about by an added initiator.104 The observation that alcohol regeneration occurs in the absence of an initiator in the reaction shown in eq 18, supports the idea that a radical reaction may not be involved in the alcohol-reforming process.4
When the Barton-McCombie reaction is initiated by Et3B–O2,105,106 it can take place at temperatures much lower than those required for AIBN initiation (eq 19).21 Because it was originally thought that reducing the temperature of a Barton-McCombie reaction to about 80 oC caused alcohol regeneration to begin to become important,12 further lowering the reaction temperature would be expected to increase alcohol formation. When reactions were conducted at lower temperature using Et3B–O2 initiation, Barton-McCombie reaction took place with little or no alcohol regeneration. This finding was not consistent with the idea that 10 (Scheme 7) was increasingly likely to abstract a hydrogen atom from Bu3SnH as the reaction temperature decreased.13 If the conversion of 10 into 32 by hydrogen-atom abstraction were not taking place, it raised the possibility that 32 was formed in a different, perhaps nonradical, reaction (Scheme 8).
Several investigators have proposed that hydride transfer may be responsible for alcohol regeneration.13,107,108 As a part of the most detailed of these proposals, it was suggested that nonradical addition of tri-n-butyltin hydride to an O‑thiocarbonyl group occurs in the first step in alcohol regeneration, and it occurs later in the reaction sequence when the thionoformate 33 is converted into the thioacetal 34 (Scheme 8).108
b. Alcohols from Dimeric Esters
The observation that minor amounts of “dimeric’ esters were formed during the phenyl thionocarbonate synthesis shown in eq 20 raised the possibility that if these esters were not removed prior to Barton-McCombie reaction, an alcohol could be formed because each dimeric ester reasonably could be expected to react with tri-n-butyltin hydride to give a molecule of a deoxygenated compound and one of the starting alcohol.103 This expectation was confirmed when the dimeric ester 37 was found to undergo the reaction shown in eq 21.109
c. Minimizing Alcohol Regeneration
The rate determining step in the Barton-McCombie reaction is believed to be the unimolecular fragmentation of the radical 10 to give the carbon-centered radical R· (Scheme 9).9 For alcohol regeneration, however, the rate is more likely to depend upon a bimolecular reaction involving Bu3SnH. This analysis (Scheme 9) is consistent with the finding that keeping the concentration of Bu3SnH at a low level during reaction (i.e., adding the tin hydride slowly as the reaction proceeds) is associated with maximizing deoxygenation; for example, the xanthate shown in eq 22 reacts to give a deoxy sugar in 54% yield when all the Bu3SnH is present at the beginning of the reaction, but the yield rises to 85% when Bu3SnH is added over a period of 1.5 h.36 Part of the reduced deoxy sugar yield in the reaction where all the Bu3SnH was added at the beginning is due to recovery of the alcohol from which the xanthate was synthesized.36
2. O-Thionocarbonyl Group Conversions and Rearrangements
a. Conversion of a Xanthate into a Dithiocarbonate
When Bu3SnH is the hydrogen-atom donor in the reaction shown in eq 23,1 a deoxy sugar forms in the normal manner, but if a less effective donor is used, xanthate conversion to a dithiocarbonate competes with the deoxygenation process. This conversion becomes significant when reaction is conducted with 2‑propanol serving as both solvent and hydrogen-atom donor. When benzene is the solvent, dithiocarbonate formation is the major reaction pathway.110 Benzene is, in fact, such a poor hydrogen-atom donor that any carbohydrate present in solution is a more likely hydrogen-atom source for the small amount of deoxy sugar formed.
The reaction shown in eq 23 begins with formation of the carbohydrate radical R·. As pictured in Scheme 10, this radical (R·) then either adds to a molecule of starting material, leading to a dithiocarbonate, or abstracts a hydrogen atom, producing a deoxy sugar. The data presented in eq 23 confirm the expectation from the proposed mechanism (Scheme 10) that deoxy sugar formation is favored when effective hydrogen-atom donors are used, and dithiocarbonates form more easily in reactions run at high xanthate concentrations in the presence of poor hydrogen-atom donors.110
b. Thionocarbonate-Thiocarbonate Rearrangement
A reaction closely related to the xanthate-dithiocarbonate rearrangement just discussed is the conversion of a thionocarbonate into a thiocarbonate. When a hydrogen-atom donor is present in a reaction mixture in an amount less than that needed to supply a hydrogen atom to each carbohydrate radical, thionocarbonate to thiocarbonate rearrangement can take place.82,88,89,111,112 In the reaction shown in Scheme 11 this rearrangement represents the major reaction pathway when the amount of tri-n-butyltin hydride is significantly less than that required for complete reduction.111 Rearrangement is, of course, undesirable when simple reduction is the goal of a reaction, but it can be useful if the purpose of the reaction is to convert a sugar into a thiosugar.112
c. Conversion of an S-Alkyl Xanthate into an O-Alkyl Xanthate
Reaction of the carbohydrate xanthate 39 (an S-alkyl xanthate) with the triphenyltin radical is the first step in a propagation sequence (Scheme 12) that converts 39 into a xanthate with the carbohydrate portion of the molecule bonded to sulfur (40, an O-alkyl xanthate).113,114 The second step in this sequence is reaction of the carbohydrate radical with triphenyltin xanthate to produce 40 and the chain-carrying, triphenyltin radical. This second step must be reversible in order to account for the epimers 41 and 42 being interconverted under the reaction conditions (eq 24).114 Effective hydrogen-atom donors, such as triphenyltin hydride, must be excluded to prevent simple reduction. Excluding triphenyltin hydride but still having the triphenyltin radical needed to initiate the reaction is accomplished by photolysis of bis(triphenyltin) (Scheme 12).
3. Reaction With Molecular Oxygen
Reaction conducted in the presence of molecular oxygen leads to rapid capture of O2 by carbon-centered radicals. This capture is faster than hydrogen-atom abstraction from Bu3SnH. Radical capture of O2 is suppressed by the normal procedure of excluding oxygen from the reaction mixture, but when O2 is deliberately added, its combination with a carbohydrate radical becomes a major or even the exclusive reaction pathway (eq 25).89
4. Elimination Reactions
a. Reactions of Compounds with Two O-Thiocarbonyl Groups
If there are two O-thiocarbonyl groups in the same molecule, their physical separation affects whether or not Barton-McCombie reaction will take place.1,115–134 In the reaction of compound 43, for example, the substituents are sufficiently well separated to allow replacement of each group by a hydrogen atom to proceed in the normal manner (eq 26115).115,130 When O‑thiocarbonyl groups are attached to adjacent carbon atoms, reaction of one of these groups produces a carbon-centered radical that forms a double bond by elimination of the second group.117–129,131,133 An example is shown in eq 27.117–119 As indicated in Scheme 13, elimination takes place when R = OC6H5 or SCH3, but in the rare event that R = C6H5, the elimination pathway leads to the unstable phenyl radical; as a consequence, radical cyclization takes place instead of elimination (eq 28).1 If two O-thiocarbonyl groups are not on adjacent carbon atoms, but the radical produced by reaction of the first is centered on an atom in close proximity to the second, internal addition will take place.115,116,132,134 Because the new radical formed by this reaction does not have a clear path to an elimination product, more complex reaction, such as that shown in eq 29,115 is likely to take place.
b. Reactions of Compounds with a Single O-Thiocarbonyl Group
As described in the previous section, elimination reactions take place when compounds with O-thiocarbonyl groups on adjacent carbon atoms react with tin or silicon hydrides. Similar reaction takes place when an O-thiocarbonyl group is attached to a carbon atom that has an azido,135 bromo,135–137 chloro,135,138,139 iodo,135 isocyano,140 methylthio,135 or phenylthio,141 substituent bonded to an adjacent carbon atom. An example is given in eq 30.135 This process begins with formation of a carbon-centered radical and ends with radical expulsion from an adjacent carbon atom (Scheme 14). In some instances it is reaction of the O‑thiocarbonyl group that generates the carbon-centered radical, but in others, particularly those involving compounds containing bromine and iodine, halogen-atom abstraction is the first reaction to take place.
c. Use of Sacrificial Olefins
One of the difficulties associated with the synthesis of unsaturated compounds by radical reaction is that a radical intermediate may add to a recently formed double bond in a product molecule. When such an unwanted addition occurs, it may be reduced in importance to an acceptable level by adding an alkene such as 1-dodecene to the reaction mixture. This alkene, a compound described as a “sacrificial olefin”, protects the elimination product by scavenging radicals before they add to product molecules.142
d. Cyclic Thionocarbonates
One way to view cyclic thionocarbonates is as compounds that have adjacent O-thiocarbonyl groups and, consequently, might undergo radical elimination. Since Barton-McCombie reactions of cyclic thionocarbonates generally give good yields of deoxy compounds, an elimination reaction can be expected only if such a reaction benefits from a special driving force. This driving force can come from reaction producing an unsaturated compound with a double bond stable enough that its formation significantly lowers transition-state energy. Consistent with this idea is formation of the glycal 46 as the only product from reaction of the cyclic thionocarbonate 45 with tri-n-butyltin hydride (eq 31).143 2',3'-O-Thiocarbonyl nucleoside derivatives undergo similar reaction to give unsaturated nucleosides, but only as minor products.144–146
e. Preventing Thermal Elimination
The Barton-McCombie reaction of xanthates is most successful when these compounds are prepared from secondary alcohols. Xanthates formed from tertiary alcohols are prone to thermal elimination (eq 32, Chugaev elimination147) at temperatures normally used in the Barton-McCombie reaction. When reaction is initiated by triethylboron–oxygen, however, it can be conducted at room temperature, where reaction of tertiary xanthates occurs without competing elimination.106,148
5. Reversible Addition to an O-Thiocarbonyl Group
Xanthates synthesized from primary alcohols usually require higher temperatures in the Barton-McCombie reaction and often give lower product yields.149,150 These problems can be overcome in some instances by changing the hydrogen-atom donor from Bu3SnH to (Me3Si)3SiH. Tri-n-butyltin hydride is a less effective donor than tris(trimethylsilyl)silane in these reactions due to greater reversibility of Bu3Sn· addition to a thiocarbonyl group (Scheme 15).149 Because the S–Si bond [90 kcal mol-1 (377 kJ mol-1)]151 is stronger than the S–Sn bond [65 kcal mol-1 (272 kJ mol‑1)],151 reversal of addition of (Me3Si)3Si· is less likely to occur. Reduced reversibility means that once a silyl radical has added to an O-thiocarbonyl group, a difficult forward reaction (e.g., one producing a primary radical149 and, sometimes, one producing a secondary radical152,153 ) can compete more effectively with reverse reaction to give the starting materials. Limiting reversibility is even more effective when reaction take place under the low temperature conditions made possible by Et3B–O2 initiation.33
6. Reduction of a Thiocarbonyl Group to a Methylene Group
Conversion of a thiocarbonyl group into a methylene group represents a rare type of competition for the Barton-McCombie reaction. This transformation changes the O‑phenylthiocarbonyl group in 47 into an O-benzyl group (eq 33),28 and it is responsible for a similar change in the cyclic thionocarbonate 38 (eq 34).89,111 The reaction shown in eq 34 occurs in much higher yield when a large excess of tri-n-butyltin hydride is present. One effect of a large excess of Bu3SnH is to increase the rate of hydrogen-atom abstraction by the radical 48 (Scheme 16) to the point that ring opening by this radical is too slow to be detected.
7. Phenyl-Group Migration
Compound 49, a xanthate with an O-[(alkylthio)thiocarbonyl] group at C-4 and 2,3-O-diphenylmethylene protection, undergoes phenyl group migration and ring opening during reaction (eq 35).154 According to the mechanism proposed in Scheme 17, this reaction depends upon the presence of substituents in the protecting group that can undergo migration by an addition-elimination process. Consistent with this mechanistic proposal is the observation that if the 2,3-O-diphenylmethylene group is replaced by a 2,3-O-isopropylidene group, normal reaction occurs (eq 36).155
F. Influence of Steric Effects on Reactivity
Reaction of the epimeric thionocarbamates 51 and 52 with Bu3SnH illustrates the role that steric effects play in the Barton-McCombie reaction (eq 37).156 The dramatically different steric environment for the C-3 substituents in 51 and 52 does not have a substantial impact on their reactivity; specifically, the difference of a factor of four in reaction times (reaction rates were not measured) is consistent with only minor influence by steric effects. A possible explanation for this modest difference in reactivity is that although the steric congestion around C-3 in compounds 51 and 52 is substantially different, it is not dramatically different in the area of the sulfur atom, where the reaction is taking place. Comparing the reactivity of the xanthates 53 and 54 provides another example of the modest role of steric effects in the Barton-McCombie reaction. The decidedly more hindered O-[(methylthio)thiocarbonyl] group in 54 renders this xanthate (54) only about five times less reactive than its epimer 53.157
G. Regioselectivity
One of the characteristics of the Barton-McCombie reaction is that when it is conducted in the normal manner (i.e., with Bu3SnH as the hydrogen-atom donor and AIBN as the initiator), higher temperature is required for reaction of a primary O-phenoxythiocarbonyl group than a secondary one.76,158 This difference in reactivity can become the basis for regioselective reaction; thus, as pictured in Scheme 18, group replacement takes place only at the 3'‑position in the nucleoside 55.76
H. Chemoselectivity
Chemoselectivity in reactions of compounds containing O-thiocarbonyl groups is discussed in Section II.B of Chapter 9 in Volume I.
I. The β-Oxygen Effect
An observation about the Barton-McCombie reaction is that some carbohydrates react more easily than would be expected on the basis of model compound behavior. This greater reactivity is observed when the carbon atom in a C–O bond is β-related to a carbon atom bearing an O-thiocarbonyl group.93 Comparison of the reactivities of 56 and 57 illustrates this difference. Compound 56 does not react with tri-n-butyltin hydride at 110 oC (it does at 130 oC),93 but 57, which also is a primary (thiocarbonyl)imidazolide, does react under these conditions42,93 This and similar observations led to the proposal that "oxygen bonded to the β-carbon of a carbon radical has a marked stabilizing effect; this permits radical reactions not seen, except at much higher temperatures, in non-oxygenated model compounds."93 When first proposed, this "β-oxygen effect" represented a potentially important factor in carbohydrate chemistry because many carbon-centered radicals would be stabilized by its existence.
Considerable effort has been invested in studying the β-oxygen effect. This work has established that the presence of a β-related, carbon–oxygen bond does not necessarily increase the rate of formation of a developing, carbon-centered radical. For example, reaction of a mixture of the xanthates 58 and 59 with a limited amount of tri-n-butyltin hydride gave products 60 and 61 in approximately equal amounts (Scheme 19).159,160 Since this result meant that xanthates 58 and 59 were reacting at essentially the same rate, a β-related, carbon–oxygen bond was providing little, if any, transition-state stabilization for the developing radical 63.160
A related experiment involved the thionocarbonate 64, which could react along either of two competing pathways (Scheme 20). If the β-oxygen effect existed, reaction by path b would be favored. Experimentation showed that reaction along each pathway was equally likely; consequently, the conclusion again was that no evidence existed for a β-oxygen effect.149
Although these experiments dispelled the idea that a β-related oxygen atom generally provides stabilization to a radical center, they did not explain why the carbohydrate derivative 57 is more reactive than the model compound 56. An explanation for this behavior, however, does come from study of the equatorial and axial isomers 65 and 66, respectively. The axial epimer (66) with its gauche (synclinal) dipoles is more reactive than the equatorial epimer (65) with trans dipoles (Figure 2). The greater reactivity of compound 66 is attributed to eliminating an unfavorable dipole-dipole interaction during bond breaking. Removing such an interaction then appears to be the primary reason for the rate differences that originally were attributed to the β-oxygen effect.149 | textbooks/chem/Organic_Chemistry/Radical_Reactions_of_Carbohydrates_(Binkley)/Radical_Reactions_of_Carbohydrates_II%3A_Radical_Reactions_of_Carbohydrates/12%3A_Reactions_of_O-Thiocarbonyl_Compounds/II._Deoxygenation%3A_The_Barton-McCombi.txt |
Addition reactions can be organized on the basis of the process that follows the initial reaction of a carbon-centered radical with a multiple bond to form an adduct radical. For an adduct radical that traces it beginning back to an O‑thiocarbonyl compound, observed further reactions include atom abstraction and radical elimination. Chapter 18 contains a more general discussion of radical addition reactions, one that focuses more on the compound to which addition is occurring rather than on the radical precursor.
IV. Radical Cyclization
It is convenient to organize cyclization reactions on the basis of the locations of the reacting multiple bond and the radical center. These two can be either in the carbohydrate framework or in a substituent group. More information about radical cyclization is found in Chapter 19, which is devoted entirely to ring-forming reactions.
V. Comparing the Reactivity of O-Thioca
When one considers the success of O-thiocarbonyl compounds as substrates in the Barton- McCombie reaction, a reasonable question to ask is “Why don’t O-carbonyl carbohydrate derivatives (in particular, O-acylated compounds) exhibit similar reactivity?” An answer to this question can be framed in terms of the reactions shown in Scheme 28.223–225 According to this explanation, the equilibrium involving addition of Bu3Sn· to a compound with an O-thiocarbonyl group produces a far greater concentration of the adduct radical 77 than the concentration of the radical 76 produced by addition of Bu3Sn· to the corresponding O-carbonyl carbohydrate. The dramatically greater equilibrium concentration of 77 leads to a corresponding increased rate of carbohydrate radical (CARB·) formation.223–225
Detectable reaction of simple esters with Bu3Sn· becomes possible only if the low equilibrium concentration of 76 can be increased in some manner or compensated for by rapid further reaction of this radical (k1 large). Due to these requirements, no reaction takes place under normal Barton-McCombie conditions, but replacement of an acyloxy group with a hydrogen atom does occur when acylated carbohydrates react with (C6H5)3Si· under vigorous conditions [(C6H5)3SiH, 140 oC, 12 h, benzoyl peroxide].226 Also, under quite different conditions (HMPA, H2O, UV light, room temperature) photochemical electron transfer leads to reduction of acylated carbohydrates to the corresponding deoxy compounds.227 These and other reactions of esterified carbohydrates are discussed in detail in Chapter 8.
VI. Summary
An effective procedure for deoxygenation begins with the conversion of a hydroxyl group in a carbohydrate into an O‑thiocarbonyl group and ends with a radical reaction (the Barton- McCombie reaction) that replaces the O‑thiocarbonyl group with a hydrogen atom. Xanthates, (thiocarbonyl)imidazolides, and thionocarbonates are the most common substrates for the Barton- McCombie reaction. Although a number of hydrogen-atom donors can be used in this reaction, the usual choice is tri-n-butyltin hydride. Safety concerns about tin hydrides and problems with product purification have caused chemists to turn increasingly to other hydrogen atom sources, in particular, tris(trimethylsilyl)silane. Barton-McCombie reaction is sometimes complicated by competing reactions, the most common of which regenerates the partially protected carbohydrate from which the O‑thiocarbonyl-containing substrate was synthesized.
Differences in reactivity sometimes favor selecting a particular type of O-thiocarbonyl compound. Phenyl thionocarbonates are particularly valuable for hydroxyl group replacement by a hydrogen atom during nucleoside synthesis, and they are the least likely O-thiocarbonyl derivative to undergo alcohol regeneration. If reaction of a xanthate is attempted with a hydrogen-atom donor much less effective than tri-n-butyltin hydride, xanthate-dithiocarbonate rearrangement can take place. Reactions of tertiary xanthates can be complicated by Chugaev elimination unless these reactions are conducted at low temperature.
Deoxygenation involving cyclic thionocarbonates differs from that of other O-thiocarbonyl compounds because reaction involves ring opening. The direction of ring opening determines which of the two carbon atoms in the ring system will become the radical center. A high yield of a single product, therefore, depends on high regioselectivity in the ring-opening process. Cyclic thionocarbonates generally react to give deoxy compounds resulting from formation of the more stable intermediate radical. In some instances release of ring strain during ring opening becomes a factor in determining where the radical center will be located. Mixtures of products are a common result when the two radicals produced by ring opening are comparable in stability.
Radicals produced by reaction of O-thiocarbonyl compounds undergo addition reactions when a compound with a reactive multiple bond is present, and they undergo cyclization when the radical itself has a properly positioned multiple bond. In either situation the reactions that take place are of the addition-abstraction or addition-elimination type. The latter includes reactions with double bond migration and those without. Observed radical cyclizations are all addition- abstraction reactions. Most of these involve addition of a framework radical to either a framework multiple bond or to a substituent multiple bond. | textbooks/chem/Organic_Chemistry/Radical_Reactions_of_Carbohydrates_(Binkley)/Radical_Reactions_of_Carbohydrates_II%3A_Radical_Reactions_of_Carbohydrates/12%3A_Reactions_of_O-Thiocarbonyl_Compounds/III._Radical_Addition.txt |
The discussion in Chapter 12 focused on reactions of O-thiocarbonyl compounds prepared by derivatization of hydroxyl groups in partially protected carbohydrates. The current chapter is concerned with reactions of another type of thiocarbonyl compound, one prepared by carboxyl group esterification. The compounds of interest are esters of N‑hydroxypyridine-2-thione, sometimes referred to as O-acyl-N-hydroxy-2-thiopyridones or “Barton esters”.
Barton esters generate carbon-centered radicals in photochemically initiated reactions in which the esters themselves produce the chain-carrying radicals needed for the chain reaction to continue (Scheme 1).1–4 Radical formation from these esters is followed by loss of carbon dioxide and, in the absence of additional reactants (e.g., a compound with an electron-deficient double bond), formation of a product with a sulfur atom bonded to the carbon atom in the carbohydrate framework where the radical was centered (eq 1).5 The carbon–sulfur bond in the product from Barton ester reaction provides greater flexibility in further synthetic transformation than does the carbon–hydrogen bond that forms when tin or silicon hydrides are present. The synthetic potential of this type of reaction is greater than that indicated in eq 1 because the intermediate, carbon- centered radical also can add to a multiple bond (eq 2),6,7 abstract a hydrogen atom from a suitable donor (eq 3),8 or undergo other radical reactions. In effect, through formation and reaction of N-hydroxypyridine-2-thione esters it is possible to replace a carboxyl group in a molecule with a carbon-atom chain or with one of a variety of functional groups.
Since relatively few carbohydrates contain the carboxyl group needed for radical formation by Barton ester photolysis, the requirement that a carbohydrate first be converted into a carboxylic acid places a barrier to the general usefulness of this procedure. In spite of this limitation a number of addition and group replacement reactions exist that are based on radical formation from N‑hydroxypyridine-2-thione esters of carbohydrates that have been modified to contain a carboxyl group.
13: Carboxylic Acid Esters of N-Hydroxypyridine-2-thione
The mechanism proposed in Scheme 1 for reaction of an N-hydroxypyridine-2-thione ester is supported by a number of experimental observations. The carbon-centered radical R· is detectable by ESR spectroscopy,9 and flash photolysis experiments identify the 2-pyridylthiyl radical (PyS·) as one of the transients formed by ester photolysis.10,11 Also, the radical-chain nature of the reaction is attested to by quantum yields that range between 6 to 35, depending upon the reaction conditions.12
There are several characteristics of reactions of N-hydroxypyridine-2-thione esters that have “come to light” as a result of mechanistic studies. One of these is that addition of R· to the carbon–sulfur double bond is reversible (Scheme 1).13 Another is that the 2-pyridylthiyl radical, produced by photolysis in the first initiation step (Scheme 1), can add to a molecule of the starting ester in the second initiation step to provide another pathway for acyloxy radical formation.14
Several factors contribute to the driving force for the rate-determining step in the reaction shown in Scheme 1. One of these is conversion of a nonaromatic starting material into an aromatic product.15,16 Another is that a weak N–O bond (BDE = 43 kcal mole-1)17 in the substrate is being replaced with a stronger N–C bond (BDE \(\cong\) 76 kcal mole-1 for the second bond between carbon and nitrogen atoms)18 in the product.
III. Group Replacement R
Since the reaction pictured in Scheme 1 depends upon R· adding to a molecule of the starting ester, one way to change the course of this reaction is to introduce a compound that will react more rapidly with R· than does the ester. Thiols meet this requirement.8,19–25 Hydrogen-atom abstraction by a carbon-centered radical from a thiol is rapid enough (kH = 1.4 x 108 M-1s-1 at 25 oC for abstraction from C6H5SH by Bu·)26 to occur in preference to radical reaction with the starting ester. | textbooks/chem/Organic_Chemistry/Radical_Reactions_of_Carbohydrates_(Binkley)/Radical_Reactions_of_Carbohydrates_II%3A_Radical_Reactions_of_Carbohydrates/13%3A_Carboxylic_Acid_Esters_of_N-Hydroxypyridine-2-thione/II._Reaction_Mechanism.txt |
A. A Competition Always Present
The rate constant for addition of a typical carbon-centered radical to an ester of N-hydroxypyridine-2-thione (kr \(\cong\) 2 x 106 M-1s-1 at 50 oC)29,30 is large enough that competing addition of such a radical to a reactant with a carbon–carbon multiple bond occurs only for compounds with the most reactive bonds (i.e., those with electron-withdrawing substituents attached.) (The rate constants for addition to these unsaturated reactants are approximately 1 x 106 M-1s-1 at 20 oC31) In practice this means that unless the multiple bond has an electron-withdrawing substituent, the rate of addition of a carbon-centered radical will be too slow to compete effectively with addition to a Barton ester. As is emphasized in Scheme 4, a similarity in rate constants means that the competition between addition of a radical to the unsaturated compound 5, as opposed to addition to a second molecule of the Barton ester 4, depends heavily upon the relative concentrations of these two reactants (4 and 5).
Examples of the competition between radical addition to a molecule with an electron deficient double bond or to a molecule of unreacted starting ester are found in the reactions shown in equations 4-7. In the first of these (eq 4) acrylamide is present in two-fold excess; yet, products from addition of R· to the amide and to the starting ester are formed in essentially equal amounts.32 In the second reaction (eq 5) a good yield of the product from radical addition to methyl acrylate requires a five-fold excess of the unsaturated ester.33 In the reaction shown in eq 6 even a six-fold excess of phenyl vinyl sulfone does not suppress completely addition to the starting ester of some of the carbohydrate radicals.34 Radical addition to 2‑nitropropene is similar to addition to other unsaturated compounds (eq 7).35,36
B. Compounds With Electron-Deficient Multiple Bonds
Generalizing the information in equations 4-7 leads to the conclusion that carbohydrate radicals formed from esters of N-hydroxypyridine-2-thione add to compounds with electron- deficient multiple bonds; specifically, these reactions include radicals combining with α,β-unsaturated amides32,37,38 and esters33,35,39–41 as well as with compounds in which a double bond has either a phenylsulfonyl 34,39,42 or nitro35,36 substituent. Among reactive compounds addition to the carbon–carbon double bond in ethyl α‑(trifluoroacetoxy)acrylate (6)32,35,43–47 is reported more often than addition to any other compound (Scheme 5). One reason for this is that the resulting adduct (7) hydrolyzes under very mild conditions to give an α‑keto ester; thus, radical addition extends the carbon chain by two atoms and introduces easily manipulated functional groups (Scheme 5).47
C. Heteroaromatic Compounds
As the reaction in eq 8 shows, radicals generated from N-hydroxypyridine-2-thione esters also add to aromatic amines.34,42,48,49 Since this type of reaction occurs only when an acid such at trifluoroacetic acid is present in the reaction mixture, the assumption is that the radical is adding to the protonated amine. [In the absence of an acid only RSPy (eq 8) is formed.] Such an assumption is reasonable because a nucleophilic radical would be expected to react more rapidly with an electron-deficient π system.
D. Carbohydrates as Chiral Auxiliaries
One synthetic application of carbohydrate-containing esters of N‑hydroxypyridine-2-thione uses the carbohydrate moiety as a chiral auxiliary during radical addition.35,36,40,41,43 The reaction shown in eq 9 illustrates the diastereoselectivity possible in this type of process.35 Selectivity in this case results primarily from steric interactions between methyl acrylate and the C-6 substituent in the reactant sugar.35
V. Cyclization Reactions
A radical generated from a Barton ester undergoes cyclization if it contains a properly positioned multiple bond (eq 1050).50,51 Such reactions are not common because unless the needed carboxylic acid is readily available, the steps involved in its synthesis often make this process less attractive than others for generating the carbohydrate radical needed for ring formation.
VI. Generating Methyl Ra
Photolysis of acetylated N-hydroxypyridine-2-thione produces a methyl radical (Scheme 6).52 Methyl radicals generated in this way react with tellurium-containing nucleosides to generate the corresponding, carbon-centered radicals.52–56 In the reaction shown in Scheme 6, radical formation from photolysis of O-acetyl-N-hydroxypyridine-2-thione is the first step in producing a new ring system.52
VII. Summary
Radical reaction of a carboxylic acid ester of N-hydroxypyridine-2-thione causes the weak N–O bond to fragment to produce acyloxy and 2-pyridylthiyl radicals. The acyloxy radical then loses carbon dioxide to form a carbon-centered, carbohydrate radical. Carbon-centered radicals produced in this way abstract hydrogen atoms from highly reactive donors, such as thiols, and they combine with 2-pyridylthiyl radicals formed in the initial N–O bond homolysis. These carbohydrate radicals also can add to α,β-unsaturated esters and amides, to compounds with phenylsulfonyl- or nitro-substituted multiple bonds, and to protonated aromatic amines. Photolysis of acetylated N-hydroxypyridine-2-thione provides a source of methyl radicals that react readily with tellurium-containing carbohydrates to produce carbohydrate radicals. | textbooks/chem/Organic_Chemistry/Radical_Reactions_of_Carbohydrates_(Binkley)/Radical_Reactions_of_Carbohydrates_II%3A_Radical_Reactions_of_Carbohydrates/13%3A_Carboxylic_Acid_Esters_of_N-Hydroxypyridine-2-thione/IV._Addition_Reactions.txt |
Carbohydrates contain two types of nitro groups. The first type has the nitrogen atom in this group attached to a carbon atom in the carbohydrate framework (creating a deoxynitro or C-nitro carbohydrate) and the second has the nitrogen atom bonded to an oxygen atom in the carbohydrate structure (making an O-nitro carbohydrate or a carbohydrate nitrate). Radical reactions of nitro compounds are highly dependent upon the atom to which the nitro group is bonded.
14: Nitro Compounds
A. Addition-Elimination Reaction
The first step in the reaction of the tri-n-butyltin radical with a nitro compound is the addition of the radical to one of the oxygen atoms in the nitro group. The reaction that takes place after this initial addition depends upon whether the reactant is an O-nitro or a C-nitro compound. O‑Nitro carbohydrates fragment to give alkoxy radicals (Scheme 1). C-Nitro carbohydrates have more varied possibilities. The adduct radical either breaks the C–N bond to give a carbon-centered radical, cleaves an O–N bond to form a nitroso compound, or abstracts a hydrogen atom from an available donor, usually Bu3SnH (Scheme 2).
B. Electron Transfer
Early investigations raised the possibility that reaction of a C-nitro compound with the tri-n-butyltin radical could involve electron transfer (Scheme 3).1–3 Later investigation, however, did not support this possibility because the electron transfer between Bu3Sn· and (CH3)2CHNO2 (Scheme 3) was found to be endothermic by at least 12 kcal mol-1 . This endothermic electron transfer was inconsistent with the large rate constant (k = 9.5 x 107 M-1s-1) observed for reaction between this pair [Bu3Sn· and (CH3)2CHNO2].4 The conclusion from this latter study was that the addition-elimination mechanism (Scheme 2)5–8 provided a better explanation for the reaction between Bu3Sn· and a C‑nitro compound. The possibility that electron transfer could be involved in reaction of O-nitro compounds (Scheme 4) has not been addressed and, thus, remains open.
C. Photochemical Reaction
Photolysis of an O-nitrocarbohydrate fragments the N–O bond in the nitro group to produce nitrogen dioxide and the corresponding alkoxy radical (Scheme 1).
III. C-Nitro Carbohydrates
A. Group Replacement
Since in the reaction of a C-nitro compound the stability of the developing radical (R·) affects the ease of cleavage of the carbon–nitrogen bond (Scheme 2), group replacement occurs more easily for tertiary nitro compounds9–14 than for secondary15–19 and, especially, primary ones.20–22 Reaction of Bu3SnH with a compound containing a tertiary nitro group is given in eq 1.9 If the nitro group in the substrate is secondary, reaction usually follows the same pathway and replaces this group with a hydrogen atom,15–18 but sometimes breaking a C–O bond leading to a nitroso compound (which isomerizes to an oxime) offers significant competition (Schemes 2 and 5).19 When a nitro group is primary, the elimination phase of the reaction is more likely to produce only the nitroso compound (eq 2),20,21 even though replacement of a primary nitro group with a hydrogen atom has been observed (eq 3).22
B. Addition Reactions
A nitro group in a reactant molecule can be involved in radical addition in several ways. First, denitration can produce a carbon-centered radical that undergoes typical addition to an electron-deficient multiple bond (eq 4).3 In a different role, nitro groups activate multiple bonds toward addition by nucleophilic radicals and affect the regioselectivity of such reactions (eq 5).23 Deprotonation of a carbon atom bearing a nitro group creates an unsaturated system to which a carbon-centered radical can add to form a new, C–C bond (Scheme 624).24–26 Finally, the electron-withdrawing character of a nitro group can contribute to turning a normally nucleophilic radical into one that is electrophilic; thus, the philicity of the radical 1, which has both nitro and ethoxycarbonyl groups attached to the radical center, is reflected in its ability to add to the electron-rich double bond in the glycal 2 (Scheme 7).27
C. Cyclization Reactions
A cyclization reaction that begins with a C‑nitro carbohydrate is shown in eq 6.28 The high yield of this reaction, which involves a radical centered on a tertiary carbon atom adding to a multiple bond, illustrates that radical addition can be relatively insensitive to steric congestion at the radical center.28,29
D. Elimination Reactions
A deoxynitro sugar can undergo an elimination reaction, if a radical center develops on a carbon atom adjacent to that bearing the nitro group.30–32 In the reaction shown in Scheme 8,30 such a radical forms and then eliminates nitrogen dioxide to give an unsaturated compound.
IV. O-Nitro Carbohydrates
Reaction of an O-nitro carbohydrate with a tri-n-butyltin radical or with ultraviolet light produces the corresponding alkoxy radical, an intermediate that is reactive enough to abstract a hydrogen atom from most C–H bonds. Also characterizing the reactivity of alkoxy radicals is rapid carbon–carbon bond cleavage to produce both a compound with a carbonyl group and a carbon-centered radical. Examples of these reactions are discussed in the next several sections.
V. Reactions of Nitro Compounds with Silanes
Concerns with the toxicity of tri-n-butyltin hydride and the purification problems that accompany its use have spawned a variety of attempts to replace this reagent with a less troublesome one (see Appendix I). Tris(trimethylsilyl)silane normally is an attractive alternative to Bu3SnH, but it fails completely in this role in group replacement reactions in nitro compounds. The reason for failure is that the radical formed by addition of (Me3Si)3Si· to a nitro group does not break the carbon–nitrogen bond required for group replacement but rather cleaves a nitrogen–oxygen bond to begin a sequence of reactions leading to a complex reaction mixture (Scheme 13).48
A more effective procedure for reducing the amount of tri-n-butyltin hydride needed for nitro-group replacement with a hydrogen atom consists of regenerating Bu3SnH from the Bu3SnNO2 formed during the substitution process. A pair of reactions that achieve group replacement and regenerate Bu3SnH are given in equations 11 and 12. This alternative method requires only 10% of the tri-n-butyltin hydride needed in the standard procedure.49
VI. Summary
The type of radical produced by reaction of a nitro group in a carbohydrate depends upon whether this group is bonded to a carbon or an oxygen atom. A nitro group bonded to an oxygen atom invariably fragments to produce an alkoxy radical, but a nitro group attached to a carbon atom either fragments to generate a carbon-centered radical, expels an oxygen-centered radical to form a nitroso compound, or abstracts a hydrogen atom.
The alkoxy radicals produced from O-nitro carbohydrates rapidly abstract hydrogen atoms either internally or from molecules present in solution. Competing with or, in some instances, superseding hydrogen-atom abstraction is carbon–carbon bond cleavage to give a carbonyl group and a carbon-centered radical.
The carbon-centered radicals derived from reaction of C-nitro carbohydrates form most readily if the radical being produced is tertiary. These radicals undergo typical replacement, addition, and cyclization reactions. Primary radicals are much less likely to form from C-nitro carbohydrates; instead, these carbohydrate derivatives usually produce nitroso compounds. | textbooks/chem/Organic_Chemistry/Radical_Reactions_of_Carbohydrates_(Binkley)/Radical_Reactions_of_Carbohydrates_II%3A_Radical_Reactions_of_Carbohydrates/14%3A_Nitro_Compounds/II._Reaction_Mechanisms.txt |
Radicals are involved in both the synthesis and reactions of carbohydrate azides and, to a much lesser extent, azo compounds. The primary contribution of radicals to azide synthesis is in the formation of 2-azido-2-deoxy sugars by addition of azide radicals to glycals. The principal radical reaction of azides is their conversion to amines by reduction with an organotin hydride, often tri-n-butyltin hydride. The most important contribution of azo compounds to radical chemistry is in their role as reaction initiators.
15: Azides and Azo Compounds
A. Reactions
1. Reduction
Azides are reduced to tin-substituted amines by reaction with tin hydrides.1–9 These tin-containing compounds generally are not isolated; rather, each is converted into either the corresponding amine or an amine derivative. Amine derivatives include compounds formed by reaction of a tin-containing product with an acid halide,1,2 anhydride,3 or ester.4 Free amines are liberated from tin-containing products by a hydrolysis that often takes place during chromatographic purification.5–9 Examples of reactions producing amine and amine derivatives are found in equations 17 and 2,1 respectively. Conversion of an azide into the corresponding amine also can take place when tri-n-butyltin hydride is replaced by tris(trimethylsilyl)silane.10
Treatment of azides with tri-n-butyltin hydride does not always cause reaction of the azido group.11–13 Carbohydrates containing O‑thiocarbonyl substituents (eq 3)11,12 or iodine atoms (eq 4)13 undergo chemoselective reaction that leaves the azido group intact. If the amount of Bu3SnH is sufficient and the conditions are conducive, azido groups will react after the replacement of more reactive groups is complete.13
The tri-n-butyltin radical potentially can add to either Nα or Nγ in an azido group (Scheme 1), but since the intermediate radical 1 is thought to be a precursor to the tin-containing product 2, it is assumed that the initial addition of Bu3Sn· is to Nα.14–16 As mentioned above and indicated in Scheme 1, normal procedure calls for either hydrolysis or derivatization prior to product isolation.
2. Bromination
Free-radical bromination of a glycosyl azide produces the corresponding N-bromoglycosylimine.17,18 The reaction shown in Scheme 2 begins such a process when H-1 is abstracted regioselectively to give the radical 3, an intermediate that is stabilized by both the ring oxygen atom and the azido group. Loss of molecular nitrogen, followed by bromine-atom abstraction, completes the reaction.
3. Nitrile Formation
If an oxygen-centered radical and an azido group are attached to adjacent carbon atoms, the bond between the carbon atoms cleaves in a reaction leading to a nitrile.19 In the example shown in Scheme 3, the alkoxy radical 4 fragments to open a six-membered ring; a sequence of steps then leads to the nitrile 5. Similar ring opening occurs in compounds with five-membered rings.19
B. Synthesis
1. Carbohydrate Radical Addition to Ethanesulfonyl and Benzenesulfonyl Azides
It is possible to synthesize a carbohydrate azide by reaction of the corresponding xanthate with ethanesulfonyl azide20 or benzenesulfonyl azide20,21 (eq 521). The propagation steps for this type of reaction are outlined in Scheme 4.21 As shown in eq 6, the presence of an acetoxy group at C-2 reduces product yield when compared to reaction in which such a group is absent (eq 5). The intermediate pyranos-1-yl radical 6, which adopts a B2,5 boat conformation, is more hindered at C-1 than the pyranos-1-yl radical 7, which lacks a C-2 substituent and has a chair conformation. (Section IV of Chapter 6 in Volume I contains more information about the conformation of pyranos-1-yl radicals.) In addition to being more hindered at C-1, the radical 6 is less nucleophilic due to the electron-withdrawing nature of the C-2 acetoxy group.
2. Azide Radical Addition to an Unsaturated Carbohydrate
a. Azide Radicals
Azide ion is converted into azide radical by reaction with oxidizing agents that include ammonium cerium(IV) nitrate (eq 7)22 and (diacetoxyiodo)benzene (eq 8).23 The electrophilic nature of the azide radical is attested to by its addition to electron-rich double bonds such as those found in glycals.24–47 Atomic orbital coefficients can be critical in determining regioselectivity in a reaction with an early transition state because the rate constant for the bond-forming process between two atoms depends in its early stages on the magnitude of the coefficients of the interacting frontier orbitals.28,29 In the azide radical addition pictured in Scheme 5, the frontier orbitals are the SOMO of the azide radical and the HOMO of the D-galactal 8. The more reactive position for addition to the double bond in 8 is at C-2 because, based on simple model systems,30 the atomic orbital coefficient for the HOMO is larger at C-2 than at C‑1 (Figure 1).
b. Azidonitration
Azidonitration takes place when the azide radical is produced by oxidation of the azide ion by ammonium cerium(IV) nitrate in the presence of a glycal.24–27,31–39 The stereoselectivity of azide radical addition is dependent on glycal structure; thus, N3· adds in a highly selective fashion to the α face of the D-galactal 8 because the β face is well protected by ring substituents (eq 9).24 Less effective protection of the β face, which occurs when 8 is replaced by the D‑glucal 9, leads to stereoselectivity that varies as the reaction conditions change (eq 10).37 When a 4,6-O-isopropylidene group is incorporated into the D-glucal structure, it creates a compound that is conformationally less mobile. With this reduction in conformational mobility comes greater stereoselectivity in addition of the azide radical to the D-glucal derivative (eq 11).37
c. Azidophenylselenylation
When azide ion is oxidized by (diacetoxyiodo)benzene, the resulting azide radical will add to a glycal to produce a phenyl selenide if diphenyldiselenide is present in the reaction mixture (eq 1240).39–48 The normal solvent for this reaction is dichloromethane, but due to limited solubility of sodium azide in this solvent, reaction usually is heterogeneous, a situation that causes product yields to suffer except in quite dilute solutions. If, however, the azide radical is generated from trimethylsilyl azide, solutions are homogeneous and good product yields are realized (eq 13).49
d. Azidohalogenation
The azide radical also forms by photolysis of chloroazide (eq 14). When this reaction takes place in the presence of a glycal, azidochlorination results (eq 15).50 The stereoselectivity of glycal addition by the azide radical formed from photolysis is similar to that observed when this radical is generated by azide ion oxidation. The photochemical reaction is different, however, in that a small amount (13%) of addition occurs in which the positions of the azido and chloro substituents are interchanged. Since photolysis of chloroazide also produces the electrophilic chlorine atom (eq 14), it is reasonable to expect that sometimes a chlorine atom would add to a glycal at C-2.
3. Atom Replacement by an Azido Group
Reaction between iodoazide and a carbohydrate containing a benzyloxy group leads to hydrogen atom replacement by an azido group (eq 16).51 In this reaction the azide radical abstracts one of the reactive hydrogen atoms from the benzyloxy group to form a benzylic radical. The azido group then can be introduced by one of the proposed pathways shown in Scheme 6.
III. Azo Compounds
Azo compounds, in particular 2,2'-azobis(isobutyronitrile), are ubiquitous initiators in radical reactions, but they rarely participate in these reactions in other ways. One reaction that includes an azo compound in a role other than that as an initiator is shown in eq 17, where the imine 11 is produced by reaction of a carbohydrate radical with an azo compound.52,53 The dimer 12 is a suggested intermediate in the proposed mechanism for this reaction, which is pictured in Scheme 7.52,53 After radical reaction is complete, the imine 11 can be hydrolyzed to produce an aminodeoxy sugar.
It is also possible to synthesize an azo compound in a radical reaction. Such a compound is formed when a carbohydrate radical, generated from a deoxyiodo sugar, adds to a diazonium salt (eq 18).54 The propagation steps for a proposed mechanism for this reaction are shown in Scheme 8.54
IV. Summary
The principal radical reaction of carbohydrate azides is reduction to tin-substituted amines by reaction with tri-n-butyltin hydride. Products from this reaction are isolated either as free amines or amine derivatives. Three radical reactions for synthesis of carbohydrate azides are known. One of these involves the addition of a carbohydrate radical to ethanesulfonyl azide. A second is the azidonitration that takes place when an azide radical adds to a glycal in the presence of ammonium cerium(IV) nitrate. Finally, reaction of a glycal with an azide radical and diphenyldiselenide, results in azidophenylselenylation. | textbooks/chem/Organic_Chemistry/Radical_Reactions_of_Carbohydrates_(Binkley)/Radical_Reactions_of_Carbohydrates_II%3A_Radical_Reactions_of_Carbohydrates/15%3A_Azides_and_Azo_Compounds/II._Azides.txt |
Radical reactions of carbohydrate nitriles and isonitriles are not widespread. Isonitrile reactions are rare because these compounds themselves are not common. Their primary reaction is replacement of the isocyano group with a hydrogen atom. Nitriles are more plentiful than isonitriles but less reactive in radical reactions. They typically are involved in radical cyclization. Both nitriles and isonitriles can be synthesized by radical reaction.
16: Nitriles and Isonitriles
A. Reactions
1. Group Replacement
a. Isocyano Groups
Reaction of tri-n-butyltin hydride with carbohydrates containing isocyano groups replaces each of these groups with a hydrogen atom. Such replacement is known to occur when isocyano groups are attached to anomeric,1,2 secondary,3–9 and primary7–9 carbon atoms. An example of replacement at an anomeric carbon atom is shown in eq 1,5 while both primary and secondary groups are replaced in the reaction described in Scheme 1.7,9
Isocyano group replacement is remarkably temperature sensitive. Reaction of the secondary groups in 1 takes place at 70 oC, but the primary isocyano group is unreactive (Scheme 1).7,9 When the temperature of the reaction mixture is raised to 80 oC, both groups are replaced. This temperature dependence provides a basis for regioselective reaction.
A mechanism for isocyano group replacement with a hydrogen atom is pictured in Scheme 2.10 In the first step of this process the tri-n-butyltin radical adds to the carbon atom of the isocyano group to produce an imidoyl radical (2). Fragmentation of this radical (2) then generates the carbon-centered radical R·, which abstracts a hydrogen atom from Bu3SnH to complete the reaction sequence. If R represents a phenyl or substituted-phenyl group, fragmentation to give an aryl radical does not occur; rather, an addition reaction takes place.11 When tris(trimethylsilyl)silane replaces tri-n-butyltin hydride in reduction of isonitriles, compounds containing primary, secondary, or tertiary isocyano groups all are reactive.12
b. Sulfhydryl Groups
An isonitrile can participate in replacement of a sulfhydryl group by a hydrogen atom.13 Such a reaction is pictured in Scheme 3, where replacement begins when the sulfur-centered radical 4 forms from the thiol 3 by hydrogen-atom abstraction. Addition of 4 to t-butyl isocyanide gives the adduct radical 5, which then fragments to produce the pyranos-1‑yl radical 6. Hydrogen-atom abstraction by 6 from another molecule of the starting thiol (3) completes the cycle and begins a new reaction sequence. Sulfhydryl group replacement represents another pathway for generating carbon-centered, carbohydrate radicals.
2. Elimination Reactions
Reaction of tri-n-butyltin hydride with a carbohydrate that has adjacent isocyano and O-thiocarbonyl groups generates a product with a C–C double bond (Scheme 4).4 In this reaction radicals 7 and 8 are both possible intermediates. Study of the diisonitrile 9 provides information helpful in choosing between 7 and 8. Reaction of 9 with Bu3Sn· produces a carbon-centered radical (10) with an isocyano group attached to the carbon atom adjacent to the radical center (Scheme 5).8 The intermediate 10 does not expel a cyano radical to form a multiple bond but rather abstracts a hydrogen atom from tri-n-butyltin hydride. Extrapolating the behavior of 10 to the reaction shown in Scheme 4 leads to the conclusion that the radical 8 is an unlikely intermediate in this process.
3. Addition Reactions
As a part of the replacement process shown in Scheme 2, an isonitrile reacts with Bu3Sn· to produce an intermediate, carbon-centered radical R·. Normal completion of this reaction involves hydrogen-atom abstraction by R· from Bu3SnH; however, if R· is formed without a hydrogen-atom transfer present, it will add to a molecule of isonitrile (Scheme 6). A specific example of this type of reaction is found in eq 2, which describes the α addition of a pyranos-1-yl radical, formed from a carbohydrate telluride, to an aromatic isonitrile.14
B. Synthesis
It is possible to produce isonitriles from isothiocyanates by radical reaction (eq 3).2 A proposed mechanism for such a structural change is shown in Scheme 7. Isonitrile formation results when reaction is conducted at room temperature (eq 3), but if the reaction temperature is raised to 110 oC, the isonitrile is not isolated because it undergoes isocyano group replacement by a hydrogen atom (eq 4).2,15
III. Nitriles
A. Synthesis
When t-butyl isonitrile reacts with a carbon-centered radical, an addition-elimination process takes place that results in the formation of a nitrile (Scheme 8).16–23 This type of nitrile synthesis is illustrated by the reaction shown in eq 5.20 (Scheme 10 in Chapter 2 describes another example of this type of reaction.22)
B. Reactions
1. Addition to a Cyano Group
Radical reactions of carbohydrate nitriles usually involve internal addition in which a carbon-centered radical generated in close proximity to a cyano group forms a new ring system.24–31 An example is shown in Scheme 9, where the radical centered on C-6 in 11 adds to the cyano group as a part of this sequential process.24 Carbohydrate radical formation in this type of reaction typically begins with the tri-n-butyltin radical abstracting a halogen atom or an O-thiocarbonyl group. After cyclization, the radical abstracts a hydrogen atom from Bu3SnH to produce an imine, which in some cases is isolated and in others is hydrolyzed to the corresponding carbonyl compound before isolation.
2. Cyano Group Migration
When a cyclic imino radical forms during carbon-centered radical addition to a cyano group, the possibility exists that ring opening will lead to cyano group migration (Scheme 10).32–34 Such a reaction is shown in Scheme 11, where migration accompanies ring opening of a benzylidene acetal.32
IV. Summary
When compared to reactions of halogenated carbohydrates or O-thiocarbonyl compounds, radical reactions of nitriles and isonitriles are few in number. The primary reaction for isonitriles is isocyano group replacement by a hydrogen atom. For nitriles the only reaction of significance is internal addition of a carbon-centered radical to a cyano group to form a cyclic imino radical.
Radical reactions can be involved in nitrile and isonitrile synthesis. Isonitriles are formed from reaction of isothiocyanates with tri-n-butyltin hydride. Nitriles are produced by addition of a carbon-centered radical to t‑butyl isonitrile and subsequent elimination of the t-butyl radical. | textbooks/chem/Organic_Chemistry/Radical_Reactions_of_Carbohydrates_(Binkley)/Radical_Reactions_of_Carbohydrates_II%3A_Radical_Reactions_of_Carbohydrates/16%3A_Nitriles_and_Isonitriles/II._Isonitriles.txt |
Carbon-centered radicals add to compounds containing carbon–nitrogen double bonds. Most reactions of this type are internal additions involving oxime ethers, but similar transformations of hydrazones, ketonitrones, and protonated heteroaromatics also take place.1,2 When compared to addition to carbon–carbon double bonds, reactions of their carbon–nitrogen counterparts are similar in that they are rarely reversible1 but different in that the rate constants for addition are larger. Also, the final products from addition to carbon–nitrogen double bonds contain functional groups that more easily undergo further modification.1
17: Oxime Ethers and Related Compounds
A. Addition Reactions
A carbon-centered radical adds preferentially to the carbon atom in the carbon–nitrogen double bond of an oxime ether (Scheme 1). The regioselectivity of this reaction is consistent with a radical adding to the oxime ether to produce the more stable of the two possible adduct radicals; that is, the radical stabilized by an oxygen atom attached to the radical center. Examples of this type of reaction are found in the addition of a pyranos-1-yl radical to O‑benzylformaldoxime (eq 1),3 and reaction of a nucleoside radical centered at C-3' to a nucleoside-containing oxime ether (eq 2).4,5
The carbon-centered radicals that add to oxime ethers typically are formed from the corresponding iodides and bromides by reaction with Bu3Sn·. Because simple reduction offers significant competition to radical addition, Bu3SnH is not the best source for the tin-centered radicals participating in these reactions. Addition is more successful when Bu3Sn· is generated from bis(tri-n-butyltin)benzpinacolate (2), a compound that produces Bu3Sn· but suppresses simple reduction by not introducing an effective hydrogen-atom transfer into the reaction mixture (Scheme 2).3 In the absence of such a donor the radical phase of this reaction is thought to produce the tin-containing compound 4, which then is hydrolyzed to the benzyloxyamine 5 in a nonradical reaction (Scheme 2).
B. Cyclization Reactions
1. Regioselectivity
An example of regioselective radical cyclization involving an oxime ether is shown in Scheme 3.6 One factor important in determining regioselectivity in this type of reaction is the energetically favored addition of a radical to the carbon atom of the carbon–nitrogen double bond (Scheme 1). Also of importance is the preference during radical cyclization of oxime ethers for forming five-membered rather than six-membered rings when both are possible and six-membered rather than seven-membered rings when either one could be formed.7
2. Stereoselectivity
Radical cyclization converts open-chain oxime ethers into highly substituted, alkoxyamino cyclopentanes6,8–14 and cyclohexanes.7,15–17 As is often the case in radical cyclization reactions, stereoselectivity can be understood, if one assumes that the lowest energy transition state is a chair-like structure that maximizes the number of pseudoequatorial substituents. The reaction shown in Scheme 3, for example, generates compounds with two new chiral centers but produces only two (9 and 10) of the four possible stereoisomers.6 These two isomers arise from the chair-like transition states 7 and 8, respectively. The transition state 8, leading to the major product 10 has the greater number of pseudoequatorial substituents.
In light of the reaction of the oxime ether 6 (Scheme 3) one might expect the structurally similar oxime ether 11 (eq 3) also to produce only the two stereoisomers 14 and 15, that is, the products expected from reaction via chair-like transition states. Actually, 11 produces all four possible stereoisomers (eq 3), and neither 14 nor 15 is the major product.11 The major product (12) arises from reaction involving the boat-like transition state 16 (Scheme 4). Similar amounts of energy often are required to reach chair-like and boat-like transition states in radical cyclization reactions (Section IV.A in Chapter 11 of Volume I); consequently, differences in structure near the reactive centers can affect product distribution significantly.
One difference between the substrates 6 and 11 is that 6 has a benzyloxy group at C-2 (Scheme 3), and 11 has an acetamido group at this position (Scheme 4). The possibility exists that hydrogen bonding involving the acetamido group stabilizes the transition state 16 to the point that the reaction pathway containing this boat-like structure (16) becomes the lowest energy one (Scheme 4).
An O-isopropylidene group that is close to the reactive centers in a cyclization reaction has a pronounced effect of stereoselectivity.17–24 This effect is illustrated by the reaction shown in Scheme 5, where the sole product has the NHOBn substituent on the side of the newly formed ring that is opposite to the O-isopropylidene group.18 In this reaction the chair-like conformation (17) of the intermediate radical is greatly favored over its boat-like counterpart (18) due, in large measure, to destabilizing steric interactions that involve the O-isopropylidene group.
3. Radical Forming Reactions
Carbohydrate radicals that add internally to carbon–nitrogen double bonds come from a variety of sources. Such radicals often form by reaction of O-thiocarbonyl compounds6,8–11,20,21,23 or bromides15–19 with tri-n-butyltin hydride. Reactions of Bu3SnH with carbonyl compounds12 and dithioacetals13 represent two less common ways for generating radicals that then undergo ring formation. The necessary radicals also can be formed by electron-transfer reaction.
4. Preserving the Double Bond
Normally, the double bond in an oxime ether is converted into a single bond during ring formation, but when an O-trityl oxime undergoes cyclization, the carbon–nitrogen double bond is preserved in the product. A proposed mechanism for this type of reaction is shown in Scheme 6, and the preferred reagents, all of which are critical to the success of the reaction, are given in eq 4.24 Slow addition of Bu3SnH and the initiator ABC also are necessary for synthetically useful reaction. The persistent radical (C6H5)3C· hinders chain propagation by its slow reaction with Bu3SnH; consequently, to avoid buildup of (C6H5)3C·, diphenyl diselenide, a convenient precursor for the very reactive hydrogen-atom transfer C6H5SeH, can be added to the reaction mixture. This addition replaces a slow reaction (eq 5) with two rapid ones (eq 6 and eq 7).24 Although the carbohydrates that have been studied thus far give good product yields without addition of diphenyl diselenide, other compounds do not; consequently, the selenide inclusion in the reaction mixture is now part of the recommended procedure.
5. Enhanced Reactivity of Vinylic Radicals
Attempted cyclization of the radical generated from the bromide 21 produces only the simple-reduction product 22 (eq 8), but reaction of the related vinylic bromide 23 gives both a cyclic product and a simple-reduction product (eq 9).25 One factor that may contribute to the difference in ability of compounds 21 and 23 to undergo radical cyclization is the considerable reactivity of vinylic radicals. This reactivity may be sufficient to cause new ring formation during reaction of 23.
In addition to halogen-atom abstraction, vinylic radicals also form by reaction of a tin-centered radical with a compound containing a carbon–carbon triple bond.26–30 When a radical formed in this way adds internally to a carbon–nitrogen double bond, it produces a cyclic, tin-containing product (eq 10).26 The triple bond in such reactions usually is a terminal one, but internal bonds, particularly those in conjugation with radical-stabilizing groups, also undergo reaction.30
C. Electron-Transfer Reaction
Electron transfer provides another means for generating a ring-forming radical from an oxime ether. The electron donor in such reactions typically is samarium(II) iodide and the electron acceptor often is an oxime ether containing a halogen atom. A typical reaction is shown in Scheme 7,31 where dissociative electron transfer involving the carbon–iodine bond in 24 leads to formation of the carbon-centered radical 25. Cyclization and hydrogen-atom abstraction complete this reaction by forming the cyclopentylamine derivative 27 in a process that passes through the boat-like transition state 26. Although chair-like transition states typically are favored in cyclization reactions, in this case chair-like structures have destabilizing interaction between C-1 and C-2 substituents; furthermore, the other boat-like transition state (28) has reduced stability due to the interaction between the C-1 and C-6 substituents (Scheme 7). If excess SmI2 is present during reaction, dissociative electron transfer breaks the N–O bond in 27 to produce, after hydrogen-atom abstraction, the corresponding cyclopentylamine 29 (Scheme 8). Nonradical migration of the acyl group from oxygen to nitrogen then forms the final product (30).31 (Samarium(II) iodide also cleaves N–O bonds that are not formed from reaction of oxime ethers.32)
Electron transfer to an aldehydo or keto group in a oxime ether also can be the initial step in ring formation.33–38 When SmI2 is the electron-donor, the first reactive intermediate is a samarium ketyl formed by one-electron reduction of the carbonyl group.1 In the reaction shown in Scheme 9, the carbon atom on which the radical in the samarium ketyl 31 is centered adds internally to the carbon–nitrogen double bond leading to the cyclic product 32. When excess SmI2 is present, electron transfer to 32 causes N–O bond cleavage and ultimate replacement of the O-benzyl group attached to nitrogen with a hydrogen atom (Scheme 9).36 Similar cyclization has been observed when the tri-n-butyltin radical reacts with a oxime ether containing an aldehydo or keto group.39
D. Oxime Esters as Radical Sources
The signature event of the cyclization reactions discussed thus far is a carbon-centered radical adding to a carbon–nitrogen double bond in an oxime ether. The majority of ring-forming reactions involving oxime derivatives take place in this way, but a change occurs in the reaction shown in Scheme 10, where the adding radical (34) is generated from an O-benzoyloxime.40 In general, adding Bu3Sn· to an O-benzoyl group is an inauspicious beginning for a radical reaction because such addition reverses rapidly, nearly always before any other reaction can occur. Further reaction is possible in this case (Scheme 10) due primarily to rapid fragmentation of the weak N‑O bond in the initially formed radical 33.40
E. Oxime Synthesis
Oximes can be the products of radical reaction.41,42 When the cobalt-containing, D‑glucopyranosyl derivative 35 is photolyzed, the radical produced adds to a molecule of nitric oxide to give a nitroso compound. This compound then tautomerizes to form the corresponding oxime 36 (eq 11).41 | textbooks/chem/Organic_Chemistry/Radical_Reactions_of_Carbohydrates_(Binkley)/Radical_Reactions_of_Carbohydrates_II%3A_Radical_Reactions_of_Carbohydrates/17%3A_Oxime_Ethers_and_Related_Compounds/II._Oximes.txt |
Radical reactions of carbohydrate hydrazones2,43–46 are less common than those of oxime ethers; reactions of imines43 are still more rare. The reported reactions of hydrazones, such as that shown in eq 12,43 all involve radical cyclization. The substrates in most of these reactions are esters derived from (2,2-diphenylhydrazono)acetic acid (eq 13).44,45
The reaction shown in eq 14 pictures a highly stereoselective cyclization involving a carbohydrate hydrazone.46 Stereoselectivity in this reaction is determined by the preferred conformation (37, Figure 1) of the intermediate produced by a phenylthiyl radical adding to the carbon–carbon double bond in the substrate. Conformation 37 has the carbon–nitrogen bond anti to the adjacent carbon–oxygen bond. The conformation 38, expected to be more stable because it has more pseudoequatorial substituents, is destabilized by dipole-dipole interactions arising from a gauche relation between the neighboring C=N and C–O bonds.46
IV. Ketonitrones
The ketonitrone 39 undergoes radical cyclization when treated with excess SmI2 (Scheme 11).47 This reaction is similar to that shown in Scheme 9 for a related ketooxime. Both reactions depend upon electron transfer from SmI2, and each produces a new five-membered ring. These reactions are stereoselective but their selectivity is controlled in different ways. In the reaction of the ketonitrone, samarium remains coordinated with the oxygen atoms in the two developing substituent groups during reaction. This coordination insures the stereoselective formation of a cyclopentane ring in which the OH and N(OH)Bn groups are cis-related (Scheme 11). Since similar coordination does not occur during ketooxime reaction (Scheme 9), the emerging OH and NHOBn groups are not held on the same side of the ring; in fact, due to the steric size of the groups attached to the bonding carbon atoms in the radical 31, OH and NHOBn groups become trans-related in the product 32.
V. Protonated Heteroaromatics
Pyranos-1-yl and furanos-1-yl radicals add to protonated heteroaromatics to produce C-nucleoside derivatives.48–54 A proposed mechanism for this type of reaction is given in Scheme 12.50 Protonation dramatically increases the rate of addition of a carbon-centered radical to a heteroaromatic compound; in fact, the rate is so fast that it is not necessary to conduct the reaction in an inert atmosphere. Not only does reaction take place in the presence of molecular oxygen but oxygen is a likely participant in the rearomatization stage of this process (Scheme 12). In reactions of this type the initially formed radical (R·) usually is generated by photolysis of an O‑acyl-N-hydroxy-2-thiopyridone (eq 15).48 (Reactions of O‑acyl-N-hydroxy-2-thiopyridones are discussed in Chapter 13.)
VI. Protected Amines
The benzyloxycarbonyl protected amine 40 reacts with (diacetoxyiodo)benzene–iodine to give a product (41) that contains a new ring system (eq 16).55 The proposed mechanism for this reaction (Scheme 13) involves internal hydrogen-atom abstraction by a nitrogen-centered radical. Similar internal hydrogen-atom abstraction takes place when a nitrogen-centered radical is generated from a sulfonamidate.56 Such reactions are reminiscent of the hydrogen-atom abstraction by alkoxy radicals described in Chapter 6.
VII. Summary
Carbon-centered radicals add to carbon–nitrogen double bonds in oxime ethers, hydrazones, ketonitrones, and protonated heteroaromatics. These reactions are regiospecific with addition occurring to the carbon atom in the double bond. The majority of such reactions involve oxime ethers and usually result in formation of a new ring. These cyclization reactions often are quite stereoselective. This is particularly true if an O-isopropylidene group is near the reactive center, in which case the nitrogen-containing substituent and the O-isopropylidene group in the product are on opposite sides of the new ring. Radical addition to protonated heteroaromatics is different from addition to other compounds containing carbon–nitrogen multiple bonds because addition is always followed by hydrogen-atom abstraction that rearomatizes the ring. | textbooks/chem/Organic_Chemistry/Radical_Reactions_of_Carbohydrates_(Binkley)/Radical_Reactions_of_Carbohydrates_II%3A_Radical_Reactions_of_Carbohydrates/17%3A_Oxime_Ethers_and_Related_Compounds/III._Hydrazones_and_Imines.txt |
Earlier chapters in this book describe radical formation from reaction of various carbohydrate derivatives. Among the reactions of these radicals is addition to unsaturated compounds. The current chapter builds on earlier ones in that the discussion of radical addition reactions continues, but the focus in the present chapter is less on how radicals are formed and more on the process of their addition to compounds with carbon–carbon multiple bonds.
18: Compounds with Carbon–Carbon Multiple Bonds I: Addition Reactions
A. Reaction Mechanisms
Radical addition can take place by either chain or nonchain reaction. For each of these there are two variations on the basic reaction mechanism. For chain reactions (Schemes 1 and 2) each variation has a different type of chain-transfer step. For nonchain reactions both mechanisms involve electron transfer. They differ in that the transfer is either from (Scheme 3) or to (Scheme 4) an organometallic complex.
1. Chain Reactions
Both mechanisms for radical addition by chain reaction have a propagation phase that begins with group or atom abstraction (Schemes 1 and 2). In each of these mechanisms Bu3Sn· is shown as the abstracting radical, although other radicals [e.g., (Me3Si)3Si·] are capable of filling this role. The defining difference between these two mechanisms is the chain-transfer step. In the reaction shown in Scheme 1 it is bimolecular, and in that in Scheme 2 it is unimolecular.
a. Bimolecular Chain Transfer
The distinguishing feature of a chain reaction that takes place by bimolecular chain-transfer is an elementary reaction between a radical and a nonradical that ends one propagation sequence and creates the radical that begins a new sequence. In the reaction shown in Scheme 1, chain transfer occurs when the adduct radical abstracts a hydrogen atom from Bu3SnH to form the addition product RCH2CH2E and generate the chain-carrying radical Bu3Sn·.
b. Unimolecular Chain Transfer
Unimolecular chain transfer describes a reaction in which the chain-transfer step in a propagation sequence is an elementary reaction with a single reactant. The propagation steps for a typical, unimolecular, chain-transfer reaction are shown in Scheme 2, where the transfer step is a β-fragmentation that produces an unsaturated compound and a chain-carrying radical.
2. Nonchain Reactions
Schemes 3 and 4 each describe a mechanism for a reaction that has no repeating cycle. In the first of these (Scheme 3) the carbohydrate radical R· forms by electron transfer from a transition-metal complex (Cp2TiCl) to a carbohydrate derivative. (Cp2TiCl is one of several transition-metal complexes known to function as an electron donor in this type of reaction.) Radical reaction ends when a second molecule of Cp2TiCl combines with an adduct radical to produce an unstable, carbometallic product. Products of this type undergo rapid, nonradical reaction (e.g., elimination of the elements of Cp2TiClH to form a double bond).
In the second type of nonchain reaction (Scheme 4) radical formation occurs when the transition-metal complex donates an electron to a CH-acidic compound. The radical phase of the reaction ends with a second electron transfer, one that produces a carbocation. This cation undergoes rapid, nonradical reaction, such as capture by a molecule of solvent.
B. Selectivity in Addition Reactions
1. Chemoselectivity
Chemoselectivity is of consequence at two stages in a radical addition reaction. The first is in the radical forming step, where selectivity is determined by the reactivity of the functional groups present in a molecule of substrate. Forming the desired radical is accomplished at this stage by insuring that the radical precursor has the most reactive substituent attached to the carbon atom where the radical center is to be located. The next place at which selectivity potentially is of significance is during radical addition to the multiple bond. Chemoselectivity is meaningful at this time if there are two or more multiple bonds to which addition can occur.
2. Regioselectivity
Most radical addition reactions involving carbohydrates are regiospecific. Addition occurs exclusively at the less substituted carbon atom in a multiple bond. The way in which the pyranos-1-yl radical 2 adds to acrylonitrile is typical (Scheme 5).1,2 The reactions in Tables 1 and 2 document a similar regiospecificity in the addition of 2 to other unsaturated compounds. The data in Tables 3-5 show that reactions of other pyranos-1-yl radicals exhibit similar reactivity. (All of these tables are located at the end of this chapter.)
Both steric and polar effects have a role in determining regiospecificity in addition reactions. Steric effects become progressively more important as the effective size of the substituents near a multiple bond in an unsaturated compound increases and as the effective size of the adding radical increases. Polar effects exert themselves whenever a nucleophilic radical adds to an electron-deficient multiple bond, or an electrophilic radical adds to an electron-rich multiple bond. The extent of the influence of each effect depends upon the point at which the transition state is reached as a reaction progresses. For example, because addition of a nucleophilic radical to an electron-deficient multiple bond (the most common type of addition reaction) is exothermic, such a reaction should have an early (reactant-like) transition state11 with minimal, new bond formation. When there is little new bond formation at the transition state, there is a diminished opportunity for steric interactions to affect regioselectivity. The situation with polar effects is different because, as described in the next paragraph, they can exert considerable regioselective control in a reaction that has an early transition state.
For a reaction with an early transition state the molecular orbitals in the reactants do not change greatly by the time the transition state is reached. In such a situation frontier-orbital interactions are helpful not only in understanding why a reaction takes place but also in explaining reaction regioselectivity. Briefly, reaction occurs readily because the SOMO of the carbohydrate radical and the LUMO of the unsaturated reactant are energetically close enough for significant stabilizing interaction between the two at the transition state (Figure 1).12,13 The extent of early bonding between a radical and the carbon atoms in a reacting multiple bond is a function of the magnitude of the atomic orbital coefficient at each carbon atom in the LUMO of the unsaturated reactant. Attaching an electron-withdrawing or electron-donating substituent to a multiple bond causes these coefficients to be quite unequal, a situation that leads to regiospecific reaction. For a polarized multiple bond such as that found in an α,β-unsaturated nitrile or carbonyl compound, regiospecific addition to the β‑carbon atom reflects the greater magnitude of the atomic orbital coefficient at this atom in the LUMO when compared to the magnitude of the coefficient at the α carbon atom (Figure 1).
3. Stereoselectivity
As described in the next several sections, stereoselectivity in radical addition reactions is determined by a combination of steric and stereoelectronic effects. The stereoelectronic effect of primary importance is the kinetic anomeric effect. Steric effects that have an impact on reaction stereoselectivity have various names, but they all depend on steric interactions favoring a particular approach of an unsaturated compound or a hydrogen-atom transfer to a radical center.
a. The Kinetic Anomeric Effect
The reaction between the pyranos-1-yl radical 2 and an unsaturated compound with an electron-deficient double bond is highly stereoselective (Scheme 5). Preferential reaction on the α face of the pyranoid ring in 2 is due, in large part, to the kinetic anomeric effect (discussed in Chapter 11 of Volume I). Stereoselectivity in the addition reactions of 2 also is affected by reaction conditions. The information in Table 1 includes several sets of conditions suitable for highly stereoselective reaction with little, competing simple reduction.
The selectivity observed in reactions of 2 extends to other D‑hexopyranos-1-yl radicals. Reactions of the D-galacto- and D-mannopyranosyl bromides 8 (eq 1) and 11 (eq 2), respectively, are at least as stereoselective as those of the corresponding D-glucopyranosyl bromide 1. The data in Tables 3 and 4 confirm that reactions of 8 and 11 occur preferentially on the α face of the pyranoid ring.
b. Steric Effects
(1). Group Shielding
Group shielding causes preferential addition to the less-hindered face of a ring system. In the reaction shown in eq 3 the 2,3-O-isopropylidene group shields the α face of the radical centered at C-4 in the pyranoid ring from approach by the unsaturated reactant.14
(2). Size of the Unsaturated Reactant
The reaction shown in Scheme 6 illustrates the combined effect of steric size of the unsaturated reactant and group shielding of a radical by ring substituents.15 In this reaction the amount of addition to the better shielded, β face of the pyranoid ring in the radical 13 decreases as the effective size of the unsaturated reactant increases. The data in Scheme 6 show that, even in a reaction with an early transition state, steric effects can have a significant role in determining reaction stereoselectivity if these effects are large enough.
(3). Size of the Hydrogen-atom transfer
Sometimes a reaction forms the less stable of two possible stereoisomers due to restricted approach of a hydrogen-atom transfer to a radical center in the product-forming step. In the reaction shown in Scheme 7, for example, the less hindered approach of sodium hypophosphite to the β face of the furanoid ring produces the less stable stereoisomer in greater the 95% yield.16
V
Radical addition to compounds with carbon–carbon multiple bonds can take place by several reaction mechanisms. The most common of these is a chain reaction that is characterized by a bimolecular, chain-transfer step. This type of reaction governs addition to unsaturated compounds of radicals centered on various carbon atoms in pyranoid and furanoid rings. These reactions are regiospecific and often highly stereoselective. Since most carbon-centered radicals, including those derived from carbohydrates, are nucleophilic, successful addition requires an electron-withdrawing substituent attached to one of the carbon atoms of the multiple bond. The primary limitation placed on this type of addition reaction is that it is in competition with a hydrogen-atom abstraction reaction that causes simple reduction of the carbohydrate derivative.
Addition reactions also can occur by unimolecular chain-transfer. This type of reaction also takes place with radicals centered on various carbon atoms in pyranoid and furanoid rings. Most of these reactions involve addition of a carbohydrate radical to an allylic or vinylic stannane followed by loss of a tin-containing substituent. An advantage to this type of reaction is that competition with simple reduction can be largely avoided. A disadvantage is that 1,3-tin migration can occur in allylic stannanes; such migration places a synthetic limitation on the usefulness of the reaction.
Although the unsaturated compound in a typical addition reaction is a noncarbohydrate, addition also occurs to unsaturated carbohydrates. The carbon–carbon double bond in such reactions usually is part of an enol ether or an α,β-unsaturated ketone or lactone and the product is often a C-disaccharide.
Most addition reactions to carbon–carbon triple bonds involve a tri-n-butyltin radical adding to a triple bond in a carbohydrate. The structures of the carbohydrates are such that the vinylic radicals produced often undergo a cyclization reaction. | textbooks/chem/Organic_Chemistry/Radical_Reactions_of_Carbohydrates_(Binkley)/Radical_Reactions_of_Carbohydrates_II%3A_Radical_Reactions_of_Carbohydrates/18%3A_Compounds_with_Carbon%E2%80%93Carbon_Multiple_Bonds_I%3A_Addition_Reactions/I.txt |
The structural requirements for a molecule destined to undergo radical cyclization are that it contain a substituent from which a radical (almost always a carbon-centered one) can be generated and that it have a properly positioned multiple bond. Carbohydrates that meet these requirements include unsaturated iodides, bromides, thionocarbonates, cyclic thionocarbonates, xanthates, and phenyl selenides. Ring formation in the reactions of these compounds usually is regiospecific and often highly stereoselective.
II. Radical Fo
Samarium(II) iodide (SmI2) reacts with a variety of carbohydrate derivatives (including halides, sulfones, aldehydes, ketones, α-acyloxy esters, and α-acyloxy lactones) to generate carbon-centered radicals by nonchain, electron-transfer reaction.1–9 These radicals undergo reactions that include hydrogen-atom abstraction and ring formation, and they combine with SmI2 to produce organosamarium compounds (Scheme 1). Organosamarium compounds are quite reactive and easily undergo elimination and protonation reactions, as well as addition to aldehydes and ketones.
20: Reactions of Samarium(II) Iodide With Carbohydrate Derivatives
A. Reaction Mechanism
Radical formation begins when SmI2 coordinates with a substituent in a carbohydrate derivative (Scheme 1), that is, when a carbohydrate derivative replaces a solvent molecule within the coordination sphere of samarium(II) iodide. Within this new complex an electron is transferred from SmI2 to the carbohydrate derivative to produce a radical anion. This radical anion dissociates rapidly to give a carbohydrate radical and an anion complexed with SmI2. It is possible in some instances that the radical anion never actually forms; instead, the bond between the carbohydrate and the functional group breaks during electron transfer.3 [Section II.C.3 of Chapter 3 in Volume I contains additional information about samarium(II) iodide and the complexes it forms.]
B. Effect of HMPA
Reaction with SmI2 typically is conducted in tetrahydrofuran (THF). Adding the cosolvent hexamethylphosphoramide (HMPA) to the reaction mixture dramatically increases the rate constant for samarium(II) iodide reaction.10,11 Since the redox potential (Eo) of Sm2+/Sm3+ increases from -1.33 V to -2.05 V with the addition of four equiv of HMPA to a THF solution of SmI2,12 the rate enhancement brought about by added HMPA can be attributed to the substantially increased ability of SmI2 to donate an electron. (Addition of HMPA beyond four equivalents does not further increase reaction rates.10)
One explanation for the effect of HMPA on the reactivity of SmI2 is based on the energies of the highest occupied (HOMO) and lowest unoccupied (LUMO) molecular orbitals pictured in Figure 1.13 (In the reaction represented in this diagram it is assumed that the substrate is a phenyl sulfone.) When HMPA complexes with SmI2, it raises the HOMO energy of the resulting complex and, in so doing, reduces the energy required for electron transfer to the σ* orbital (LUMO) of the sulfone (Figure 1). This energy reduction translates into a larger rate constant for reaction. HMPA also increases the rate of reaction of SmI2 with halogenated compounds by elongating the carbon–halogen bond.11b
Radical formation by reaction of samarium(II) iodide with carbohydrate derivatives has been conducted under a variety of conditions.14–18 In addition to HMPA, other additives used are DMPU (1),14 ethylene glycol,19 and visible light.17 Alternative conditions also include reaction with HMPA in the presence of a proton donor14–18 or a catalytic amount of nickel(II) halide.9,17 Motivation for trying new reaction conditions comes from the possibilities of gaining greater understanding of the reaction mechanism, improving product yields, developing greater stereoselectivity, and finding a promoter for SmI2 reaction that is safer than HMPA.
III. Formation
At some point after its formation a carbohydrate radical will combine with a molecule of SmI2 to produce an organosamarium compound. The radical combining with SmI2 sometimes is the one initially formed and other times is one produced by reaction of the initially formed radical. Because carbon-centered radicals react rapidly with SmI2 (k = 7.0 x 106 M-1s-1 for reaction of the 5-hexenyl radical with SmI2 at 25 oC in the presence of five equiv of HMPA),10 only a limited number of radical reactions are fast enough to take place before an organosamarium compound forms. (Even those that are fast enough produce new radicals that are destined to be captured by SmI2.) Organosamarium compounds are quite reactive and, consequently, rarely isolated. Evidence for their existence takes the form of characteristic reactions (e.g., proton transfer, β elimination, and addition to a carbonyl compound). To understand the outcome of reactions begun by transfer of an electron from SmI2 to a carbohydrate derivative, it is necessary to be familiar with the reactions of organosamarium compounds. | textbooks/chem/Organic_Chemistry/Radical_Reactions_of_Carbohydrates_(Binkley)/Radical_Reactions_of_Carbohydrates_II%3A_Radical_Reactions_of_Carbohydrates/19%3A_Compounds_With_Carbon%E2%80%93Carbon_Multiple_Bonds_II%3A_Cyclization_Reactio.txt |
A. Protonation
Accepting a proton from a suitable donor is a characteristic reaction of an organosamarium compound.19–28 Lactones with α substituents19–23 and esters that are similarly substituted22,25 are common substrates in this type of reaction. Equation 1 describes a typical example.22 A mechanism for the reaction shown in eq 1 is proposed in Scheme 2. Tosylates26 and sulfones27 also form organosamarium compounds that readily protonate.
B. β Elimination
Another characteristic reaction of organosamarium compounds is eliminating a samarium-containing group along with a substituent on a neighboring carbon atom to form a compound with a carbon–carbon double bond.19,20,27,29–32 An example of a reaction in which this happens is shown in Scheme 3, where the glycosyl phenyl sulfone 2 reacts with SmI2 to give the organosamarium intermediate 3, from which β elimination produces the corresponding glycal.27
In the reactions of glycosyl phenyl sulfones with SmI2 the amount of glycal formed depends upon how well the departing, C-2 substituent supports a negative charge. In the reaction shown in eq 2 the compound with the O‑acetyl group at C-2 gives a far higher yield of glycal than does the substrate with the O-benzyl group in the 2-position.27,32 The primary process competing with glycal formation in this reaction is proton transfer to the organosamarium intermediate from the trace amount of water present in the reaction mixture. For the substrate with the O-acetyl group, proton transfer from water is too slow to be of consequence, but for that with the O‑benzyl group proton transfer is significant and becomes the major reaction pathway when greater than trace amounts of water are present (eq 2).
C. Addition to a Carbonyl Compound
Scheme 4 describes a reaction between samarium(II) iodide and a carbohydrate with an arylsulfonyl group to give an organosamarium compound that then adds to cyclohexanone.33 Addition to aldehydes and ketones is another characteristic reaction of an organosamarium intermediate. In Scheme 4 the overall reaction is given first, and then a mechanism for the radical and nonradical phases of the reaction is proposed.
1. The Samarium-Barbier Reaction
Because in the reaction shown in Scheme 4 the sulfone 4 and cyclohexanone both are present in the reaction mixture from the outset, the reaction is described as a Barbier-type2 or samarium-Barbier3,34 reaction. The mechanism pictured in Scheme 4 is a widely accepted one for this type of process.1–9,34–37 The carbohydrate reactant frequently is a glycosyl sulfone,28‑30,32,33,37‑43 but it also can be a glycosyl halide31,32,37,44-46 or phosphate.47 Possibilities for the carbonyl compound include ketones,2,28,32,39,42,43,46 aldehydes,2,28–30,32,38–46, and lactones.45, 48–50 Usually the carbonyl compound is a simple organic molecule, but sometimes the carbonyl group is part of the more complex structure found in a carbohydrate.38–40,42–44
2. The Samarium-Grignard Reaction
The defining characteristic of the samarium-Barbier reaction is that all of the reactants are present in the reaction mixture at the outset. If an intermediate organosamarium compound is sufficiently stable, it can be formed prior to adding the carbonyl compound. When reaction takes place using such a procedure, it is described as a Grignard-type or samarium-Grignard reaction.3,34 Many organosamarium compounds are not stable enough to undergo reaction in this way; in particular, the reaction shown in Scheme 4 is only successful when run under samarium-Barbier conditions.33
D. Formation and Reaction of Samarium Ketyls
Reaction of samarium(II) iodide with aldehydes and ketones produces ketyl radical anions, sometimes referred to as samarium ketyls (eq 3). These intermediates, each of which has considerable radical character on the former carbonyl carbon atom, form reversibly and have longer lifetimes than typical radical anions and most carbon-centered radicals.
1. Internal Addition to a Carbon–Carbon Multiple Bond
A samarium ketyl that contains a properly positioned multiple bond readily forms a new ring system.51–62 Examples indicating the range of reactivity of these ketyl intermediates are found in the reactions shown in equations 4 and 5 and Scheme 5. In the reaction pictured in eq 4, an unsaturated aldehyde forms a samarium ketyl that cyclizes and then reacts with cyclohexanone.60 Eq 5 describes the reaction of an unsaturated carbonyl compound that has a substituent on the carbon atom α to the carbonyl group. If the substituent is a poor leaving group [e.g., (C6H5)3CO], ring formation takes place, but when a better leaving group [e.g., (CH3)3CCO2] is present, cyclization is replaced by elimination of the corresponding anion [e.g., (CH3)3CCO2-] followed by hydrogen-atom abstraction.62 The highly stereoselective cyclization shown in Scheme 5 is an internal addition of a samarium ketyl to a triple bond.51 The resulting cyclic intermediate (7) either can react with another molecule of samarium(II) iodide or, since 7 is a highly reactive radical, abstract a hydrogen atom from the solvent (THF). Either reaction can be part of a two-step sequence leading to the final product (Scheme 5).
2. Internal Addition to a Carbon–Oxygen Double Bond (Pinacol Formation)
Reaction of samarium(II) iodide with a compound that has 1,4-,63 1,5‑,64–68 or 1,6‑69–77 related aldehydo or keto groups produces a samarium ketyl that then forms a cyclic pinacol. A typical example of such a reaction is shown in eq 6,64 and a general mechanism for pinacol formation is proposed in Scheme 6.78 Based on this proposal, one would expect that the two hydroxyl groups in a pinacol should be found on the same side of the newly formed ring system because during reaction the oxygen atoms in these two groups interact simultaneously with a single samarium ion. Further, one also would anticipate that reaction should place the hydroxyl groups stereoselectively on the less-hindered face of the new ring. Both of these expectations are realized not only in the reaction shown in eq 6 but in other, similar reactions, where the major products always are cis diols formed by minimizing steric interactions during ring construction.63,65–77
3. Internal Addition to a Carbon–Nitrogen Double Bond
Reaction analogous to pinacol formation occurs when one of the carbonyl groups in a reactant molecule is replaced by a group with a C–N double bond (eq 779).79–83 A significant stereochemical difference between this type of reaction and pinacol formation is that the hydroxyl and substituted amino groups produced during cyclization are on opposite faces of the newly formed ring. This result indicates that complexation between the carbonyl groups and the samarium ion during pinacol formation has no analogous interaction in reactions of keto-oximes.
Sometimes the cyclization of a keto-oxime produces an amine rather than a substituted amine (eq 7).79,84 This occurs when samarium(II) iodide, in excess of that needed for cyclization, transfers an electron to the N–O bond in the cyclic product leading to replacement of the amine substituent with a hydrogen atom. This reaction is accelerated by addition of water to the reaction mixture.
4. Ring-Contraction Reactions
Scheme 7 describes a reaction in which a samarium ketyl is involved in ring contraction. This process begins with electron transfer from SmI2 to the carbohydrate iodide 8 to generate the radical 9.85 Reaction of 9 with a second molecule of SmI2 produces the organosamarium compound 10. Elimination of the elements of MeOSmI2 from 10 causes the pyranoid ring to open to give the unsaturated aldehyde 11, which reacts with SmI2 to form a samarium ketyl that then cyclizes to give the substituted cyclopentanes 12 and 13. Similar ring contractions occur when 6-aldehydo hexopyranosyl derivatives react with SmI2.86,87 | textbooks/chem/Organic_Chemistry/Radical_Reactions_of_Carbohydrates_(Binkley)/Radical_Reactions_of_Carbohydrates_II%3A_Radical_Reactions_of_Carbohydrates/20%3A_Reactions_of_Samarium(II)_Iodide_With_Carbohydrate_Derivatives/IV._Reactions_.txt |
A. Substrates for Radical Cyclization
Radicals capable of cyclization can be generated from reaction of SmI2 with unsaturated carbohydrate sulfones13,27,88–91 or halides.14–18,92,93 Internal addition is possible to either a C–C14–18 (Scheme 8)14 or C–N92,93 (eq 8)92 double bond. Glycosyl phenyl sulfones are often the starting materials of choice for forming pyranos-1-yl radicals because not only are such sulfones more stable32 than the corresponding iodides and bromides, but they also produce radicals readily upon reaction with SmI2 in the presence of HMPA (Scheme 3). HMPA is critical to phenyl sulfone reactivity because in the absence of this cosolvent these sulfones are unreactive.13,91
In contrast to phenyl sulfones, HMPA is not required for reaction of 2‑pyridyl sulfones. This contrasting behavior is attributed to the effect of the 2-pyridyl group on sulfone MO energy levels. Because the LUMO energy of a 2‑pyridyl sulfone is lower than that of a phenyl sulfone, transfer of an electron to the 2‑pyridyl derivative occurs more easily (Figure 2); as a result, reaction can take place without HMPA being present (Scheme 9).13,91
B. Radical Cyclization Versus Cyclization of an Organosamarium Compound
When single-electron transfer takes place from SmI2 to a substrate molecule, it often is not clear whether the reactive species is a radical, an organosamarium compound, or even an anion.94 In the reaction shown in Scheme 10, for example, radical cyclization and organosamarium compound formation are both possible from the radical 19.27 Since neither protonation nor β elimination, characteristic reactions of an organosamarium intermediate, is observed, the indication is that the radical 19 undergoes cyclization before formation of the organosamarium compound 20 can take place.
The reaction pictured in Scheme 814 is similar to the one shown in Scheme 10 in that ring formation occurs without the simple reduction or β elimination that characterize organosamarium intermediates. In this reaction (Scheme 8) stereoselectivity is highly dependent on the reaction conditions. For the AIBN initiated reaction of 14 with tri-n-butyltin hydride, there is little doubt that radical cyclization is taking place. The similarity in product ratios between this reaction and that caused by SmI2 (in the absence of HMPA) supports the idea that both reactions involve radical cyclization.
There is a dramatic change in stereoselectivity when HMPA is added to the reaction shown in Scheme 8.14 This change has been attributed to an HMPA-complexed samarium ion becoming associated with the carbonyl group in 15. The size of this group is believed to be sufficient to create severe steric interaction with the isopropylidene group, an interaction that forces these two groups to opposite faces of the newly formed ring.14 It is also possible, however, that the large change in stereoselectivity signals a new reaction mechanism. Cyclization may occur from the organosamarium intermediate 16. For this to happen, however, it would require 16 to be formed faster than internal radical addition to an activated double bond. It also would require cyclization of the organosamarium compound to be faster than protonation or β elimination from 16. The available information does not provide a definitive, mechanistic choice for this reaction, that is, the radical cyclization shown in Scheme 8 when HMPA is present.14 | textbooks/chem/Organic_Chemistry/Radical_Reactions_of_Carbohydrates_(Binkley)/Radical_Reactions_of_Carbohydrates_II%3A_Radical_Reactions_of_Carbohydrates/20%3A_Reactions_of_Samarium(II)_Iodide_With_Carbohydrate_Derivatives/V._Cyclization.txt |
The reaction shown in Scheme 11 describes the formation of the C‑glycoside 25 by addition of the oxygen-stabilized radical 21 or the organosamarium compound 22 (or both) to a molecule of acetone.95 There is evidence for participation of both of these intermediates at some stage in this reaction. Conducting the reaction in the presence of t-BuSH quenches the addition process and dramatically increases the yield of the reduction product 23. Such a change would be expected from hydrogen-atom abstraction by the radical 21. In the absence of t-BuSH, formation of 23 and the elimination product 24 provide evidence for the organosamarium compound 22 also being present in the reaction mixture. Since conducting the reaction in the presence of D2O decreases the yield of the C-glycoside 25 in favor of the reduction and elimination products 23 and 24, respectively, the organosamarium compound 22 appears to be a likely intermediate in the addition process, but since a large excess of D2O only modestly reduces the yield of 25, radical addition remains a possible (perhaps major) pathway to C-glycoside formation.
The radical-addition pathway shown in Scheme 11 involves the nucleophilic, carbon-centered radical 21 adding to the carbonyl carbon atom in acetone. The carbonyl carbon atom is rendered quite electron deficient by complexation of acetone with SmI2. This combination of a reactive radical adding to a double bond with a decidedly electron-deficient atom is found in other reactions promoted by SmI2.96,97
VII. Compariso
Chromium(II) reagents participate in radical reactions98-104 that are similar both mechanistically and in terms of product formation to those occurring when samarium(II) iodide reacts with carbohydrate derivatives. Radical formation from reaction of chromium compounds with carbohydrate derivatives is far less common than radical formation from reaction with samarium(II) iodide. An example of a reaction involving a chromium(II) complex is given in Scheme 12 where [CrII(EDTA)]2- reacts with a glycosyl halide to produce a pyranos-1-yl radical that then combines with additional [CrII(EDTA)]2- to generate a glycosylchromium complex.98,99 This complex undergoes β elimination to give a glycal.
Radicals generated by chromium(II) reagents also undergo cyclization reactions such as that occurring when the bromide 26 reacts with chromium(II) acetate (eq 9).103 The presence of a carbon–carbon double bond in the final product (27) indicates that a transient organochromium complex forms during this reaction but then reacts to give the unsaturated, bicyclic carbohydrate 27.
VIII. Summary
Transfer of an electron from samarium(II) iodide to a carbohydrate with an electron-accepting substituent produces a radical anion. Dissociation of this radical anion generates a carbohydrate radical along with an anion derived from the substituent group. A carbohydrate radical formed in this way reacts rapidly with a second molecule of SmI2 to produce an organosamarium compound. This organometallic compound can undergo reactions that include addition to a compound containing a carbonyl group, proton capture, and elimination of a samarium-containing leaving group. In order for a radical formed from reaction of a carbohydrate derivative with SmI2 to avoid immediate reaction with a second molecule of SmI2, a rapid, radical process must intervene. The one of greatest interest is radical cyclization; thus, a carbohydrate derivative that has a properly placed multiple bond and an electron-accepting substituent reacts with samarium(II) iodide to a form radical that cyclizes. Carbohydrates that contain a pair of properly positioned aldehydo or keto groups cyclize to form pinacols. The intermediate in the cyclization step leading to a pinacol is a radical anion. Organochromium complexes form and react in a manner similar to organosamarium compounds.
21: Reactions of Radicals Produced by Electron Transfer to Manganese(III) Acetate
Manganese(III) acetate and ammonium cerium(IV) nitrate each react with CH-acidic compounds to produce carbon-centered radicals.1–12 These radicals add preferentially to compounds with electron-rich multiple bonds. The role of a carbohydrate in a reaction of this type is to provide the multiple bond to which addition occurs. | textbooks/chem/Organic_Chemistry/Radical_Reactions_of_Carbohydrates_(Binkley)/Radical_Reactions_of_Carbohydrates_II%3A_Radical_Reactions_of_Carbohydrates/20%3A_Reactions_of_Samarium(II)_Iodide_With_Carbohydrate_Derivatives/VI._Radical_Ad.txt |
Titanocene(III) chloride [Cp2TiCl, bis(cyclopentadienyl)titanium(III) chloride] is an oxygen-sensitive compound that is prepared by reaction of Cp2TiCl2 with metals such as zinc, aluminum, or manganese. Cp2TiCl exists as a dimer in the solid state, but coordinating solvents (e.g., tetrahydrofuran) dissociate the dimer into a reactive monomer (eq 1).1,2 (Although the monomer is coordinated with a solvent molecule, it usually is represented simply as Cp2TiCl; more generally, Cp2TiCl can be looked upon as representing all the Ti(III) species present in a solution of titanocene(III) chloride.1–3)
22: Reactions of Carbohydrate Derivatives With Titanocene(III) Chloride
Three types of carbohydrate derivatives form carbon-centered radicals upon reaction with Cp2TiCl. Halogen-atom abstraction from glycosyl halides produces furanos-1-yl and pyranos-1-yl radicals.1,4–11 Radicals also can be generated by abstractive ring opening of epoxides.12–19 Finally, Cp2TiCl produces pyranos-1-yl radicals when it reacts with glycosyl 2-pyridyl sulfones.7 An example of the first type of reaction is found in eq 2,5,6 one of the second type in eq 3,16 and one of the third in eq 4.7 These radical-forming reactions have the attractive, chemoselective feature that Cp2TiCl does not affect acetal, ester, or silyl ether protection.5,6
III. Elec
Ruthenium is a transition metal that, like titanium, can transfer an electron to a glycosyl halide. Photochemical reaction of [Ru(bpy)3]2+with a tertiary amine produces [Ru(bpy)3]+, a complex that then donates an electron to a glycosyl bromide to form a pyranos-1-yl radical (Scheme 10).29,30 The radical formed in this way from the bromide 20 is capable of adding to a variety of electron-deficient alkenes (eq 13). The role of the additive in this reaction is to improve product yield by suppressing oligmerization.29
IV. Summa
Titanocene(III) chloride reacts with glycosyl halides and with epoxides to generate carbon-centered radicals. The primary reaction of these radicals is combination with another molecule of Cp2TiCl. These radicals also can abstract hydrogen atoms from the solvent or other hydrogen-atom transfers in the reaction mixture or undergo radical addition and cyclization reactions. If a radical combines with a second molecule of titanocene(III) chloride, the resulting organotitanium compound typically undergoes a β-elimination reaction. The result of such a reaction usually is formation of a glycal.
I. Organocobalt Compounds
An organometallic complex that contains a carbon–cobalt bond can function as a radical precursor because such a bond is easily broken homolytically. Facile cleavage occurs because carbon–cobalt bonds are significantly weaker than most covalent bonds:1,2 in fact, the C–Co bond in coenzyme B12 (1, Figure 1) is one of the weakest covalent bonds known (BDE = 31.5 kcal mol‑1).3 Enzymatic reaction, mild heating, and photolysis with visible light all cause homolysis of C–Co bonds. Adding to the usefulness of organocobalt complexes as radical precursors is the fact that, despite their considerable reactivity, many of these complexes can be handled in the laboratory.
Although C–Co bond homolysis takes place at relatively low temperatures, photolysis is the method of choice for radical formation in reactions conducted outside biological settings.4–9 The reason for this choice is that C–Co bond fragmentation occurs with low-energy (visible) light at temperatures that avoid possible side reactions from even mild heating of complex, cobalt-containing compounds.
Coenzyme B12 (1, Figure 1) provided the original stimulus for using carbon–cobalt bond homolysis to form carbon-centered radicals.7–11 The enzyme-induced homolysis of the C–Co bond in 1 produced the 5‑deoxyadenosyl radical 2 and the cobalt-containing radical 3 (eq 1). The discovery that carbon-centered radicals could be produced in this way led to interest in finding simpler molecules that would mimic such behavior.
Among the several types of organocobalt complexes found to be useful in generating carbon-centered radicals, cobaloximes [bis(dimethylglyoximato)cobalt complexes] (Figure 2) are the most widely used in carbohydrate chemistry.11-15 Many reactions of cobalt-containing carbohydrates and much of the mechanistic information about reactions caused by C–Co bond homolysis come from study of cobaloximes.
II. Organomercury Compounds
There are similarities between the reactivities of carbon–cobalt and carbon–mercury bonds. Both are strong enough to exist in stable structures that can be isolated and both readily cleave upon heating or photolysis. The result in each case is formation of a metal-centered and a carbon-centered radical. Carbon-centered radicals produced by carbon–mercury bond homolysis undergo typical radical reactions, such as hydrogen-atom abstraction (Scheme 627),27,28 addition to a multiple bond (eq 9),29 and combination with molecular oxygen (eq 1030).30,31 Although organomercury compounds can be effective sources of carbon-centered radicals, their use in this role is limited by toxicity and environmental concerns.
Two basic methods exist for generating radicals from organomercury compounds. The first, photochemical homolysis of a carbon–mercury bond, is illustrated by the reaction shown in Scheme 6.27 The second is more complicated and consists of initially converting an organomercury compound into the corresponding mercury hydride by reaction with NaBH4 (Scheme 7).32 The hydride then produces a carbon-centered radical capable of reactions such as the addition to acrylonitrile shown in eq 9.29 Adventitious initiation is credited with beginning this reaction.
III. Summary
Organocobalt complexes are sources of free radicals because heating, photolysis, or enzymatic reaction cleaves a carbon–cobalt bond homolytically to produce carbon-centered and cobalt-centered radicals. Cleaving the carbon–cobalt bond in this way changes the oxidation state of cobalt from Co(III) to Co(II). Complexes with cobalt in the Co(II) oxidation state exhibit radical reactivity. Cobalt-containing carbohydrates easily undergo epimerization reactions because the radicals formed by bond fragmentation readily recombine. Carbon-centered radicals produced from organocobalt complexes also undergo the characteristic radical reactions of addition and cyclization.
Organocobalt and organomercury compounds have a similarity in reactivity because each contains a carbon-metal bond that is easily cleaved by heating or photolysis. Carbon-centered radicals produced from organomercury compounds undergo hydrogen-atom abstraction and radical addition reactions. Concern about the toxicity of organomercury compounds reduces their usefulness as radical precursors. | textbooks/chem/Organic_Chemistry/Radical_Reactions_of_Carbohydrates_(Binkley)/Radical_Reactions_of_Carbohydrates_II%3A_Radical_Reactions_of_Carbohydrates/22%3A_Reactions_of_Carbohydrate_Derivatives_With_Titanocene(III)_Chloride/II._React.txt |
The previous four chapters describe electron-transfer reactions between carbohydrate derivatives and transition-metal ions. Some ions [chromium(II), samarium(II), and titanium(III)] are electron donors and others [cobalt(III), cerium(IV), manganese(III), and mercury(II)] are electron acceptors. Another form in which a transition-metal ion can participate in a radical reaction is as a part of a redox couple. (A redox couple is a combination of a transition metal and an ion from a different transition metal that act together in donating electrons to organic compounds.) Redox couples promote the addition of halogenated carbohydrates to electron-deficient double bonds, and they participate in the conversion of glycosyl halides into glycals and simple reduction products.
24: Redox Couples
A. Direct Electron Donation
Electron donation from the metal in a redox couple to a halogenated carbohydrate can occur either directly or indirectly. With direct reaction the role of the metal ion is primarily to prepare the surface of the metal for interaction with the halogenated compound. This is believed to be the purpose of the copper ion in a zinc–copper couple, a reagent that has been described as an active form of zinc metal.1 Equation 1 pictures an addition reaction in which the adding radical is generated by reaction of a deoxyiodo sugar with a zinc–copper couple (Zn and CuI) suspended in an ethanol-water solution.2
B. Indirect Electron Donation
Indirect electron donation from the metal in a redox couple occurs when the metal ion is actively involved in the transfer process.3–5 An example of this type of participation is shown in eq 2, where Ni(I) is oxidized to Ni(II) during reaction with a halogenated carbohydrate, and Ni(II) then is reduced to Ni(I) by the manganese metal.3 Since the electrons being transferred to the carbohydrate are coming indirectly from manganese, the nickel ion performs a delivery role in the reaction and the manganese is the stoichiometric reactant (Scheme 1). In reactions of this type the metal ion need be present in only catalytic amounts; for example, in the glycal formation pictured in eq 3, the complex containing the titanium ion is added in as little as 10 mol%.4
III. Reactions with Redox Couples
A. Addition Reactions
The total number of reactions of carbohydrate derivatives with redox couples is modest; among these addition reactions are reported more often than any other type.2,3,5–9 Addition processes often involve a couple formed by combining zinc metal with a copper salt.2,6,7,9 Such a reactions is illustrated in eq 1, where a zinc–copper couple participates in the addition of a halogenated carbohydrate to a compound with an electron-deficient double bond.
Since reaction between a zinc–copper couple and a carbohydrate is a heterogeneous process that takes place on the surface of finely divided zinc, efficient mixing during reaction is essential. Sonication, which often is used during redox-couple preparation and reaction, is believed to aid electron transfer indirectly by increasing mixing and improving metal-surface cleaning and directly by promoting electron transfer through the influence of ultrasonic waves.10
B. Elimination and Hydrogen-atom abstraction Reactions
The idea that a copper ion is not directly involved in electron transfer from a zinc–copper couple garners some support from the reaction shown in eq 4, where zinc metal alone is able to act as the electron source in generating a pyranos-1-yl radical.11 After formation, this radical undergoes further reaction that leads to the D-glucal 3. Support for the intermediacy of a pyranos-1-yl radical in this reaction comes from conducting reaction in the presence of 1-dodecanethiol (C12H25SH), an excellent hydrogen-atom transfer. When this thiol is present, a substantial amount of hydrogen-atom abstraction by the pyranos-1-yl radical takes place to produce the simple-reduction product 4. Proton transfer (the competing, nonradical possibility) does not appear to be involved in formation of 4 because when methanol replaces 1-dodecanethiol in the reaction mixture, none of this simple-reduction product is formed (eq 4).11
IV. Reaction Mechanism
Although the radical mechanism shown in Scheme 1 offers a reasonable explanation for the reactions pictured in eq 2, there is uncertainty in some reactions involving redox couples about whether a free-radical is ever produced. This uncertainty is reflected in the reaction mechanism shown in Scheme 2, which describes two possible pathways for participation of a zinc–copper couple in an addition reaction. One pathway involves radical formation by electron transfer, and the second describes formation of an organozinc intermediate. The stereochemical evidence and solvent effects described in the next two sections offer insight into the nature of the reactive species generated by a typical redox couple.
V. Summary
A redox couple is a combination of a transition metal with an ion from another transition metal. These couples serve as electron donors in addition of halogenated carbohydrates to compounds with electron-deficient double bonds. There is some uncertainty as to whether a free radical or an organometallic compound is the intermediate in this type of reaction. Stereochemical evidence supports the radical pathway, but solvent effects indicate a more complicated situation. | textbooks/chem/Organic_Chemistry/Radical_Reactions_of_Carbohydrates_(Binkley)/Radical_Reactions_of_Carbohydrates_II%3A_Radical_Reactions_of_Carbohydrates/24%3A_Redox_Couples/II._Electron_Transfer_from_a_Redox_Couple.txt |
Hydrogen-atom donors are widely used in radical reactions because hydrogen-atom abstraction is the final step in most radical chain processes. Donors can have a hydrogen atom bonded to a tin, silicon, sulfur, selenium, boron, phosphorous, or carbon atom. Most reactions involve organotin compounds, usually tri-n-butyltin hydride (Bu3SnH). Some organosilanes, in particular tris(trimethylsilyl) silane [(Me3Si)3SiH], are effective enough as hydrogen-atom transfers to serve as replacements for organotin hydrides. Most other hydrogen-atom transfers are either so reactive or so unreactive that they typically are used only in special situations.
Appendix I: Hydrogen-Atom Donors
Organotin hydrides are the most frequently employed hydrogen-atom donors in radical reactions of carbohydrates. Clearly, the compound of choice is tri-n-butyltin hydride (Section II.A). Phenyl-substituted compounds, such as triphenyltin hydride, can serve in the same role, but they offer no advantage and are rarely used. Polymer-supported (Section II.B) and fluorous (Section II.C) tin hydrides have been used as replacements that avoid some of the difficulties inherent in the use of tri-n-butyltin hydride.
III. Organosilanes
Difficulties associated with use of tri-n-butyltin hydride have prompted chemists to search for alternative, hydrogen-atom sources, ones that avoid the problems associated with organotin compounds. Most attention has focused on organosilanes, compounds that do not have the toxicity associated with organotin reagents.4 Initially, the outlook was not promising because simple organosilanes are poor hydrogen-atom transfers when reacting with alkyl radicals and do not support chain reactions under normal conditions.4 Innovative ideas, however, have overcome these problems.
IV. Compounds with Phosphorous–Hydrogen
The search for less problematic hydrogen-atom transfers for use in the Barton-McCombie reaction has led to compounds with phosphorus–hydrogen bonds. These include dialkylphosphine oxides (11), alkyl phosphites (12), hypophosphorous acid (13), and salts of hypophosphorous acid (14) (Figure 2). All of these compounds can function as inexpensive, nontoxic hydrogen-atom transfers that form the chain-carrying radicals needed for reaction and do not produce byproducts difficult to remove.3,10,57 An example of a reaction in which hydrogen donation is from a P–H bond is shown in eq 13.58
Alkyl phosphites (12) are excellent hydrogen-atom transfers, but reactions involving these compounds have the disadvantage of not being able to be initiated by 2,2'-azobis(isobutyronitrile); benzoyl peroxide usually is the initiator.3,10 Reactions in which the hydrogen-atom transfer is a dialkylphosphine oxide (11), hypophosphorous acid (13), or a salt of hypophosphorous acid (14) can be initiated by AIBN.9,57 Because it is difficult to completely remove water from hypophosphorous acid and its salts, these donors are less attractive choices when moisture sensitive compounds are reacting.9
V. Compounds with Boron–Hydrogen Bonds
Phosphine-boranes (15) (Figure 2) are a group of compounds that have the ability to react selectively with xanthates in the presence of compounds containing bromine or chlorine (but not iodine).11 For example, cyclohexyl bromide is recovered without change when it is added to the reaction shown in eq 14; in contrast, tri-n-butyltin hydride and most other hydrogen-atom transfers used in radical reactions readily dehalogenate bromides. If this lack of reactivity between alkyl bromides and phosphine-boranes extends to halogenated carbohydrates, it will make possible their chemoselective deoxygenation without dehalogenation.
VI. Compounds with Carbon–Hydrogen Bonds
A. 2-Propanol
Few compounds in which a carbon–hydrogen bond must serve as the hydrogen-atom source are reactive enough to function as hydrogen-atom transfers in radical reactions of carbohydrates. The reason for this is that when less reactive donors are used, other reactions become competitive. Even compounds with quite reactive C–H bonds are poor hydrogen-atom transfers when compared to tri-n-butyltin hydride or tris(trimethylsilyl)silane. One compound that does have the necessary reactivity, but just barely, is 2‑propanol. When reaction of the xanthate 16 is conducted with 2‑propanol as the solvent, hydrogen-atom abstraction is in spirited competition with xanthate-dithiocarbonate rearrangement (eq 15).59 This competition exists because hydrogen-atom abstraction by the carbohydrate radical R· is slow enough that addition of R· to another molecule of the xanthate 16 has a comparable rate (Scheme 5). The adduct radical formed by this addition fragments to give the dithiocarbonate 18 and a carbohydrate radical (R·).
B. Cyclohexane
The xanthate 19 reacts to form the corresponding deoxy sugar in 85% yield (eq 16).60 In this reaction cyclohexane functions as the hydrogen-atom transfer. Since cyclohexane is not a noticeably better hydrogen-atom transfer than 2-propanol, it is initially surprising that no dithiocarbonate is formed from 19 even though (as described in the previous section) dithiocarbonate formation is significant in reaction of the xanthate 16 (eq 15). The structural difference between the starting materials (16 and 19) in these two reactions accounts for their difference in reactivity. Unlike 16, the xanthate 19 has a sulfur atom directly attached to the carbohydrate portion of the molecule. This means that when the carbohydrate radical R· adds to 19, the options available to the adduct radical 20 are either regenerating the starting materials or expelling an unstabilized, primary radical (Scheme 6). Not surprisingly, no dithiocarbonate from primary radical expulsion is observed; therefore, the only operative pathway for the radical 20 is reforming of R· and the xanthate 19. Each regeneration of R· creates a new opportunity for it to abstract a hydrogen atom. With these multiple opportunities even a marginally effective hydrogen-atom transfer eventually is able to react with R· to produce the hydrogen-abstraction product RH.
Even though the yield is good, the reaction shown in eq 16 is not an attractive option for deoxy sugar synthesis because it requires reaction of the carbohydrate to replace a C–O bond with a C–S bond before conducting the Barton-McCombie reaction. The additional steps necessary for this conversion add to the effort required for deoxygenation.
C. Silylated Cyclohexadienes
Silylated cyclohexadienes, such as 21, are effective hydrogen-atom transfers in Barton-McCombie reactions (eq 17).61 Compound 21 has the advantage of being a solid material that can be easily stored and handled. Although this compound (21) is an order of magnitude less reactive than (Me3Si)3SiH (3), it is able to support chain reactions. The propagation steps in a proposed mechanism for replacement of an O-phenoxythiocarbonyl group with a hydrogen atom supplied by 21 are given in Scheme 7.
VII. Compounds with Sulfur–Hydrogen or S
The rate constants for hydrogen-atom abstraction from sulfur–hydrogen and selenium–hydrogen bonds are so rapid [kSH = 1.5 x 108 M-1s-1 (from C6H5SH) and kSeH = 2.1 x 109 M-1s-1 (from C6H5SeH)] that abstraction typically will take place before other radical reactions (e.g., addition, cyclization, and rearrangement) can occur. (Rate constants for hydrogen-atom abstraction from various, hydrogen-atom donors, as well as rate constants for other radical reactions are given in Chapter 8 of Volume I.)
VIII. Summary
Since some tin-containing compounds are toxic and can cause purification problems, procedures have been developed both to minimize the amount of these materials needed for successful reaction and to make their removal easier and more complete. Another solution to toxicity and purification problems created by tin-containing compounds is to replace them with less offensive reagents. An effective replacement is tris(trimethylsilyl)silane. For reactive carbohydrate iodides cyclohexane also can be used. | textbooks/chem/Organic_Chemistry/Radical_Reactions_of_Carbohydrates_(Binkley)/Radical_Reactions_of_Carbohydrates_II%3A_Radical_Reactions_of_Carbohydrates/Appendix_I%3A_Hydrogen-Atom_Donors/II._Organotin_Hydrides.txt |
The nomenclature of acid halides starts with the name of the corresponding carboxylic acid. The –ic acid ending is removed and replaced with the ending -yl followed by the name of the halogen with an –ide ending. This is true for both common and IUPAC nomenclature. The carbonyl carbon is given the #1 location number. It is not necessary to include the location number in the name because it is assumed that the functional group will be on the end of the parent chain.
Contributors
Prof. Steven Farmer (Sonoma State University)
Properties of Acyl Halides
This page defines acyl halides and discusses their simple physical properties, introducing chemical reactivity in a general way.
Acyl halides as "acid derivatives"
A carboxylic acid such as ethanoic acid has the following structure:
There are a number of related compounds in which the -OH group in the acid is replaced by something else that leaves the acyl carbon in a +3 oxidation state. Compounds like this are described as acid derivatives. Acyl halides (also known as acid halides) are one example of an acid derivative. In this example, the -OH group has been replaced by a chlorine atom; chlorine is the most commonly used acid halide.
The acyl group
The acyl group is a hydrocarbon group attached to a carbon-oxygen double bond:
The "R" group is normally restricted to an alkyl group. It could, however, be a group based on a benzene ring.
Naming acyl halides
The easiest way to name an acyl halide is to consider the relationship with the corresponding carboxylic acid:
carboxylic acid name acyl halide name acyl halide formula
ethanoic acid ethanoyl chloride CH3COCl
propanoic acid propanoyl chloride CH3CH2COCl
butanoic acid butanoyl chloride CH3CH2CH2COCl
butanoic acid butanoyl bromide CH3CH2CH2COBr
butanoic acid butanoyl iodide CH3CH2CH2COI
If something is substituted into the hydrocarbon chain, the carbon in the -COX (X = halide) group is the number 1 carbon.
For example, 2-methylbutanoyl chloride is named as follows:
Physical properties of acyl halides
An acyl halide such as ethanoyl chloride is a colorless, fuming liquid. The strong smell of ethanoyl chloride is a mixture of the smell of vinegar (ethanoic acid) and the acrid smell of hydrogen chloride gas. The smell and the fumes originate from the reactions between ethanoyl chloride and water vapor in the air.
Solubility in water
Acyl halides do not dissolve in water because they react (often violently) with it to produce carboxylic acids and hydrogen halides (e.g. HCl). The strong reaction makes it impossible to obtain a simple aqueous solution of an acyl halide.
Boiling points
Taking ethanoyl chloride as a typical example: ethanoyl chloride boils at 51°C and is a polar molecule. Therefore, it has dipole-dipole attractions between its molecules as well as van der Waals dispersion forces. However, it does not form hydrogen bonds. Its boiling point is, therefore, higher than an alkane of similar size (which has no permanent dipoles) such as ethane, which boils at -88.5°C, but not as high as a similarly sized alcohol (which forms hydrogen bonds in addition to everything else) such as ethanol, which boils at 78.37°C.
Reactivity
Acyl halides are extremely reactive, and in each of their reactions the halogen atom is replaced by another functional group. In each case, in the first step, hydrogen halide (normally hydrogen chloride) gas is produced as steamy acidic fumes. However, in some cases the hydrogen halide goes on to react with one of the substances in the reaction mixture. Taking ethanoyl chloride as typical, the initial reaction is the following:
These reactions involve water, alcohols and phenols, or ammonia and amines as nucleophiles. Each of these particular cases contains a very electronegative element with an active lone pair of electrons—either oxygen or nitrogen. The most commonly performed reaction with acyl halides and the example reaction is known as nucleophilic acyl substitution.
Contributors
Jim Clark (Chemguide.co.uk)
• Mark Tye (Diablo Valley College) | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Acid_Halides/Nomenclature_of_Acid_Halides.txt |
Introduction
Acid chlorides react with carboxylic acids to form anhydrides.
Example 1:
Mechanism
1) Nucleophilic attack by the alcohol
2) Leaving group is removed
3) Deprotonation
Contributors
Prof. Steven Farmer (Sonoma State University)
Acid chlorides can be converted to aldehydes using LiAlH(Ot-Bu)3
Acid chlorides can be converted to aldehydes using lithium tri-tert-butoxyaluminum hydride (LiAlH(Ot-Bu)3). The hydride source (LiAlH(Ot-Bu)3) is a weaker reducing agent than lithium aluminum hydride. Because acid chlorides are highly activated they still react with the hydride source; however, the formed aldehyde will react slowly, which allows for its isolation.
General Reaction:
Acid chlorides react with alcohols to form esters
Acid chlorides react with alcohols to form esters
Mechanism
1) Nucleophilic attack by the alcohol
2) Leaving group is removed
3) Deprotonation
Contributors
Prof. Steven Farmer (Sonoma State University)
Acid chlorides react with ammonia 1 amines and 2 amines to form amides
Acid chlorides react with ammonia, 1o amines and 2o amines to form amides.
Mechanism
1) Nucleophilic attack by the amine
2) Leaving group is removed
3) Deprotonation
Contributors
Prof. Steven Farmer (Sonoma State University)
Acid chlorides react with water to form carboxylic acids.
Introduction
Acid chlorides react with water to form carboxylic acids.
Example 1:
Mechanism
1) Nucleophilic attack by water
2) Leaving group is removed
3) Deprotonation
Contributors
Prof. Steven Farmer (Sonoma State University)
Addition and Elimination Reactions in Acyl Chlorides
Acyl chlorides (also known as acid chlorides) are one of a number of types of compounds known as "acid derivatives". This is ethanoic acid:
If you remove the -OH group and replace it by a -Cl, you have produced an acyl chloride.
This molecule is known as ethanoyl chloride and for the rest of this topic will be taken as typical of acyl chlorides in general. Acyl chlorides are extremely reactive. They are open to attack by nucleophiles - with the overall result being a replacement of the chlorine by something else.
Nucleophiles
A nucleophile is a species (an ion or a molecule) which is strongly attracted to a region of positive charge in something else.Nucleophiles are either fully negative ions, or else have a strongly $\delta^-$ charge somewhere on a molecule. The nucleophiles that we shall be looking at all depend on lone pairs on either an oxygen atom or a nitrogen atom.
Nucleophiles based on oxygen atoms
We shall be talking about water and alcohols, taking ethanol as a typical alcohol.
Notice how similar these two molecules are around the oxygen atom. That's what turns out to be important.
Nucleophiles based on nitrogen atoms
We shall be considering ammonia and primary amines, taking ethylamine as a typical primary amine. A primary amine contains the -NH2 group attached to either an alkyl group (as it is here) or a benzene ring. As far as these reactions are concerned, the nature of any hydrocarbon attached to the nitrogen makes no difference. The nitrogen atom is the important bit.
Again, notice how similar these two molecules are around the nitrogen atom - and also how similar they are to the previous ones containing oxygen. Both types of molecule have an active lone pair of electrons attached to one of the most electronegative elements. All of these molecules also have at least one hydrogen atom attached to the oxygen or nitrogen.
Why are acyl chlorides attacked by nucleophiles?
The carbon atom in the -COCl group has both an oxygen atom and a chlorine atom attached to it. Both of these are very electronegative. They both pull electrons towards themselves, leaving the carbon atom quite positively charged.
The Overall Reaction
We are going to generalize this for the moment by writing the reacting molecule as "Nu-H". Nu is the bit of the molecule which contains the nucleophilic oxygen or nitrogen atom. The attached hydrogen turns out to be essential to the reaction. The general equation for the reaction is:
In each case, the net effect is that you replace the -Cl by -Nu, and hydrogen chloride is formed as well.
Since the initial attack is by a nucleophile, and the overall result is substitution, it would seem reasonable to describe the reaction as nucleophilic substitution. However, the reaction happens in two distinct stages. The first involves an addition reaction, which is followed by an elimination reaction where HCl is produced. So the mechanism is also known as nucleophilic addition / elimination.
The General Mechanism
Step 1: Addition
As the lone pair on the nucleophile approaches the fairly positive carbon in the acyl chloride, it moves to form a bond with it. In the process, the two electrons in one of the carbon-oxygen bonds are repelled entirely onto the oxygen, leaving that oxygen negatively charged.
Notice the positive charge that forms on the nucleophile. Just accept this for the moment. It's much easier to explain why that charge must be there if you have a real example in front of you. This is fully explained in the pages on the reactions involving water, ammonia and so on.
Step 2: Elimination
The elimination stage occurs in two steps. In the first step, the carbon-oxygen double bond reforms. To make room for it, the electrons in the carbon-chlorine bond are repelled until they are entirely on the chlorine atom - forming a chloride ion.
Finally, the chloride ion plucks the hydrogen off the original nucleophile. It removes it as a hydrogen ion, leaving the pair of electrons behind on the oxygen or nitrogen atom in that nucleophile. That cancels the positive charge.
Contributors
Jim Clark (Chemguide.co.uk)
General reaction mechanism of acid chlorides
Introduction
Carboxylic acid derivatives are functional groups whose chemistry is closely related. The main difference is the presence of an electronegative substituent that can act as a leaving group during a nucleophile substitution reaction. Although there are many types of carboxylic acid derivatives known, this article focuses on four: acid halides, acid anhydrides, esters, and amides.
General mechanism
1) Nucleophilic attack on the carbonyl
2) Leaving group is removed
Although aldehydes and ketones also contain carbonyls, their chemistry is distinctly different because they do not contain suitable leaving groups. Once a tetrahedral intermediate is formed, aldehydes and ketones cannot reform their carbonyls. Because of this, aldehydes and ketones typically undergo nucleophilic additions and not substitutions.
The relative reactivity of carboxylic acid derivatives toward nucleophile substitutions is related to the electronegative leaving group’s ability to activate the carbonyl. The more electronegative leaving groups withdraw electron density from the carbonyl, thereby increasing its electrophilicity.
Contributors
Prof. Steven Farmer (Sonoma State University)
Organocuprate reagents convert acid chlorides to ketones
Introduction
Organocuprate reagents convert acid chlorides to ketones.
Example 1:
Contributors
Prof. Steven Farmer (Sonoma State University) | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Acid_Halides/Reactions_of_Acid_Halides/Acid_Chlorides_react_with_carboxylic_acids_to_form_anhydrides.txt |
This page discusses the reactions of acyl chlorides (acid chlorides) with ammonia and primary amines. These reactions are considered together because their chemistry is so similar.
Comparing the structures of ammonia and primary amines
Each of these substances contain an -NH2 group. In ammonia, this is attached to a hydrogen atom. In a primary amine, it is attached to an alkyl group (shown by "R" in the diagram below) or a benzene ring.
Reactions with acyl chlorides
Taking a general case of a reaction between ethanoyl chloride and a compound XNH2 (where X is hydrogen, an alkyl group, or a benzene ring). The reaction happens in two stages:
First:
In each case, hydrogen chloride gas is initially formed, the hydrogen coming from the -NH2 group, and the chlorine from the ethanoyl chloride. The remaining species join together. However, ammonia and amines are basic, and react with hydrogen chloride to produce a salt. Therefore, the second stage of the reaction is the following:
$XNH_2 + HCl \rightarrow XNH_3^+ Cl^-$
This is illustrated with specific compounds below.
The individual reactions
The reaction with ammonia
In this case, the "X" in the equations above is a hydrogen atom. In the first instance, hydrogen chloride gas and an organic compound called an amide are formed. An amide contains the group -CONH2. In the reaction between ethanoyl chloride and ammonia, the amide formed is called ethanamide:
This reaction is more commonly written as follows:
The hydrogen chloride produced reacts with excess ammonia to form ammonium chloride:
These equations can be combined to give the overall equation:
The ethanoyl chloride is typically added to a concentrated solution of ammonia in water. This is a violent reaction producing large amounts of white smoke, a mixture of solid ammonium chloride and ethanamide. Some of the mixture remains dissolved in water as a colorless solution.
The reaction with primary amines
The reaction with methylamine
Consider methylamine as a typical primary amine, in which the -NH2 is attached to an alkyl group. The initial equation would be the following:
The organic product in this case is called an N-substituted amide. Compare the structure with the amide produced in the reaction with ammonia, the only difference is that one of the hydrogen atoms on the nitrogen has been substituted for a methyl group. This particular compound is N-methylethanamide. The "N" indicates that the substitution is on the nitrogen atom, and not elsewhere in the molecule.
This reaction equation is typically be written the following way:
$CH_3COCl + CH_3NH_2 \rightarrow CH_3CONHCH_3 + HCl$
Primary amines can be thought of as modified ammonia molecules. If ammonia is basic and forms a salt with the hydrogen chloride, excess methylamine does exactly the same thing:
$CH_3NH_2 + HCl \rightarrow CH_3NH_3^+ \; Cl^-$
The salt is called methylammonium chloride. It is like ammonium chloride, except that one of the hydrogens has been replaced by a methyl group. These two equations are combined into one overall equation for the reaction, as shown below:
$CH_3COCl + 2CH_3NH_2 \rightarrow CH_3CONHCH_3 + CH_3NH_3Cl$
The reaction looks identical to the ammonia reaction. The methylamine is again used as a concentrated aqueous solution. There is a violent reaction producing a white solid mixture of N-methylethanamide and methylammonium chloride.
The reaction with phenylamine (aniline)
Phenylamine is the simplest primary amine in which the -NH2 group is attached directly to a benzene ring. Its old name is aniline. In phenylamine, no other groups are attached to the ring. The formula of phenylamine is C6H5NH2. There is no essential difference between this reaction and the methylamine reaction, except that phenylamine is a brown liquid, and the solid products tend to be stained brown.
The overall equation for the reaction is given below:
$CH_3COCl + 2C_6H_5NH_2 \rightarrow CH_3CONHC_6H_5 + C_6H_5NH_3Cl$
The products are N-phenylethanamide and phenylammonium chloride. This reaction is often confusing if the phenylamine is drawn showing the benzene ring, and especially if the reaction is examined from the point of view of the phenylamine. For example, the product molecule might be drawn looking like this:
This is obviously the same molecule as in the equation above, but it emphasizes the phenylamine part. Notice that one of the hydrogen atoms of the -NH2 group has been replaced by an acyl group, an alkyl group attached to a carbon-oxygen double bond. Phenylamine can be said to have been acylated or undergone acylation. Because of the nature of this particular acyl group, it is also described as ethanoylation. The hydrogen atom is replaced by an ethanoyl group, CH3CO-.
Contributors
Jim Clark (Chemguide.co.uk) | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Acid_Halides/Reactions_of_Acid_Halides/Reactions_of_Acyl_Chlorides_Involving_Nitrogen_Compounds.txt |
This page discusses the reactions of acyl chlorides (acid chlorides) with water, alcohols and phenol. These reactions are considered together because their chemistry is similar.
Comparing the structures of water, ethanol and phenol
Each of these substances contains an -OH group. In water, this is attached to a hydrogen atom. In an alcohol, it is attached to an alkyl group, indicated in the diagrams below as "R". In phenol, it is attached to a benzene ring (phenol is C6H5OH).
General Reactions with acyl chlorides
Consider ethanoyl chloride as a typical example of the acyl chlorides. Taking a general case of a reaction between ethanoyl chloride and a compound X-O-H (where X is hydrogen, or an alkyl group or a benzene ring):
In each case, hydrogen chloride gas is produced, the hydrogen coming from the -OH group, and the chlorine from the ethanoyl chloride. The remaining species are joined together.
The reaction with water
Modifying the general equation above, ethanoic acid is produced together with hydrogen chloride gas.
This is more often written as follows:
$3CH_3COCl + H_2O \rightarrow CH_3COOH + HCl$
Adding an acyl chloride to water produces the corresponding carboxylic acid together with steamy acidic fumes of hydrogen chloride. The reaction is usually extremely vigorous at room temperature.
The reaction with alcohols
The general equation for any alcohol reacting with ethanoyl chloride is the following:
The organic product in this case is an ester. For example, ethanol produces the ester ethyl ethanoate:
The equation is more commonly written as follows:
$CH_3COCl + CH_3CH_2OH \rightarrow CH_3COOCH_2CH_3 + HCl$
This is an easy way of producing an ester from an alcohol because the reaction occurs at room temperature, and is irreversible. Making an ester from an alcohol and a carboxylic acid (the usual alternative method) requires heat and a catalyst, and is reversible, making it difficult to obtain a 100% conversion.
The reaction with phenols
A phenol has an -OH group attached directly to a benzene ring. In the substance normally called "phenol", there is nothing else attached to the ring; this simplest case is considered first. The reaction between phenol and ethanoyl chloride is not quite as vigorous as that between alcohols and ethanoyl chloride; the reactivity of the -OH group is modified by the benzene ring.
Aside from this, the reaction is just the same as with an alcohol:
Written more simply:
$CH_3COCl + C_6H_5OH \rightarrow CH_3COOC_6H_5 + HCl$
Particularly with equation in this second form, it is obvious that another ester is produced, in this case phenyl ethanoate. However, it this product is often drawn a variety of ways that make it appear as a derivative of phenol (which it is).
For example:
This form is useful when concentrating on the reactions of the phenol rather than the acyl chloride. In this form, it is apparent that the hydrogen of the phenol -OH group has been replaced by an acyl group, an alkyl group attached to a carbon-oxygen double bond. The phenol is said to have been acylated or to have undergone acylation. Because of the nature of this particular acyl group, it is also described as ethanoylation. The hydrogen is replaced by an ethanoyl group, CH3CO-.
Using a similar reaction to make aspirin
The reaction with phenol itself is not very important, but aspirin can be made in a similar reaction. Below is shown 2-hydroxybenzoic acid (also known as 2-hydroxybenzenecarboxylic acid). The old name for this compound is salicylic acid. This may be written in either of the following two ways. They are the same structure with the molecule flipped in space.
When this compound reacts with ethanoyl chloride, it is ethanoylated (or acylated, to use the more general term) to give,
This molecule is aspirin.
Contributors
Jim Clark (Chemguide.co.uk) | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Acid_Halides/Reactions_of_Acid_Halides/Reactions_of_Acyl_Chlorides_Involving_Oxygen_Compounds.txt |
This page gives you the facts and a simple, uncluttered mechanism for the nucleophilic addition / elimination reaction between acyl chlorides (acid chlorides) and alcohols.
Ethanoyl chloride is taken as a typical acyl chloride. Any other acyl chloride will behave in the same way. Simply replace the \(CH_3\) group in what follows by anything else you want. Similarly, ethanol is taken as a typical alcohol. If you are interested in another alcohol, you can replace the \(CH_3CH_2\) group by any other alkyl group.
The reaction between ethanoyl chloride and ethanol
Ethanoyl chloride reacts instantly with cold ethanol. There is a very exothermic reaction in which a steamy acidic gas is given off (hydrogen chloride). Ethyl ethanoate (an ester) is formed.
The mechanism
The first stage (the addition stage of the reaction) involves a nucleophilic attack on the fairly positive carbon atom by one of the lone pairs on the oxygen of an ethanol molecule.
The second stage (the elimination stage) happens in two steps. In the first, the carbon-oxygen double bond reforms and a chloride ion is pushed off.
That is followed by removal of a hydrogen ion by the chloride ion to give ethyl ethanoate and hydrogen chloride.
Contributors
Jim Clark (Chemguide.co.uk)
Reactions of Acyl Chlorides with Ammonia
This page gives you the facts and a simple, uncluttered mechanism for the nucleophilic addition / elimination reaction between acyl chlorides (acid chlorides) and ammonia. Ethanoyl chloride is taken as a typical acyl chloride. Any other acyl chloride will behave in the same way. Simply replace the CH3 group in what follows by anything else you want.
The reaction between ethanoyl chloride and ammonia
Ethanoyl chloride reacts violently with a cold concentrated solution of ammonia. A white solid product is formed which is a mixture of ethanamide (an amide) and ammonium chloride.
Notice that, unlike the reactions between ethanoyl chloride and water or ethanol, hydrogen chloride isn't produced - at least, not in any quantity. Any hydrogen chloride formed would immediately react with excess ammonia to give ammonium chloride.
$NH_3 + HCl \rightarrow NH_4Cl$
The Mechanism
The first stage (the addition stage of the reaction) involves a nucleophilic attack on the fairly positive carbon atom by the lone pair on the nitrogen atom in the ammonia.
The second stage (the elimination stage) happens in two steps. In the first, the carbon-oxygen double bond reforms and a chloride ion is pushed off.
That is followed by removal of a hydrogen ion from the nitrogen. This might happen in one of two ways:
It might be removed by a chloride ion, producing HCl (which would immediately react with excess ammonia to give ammonium chloride as above) . . .
and
$NH_3 +HCl \rightarrow NH_4Cl$
. . . or it might be removed directly by an ammonia molecule.
The ammonium ion, together with the chloride ion already there, makes up the ammonium chloride formed in the reaction.
Contributors
Jim Clark (Chemguide.co.uk)
Reactions of Acyl Chlorides with Primary Amines
This page gives you the facts and a simple, uncluttered mechanism for the nucleophilic addition / elimination reaction between acyl chlorides (acid chlorides) and amines.
• Ethanoyl chloride is taken as a typical acyl chloride. Any other acyl chloride will behave in the same way. Simply replace the CH3 group in what follows by anything else you want.
• Similarly, ethylamine is taken as a typical amine. Any other amine will behave in the same way. Replacing the CH3CH2 group by any other hydrocarbon group won't affect the mechanism in any way.
The reaction between ethanoyl chloride and ethylamine
Ethanoyl chloride reacts violently with a cold concentrated solution of ethylamine. A white solid product is formed which is a mixture of N-ethylethanamide (an N-substituted amide) and ethylammonium chloride.
Notice that, unlike the reactions between ethanoyl chloride and water or ethanol, hydrogen chloride isn't produced - at least, not in any quantity. Any hydrogen chloride formed would immediately react with excess ethylamine to give ethylammonium chloride.
$CH_3CH_2NH_2 + HCl \rightarrow CH_3CH_2NH_3Cl$
The mechanism
The first stage (the addition stage of the reaction) involves a nucleophilic attack on the fairly positive carbon atom by the lone pair on the nitrogen atom in the ethylamine.
The second stage (the elimination stage) happens in two steps. In the first, the carbon-oxygen double bond reforms and a chloride ion is pushed off.
That is followed by removal of a hydrogen ion from the nitrogen. This might happen in one of two ways:
It might be removed by a chloride ion, producing HCl (which would immediately react with excess ethylamine to give ethylammonium chloride as above) . . .
and
$CH_3CH_2NH_2 + HCl \rightrrow CH_3CH_2NH_3Cl$
. . . or it might be removed directly by an ethylamine molecule.
The ethylammonium ion, together with the chloride ion already there, makes up the ethylammonium chloride formed in the reaction.
Contributors
Jim Clark (Chemguide.co.uk)
Reactions of Acyl Chlorides with Water
This page gives you the facts and a simple, uncluttered mechanism for the nucleophilic addition / elimination reaction between acyl chlorides (acid chlorides) and water. Ethanoyl chloride is taken as a typical acyl chloride. Any other acyl chloride will behave in the same way. Simply replace the CH3 group in what follows by anything else you want.
The facts
Ethanoyl chloride reacts instantly with cold water. There is a very exothermic reaction in which a steamy acidic gas is given off (hydrogen chloride) and ethanoic acid is formed.
The mechanism
The first stage (the addition stage of the reaction) involves a nucleophilic attack on the fairly positive carbon atom by one of the lone pairs on the oxygen of a water molecule.
The second stage (the elimination stage) happens in two steps. In the first, the carbon-oxygen double bond reforms and a chloride ion is pushed off.
That is followed by removal of a hydrogen ion by the chloride ion to give ethanoic acid and hydrogen chloride.
Contributors
Jim Clark (Chemguide.co.uk) | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Acid_Halides/Reactions_of_Acid_Halides/Reactions_of_Acyl_Chlorides_with_Alcohols.txt |
This page discusses the reaction of acyl chlorides (acid chlorides) with benzene in the presence of an aluminum chloride catalyst. This is known as a Friedel-Crafts acylation.
Friedel-Crafts acylation of benzene
Acylation is the term given to substituting an acyl group such as CH3CO- onto another molecule. An acyl group is a hydrocarbon group attached to a carbon-oxygen double bond. The most common example of an acyl group is the ethanoyl group, CH3CO-, and this group is used throughout this article. If benzene is reacted with ethanoyl chloride in the presence of an aluminium chloride catalyst, the equation for the reaction is as follows:
This equation can be simplified as follows:
$C_6H_6 + CH_3COCl \rightarrow C_6H_6COCH_3 + HCl$
In the simplified formula for the product, the phenyl group is usually written on the left side and the alkyl group to the right of the carbon-oxygen double bond. The aluminum chloride is not written into these equations because it acts as a catalyst. It could be included as an AlCl3 over the top of the arrow. The product is called phenylethanone (old name: acetophenone).
Reaction conditions
Ethanoyl chloride is added carefully to a mixture of benzene and solid aluminium chloride cold conditions. Hydrogen chloride gas is given off. When all the ethanoyl chloride has been added, the mixture is heated under reflux at a temperature of 60°C for about 30 minutes to complete the reaction.
The importance of this reaction
Friedel-Crafts acylation is a very effective way of attaching a hydrocarbon-based group to a benzene ring. Although the product is a ketone (a compound containing a carbon-oxygen double bond with a hydrocarbon group either side), it is easily converted into other species. For example, the carbon-oxygen double bond can be reduced to give a secondary alcohol, which in turn can undergo a many other reactions.
Reduction of phenylethanone to produce ethylbenzene
This is known as the Clemmensen reduction, and involves heating the ketone with amalgamated zinc (a mixture of zinc and mercury) and concentrated hydrochloric acid for a long time.
This indirect route is the best way to attach an alkyl group to a benzene ring. It is possible to attach an alkyl group directly to the ring, but it is impossible to limit the reaction to one substitution. An alkyl group attached to the ring makes the ring more reactive than the original benzene, meaning like ethylbenzene (for example) reacts faster than benzene itself.
The result is that several ethyl groups are substituted around the ring, rather than just one. Acyl groups, however, have the opposite effect. Attaching an acyl group to the ring makes the ring so unreactive that a second one cannot be substituted.
Contributors
Jim Clark (Chemguide.co.uk)
Preparation of Acyl Chlorides
This page discusses the methods of swapping the -OH group in the -COOH group of a carboxylic acid for a chlorine atom to make acyl chlorides (acid chlorides). This section covers the use of phosphorus(V) chloride, phosphorus(III) chloride and sulfur dichloride oxide (thionyl chloride). In the examples below, consider the conversion of ethanoic acid to ethanoyl chloride to be typical of these types of reactions.
Replacing the -OH group using phosphorus(V) chloride, PCl5
Phosphorus(V) chloride is a solid that reacts with carboxylic acids under cold conditions to produce hydrogen chloride fumes. The reaction leaves a liquid mixture of acyl chloride and a phosphorus compound, phosphorus trichloride oxide (phosphorus oxychloride, POCl3). The acyl chloride can be separated by fractional distillation. The equation for this reaction is given below:
$3CH_3COOH + PCl_5 \rightarrow 3CH_3COCl + POCl_3 + HCl$
Replacing the -OH group using Phosphorous(III) chloride, PCl3
Phosphorus(III) chloride is a liquid at room temperature. Its reaction with a carboxylic acid is less dramatic than that of phosphorus(V) chloride because there is no hydrogen chloride produced. A mixture of the acyl chloride and phosphoric(III) acid is produced (old names: phosphorous acid or orthophosphorous acid), $H_3PO_3$. For example:
$3CH_3COOH + PCl_3 \rightarrow 3CH_3COCl + H_3PO_3$
As before, ethanoyl chloride can be separated by fractional distillation.
Replacing the -OH group using sulfur dichloride oxide (thionyl chloride)
Sulfur dichloride oxide (thionyl chloride) is a liquid at room temperature and has the formula $SOCl_2$. Traditionally, the formula is written as shown, although the modern name writes the chlorine before the oxygen. Sulfur dichloride oxide reacts with carboxylic acids to produce an acyl chloride, giving off sulfur dioxide and hydrogen chloride gases. For example:
$CH_3COOH + SOCl_2 \rightarrow CH_3COCl +SO_2 + HCl$
The separation is simplified to an extent because the by-products are both gases. Fractional distillation is still required to separate the acyl chloride from any excess acid or sulfur dichloride oxide.
Contributors
Jim Clark (Chemguide.co.uk) | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Acid_Halides/Reactions_of_Acid_Halides/Using_Acyl_Chlorides_in_Friedel-Crafts_Reactions.txt |
In the IUPAC system of nomenclature, functional groups are normally designated in one of two ways. The presence of the function may be indicated by a characteristic suffix and a location number. This is common for the carbon-carbon double and triple bonds which have the respective suffixes -ene and -yne. Halogens, on the other hand, do not have a suffix and are named as substituents, for example: (CH3)2C=CHCHClCH3 is 4-chloro-2-methyl-2-pentene.
Alcohols are usually named by the first procedure and are designated by an -ol suffix, as in ethanol, CH3CH2OH (note that a locator number is unnecessary on a two-carbon chain). On longer chains the location of the hydroxyl group determines chain numbering. For example: (CH3)2C=CHCH(OH)CH3 is 4-methyl-3-penten-2-ol. Other examples of IUPAC nomenclature are shown below, together with the common names often used for some of the simpler compounds. For the mono-functional alcohols, this common system consists of naming the alkyl group followed by the word alcohol. Alcohols may also be classified as primary, , secondary, , and tertiary, , in the same manner as alkyl halides. This terminology refers to alkyl substitution of the carbon atom bearing the hydroxyl group (colored blue in the illustration).
Many functional groups have a characteristic suffix designator, and only one such suffix (other than "-ene" and "-yne") may be used in a name. When the hydroxyl functional group is present together with a function of higher nomenclature priority, it must be cited and located by the prefix hydroxy and an appropriate number. For example, lactic acid has the IUPAC name 2-hydroxypropanoic acid.
Nomenclature of Alcohols
Alcohols are one of the most important functional groups in organic chemistry. Alcohols are a good source of reagents for synthesis reactions. The ability to identify alcohols is important especially when looking at IR and NMR spectra. The alcohol signal is very easy to spot on IR graphs, because they have a strong signal near the 3200 cm-1 region.
Introduction
The following is list of some common primary alcohols based on the IUPAC naming system.
Name
Molecular Formula
Methanol (methyl alcohol)
CH3OH
Ethanol (ethyl alcohol)
C2H5OH
Propanol
C3H7OH
Butanol
C4H9OH
Pentanol
C5H11OH
Hexanol
C6H13OH
Heptanol
C7H15OH
Octanol
C8H17OH
Rules for naming the alcohols
1. Find the longest chain containing the hydroxy group (OH). If there is a chain with more carbons than the one containing the OH group it will be named as a subsitutent.
2. Place the OH on the lowest possible number for the chain. With the exception of carbonyl groups such as ketones and aldehydes, the alcohol or hydroxy groups have first priority for naming.
3. When naming a cyclic structure, the -OH is assumed to be on the first carbon unless the carbonyl group is present, in which case the later will get priority at the first carbon.
4. When multiple -OH groups are on the cyclic structure, number the carbons on which the -OH groups reside.
5. Remove the final e from the parent alkane chain and add -ol. When multiple alcohols are present use di, tri, et.c before the ol, after the parent name. ex. 2,3-hexandiol. If a carbonyl group is present, the -OH group is named with the prefix "hydroxy," with the carbonyl group attached to the parent chain name so that it ends with -al or -one.
Examples
Ethane: CH3CH3 ----->Ethanol: (the alcohol found in beer, wine and other consumed sprits)
Secondary alcohol: 2-propanol
Other functional groups on an alcohol: 3-bromo-2-pentanol
Cyclic alcohol (two -OH groups): cyclohexan-1,4-diol
Other functional group on the cyclic structure: 3-hexeneol (the alkene is in bold and indicated by numbering the carbon closest to the alcohol)
A complex alcohol: 4-ethyl-3hexanol (the parent chain is in red and the substituent is in blue)
Problems
Name the following alchols:
1.
Contributors
• Abhiram Kondajji (UCD) | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alcohols/Nomenclature_of_Alcohols/Naming_Alcohols.txt |
Alcohols are some of the most important molecules in organic chemistry. They can be prepared from and converted into many different types of compounds. Alcohols contain the hydroxy functional group (-OH), bonded to a carbon atom of an alkyl or substituted alkyl group. The functional group of an alcohol is the hydroxyl group, –OH. Unlike the alkyl halides, this group has two reactive covalent bonds, the C–O bond and the O–H bond. The electronegativity of oxygen is substantially greater than that of carbon and hydrogen. Consequently, the covalent bonds of this functional group are polarized so that oxygen is electron rich and both carbon and hydrogen are electrophilic, as shown in the figure below.
Indeed, the dipolar nature of the O–H bond is such that alcohols are much stronger acids than alkanes (by roughly 1030 times), and nearly that much stronger than ethers. The most reactive site in an alcohol molecule is the hydroxyl group, despite the fact that the O–H bond strength is significantly greater than that of the C–C, C–H and C–O bonds, demonstrating again the difference between thermodynamic and chemical stability.
Properties of Alcohols
Alcohols are very weak Brønsted acids with pKa values generally in the range of 15 - 20. Because the hydroxyl proton is the most electrophilic site, proton transfer is the most important reaction to consider with nucleophiles. There are small differences in the acidities of aliphatic alcohols in aqueous solution, which are due to differences in structure and, more importantly, solvation.
General Assessment of Acidities
When considering alcohols as organic reagents, pKas are often used because they reflect reactivity in aqueous solution. In general, alcohols in aqueous solution are slightly less acidic than water. However, the differences among the pKas of the alcohols are not large. This is not surprising because all alcohols are oxy-acids (OH), and the differences in acidities are due to the effect of substituents in the 1-position removed from the acidic site. Moreover, the more highly substituted alcohols vary only in the structure two positions removed from the acidic site. The marginal effects of additional substituents at the carbon tow positions removed from the acidic site are even evident in the gas-phase enthalpies of reaction for the reaction
$ROH \rightarrow RO^- + H^+ \tag{1}$
The pKas and gas-phase enthalpies of reaction for various alcohols, ROH, with various substituents (R) are shown in Table 1 below.
R Name pKa1 $\Delta H_{acid}\; kJ/mol^2$
Table 1: pKas and gas-phase enthalpies of reaction
H water 14.0 1633.1
CH3 methanol 15.5 1597 ± 6
CH3CH2 ethanol 15.9 1587 ± 4
(CH3)2CH propan-2-ol (isopropyl alcohol) 16.5 1569 ± 4
(CH3)3C 2-methylpropan-2-ol (t-butanol) 17 1568 ± 4
C6H5 (phenyl) phenol 9.95 1462 ± 10
Interpretation of the Relative Acidities of Alcohols
There are many sites on the internet with explanations of the relative ordering of alcohol acidities in aqueous solution. The general explanation is that the larger substituents are better electron donors, which destabilize the resulting alkoxide anions. Because hydrogen is least donating of the substituents, water is the strongest acid. Unfortunately, although this belief persists, it is incomplete because it does not account for the gas-phase results. The problem with the electron donation explanation is that it suggests that the order of acidity is due solely to the intrinsic electronic effects of the substituents. However, if that were the case, the electron donating effect should also be evident in the gas-phase data. However, the relative acidities in the gas phase are opposite to those in aqueous solution. Consequently, any interpretation of the acidities of alcohols must take the gas phase data into account.
The inversion of the acidities of alcohols between the gas phase and aqueous solution was pointed out by Brauman and Blair in 1968.3 They proposed that the ordering of acidities of alcohols in solution is predominantly due to the combination of a) polarizibility and b) solvation, and that the electron donating ability of the substituent does not play a significant role.4
Polarizibility almost completely accounts for the trend in gas-phase acidities. As the size of the substituent increases, the acid becomes stronger due to the ability for the charge to be distributed over a larger volume, thereby reducing the charge density and, consequently, the Coulombic repulsion. Therefore, in the gas-phase, t-butanol is the most acidic alcohol, more acidic than isopropanol, followed by ethanol and methanol. In the gas phase, water is much less acidic than methanol, which is consistent with the difference in polarizibility between a proton and a methyl group. As before, the fact that water is less acidic than methanol in the gas phase is not consistent with the expected electronic donating capabilities of the two substituents. Given the absence of a solvent, the gas-phase properties reflect the instrinsic effects on the acidities.
In solution, however, the ions can be stabilized by solvation, and this is what leads to the inversion of acidity ordering. Brauman and Blair3 showed that smaller ions are better stabilized by solvation, which is consistent with the Born equation. Therefore, methanol is more acidic than t-butanol because the smaller methoxide ion has a shorter radius of solvation, leading to a larger solvation energy, which overcomes the stabilization that results from polarization of the charge. Because the solvation energy of hydroxide is even larger than that of methoxide, water is more acidic than methanol.
Note: Phenol
Discussions of acidities of alcohols usually include phenol in which the enhanced acidity is generally attributed to stabilization of the phenoxide ion by resonance delocalization. In this case, the gas-phase results agree with the solution trend that phenol is a much stronger acid than the aliphatic alcohols, and the difference is certainly due to electronic effects. However, this commonly encountered explanation is incomplete because it ignores the role that inductive effects have on acidities of oxy-acids. However, it is true that the acidity of phenol is much more a result of resonance stabilization than, for example, the acidities of carboxylic acids.
Practical Considerations
With pKa values in the 15.5 - 20.0 range, useful concentrations of alkoxides cannot be formed by proton transfer with hydroxide:
$ROH + OH^- \rightarrow RO^- + H_2O \tag{2}$
The equilibrium constant for the proton transfer reaction is on the order of 10-2-10-5. Phenoxide can be formed almost completely by reaction with aqueous alkaline base because the value of the equilibrium constant is roughly 104. The acidity of alcohols also indicates that it will react by proton transfer with any base more basic than hydroxide, which includes most organic bases, such as acetylide ions, cyanide, and vinyl/phenyl/alkyl anions. Therefore, alcohols will protonate most organic nucleophiles and effectively destroy most organometallic reagents, including Grignard or organolithium reagents.
Formation of alkoxide ions
Alkoxide ions can be formed by deprotonating alcohols with an extremely strong base such as an amide ion, NH2-. However, this method is rarely used. Alkoxides are more often formed by reaction of an alkali metal such as sodium with the pure alcohol:
$2ROH + 2Na \rightarrow 2RO^- + 2Na^+ + H_2 \tag{3}$
Conclusions
Relative acidities of all acids depend on many factors, including intrinsic electronic factors such as electronegativity, inductive and resonance effects, and polarizibility, as well as extrinsic factors such as solvation. For aliphatic alcohols, the most important effects are polarizibility and solvation, not electronic donation, as is generally assumed. In systems where the intrinsic factors are large, their effects are manifested in the overall properties, regardless of the medium. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alcohols/Properties_of_Alcohols/Acidities_of_Alcohols.txt |
This page defines an alcohol, and explains the differences between primary, secondary and tertiary alcohols. It examines in some detail their simple physical properties such as solubility and boiling points. Alcohols are compounds in which one or more hydrogen atoms in an alkane have been replaced by an -OH group. Alcohols fall into different classes depending on how the -OH group is positioned on the chain of carbon atoms. There are some chemical differences between the various types.
Primary alcohols
In a primary (1°) alcohol, the carbon atom that carries the -OH group is only attached to one alkyl group. Some examples of primary alcohols are shown below:
Notice that the complexity of the attached alkyl group is irrelevant. In each case there is only one linkage to an alkyl group from the CH2 group holding the -OH group. There is an exception to this. Methanol, CH3OH, is counted as a primary alcohol even though there are no alkyl groups attached to the the -OH carbon atom.
Secondary alcohols
In a secondary (2°) alcohol, the carbon atom with the -OH group attached is joined directly to two alkyl groups, which may be the same or different. Examples include the following:
Tertiary alcohols
In a tertiary (3°) alcohol, the carbon atom holding the -OH group is attached directly to three alkyl groups, which may be any combination of the same or different groups. Examples of tertiary alcohols are given below:
Physical properties of alcohols
Boiling Points
The chart below shows the boiling points of the following simple primary alcohols with up to 4 carbon atoms:
These boiling points are compared with those of the equivalent alkanes (methane to butane) with the same number of carbon atoms.
Notice that:
• The boiling point of an alcohol is always significantly higher than that of the analogous alkane.
• The boiling points of the alcohols increase as the number of carbon atoms increases.
The patterns in boiling point reflect the patterns in intermolecular attractions.
Hydrogen bonding
Hydrogen bonding occurs between molecules in which a hydrogen atom is attached to a strongly electronegative element: fluorine, oxygen or nitrogen. In the case of alcohols, hydrogen bonds occur between the partially-positive hydrogen atoms and lone pairs on oxygen atoms of other molecules.
The hydrogen atoms are slightly positive because the bonding electrons are pulled toward the very electronegative oxygen atoms. In alkanes, the only intermolecular forces are van der Waals dispersion forces. Hydrogen bonds are much stronger than these; therefore, more energy is required to separate alcohol molecules than to separate alkane molecules. This is the main reason for higher boiling points in alcohols.
Compound IUPAC Name Common Name Melting Poing (oC) Boiling Point (oC) Solubility in H2O at 23oC
Physical Properties of Alchols and Selected Analogous Haloalkanes and Alkanes
CH3OH Methanol Methyl alcohol -97.8 65.0 Infinite
CH3Cl Chloromethane Methyl chloride -97.7 -24.2 0.74 g/100 mL
CH4 Methane -182.5 -161.7 3.5 mL (gas)/ 100 mL
CH3CH2OH Ethanol Ethyl alcohol -114.7 78.5 Infinite
CH3CH2Cl Chloroethane Ethyl chloride -136.4 12.3 0.447 g/100 mL
CH3CH3 Ethane -183.3 -88.6 4.7 mL (gas)/ 100 mL
CH3CH2CH2OH 1-Propanol Propyl alcohol -126.5 97.4 Infinite
CH3CH2CH3 Propane -187.7 -42.1 6.5 mL (gas)/ 100 mL
CH3CH2CH2CH2OH 1-Butanol Butyl alcohol -89.5 117.3 8.0 g/100 mL
CH3(CH2)4OH 1-Pentanol Pentyl alcohol -79 138 2.2 g/100 mL
This table shows that alcohols (in red) have higher boiling points and greater solubility in H2O than haloalkanes and alkanes with the same number of carbons. It also shows that the boiling point of alcohols increase with the number of carbon atoms.
The effect of van der Waals forces
• Boiling points of alcohols: Hydrogen bonding is not the only intermolecular force alcohols experience. They also experience van der Waals dispersion forces and dipole-dipole interactions. The hydrogen bonding and dipole-dipole interactions are similar for all alcohols, but dispersion forces increase as the size of the alcohols increase. These attractions become stronger as the molecules lengthen and contain more electrons. This increases the sizes of the temporary dipoles formed. This is why the boiling points increase as the number of carbon atoms in the chains increases. It takes more energy to overcome the dispersion forces; thus, the boiling points rise.
• Comparison between alkanes and alcohols: Even without any hydrogen bonding or dipole-dipole interactions, the boiling point of the alcohol would be higher than the corresponding alkane with the same number of carbon atoms.
Compare ethane and ethanol:
Ethanol is a longer molecule, and the oxygen atom brings with it an extra 8 electrons. Both of these increase the size of the van der Waals dispersion forces, and subsequently the boiling point. A more accurate measurement of the effect of the hydrogen bonding on boiling point would be a comparison of ethanol with propane rather than ethane. The lengths of the two molecules are more similar, and the number of electrons is exactly the same.
Solubility of alcohols in water
Small alcohols are completely soluble in water; mixing the two in any proportion generates a single solution. However, solubility decreases as the length of the hydrocarbon chain in the alcohol increases. At four carbon atoms and beyond, the decrease in solubility is noticeable; a two-layered substance may appear in a test tube when the two are mixed.
Consider ethanol as a typical small alcohol. In both pure water and pure ethanol the main intermolecular attractions are hydrogen bonds.
In order to mix the two, the hydrogen bonds between water molecules and the hydrogen bonds between ethanol molecules must be broken. Energy is required for both of these processes. However, when the molecules are mixed, new hydrogen bonds are formed between water molecules and ethanol molecules.
The energy released when these new hydrogen bonds form approximately compensates for the energy needed to break the original interactions. In addition, there is an increase in the disorder of the system, an increase in entropy. This is another factor in deciding whether chemical processes occur. Consider a hypothetical situation involving 5-carbon alcohol molecules.
The hydrocarbon chains are forced between water molecules, breaking hydrogen bonds between those water molecules. The -OH ends of the alcohol molecules can form new hydrogen bonds with water molecules, but the hydrocarbon "tail" does not form hydrogen bonds. This means that many of the original hydrogen bonds being broken are never replaced by new ones.
In place of those original hydrogen bonds are merely van der Waals dispersion forces between the water and the hydrocarbon "tails." These attractions are much weaker, and unable to furnish enough energy to compensate for the broken hydrogen bonds. Even allowing for the increase in disorder, the process becomes less feasible. As the length of the alcohol increases, this situation becomes more pronounced, and thus the solubility decreases.
Uses of Alcohols
This page briefly discusses some of the important uses of simple alcohols, such as methanol, ethanol and propan-2-ol.
Alcoholic Drinks
The word "alcohol" in alcoholic drinks refers to ethanol (CH3CH2OH).
Industrial methylated spirits
Ethanol is usually sold as industrial methylated spirits, which is ethanol with a small quantity of methanol and possibly some color added. Because methanol is poisonous, industrial methylated spirits are unfit to drink, allowing purchasers to avoid the high taxes levied on alcoholic drinks.
Use of ethanol as a fuel
Ethanol burns to produce carbon dioxide and water, as shown in the equation below, and can be used as a fuel in its own right or in mixtures with petrol (gasoline). "Gasohol" is a petrol/ethanol mixture containing approximately 10–20% ethanol. Because ethanol can be produced by fermentation, this is a useful method for countries without an oil industry to reduce the amount of petrol imports.
$CH_3CH_2OH +3O_2 \rightarrow 2CO_2 + 3H_2O$
Ethanol as a solvent
Ethanol is widely used as a solvent. It is relatively safe and can be used to dissolve many organic compounds that are insoluble in water. It is used, for example, in many perfumes and cosmetics.
Methanol as a fuel
Methanol also burns to form carbon dioxide and water:
$2CH_3OH +3O_2 \rightarrow 2CO_2 + 4H_2O$
It can be used a a petrol additive to improve combustion, and its use as a fuel in its own right is under investigation.
Methanol as an industrial feedstock
Most methanol is used to make other compounds, for example, methanal (formaldehyde), ethanoic acid, and methyl esters of various acids. In most cases, these are then converted into further products. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alcohols/Properties_of_Alcohols/Background.txt |
The functional group of the alcohols is the hydroxyl group, –OH. Unlike the alkyl halides, this group has two reactive covalent bonds, the C–O bond and the O–H bond. The electronegativity of oxygen is substantially greater than that of carbon and hydrogen. Consequently, the covalent bonds of this functional group are polarized so that oxygen is electron rich and both carbon and hydrogen are electrophilic.
Reactivity of Alcohols
This page discusses the dehydration of alcohols in the lab to make alkenes—for example, dehydrating ethanol to produce ethene.
The dehydration of ethanol to give ethene
This is a simple method of making gaseous alkenes such as ethene. If ethanol vapor is passed over heated aluminum oxide powder, the ethanol is essentially cracked to yield ethene and water vapor.
To produce a few test tubes of ethene, the following apparatus can be used:
This system can be scaled up by boiling ethanol in a flask and passing the vapor over aluminum oxide that is heated in a long tube.
Dehydration of alcohols using an acid catalyst
The acid catalysts normally used in alcohol dehydration are either concentrated sulfuric acid or concentrated phosphoric(V) acid, H3PO4. Concentrated sulfuric acid produces messy results. Because sulfuric acid is also a strong oxidizing agent, it oxidizes some of the alcohol to carbon dioxide and is simultaneously reduced itself to sulfur dioxide. Both of these gases must be removed from the alkene. Sulfuric acid also reacts with the alcohol to produce a mass of carbon. There are other side reactions as well (not discussed here).
The dehydration of ethanol to yield ethene
In this process, ethanol is heated with an excess of concentrated sulfuric acid at a temperature of 170°C. The gases produced are passed through a sodium hydroxide solution to remove the carbon dioxide and sulfur dioxide produced from side reactions. The ethene is collected over water.
The concentrated sulfuric acid is a catalyst. Therefore, it is written over the reaction arrow rather than in the equation.
The dehydration of cyclohexanol to yield cyclohexene
This is a preparation commonly used to illustrate the formation and purification of a liquid product. The fact that the carbon atoms are joined in a ring has no bearing on the chemistry of the reaction. Cyclohexanol is heated with concentrated phosphoric(V) acid, and the liquid cyclohexene distils off and can be collected and purified. Phosphoric(V) acid tends to be used instead of sulfuric acid because it is safer and facilitates a less complex reaction.
The dehydration of more complicated alcohols
With more complicated alcohols, the formation of more than one alkene is possible. Butan-2-ol is a good example of this, with three different alkenes formed when it is dehydrated.
When an alcohol is dehydrated, the -OH group and a hydrogen atom from the next carbon atom in the chain are removed. With molecules like butan-2-ol, there are two possibilities for this.
The following products are formed:
The products are but-1-ene, CH2=CHCH2CH3, and but-2-ene, CH3CH=CHCH3.
This situation is further complicated by the fact that but-2-ene exhibits geometric isomerism; thus, a mixture of two isomers is formed: cis-but-2-ene and trans-but-2-ene.
The compoundcis-but-2-ene is also known as (Z)-but-2-ene; trans-but-2-ene is also known as (E)-but-2-ene. Which isomer is formed is a matter of chance.
Hence, the dehydration of butan-2-ol leads to a mixture containing the following compounds:
• but-1-ene
• cis-but-2-ene (also known as (Z)-but-2-ene)
• trans-but-2-ene (also known as (E)-but-2-ene)
Contributors
Jim Clark (Chemguide.co.uk)
Electrophilic Substitution at Oxygen
Because of its enhanced acidity, the hydrogen atom of a hydroxyl group is easily replaced by other substituents. A simple example is the facile reaction of simple alcohols with sodium (and sodium hydride), as described in the first equation below. Another such substitution reaction is the isotopic exchange that occurs when mixing an alcohol with deuterium oxide (heavy water). This exchange, which is catalyzed by acid or base, is rapid under normal conditions because it is difficult to avoid traces of these catalysts in most experimental systems.
2 R–O–H + 2 Na 2 R–O(–)Na(+) + H2
R–O–H + D2O R–O–D + D–O–H
The mechanism by which these substitution reactions proceed is straightforward. The oxygen atom of an alcohol is nucleophilic; therefore, it is prone to attack by electrophiles. The resulting "onium" intermediate then loses a proton to a base, forming the substitution product. If a strong electrophile is not present, then the nucleophilicity of the oxygen may be enhanced by conversion to its conjugate base (an alkoxide). This powerful nucleophile then attacks the weak electrophile. These two variations of the substitution mechanism are illustrated in the following diagram.
The preparation of tert-butyl hypochlorite from tert-butyl alcohol is an example of electrophilic halogenation of oxygen, but this reaction is restricted to 3º alcohols; 1º and 2º hypochlorites lose HCl, forming aldehydes and ketones. In the following equation the electrophile may be regarded as Cl(+).
(CH3)3C–O–H + Cl2 + NaOH (CH3)3C–O–Cl + NaCl + H2O
Alkyl substitution of the hydroxyl group creates ethers. This reaction provides examples of both strong electrophilic substitution (first equation below) and weak electrophilic substitution (second equation). The latter SN2 reaction is known as the Williamson ether synthesis and is generally used only with 1º alkyl halide reactants because the strong alkoxide base leads to E2 elimination of 2º and 3º alkyl halides.
Ester Formation
One of the most important substitution reactions at oxygen is ester formation, resulting from the reaction of alcohols with electrophilic derivatives of carboxylic and sulfonic acids. The following illustration displays the general formulas of these reagents and their ester products; the R'–O– group represents the alcohol moiety. The electrophilic atoms in the acid chlorides and anhydrides are colored red. Examples of specific esterification reactions are shown below. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alcohols/Reactivity_of_Alcohols/Dehydrating_Alcohols_to_Make_Alkenes.txt |
The discussion of alkyl halide reactions noted that 2º and 3º-alkyl halides experience rapid E2 elimination when treated with strong bases such as hydroxide and alkoxides. Alcohols do not undergo such base-induced elimination reactions and are, in fact, often used as solvents for such reactions. This is yet another example of how leaving-group stability influences the rate of a reaction. When an alcohol is treated with sodium hydroxide, the following acid-base equilibrium occurs. Most alcohols are slightly weaker acids than water, so the left side is favored.
R–O–H + Na(+) OH(–) R–O(–) Na(+) + H–OH
The elimination of water from an alcohol is called dehydration. Recalling that water is a much better leaving group than hydroxide ion, it is sensible to use acid-catalysis rather than base-catalysis in such reactions. Four examples of this useful technique are shown below. Note that hydrohalic acids (HX) are not normally used as catalysts because their conjugate bases are good nucleophiles and may create substitution products. The conjugate bases of sulfuric and phosphoric acids are not good nucleophiles, and do not participate in substitution under typical conditions.
The first two examples (in the top row) are typical, and the more facile elimination of the 3º-alcohol suggests the predominant E1 character for the reaction. This agrees with the tendency of branched 1º and 2º-alcohols to yield rearrangement products, as shown in the last example. The last two reactions also demonstrate that the Zaitsev rule applies to alcohol dehydrations, as well as to alkyl halide eliminations. Therefore, the more highly-substituted double bond isomer is favored among the products.
It should be noted that the acid-catalyzed dehydrations discussed here are the reverse of the acid-catalyzed hydration reactions of alkenes. Indeed, for these types of reversible reactions, the laws of thermodynamics require that the mechanism in both directions proceed by the same reaction path. This is known as the principle of microscopic reversibility. To illustrate, the following diagram lists the three steps in each transformation. The dehydration reaction is shown by the blue arrows; the hydration reaction by magenta arrows. The intermediates in these reactions are common to both, and common transition states are involved. This can be seen clearly in the energy diagrams depicted by clicking the button beneath the equations.
Base-induced E2 eliminations of alcohols may be achieved from sulfonate ester derivatives. This has the advantage of avoiding strong acids, which may cause molecular rearrangement or double bond migration in some cases. Because 3º-sulfonate derivatives are sometimes unstable, this procedure is best used with 1º and 2º-mesylates or tosylates. Application of this reaction sequence is shown here for 2-butanol. The Zaitsev rule favors the formation of 2-butene (cis + trans) over 1-butene.
CH3CH2CH(CH3)–OH CH3SO2Cl
CH3CH2CH(CH3)–OSO2CH3 C2H5O(–)Na(+)
CH3CH=CHCH3 + CH3CH2CH=CH2 + CH3SO2O(–) Na(+) + C2H5OH
The E2 elimination of 3º-alcohols under relatively non-acidic conditions may be accomplished by treatment with phosphorous oxychloride (POCl3) in pyridine. This procedure is also effective with hindered 2º-alcohols, but for unhindered and 1º-alcohols, an SN2 chloride ion substitution of the chlorophosphate intermediate competes with elimination. Examples of these and related reactions are given in the following figure. The first equation shows the dehydration of a 3º-alcohol. The predominance of the non-Zaitsev product (less substituted double bond) is presumed due to steric hindrance of the methylene group hydrogen atoms, which interferes with the approach of the base at that site. The second example shows two elimination procedures applied to the same 2º-alcohol. The first uses the single step POCl3 method, which works well in this case because SN2 substitution is impeded by steric hindrance. The second method is another example in which an intermediate sulfonate ester confers halogen-like reactivity on an alcohol. In every case, the anionic leaving group is the conjugate base of a strong acid.
Pyrolytic syn-Eliminations
Ester derivatives of alcohols may undergo unimolecular syn-elimination upon heating. To see examples of these, Click Here. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alcohols/Reactivity_of_Alcohols/Elimination_Reactions_of_Alcohols.txt |
Nucleophilic Substitution of the Hydroxyl Group
The chemical behavior of alkyl halides can be used as a reference in discovering analogous substitution and elimination reactions of alcohols. The chief difference, of course, is a change in the leaving anion from halide to hydroxide. Because oxygen is slightly more electronegative than chlorine (3.5 vs. 2.8 on the Pauling scale), the C-O bond is expected to be more polar than a C-Cl bond. Furthermore, an independent measure of the electrophilic characteristics of carbon atoms from their NMR chemical shifts (both 13C and alpha protons) indicates that oxygen and chlorine substituents exert a similar electron-withdrawing influence when bonded to sp3 hybridized carbon atoms. Despite this promising background evidence, alcohols do not undergo the same SN2 reactions commonly observed with alkyl halides. For example, the rapid SN2 reaction of 1-bromobutane with sodium cyanide, shown below, has no parallel when 1-butanol is treated with sodium cyanide. In fact, ethyl alcohol is often used as a solvent for alkyl halide substitution reactions such as this.
CH3CH2CH2CH2–Br + Na(+) CN(–) CH3CH2CH2CH2–CN + Na(+) Br(–)
CH3CH2CH2CH2–OH + Na(+) CN(–) No Reaction
The key factor here is the stability of the leaving anion (bromide vs. hydroxide). HBr is a much stronger acid than water (by more than 18 orders of magnitude), and this difference is reflected in reactions that generate their respective conjugate bases. The weaker base, bromide, is more stable, and its release in a substitution or elimination reaction is much more favorable than that of hydroxide ion, a stronger and less stable base.
A clear step toward improving the reactivity of alcohols in SN2 reactions would be to modify the –OH functional group in a way that improves its stability as a leaving anion. One such modification is to conduct the substitution reaction in a strong acid, converting –OH to –OH2(+). Because the hydronium ion (H3O(+)) is a much stronger acid than water, its conjugate base (H2O) is a better leaving group than hydroxide ion. The only problem with this strategy is that many nucleophiles, including cyanide, are deactivated by protonation in strong acids, effectively removing the nucleophilic co-reactant required for the substitution. The strong acids HCl, HBr and HI are not subject to this difficulty because their conjugate bases are good nucleophiles and are even weaker bases than alcohols. The following equations illustrate some substitution reactions of alcohols that may be affected by these acids. As with alkyl halides, the nucleophilic substitution of 1º-alcohols proceeds by an SN2 mechanism, whereas 3º-alcohols react by an SN1 mechanism. Reactions of 2º-alcohols may occur by both mechanisms and often produce some rearranged products. The numbers in parentheses next to the mineral acid formulas represent the weight percentage of a concentrated aqueous solution, the form in which these acids are normally used.
CH3CH2CH2CH2–OH + HBr (48%) CH3CH2CH2CH2–OH2(+) Br(–) CH3CH2CH2CH2Br + H2O SN2
(CH3)3C–OH + HCl (37%) (CH3)3C–OH2(+) Cl(–) (CH3)3C(+) Cl(–) + H2O (CH3)3C–Cl + H2O SN1
Although these reactions are sometimes referred to as "acid-catalyzed," this is not strictly correct. In the overall transformation, a strong HX acid is converted to water, a very weak acid, so at least a stoichiometric quantity of HX is required for a complete conversion of alcohol to alkyl halide. The necessity of using equivalent quantities of very strong acids in this reaction limits its usefulness to simple alcohols of the type shown above. Alcohols with acid-sensitive groups do not, of course, tolerate such treatment. Nevertheless, the idea of modifying the -OH functional group to improve its stability as a leaving anion can be pursued in other directions. The following diagram shows some modifications that have proven effective. In each case the hydroxyl group is converted to an ester of a strong acid. The first two examples show the sulfonate esters described earlier. The third and fourth examples show the formation of a phosphite ester (X represents the remaining bromines or additional alcohol substituents) and a chlorosulfite ester, respectively. All of these leaving groups (colored blue) have conjugate acids that are much stronger than water (by 13 to 16 powers of ten); thus, the leaving anion is correspondingly more stable than the hydroxide ion. The mesylate and tosylate compounds are particularly useful because they may be used in substitution reactions with a wide variety of nucleophiles. The intermediates produced in reactions of alcohols with phosphorus tribromide and thionyl chloride (last two examples) are seldom isolated, and these reactions continue to produce alkyl bromide and chloride products.
The importance of sulfonate ester intermediates in general nucleophilic substitution reactions of alcohols may be illustrated by the following conversion of 1-butanol to pentanenitrile (butyl cyanide), a reaction that does not occur with the alcohol alone. The phosphorus and thionyl halides, on the other hand, only act to convert alcohols to the corresponding alkyl halides.
CH3CH2CH2CH2–OH + CH3SO2Cl pyridine
CH3CH2CH2CH2–OSO2CH3 Na(+) CN(–)
CH3CH2CH2CH2CN + CH3SO2O(–) Na(+)
Some examples of alcohol substitution reactions using this approach to activating the hydroxyl group are shown in the following diagram. The first two cases serve to reinforce the fact that sulfonate ester derivatives of alcohols may replace alkyl halides in a variety of SN2 reactions. The next two cases demonstrate the use of phosphorus tribromide in converting alcohols to bromides. This reagent may be used without added base (e.g. pyridine) because the phosphorous acid product is a weaker acid than HBr. Phosphorus tribromide is best used with 1º-alcohols because 2º-alcohols often yield rearrangement by-products resulting from competing SN1 reactions. Note that the ether oxygen in reaction 4 is not affected by this reagent, whereas the alternative synthesis using concentrated HBr cleaves ethers. Phosphorus trichloride (PCl3) converts alcohols to alkyl chlorides in a similar manner, but thionyl chloride is usually preferred for this transformation because the inorganic products are gases (SO2 & HCl). Phosphorus triiodide is not stable but may be generated in situ from a mixture of red phosphorus and iodine and acts to convert alcohols to alkyl iodides. The last example shows the reaction of thionyl chloride with a chiral 2º-alcohol. The presence of an organic base such as pyridine is important because it provides a substantial concentration of chloride ion required for the final SN2 reaction of the chlorosufite intermediate. In the absence of a base, chlorosufites decompose upon heating to yield the expected alkyl chloride with retention of configuration
Tertiary alcohols are not commonly used for substitution reactions of the type discussed here because SN1 and E1 reaction paths are dominant and are difficult to control.
The importance of sulfonate esters as intermediates in many substitution reactions cannot be overstated. A rigorous proof of the configurational inversion that occurs at the substitution site in SN2 reactions makes use of such reactions. An example of such a proof is displayed below. Abbreviations for the more commonly used sulfonyl derivatives are given in the following table.
Sulfonyl Group CH3SO2 CH3C6H4SO2 BrC6H4SO2 CF3SO2
Name & Abbrev. Mesyl or Ms Tosyl or Ts Brosyl or Bs Trifyl or Tf
Inversion Proof
For a more complete discussion of hydroxyl substitution reactions and a description of other selective methods for this transformation, Click Here.
Hydroxyl Group Substitution
The most common methods for converting 1º- and 2º-alcohols to the corresponding chloro and bromo alkanes (i.e. replacement of the hydroxyl group) are treatments with thionyl chloride and phosphorus tribromide, respectively. These reagents are generally preferred over the use of concentrated HX due to the harsh acidity of these hydrohalic acids and the carbocation rearrangements associated with their use.
Of course, it is possible to avoid such problems by first preparing a mesylate or tosylate derivative, followed by nucleophilic substitution of the sulfonate ester by the appropriate halide anion. In this two-step approach, a clean configurational inversion occurs in the first SN2 reaction; however, the resulting alkyl halide may then undergo repeated SN2 halogen exchange reactions, thus destroying any stereoisomeric identity held by the initial carbinol carbon. For these and other reasons, alternative mild and selective methods for transforming such alcohols by nucleophilic substitution of the hydroxyl group have been devised. It should be noted that 3º-alcohols are not good substrates for the new procedures.
Disadvantages to using \(PBr_3\) and \(SOCl_2\)
Despite their general usefulness, phosphorous tribromide and thionyl chloride have shortcomings. Hindered 1º- and 2º-alcohols react sluggishly with the former and may form rearrangement products, as noted in the following equation.
Below, an abbreviated mechanism for the reaction is displayed. The initially formed trialkylphosphite ester may be isolated if the HBr byproduct is scavenged by a base. In the presence of HBr, a series of acid-base and SN2 reactions occur, along with the transient formation of carbocation intermediates. Rearrangement (pink arrows) of the carbocations leads to isomeric products.
The reaction of thionyl chloride with chiral 2º-alcohols has been observed to proceed with either inversion or retention. In the presence of a base such as pyridine, the intermediate chlorosulfite ester reacts to form a "pyridinium" salt, which undergoes a relatively clean SN2 reaction to produce the inverted chloride. In ether and similar solvents, the chlorosulfite reacts with retention of configuration, presumably by way of a tight or intimate ion pair. This is classified as an SNi reaction (nucleophilic substitution internal). The carbocation partner in the ion pair may also rearrange. These reactions are illustrated by the following equations. An alternative explanation for the retention of configuration, involving an initial solvent molecule displacement of the chlorosulfite group (as SO2 and chloride anion), followed by chloride ion displacement of the solvent moiety, has been suggested. In this case, two inversions lead to retention.
Another characteristic of thionyl chloride reactions is their tendency to yield allylic rearrangement products with allylic alcohols. This fact is demonstrated by the following equations. Reactions of this type have been classified as SNi', where the prime mark indicates an allylic characteristic to the internal substitution. They may also be considered retro-ene reactions, a special class of pericyclic reactions.
A similar substitutive rearrangement also occurs with propargyl alcohols, as shown below.
Hypervalent Phosphorous Reagents
The ability of phosphorous to assume many different valencies or oxidation states was noted elsewhere. The nucleophilicity of trialkyl phosphines allows them to bond readily to electrophiles, and the resulting phosphonium ions may then bond reversibly to other nucleophiles, especially oxygen nucleophiles. The use of phosphorous ylides in the Wittig reaction is an example of this reactivity.
The reaction of triphenylphosphine with halogens further illustrates this hypervalency. As shown in the following diagram, triphenylphosphine (yellow box on the left) reacts to form a pentavalent dihalide, which is in equilibrium with its ionic components in solution.
Chemists have used these and similar reagents to achieve the mild conversion of alcohols to alkyl halides with clean inversion of configuration. As with other OH substitution reactions, an inherently poor leaving group (hydroxide anion) is modified to provide a better leaving group, the stable compound triphenylphosphine oxide. Two such reactions are shown in the following diagram. Using this approach, even sluggish alcohols that are prone to rearrangement (e.g. neopentyl alcohol) are converted to their corresponding halides.
The instability of vicinal diiodides relative to their double bond analogs is the driving force for a novel transformation of vic-glycols to their corresponding alkenes. An example is displayed below.
The allylic rearrangement observed in thionyl chloride is similarly avoided by using triphenylphosphine dichloride, or alternatively, by a two-step procedure via a sulfonate ester. This is displayed below.
The Mitsunobu Reaction
The Japanese chemist, O.Mitsunobu, devised a general and exceptionally versatile variant of hypervalent phosphorous chemistry that has been applied to a wide selection of alcohols. This method, which now carries his name, uses a reagent mixture consisting of triphenylphosphine, diethyl azodicarboxylate (DEAD) and a moderate to strong acid. The steps leading to hydroxyl substitution are outlined in the following diagram. It should be noted that the nucleophile involved in the final SN2 substitution may be the conjugate base of the acid component or a separate species.
A common use of the Mitsunobu reaction is to invert the configuration of a 2º-alcohol. This application usually employs benzoic acid or a benzoate salt, and the resulting configurationally inverted ester is then hydrolyzed to the epimeric alcohol. An example of this procedure is displayed below.
In addition to achieving configurational inversion of carbinol sites, the Mitsunobu reaction has also been used to introduce the azide precursor of amines and for the intramolecular preparation of cyclic ethers. Examples are shown below. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alcohols/Reactivity_of_Alcohols/Hydroxyl_Group_Substitution/Additional_Methods_of_Hydroxyl_Substitution.txt |
When alcohols react with a hydrogen halide, a substitution occurs, producing an alkyl halide and water:
Scope of Reaction
• The order of reactivity of alcohols is 3° > 2° > 1° methyl.
• The order of reactivity of the hydrogen halides is HI > HBr > HCl (HF is generally unreactive).
The reaction is acid catalyzed. Alcohols react with the strongly acidic hydrogen halides HCl, HBr, and HI, but they do not react with nonacidic NaCl, NaBr, or NaI. Primary and secondary alcohols can be converted to alkyl chlorides and bromides by allowing them to react with a mixture of a sodium halide and sulfuric acid:
Mechanisms of the Reactions of Alcohols with HX
Secondary, tertiary, allylic, and benzylic alcohols appear to react by a mechanism that involves the formation of a carbocation in an \(S_N1\) reaction with the protonated alcohol acting as the substrate.
The \(S_N1\) mechanism is illustrated by the reaction tert-butyl alcohol and aqueous hydrochloric acid (\(H_3O^+\), \(Cl^-\) ). The first two steps in this \(S_n1\) substitution mechanism are protonation of the alcohol to form an oxonium ion. Although the oxonium ion is formed by protonation of the alcohol, it can also be viewed as a Lewis acid-base complex between the cation (\(R^+\)) and \(H_2O\). Protonation of the alcohol converts a poor leaving group (OH-) to a good leaving group (\)H_2O\_), which makes the dissociation step of the \(S_N1\) mechanism more favorable.
In step 3, the carbocation reacts with a nucleophile (a halide ion) to complete the substitution.
When we convert an alcohol to an alkyl halide, we perform the reaction in the presence of acid and in the presence of halide ions and not at elevated temperatures. Halide ions are good nucleophiles (they are much stronger nucleophiles than water), and because halide ions are present in a high concentration, most of the carbocations react with an electron pair of a halide ion to form a more stable species, the alkyl halide product. The overall result is an \(S_n1\) reaction.
Primary Alcohols
Not all acid-catalyzed conversions of alcohols to alkyl halides proceed through the formation of carbocations. Primary alcohols and methanol react to form alkyl halides under acidic conditions by an SN2 mechanism.
In these reactions, the function of the acid is to produce a protonated alcohol. The halide ion then displaces a molecule of water (a good leaving group) from carbon; this produces an alkyl halide:
Again, acid is required. Although halide ions (particularly iodide and bromide ions) are strong nucleophiles, they are not strong enough to carry out substitution reactions with alcohols themselves. Direct displacement of the hydroxyl group does not occur because the leaving group would have to be a strongly basic hydroxide ion:
We can see now why the reactions of alcohols with hydrogen halides are acid-promoted.
The role of acid catalysis
Acid protonates the alcohol hydroxyl group, making it a good leaving group. However, other strong Lewis acids can be used instead of hydrohalic acids. Because the chloride ion is a weaker nucleophile than bromide or iodide ions, hydrogen chloride does not react with primary or secondary alcohols unless zinc chloride or a similar Lewis acid is added to the reaction mixture as well. Zinc chloride, a good Lewis acid, forms a complex with the alcohol through association with an unshared pair of electrons on the oxygen atom. This enhances the hydroxyl’s leaving group potential sufficiently so that chloride can displace it.
Rearrangement
As we might expect, many reactions of alcohols with hydrogen halides, particularly those in which carbocations are formed, are accompanied by rearrangements. The general rule is that if rearrangement CAN OCCUR (to form more stable or equally stable cations), it will! In these reactions, mixtures of products can be formed.
Reduction of Alcohols
Because the most electrophilic site of an alcohol is the hydroxyl proton and because OH- is a poor leaving group, alcohols do not undergo substitution reactions with nucleophiles. However, they can be converted into tosylates or other sulfate esters, which have very good (sulfate) leaving groups that are readily displaced in substitution reactions.
Nucleophiles that displace tosylate groups include hydride ions, H-. Therefore, tosylates formed from alcohols undergo nucleophilic substitution reacctions with hydride sources, such as lithium aluminum hydride ($LiAlH_4$, aka $LAH$). The net result of the process is the reduction of alcohols to alkanes.
$ROH + TsCl \rightarrow ROTs$
$ROTs + LiAlH_4 \rightarrow RH$ | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alcohols/Reactivity_of_Alcohols/Reactions_of_alcohols_with_hydrohalic_acids_%28HX%29.txt |
This page looks at reactions in which the -OH group in an alcohol is replaced by a halogen such as chlorine or bromine. It includes a simple test for an -OH group using phosphorus(V) chloride. The general reaction looks like this:
$ROH + HX \rightarrow RX + H_2O$
Reaction with hydrogen chloride
Tertiary alcohols react reasonably rapidly with concentrated hydrochloric acid, but for primary or secondary alcohols the reaction rates are too slow for the reaction to be of much importance. A tertiary alcohol reacts if it is shaken with with concentrated hydrochloric acid at room temperature. A tertiary halogenoalkane (haloalkane or alkyl halide) is formed.
Replacing -OH by bromine
Rather than using hydrobromic acid,the alcohol is typically treated with a mixture of sodium or potassium bromide and concentrated sulfuric acid. This produces hydrogen bromide, which reacts with the alcohol. The mixture is warmed to distil off the bromoalkane.
$CH_3CH_2OH + HBr \rightarrow CH_3CH_2Br + H_2O \label{1.1.2}$
Replacing -OH by iodine
In this case, the alcohol is reacted with a mixture of sodium or potassium iodide and concentrated phosphoric(V) acid, H3PO4, and the iodoalkane is distilled off. The mixture of the iodide and phosphoric(V) acid produces hydrogen iodide, which reacts with the alcohol.
$CH_3CH_2OH + HI \rightarrow CH_3CH_2I + H_2O \label{1.1.3}$
Phosphoric(V) acid is used instead of concentrated sulfuric acid because sulfuric acid oxidizes iodide ions to iodine and produces hardly any hydrogen iodide. A similar phenomenon occurs to some extent with bromide ions in the preparation of bromoalkanes but not enough to interfere with the main reaction. There is no reason why you could not use phosphoric(V) acid in the bromide case instead of sulfuric acid if desired.
Reacting Alcohols with Phosphorus Halides
Alcohols react with liquid phosphorus(III) chloride (also called phosphorus trichloride) to yield chloroalkanes.
$3CH_3CH_2CH_2OH + PCl_3 \rightarrow 3CH_3CH_2CH_2Cl + H_3PO_3 \label{1.1.3a}$
Alcohols also violently react with solid phosphorus(V) chloride (phosphorus pentachloride) at room temperature, producing clouds of hydrogen chloride gas. While it is not a good approach to make chloroalkanes, it is a good test for the presence of -OH groups. To show that a substance was an alcohol, you would first have to eliminate all the other groups that also react with phosphorus(V) chloride. For example, carboxylic acids (containing the -COOH group) also react with it (because of the -OH in -COOH) as does water (H-OH).
If you have a neutral liquid not contaminated with water, and clouds of hydrogen chloride are produced when you add phosphorus(V) chloride, then you have an alcohol group present.
$CH_3CH_2CH_2OH + PCl_5 \rightarrow CH_3CH_2CH_2Cl + POCl_3 + HCl \label{1.1.4}$
There are also side reactions involving the $POCl_3$ reacting with the alcohol.
Other reactions involving phosphorus halides
Instead of using phosphorus(III) bromide or iodide, the alcohol is usually heated under reflux with a mixture of red phosphorus and either bromine or iodine. The phosphorus first reacts with the bromine or iodine to give the phosphorus(III) halide.
$2P_{(s)} + 3Br_2 \rightarrow 2PBr_3\label{1.1.5}$
$2P_{(s)} + 3I_2 \rightarrow 2PI_3 \label{1.1.6}$
These then react with the alcohol to give the corresponding halogenoalkane, which can be distilled off.
$3CH_3CH_2CH_2OH + PBr_3 \rightarrow 3CH_3CH_2CH_2Br + H_3PO_3 \label{1.1.7}$
$3CH_3CH_2CH_2OH + PI_3 \rightarrow 3CH_3CH_2CH_2I + H_3PO_3 \label{1.1.8}$
Reacting alcohols with Thionyl Chloride
Sulfur dichloride oxide (thionyl chloride) has the formula SOCl2. Traditionally, the formula is written as shown, despite the fact that the modern name writes the chlorine before the oxygen (alphabetical order). The sulfur dichloride oxide reacts with alcohols at room temperature to produce a chloroalkane. Sulfur dioxide and hydrogen chloride are given off. Care would have to be taken because both of these are poisonous.
$CH_3CH_2CH_2OH + SOCl_2 \rightarrow CH_3CH_2CH_2Cl + SO_2 + HCl \label{1.1.9}$
The advantage that this reaction has over the use of either of the phosphorus chlorides is that the two other products of the reaction (sulfur dioxide and HCl) are both gases. That means that they separate themselves from the reaction mixture. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alcohols/Reactivity_of_Alcohols/Replacing_the_OH_Group_by_Halogen_Atoms.txt |
This page looks at the oxidation of alcohols using acidified sodium or potassium dichromate(VI) solution. This reaction is used to make aldehydes, ketones and carboxylic acids, and as a way of distinguishing between primary, secondary and tertiary alcohols.
Oxidizing the different types of alcohols
The oxidizing agent used in these reactions is normally a solution of sodium or potassium dichromate(VI) acidified with dilute sulfuric acid. If oxidation occurs, then the orange solution containing the dichromate(VI) ions is reduced to a green solution containing chromium(III) ions. The electron-half-equation for this reaction is as follows:
$Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$
Primary alcohols
Primary alcohols can be oxidized to either aldehydes or carboxylic acids, depending on the reaction conditions. In the case of the formation of carboxylic acids, the alcohol is first oxidized to an aldehyde, which is then oxidized further to the acid.
An aldehyde is obtained if an excess amount of the alcohol is used, and the aldehyde is distilled off as soon as it forms. An excess of the alcohol means that there is not enough oxidizing agent present to carry out the second stage, and removing the aldehyde as soon as it is formed means that it is not present to be oxidized anyway!
If you used ethanol as a typical primary alcohol, you would produce the aldehyde ethanal, $CH_3CHO$. The full equation for this reaction is fairly complicated, and you need to understand the electron-half-equations in order to work it out.
$3CH_3CH_2OH + Cr_2O_7^{2-} + 8H^+ \rightarrow 3CH_3CHO + 2Cr^{3+} + 7H_2O$
In organic chemistry, simplified versions are often used that concentrate on what is happening to the organic substances. To do that, oxygen from an oxidizing agent is represented as $[O]$. That would produce the much simpler equation:
It also helps in remembering what happens. You can draw simple structures to show the relationship between the primary alcohol and the aldehyde formed.
Full oxidation to carboxylic acids
An excess of the oxidizing agent must be used, and the aldehyde formed as the half-way product should remain in the mixture. The alcohol is heated under reflux with an excess of the oxidizing agent. When the reaction is complete, the carboxylic acid is distilled off. The full equation for the oxidation of ethanol to ethanoic acid is as follows:
$3CH_3CH_2OH + 2Cr_2O_7^{2-} + 16H+ \rightarrow 3CH_3COOH + 4Cr^{3+} + 11H_2O$
The more typical simplified version looks like this:
$CH_3CH_2OH + 2[O] \rightarrow CH_3COOH + H_2O$
Alternatively, you could write separate equations for the two stages of the reaction - the formation of ethanal and then its subsequent oxidation.
$CH_3CH_2OH + [O] \rightarrow CH_3CHO + H_2O$
$CH_3CHO + [O] \rightarrow CH_3COOH$
This is what is happening in the second stage:
Secondary alcohols
Secondary alcohols are oxidized to ketones - and that's it. For example, if you heat the secondary alcohol propan-2-ol with sodium or potassium dichromate(VI) solution acidified with dilute sulfuric acid, propanone is formed. Changing the reaction conditions makes no difference to the product. Folloiwng is the simple version of the equation, showing the relationship between the structures:
If you look back at the second stage of the primary alcohol reaction, you will see that an oxygen inserted between the carbon and the hydrogen in the aldehyde group to produce the carboxylic acid. In this case, there is no such hydrogen - and the reaction has nowhere further to go.
Tertiary alcohols
Tertiary alcohols are not oxidized by acidified sodium or potassium dichromate(VI) solution - there is no reaction whatsoever. If you look at what is happening with primary and secondary alcohols, you will see that the oxidizing agent is removing the hydrogen from the -OH group, and a hydrogen from the carbon atom is attached to the -OH. Tertiary alcohols don't have a hydrogen atom attached to that carbon.
You need to be able to remove those two particular hydrogen atoms in order to set up the carbon-oxygen double bond.
Using these reactions as a test for the different types of alcohols
First, the presence of an alcohol must be confirmed by testing for the -OH group. The liquid would need to be verified as neutral, free of water and that it reacted with solid phosphorus(V) chloride to produce a burst of acidic steamy hydrogen chloride fumes. A few drops of the alcohol would be added to a test tube containing potassium dichromate(VI) solution acidified with dilute sulfuric acid. The tube would be warmed in a hot water bath.
Determining the tertiary alcohol
In the case of a primary or secondary alcohol, the orange solution turns green. With a tertiary alcohol, there is no color change. After heating, the following colors are observed:
Distinguishing between the primary and secondary alcohols
A sufficient amount of the aldehyde (from oxidation of a primary alcohol) or ketone (from a secondary alcohol) must be produced to be able to test them. There are various reactions that aldehydes undergo that ketones do not. These include the reactions with Tollens' reagent, Fehling's solution and Benedict's solution, and these reactions are covered on a separate page.
These tests can be difficult to carry out, and the results are not always as clear-cut as the books say. A much simpler but fairly reliable test is to use Schiff's reagent. Schiff's reagent is a fuchsin dye decolorized by passing sulfur dioxide through it. In the presence of even small amounts of an aldehyde, it turns bright magenta.
It must, however, be used absolutely cold, because ketones react with it very slowly to give the same color. If you heat it, obviously the change is faster - and potentially confusing. While you are warming the reaction mixture in the hot water bath, you can pass any vapors produced through some Schiff's reagent.
• If the Schiff's reagent quickly becomes magenta, then you are producing an aldehyde from a primary alcohol.
• If there is no color change in the Schiff's reagent, or only a trace of pink color within a minute or so, then you are not producing an aldehyde; therefore, no primary alcohol is present.
Because of the color change to the acidified potassium dichromate(VI) solution, you must, therefore, have a secondary alcohol. You should check the result as soon as the potassium dichromate(VI) solution turns green - if you leave it too long, the Schiff's reagent might start to change color in the secondary alcohol case as well.
The Oxidation of Alcohols
One of the reagents that is commonly used for oxidation in organic chemistry is chromic acid. This reagent is straightforward to use once deciphered. However, there are a vast number of different ways that textbooks (and instructors) show it being used in reactions.
Chromic acid, H2CrO4, is a strong acid and a reagent for oxidizing alcohols to ketones and carboxylic acids. For fairly mundane reasons owing primarily to safety and convenience, chromic acid tends to be made in the reaction vessel as needed (through addition of acid to a source of chromium), rather than being dispensed from a bottle.
And that’s where the trouble begins. Choosing a source of chromium to make H2CrO4 from is a lot like choosing a favorite brand of bottled water. Beyond the packaging, they’re pretty much all the same. Depending on which textbook or instructor you have, however, you might see several different ways to do this, and it can be very confusing.
The key point is that Na2CrO4 (sodium chromate), Na2Cr2O7 (sodium dichromate), K2CrO4(potassium chromate), K2Cr2O7 (potassium dichromate), and CrO3 (chromium trioxide) are all alike in one crucial manner: when they are combined with aqueous acid, each of them forms H2CrO4, and ultimately it’s H2CrO4 which does the important chemistry. Unfortunately I rarely see this point explained in textbooks. I remember this causing some confusion for me when I took the course. The K or Na ions present are just spectators.
Once H2CrO4 is formed, its reactions are pretty straightforward: it converts primary alcohols (and aldehydes) to carboxylic acids and secondary alcohols to ketones.
It does this through addition of the alcohol oxygen to chromium, which makes it a good leaving group; a base (water being the most likely culprit) can then remove a proton from the carbon, forming a new π bond and breaking the O-Cr bond.
Due to its high toxicity, chromic acid tends to find very little use in the organic chemistry laboratory outside of undergrad labs. There are far more useful reagents out there for performing these transformations.
Oxidation by PCC (pyridinium chlorochromate)
Pyridinium chlorochromate (PCC) is a milder version of chromic acid.
PCC oxidizes alcohols one rung up the oxidation ladder, from primary alcohols to aldehydes and from secondary alcohols to ketones. In contrast to chromic acid, PCC will not oxidize aldehydes to carboxylic acids. Similar to or the same as: \(CrO_3\) and pyridine (the Collins reagent) will also oxidize primary alcohols to aldehydes. Here are two examples of PCC in action.
• If you add one equivalent of PCC to either of these alcohols, the oxidized version will be produced. The byproducts (featured in grey) are Cr(IV) as well as pyridinium hydrochloride.
• One has to be careful with the amount of water present in the reaction. If water is present, it can add to the aldehyde to create the hydrate, which could be further oxidized by a second equivalent of PCC if it is present. This is not a concern with ketones because there is no H directly bonded to C.
How does it work? Oxidation reactions of this sort are actually a type of elimination reaction. The reaction starts with a carbon-oxygen single bond and results in a carbon-oxygen double bond. The elimination reaction can occur because of the good leaving group on the oxygen, namely the chromium, which will be displaced when the neighboring C-H bond is broken with a base.
The first step is attack of oxygen on the chromium to form the Cr-O bond. Secondly, a proton on the (now positive) OH is transferred to one of the oxygens of the chromium, possibly through the intermediacy of the pyridinium salt. A chloride ion is then displaced in a reaction reminiscent of a 1,2 elimination reaction to form what is known as a chromate ester.
The C-O double bond is formed when a base removes the proton on the carbon adjacent to the oxygen. [aside: I've drawn the base as Cl(-) although there are certainly other species which could also act as bases here (such as an alcohol). It is also possible for pyridine to be used as the base here, although only very low concentrations of the deprotonated form will be present under these acidic conditions.] The electrons from the C-H bond move to form the C-O bond, and in the process, the O-Cr bond is broken, and Cr(VI) becomes Cr(IV) (drawn here as O=Cr(OH)2 ).
Real life notes: If you end up using PCC in the lab, do not forget to add molecular sieves or Celite or some other solid to the bottom of the flask because otherwise you get a nasty brown tar that is difficult to clean up. The toxicity and mess associated with chromium has spurred the development of other alternatives, such as TPAP, IBX, DMP, and a host of other reagents you generally do not typically learn about until grad school. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alcohols/Reactivity_of_Alcohols/The_Oxidation_of_Alcohols/Oxidation_by_Chromic_Acid.txt |
This page describes the reaction between alcohols and metallic sodium,and introduces the properties of the alkoxide that is formed. We will look at the reaction between sodium and ethanol as being typical, but you could substitute any other alcohol and the reaction would be the same.
The Reaction between Sodium Metal and Ethanol
If a small piece of sodium is dropped into ethanol, it reacts steadily to give off bubbles of hydrogen gas and leaves a colorless solution of sodium ethoxide: $CH_3CH_2ONa$. The anion component is an alkoxide.
$2CH_3CH_2OH_{(l)} + 2Na_{(s)} \rightarrow 2CH_3CH_2O^-_{(aq)} + 2Na^+_{(aq)} + H_{2(g)}$
If the solution is evaporated carefully to dryness, then sodium ethoxide ($CH_3CH_2ONa$) is left behind as a white solid. Although initially this appears as something new and complicated, in fact, it is exactly the same (apart from being a more gentle reaction) as the reaction between sodium and water - something you have probably known about for years.
$2H_2O_{(l)} + 2Na_{(s)} \rightarrow 2OH^-_{(aq)} + 2Na^+_{(aq)} + H_{2(g)}$
If the solution is evaporated carefully to dryness, then the sodium hydroxide ($NaOH$) is left behind as a white solid.
We normally, of course, write the sodium hydroxide formed as $NaOH$ rather than $HONa$ - but that's the only difference. Sodium ethoxide is just like sodium hydroxide, except that the hydrogen has been replaced by an ethyl group. Sodium hydroxide contains $OH^-$ ions; sodium ethoxide contains $CH_3CH_2O^-$ ions.
Note
The reason that the ethoxide formula is written with the oxygen on the right unlike the hydroxide ion is simply a matter of clarity. If you write it the other way around, it doesn't immediately look as if it comes from ethanol. You will find the same thing happens when you write formulae for organic salts like sodium ethanoate, for example.
There are two simple uses for this reaction:
• To safely dispose of small amounts of sodium: If you spill some sodium on the bench or have a small amount left over from a reaction you cannot simply dispose of it in the sink. It tends to react explosively with the water - and comes flying back out at you again! It reacts much more gently with ethanol. Ethanol is, therefore, used to dissolve small quantities of waste sodium. The solution formed can be washed away without problems (provided you remember that sodium ethoxide is strongly alkaline - see below).
• To test for the -OH group in alcohols: Because of the dangers involved in handling sodium, this is not the best test for an alcohol at this level. Because sodium reacts violently with acids to produce a salt and hydrogen, you would first have to be sure that the liquid you were testing was neutral. You would also have to be confident that there was no trace of water present because sodium reacts with the -OH group in water even better than with the one in an alcohol. With those provisos, if you add a tiny piece of sodium to a neutral liquid free of water and get bubbles of hydrogen produced, then the liquid is an alcohol.
Ethoxide Ions are Strongly Basic
If you add water to sodium ethoxide, it dissolves to give a colorless solution with a high pH. The solution is strongly alkaline because ethoxide ions are Brønsted-Lowry bases and remove hydrogen ions from water molecules to produce hydroxide ions, which increase the pH.
$CH_3CH_2O^- + H_2O \rightarrow CH_3CH_2OH + OH^-$
Ethoxide Ions are Good Nucleophiles
A nucleophile is a chemical species that carries a negative or partial negative charge that it uses to attack positive centers in other molecules or ions. Hydroxide ions are good nucleophiles, and you may have come across the reaction between a halogenoalkane (also called a haloalkane or alkyl halide) and sodium hydroxide solution. The hydroxide ions replace the halogen atom.
$CH_3CH_2CH_2Br + OH^- \rightarrow CH_3CH_2CH_2OH + Br^-$
In this case, an alcohol is formed. The ethoxide ion behaves in exactly the same way. If you knew the mechanism for the hydroxide ion reaction, you could work out exactly what happens in the reaction between a halogenoalkane and ethoxide ion.
Compare this equation with the last one.
$CH_3CH_2CH_2OH + CH_3CH_2Br \rightarrow CH_3CH_2CH_2OCH_2CH_3 + Br^-$
The only difference is that where there was a hydrogen atom at the right-hand end of the product molecule, an alkyl group is now present. Two alkyl (or other hydrocarbon) groups bridged by an oxygen atom is called an ether. This particular one is 1-ethoxypropane or ethyl propyl ether. This reaction is known as the Williamson Ether Synthesis and is a good method of synthesizing ethers in the lab.
Contributors
• UndefinedNameError: reference to undefined name 'ContribClark' (click for details)
Callstack:
at (Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alcohols/Reactivity_of_Alcohols/The_Reaction_Between_Alcohols_and_Sodium), /content/body/div[4]/ul/li/span, line 1, column 1 | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alcohols/Reactivity_of_Alcohols/The_Reaction_Between_Alcohols_and_Sodium.txt |
The triiodomethane (iodoform) reaction can be used to identify the presence of a CH3CH(OH) group in alcohols. There are two apparently quite different mixtures of reagents that can be used to do this reaction, but are chemically equivalent.
• Iodine and sodium hydroxide solution
This is chemically the more obvious method. Iodine solution is added to a small amount of an alcohol, followed by just enough sodium hydroxide solution to remove the color of the iodine. If nothing happens in the cold, it may be necessary to warm the mixture very gently. A positive result is the appearance of a very pale yellow precipitate of triiodomethane (previously known as iodoform): CHI3, which apart from its color, can also be recognized by its faintly "medical" smell. It is used as an antiseptic on the sort of sticky plasters you put on minor cuts, for example.
• Using potassium iodide and sodium chlorate(I) solutions
Sodium chlorate(I) is also known as sodium hypochlorite. Potassium iodide solution is added to a small amount of an alcohol, followed by sodium chlorate(I) solution. Again, if no precipitate is formed in the cold, it may be necessary to warm the mixture very gently. The positive result is the same pale yellow precipitate as before.
What the Iodoform Reaction Shows
A positive result is the formation of a pale yellow precipitate of triiodomethane (iodoform) - is given by an alcohol containing the grouping:
"R" can be a hydrogen atom or a hydrocarbon group (for example, an alkyl group). If "R" is hydrogen, then you have the primary alcohol ethanol, CH3CH2OH.
1. Ethanol is the only primary alcohol to give the triiodomethane (iodoform) reaction.
2. If "R" is a hydrocarbon group, then you have a secondary alcohol. Lots of secondary alcohols give this reaction, but those that do all have a methyl group attached to the carbon with the -OH group.
3. No tertiary alcohols can contain this group because no tertiary alcohols can have a hydrogen atom attached to the carbon with the -OH group. No tertiary alcohols give the triiodomethane (iodoform) reaction.
Mechanism
We will assume the iodine/sodium hydroxide solution for the reaction:
This is being given as a flow scheme rather than full equations. The equations for the other two steps are given elsewhere.
Contributors
Jim Clark (Chemguide.co.uk)
Thionyl Chloride
If There is one thing you learn how to do well in Org 1, it’s make alcohols. Let’s count the ways: hydroboration, acid-catalyzed hydration, oxymercuration for starters, and then substitution of alkyl halides with water or \(HO^-\). If you want to extend it even further, There is dihydroxylation (to make diols) using \(OsO_4\) or cold \(KMnO_4\), and even opening of epoxides under acidic or basic conditions to give alcohols.
There is just one issue here and it comes up once you try to use alcohols in synthesis. Let’s say you want to use that alcohol in a subsequent substitution step, getting rid of the \(HO^-)\ and replacing it with something else. See any problems with that? Remember that good leaving groups are weak bases – and the hydroxide ion, being a strong base, tends to be a pretty bad leaving group.
So what can we do?
What you want to do is convert the alcohol into a better leaving group. One way is to convert the alcohol into a sulfonate ester – we talked about that with \( TsCl\) and \(MsCl\). Alternatively, alcohols can be converted into alkyl chlorides with thionyl chloride (\(SOCl_2\)). This is a useful reaction, because the resulting alkyl halides are versatile compounds that can be converted into many compounds that are not directly accessible from the alcohol itself.
If you take an alcohol and add thionyl chloride, it will be converted into an alkyl chloride. The byproducts here are hydrochloric acid (\(HCl\)) and sulfur dioxide (\(SO_2\)). Note: there are significant differences in how this reaction is taught at different schools. Consult your instructor to be 100% sure that this applies to your course). See post here
Example \(1\): Conversion of Alcohols to Alkyl Chlorides
There is one important thing to note here: see the stereochemistry? It’s been inverted.*(white lie alert – see below) That’s an important difference between \(SOCl_2\) and TsCl, which leaves the stereochemistry alone. We’ll get to the root cause of that in a moment, but in the meantime, can you think of a mechanism which results in inversion of configuration at carbon?
As an extra bonus, thionyl chloride will also convert carboxylic acids into acid chlorides (“acyl chlorides”). Like alcohols, carboxylic acids have their limitations as reactants: the hydroxyl group interferes with many of the reactions we learn for nucleophilic acyl substitution (among others). Conversion of the OH into Cl solves this problem.
Mechanism
As you might have guessed, conversion of alcohols to alkyl halides proceeds through a substitution reaction – specifically, an \(S_N2\) mechanism. The first step is attack of the oxygen upon the sulfur of \(SOCl_2\), which results in displacement of chloride ion. This has the side benefit of converting the alcohol into a good leaving group: in the next step, chloride ion attacks the carbon in \(S_N2\) fashion, resulting in cleavage of the C–O bond with inversion of configuration. The \(HOSCl\) breaks down into \(HCl\) and sulfur dioxide gas, which bubbles away.
Formation of Alkyl Chlorides
Since the reaction proceeds through a backside attack (\(S_N2\)), there is inversion of configuration at the carbon
The mechanism for formation of acid chlorides from carboxylic acids is similar. The conversion of caboxylic acids to acid chlorides is similar, but proceeds through a [1,2]-addition of chloride ion to the carbonyl carbon followed by [1,2]-elimination to give the acid chloride, \(SO_2\) and \(HCl\)
In the laboratory
Like many sulfur-containing compounds, thionyl chloride is noseworthy for its pungent smell. Thionyl chloride has a nauseating sickly-sweet odor to it that imprints itself forever upon your memory. One accident that occurred during my time as a TA involved a student dropping a flask with 5 mL of thionyl chloride into a rotovap bath outside the fume hood. The cloud of \(SO_2\) and \(HCl\) that formed cleared the teaching lab for half an hour, so you can imagine what thionyl chloride would do if exposed to the moisture in your lungs. Treat with caution, just as you would if you were working with phosgene.
*Here’s the white lie. Although it’s often taught that \(SOCl_2\) leads to 100% inversion of configuration, in reality it’s not always that simple. Inversion of configuration with \(SOCl_2\) is very solvent dependent. Depending on the choice of solvent, one can get either straight inversion, or a mixture of retention and inversion. For the purposes of beginning organic classes, most students can ignore this message. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alcohols/Reactivity_of_Alcohols/The_Triiodomethane_%28Iodoform%29_Reaction.txt |
Hydroboration-Oxidation is a two step pathway used to produce alcohols. The reaction proceeds in an Anti-Markovnikov manner, where the hydrogen (from BH3 or BHR2) attaches to the more substituted carbon and the boron attaches to the least substituted carbon in the alkene bouble bond. Furthermore, the borane acts as a lewisAnti-Markovnikov acid by accepting two electrons in its empty p orbital from an alkene that is electron rich. This process allows boron to have an electron octet. A very interesting characteristic of this process is that it does not require any activation by a catalyst. The Anti-MarkovnikovHydroboration mechanism has the elements of both hydrogenation and electrophilic addition and it is a stereospecific (syn addition), meaning that the hydroboration takes place on the same face of the double bond, this leads cis stereochemistry.
Introduction
Hydroboration-oxidation of alkenes has been a very valuable laboratory method for the stereoselectivity and regioselectivity of alkenes. An Additional feature of this reaction is that it occurs without rearrangement.
The Borane Complex
First off it is very imporatnt to understand little bit about the structure and the properties of the borane molecule. Borane exists naturally as a very toxic gas and it exists as dimer of the general formula B2H6 (diborane). Additionally, the dimer B2H6 ignites spontaneously in air. Borane is commercially available in ether and tetrahydrofuran (THF), in these solutions the borane can exist as a lewis acid-base complex, which allows boron to have an electron octet.
$2BH_3 \rightarrow B_2H_6$
The Mechanism
Step #1
• Part #1: Hydroboration of the alkene. In this first step the addittion of the borane to the alkene is initiated and prceeds as a concerted reaction because bond breaking and bond formation occurs at the same time. This part consists of the vacant 2p orbital of the boron electrophile pairing with the electron pair of the ? bondof the nucleophile.
Transition state
* Note that a carbocation is not formed. Therefore, no rearrangement takes place.
• Part #2: The Anti Markovnikov addition of Boron. The boron adds to the less substituted carbon of the alkene, which then places the hydrogen on the more substituted carbon. Both, the boron and the hydrogen add simultaneously on the same face of the double bond (syn addition).
Oxidation of the Trialkylborane by Hydrogen Peroxide
Step #2
• Part #1: the first part of this mechanism deals with the donation of a pair of electrons from the hydrogen peroxide ion. the hydrogen peroxide is the nucleophile in this reaction because it is the electron donor to the newly formed trialkylborane that resulted from hydroboration.
EpoxidationEpoxidation
• Part 2: In this second part of the mechanism, a rearrangement of an R group with its pair of bonding electrons to an adjacent oxygen results in the removal of a hydroxide ion.
Two more of these reactions with hydroperoxide will occur in order give a trialkylborate
• Part 3: This is the final part of the Oxidation process. In this part the trialkylborate reacts with aqueous NaOH to give the alcohol and sodium borate.
If you need additional visuals to aid you in understanding the mechanism, click on the outside links provided here that will take you to other pages and media that are very helpful as well.
Problems
What are the products of these following reactions?
#1.
#2.
#3.
Draw the structural formulas for the alcohols that result from hydroboration-oxidation of the alkenes shown.
#4.
#5. (E)-3-methyl-2-pentene
If you need clarification or a reminder on the nomenclature of alkenes refer to the link below on naming the alkenes.
#1.
#2.
#3.
#4.
#5.
Contributors
• Gilbert Torres (UCD) | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alcohols/Synthesis_of_Alcohols/Alcohols_from_Hydroboration-Oxidation_of_Alkenes.txt |
Electrophilic hydration is the act of adding electrophilic hydrogen from a non-nucleophilic strong acid (a reusable catalyst, examples of which include sulfuric and phosphoric acid) and applying appropriate temperatures to break the alkene's double bond. After a carbocation is formed, water bonds with the carbocation to form a 1º, 2º, or 3º alcohol on the alkane.
What Is Electrophilic Hydration?
Electrophilic hydration is the reverse dehydration of alcohols and has practical application in making alcohols for fuels and reagents for other reactions. The basic reaction under certain temperatures (given below) is the following:
The phrase "electrophilic" literally means "electron loving" (whereas "nucleophilic" means "nucleus loving"). Electrophilic hydrogen is essentially a proton: a hydrogen atom stripped of its electrons. Electrophilic hydrogen is commonly used to help break double bonds or restore catalysts (see SN2 for more details).
How Does Electrophilic Hydration Work?
Temperatures for Types of Alcohol Synthesis
Heat is used to catalyze electrophilic hydration; because the reaction is in equilibrium with the dehydration of an alcohol, which requires higher temperatures to form an alkene, lower temperatures are required to form an alcohol. The exact temperatures used are highly variable and depend on the product being formed.
• Primary Alcohol: Less than 170ºC
• Secondary Alcohol: Less than 100ºC
• Tertiary Alcohol: Less than 25ºC
But...Why Does Electrophilic Hydration Work?
• An alkene placed in an aqueous non-nucleophilic strong acid immediately "reaches out" with its double bond and attacks one of the acid's hydrogen atoms (meanwhile, the bond between oxygen and hydrogen performs heterolytic cleavage toward the oxygen—in other words, both electrons from the oxygen/hydrogen single bond move onto the oxygen atom).
• A carbocation is formed on the original alkene (now alkane) in the more-substituted position, where the oxygen end of water attacks with its 4 non-bonded valence electrons (oxygen has 6 total valence electrons because it is found in Group 6 on the periodic table and the second row down: two electrons in a 2s-orbital and four in 2p-orbitals. Oxygen donates one valence electron to each bond it forms, leaving four 4 non-bonded valence electrons).
• After the blue oxygen atom forms its third bond with the more-substituted carbon, it develops a positive charge (3 bonds and 2 valence electrons give the blue oxygen atom a formal charge of +1).
• The bond between the green hydrogen and the blue oxygen undergoes heterolytic cleavage, and both the electrons from the bond move onto the blue oxygen. The now negatively-charged strong acid picks up the green electrophilic hydrogen.
• Now that the reaction is complete, the non-nucleophilic strong acid is regenerated as a catalyst and an alcohol forms on the most substituted carbon of the current alkane. At lower temperatures, more alcohol product can be formed.
What is Regiochemistry and How Does It Apply?
Regiochemistry deals with where the substituent bonds on the product. Zaitsev's and Markovnikov's rules address regiochemistry, but Zaitsev's rule applies when synthesizing an alkene while Markovnikov's rule describes where the substituent bonds onto the product. In the case of electrophilic hydration, Markovnikov's rule is the only rule that directly applies. See the following for an in-depth explanation of regiochemistry Markovnikov explanation: Radical Additions--Anti-Markovnikov Product Formation
In the mechanism for a 3º alcohol shown above, the red H is added to the least-substituted carbon connected to the nucleophilic double bonds (it has less carbons attached to it). This means that the carbocation forms on the 3º carbon, causing it to be highly stabilized by hyperconjugationelectrons in nearby sigma (single) bonds help fill the empty p-orbital of the carbocation, which lessens the positive charge. More substitution on a carbon means more sigma bonds are available to "help out" (by using overlap) with the positive charge, which creates greater carbocation stability. In other words, carbocations form on the most substituted carbon connected to the double bond. Carbocations are also stabilized by resonance, but resonance is not a large factor in this case because any carbon-carbon double bonds are used to initiate the reaction, and other double bonded molecules can cause a completely different reaction.
If the carbocation does originally form on the less substituted part of the alkene, carbocation rearrangements occur to form more substituted products:
• Hydride shifts: a hydrogen atom bonded to a carbon atom next to the carbocation leaves that carbon to bond with the carbocation (after the hydrogen has taken both electrons from the single bond, it is known as a hydride). This changes the once neighboring carbon to a carbocation, and the former carbocation becomes a neighboring carbon atom.
• Alkyl shifts: if no hydrogen atoms are available for a hydride shift, an entire methyl group performs the same shift
The nucleophile attacks the positive charge formed on the most substituted carbon connected to the double bond, because the nucleophile is seeking that positive charge. In the mechanism for a 3º alcohol shown above, water is the nucleophile. When the green H is removed from the water molecule, the alcohol attached to the most substituted carbon. Hence, electrophilic hydration follows Markovnikov's rule.
What is Stereochemistry and How Does It Apply?
Stereochemistry deals with how the substituent bonds on the product directionally. Dashes and wedges denote stereochemistry by showing whether the molecule or atom is going into or out of the plane of the board. Whenever the bond is a simple single straight line, the molecule that is bonded is equally likely to be found going into the plane of the board as it is out of the plane of the board. This indicates that the product is a racemic mixture.
Electrophilic hydration adopts a stereochemistry wherein the substituent is equally likely to bond pointing into the plane of the board as it is pointing out of the plane of the board. The 3º alcohol product could look like either of the following products:
Note: Whenever a straight line is used along with dashes and wedges on the same molecule, it could be denoting that the straight line bond is in the same plane as the board. Practice with a molecular model kit and attempting the practice problems at the end can help eliminate any ambiguity.
Is this a Reversible Synthesis?
Electrophilic hydration is reversible because an alkene in water is in equilibrium with the alcohol product. To sway the equilibrium one way or another, the temperature or the concentration of the non-nucleophilic strong acid can be changed. For example:
• Less sulfuric or phosphoric acid and an excess of water help synthesize more alcohol product.
• Lower temperatures help synthesize more alcohol product.
Better Ways to Add Water to Synthesize Alcohols from Alkenes?
The Oxymercuration - Demercuration reaction is a more efficient reaction to make alcohols and does not allow for rearrangements. However, it requires the use of mercury, which is highly toxic. Detractions for using electrophilic hydration to make alcohols include:
• Allowing for carbocation rearrangements
• Poor yields due to the reactants and products being in equilibrium
• Allowing for product mixtures (such as an (R)-enantiomer and an (S)-enantiomer)
• Using sulfuric or phosphoric acid
Problems
Predict the product of each reaction.
1)
2) How does the cyclopropane group affect the reaction?
3) (Hint: What is different about this problem?)
4) (Hint: Consider stereochemistry.)
5) Indicate any shifts as well as the major product:
Answers to Practice Problems
1) This is a basic electrophilic hydration.
2) The answer is additional side products, but the major product formed is still the same (the product shown). Depending on the temperatures used, the cyclopropane may open up into a straight chain, which makes it unlikely that the major product will form (after the reaction, it is unlikely that the 3º carbon will remain as such).
3) A hydride shift actually occurs from the top of the 1-methylcyclopentane to where the carbocation had formed.
4) This reaction will have poor yields due to a very unstable intermediate. For a brief moment, carbocations can form on the two center carbons, which are more stable than the outer two carbons. The carbocations have an sp2 hybridization, and when the water is added on, the carbons change their hybridization to sp3. This makes the methyl and alcohol groups equally likely to be found going into or out of the plane of the paper- the product is racemic.
5) In the first picture shown below, an alkyl shift occurs but a hydride shift (which occurs faster) is possible. Why doesn't a hydride shift occur? The answer is because the alkyl shift leads to a more stable product. There is a noticeable amount of side product that forms where the two methyl groups are, but the major product shown below is still the most significant due to the hyperconjugation that occurs by being in between the two cyclohexanes.
Contributors
• Lance Peery (UCD) | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alcohols/Synthesis_of_Alcohols/Electrophilic_Hydration_to_Make_Alcohols.txt |
This page looks at the manufacture of alcohols by the direct hydration of alkenes, concentrating mainly on the hydration of ethene to make ethanol. It then compares that method with making ethanol by fermentation.
Manufacturing alcohols from alkenes
Ethanol is manufactured by reacting ethene with steam. The catalyst used is solid silicon dioxide coated with phosphoric(V) acid. The reaction is reversible.
Only 5% of the ethene is converted into ethanol at each pass through the reactor. By removing the ethanol from the equilibrium mixture and recycling the ethene, it is possible to achieve an overall 95% conversion. A flow scheme for the reaction looks like this:
The manufacture of other alcohols from alkenes
Some - but not all - other alcohols can be made by similar reactions. The catalyst used and the reaction conditions will vary from alcohol to alcohol. The reason that there is a problem with some alcohols is well illustrated with trying to make an alcohol from propene, CH3CH=CH2. In principle, there are two different alcohols which might be formed:
You might expect to get either propan-1-ol or propan-2-ol depending on which way around the water adds to the double bond. In practice what you get is propan-2-ol. If you add a molecule H-X across a carbon-carbon double bond, the hydrogen nearly always gets attached to the carbon with the most hydrogens on it already - in this case the CH2 rather than the CH. The effect of this is that there are bound to be some alcohols which it is impossible to make by reacting alkenes with steam because the addition would be the wrong way around.
Making ethanol by fermentation
This method only applies to ethanol and you cannot make any other alcohol this way. The starting material for the process varies widely, but will normally be some form of starchy plant material such as maize (US: corn), wheat, barley or potatoes. Starch is a complex carbohydrate, and other carbohydrates can also be used - for example, in the lab sucrose (sugar) is normally used to produce ethanol. Industrially, this wouldn't make sense. It would be silly to refine sugar if all you were going to use it for was fermentation. There is no reason why you should not start from the original sugar cane, though.
The first step is to break complex carbohydrates into simpler ones. For example, if you were starting from starch in grains like wheat or barley, the grain is heated with hot water to extract the starch and then warmed with malt. Malt is germinated barley which contains enzymes which break the starch into a simpler carbohydrate called maltose, \(C_{12}H_{22}O_{11}\). Maltose has the same molecular formula as sucrose but contains two glucose units joined together, whereas sucrose contains one glucose and one fructose unit.
Yeast is then added and the mixture is kept warm (say 35°C) for perhaps several days until fermentation is complete. Air is kept out of the mixture to prevent oxidation of the ethanol produced to ethanoic acid (vinegar). Enzymes in the yeast first convert carbohydrates like maltose or sucrose into even simpler ones like glucose and fructose, both \(C_6H_{12}O_6\), and then convert these in turn into ethanol and carbon dioxide. You can show these changes as simple chemical equations, but the biochemistry of the reactions is much, much more complicated than this suggests.
\[ C_{12}H_{22}O_{11} + H_2O \longrightarrow 2C_6H_{12}O_6 \]
\[ C_6H_{12}O_6 \longrightarrow 2CH_3CH_2OH + 2CO2\]
Yeast is killed by ethanol concentrations in excess of about 15%, and that limits the purity of the ethanol that can be produced. The ethanol is separated from the mixture by fractional distillation to give 96% pure ethanol. For theoretical reasons (minimum boiling point azeotrope), it is impossible to remove the last 4% of water by fractional distillation.
Fermentation Hydration of ethene
Table 1.1.1: A comparison of fermentation with the direct hydration of ethene
Type of process A batch process. Everything is put into a container and then left until fermentation is complete. That batch is then cleared out and a new reaction set up. This is inefficient. A continuous flow process. A stream of reactants is passed continuously over a catalyst. This is a more efficient way of doing things.
Rate of reaction Very slow. Very rapid.
Quality of product Produces very impure ethanol which needs further processing Produces much purer ethanol.
Reaction conditions Uses gentle temperatures and atmospheric pressure. Uses high temperatures and pressures, needing lots of energy input.
Use of resources Uses renewable resources based on plant material. Uses finite resources based on crude oil. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alcohols/Synthesis_of_Alcohols/The_Manufacture_of_Alcohols.txt |
Aldehydes and ketones contain the carbonyl group. Aldehydes are considered the most important functional group. They are often called the formyl or methanoyl group. Aldehydes derive their name from the dehydration of alcohols. Aldehydes contain the carbonyl group bonded to at least one hydrogen atom. Ketones contain the carbonyl group bonded to two carbon atoms.
Aldehydes and ketones are organic compounds which incorporate a carbonyl functional group, C=O. The carbon atom of this group has two remaining bonds that may be occupied by hydrogen, alkyl or aryl substituents. If at least one of these substituents is hydrogen, the compound is an aldehyde. If neither is hydrogen, the compound is a ketone.
Naming Aldehydes
The IUPAC system of nomenclature assigns a characteristic suffix -al to aldehydes. For example, H2C=O is methanal, more commonly called formaldehyde. Since an aldehyde carbonyl group must always lie at the end of a carbon chain, it is always is given the #1 location position in numbering and it is not necessary to include it in the name. There are several simple carbonyl containing compounds which have common names which are retained by IUPAC.
Also, there is a common method for naming aldehydes and ketones. For aldehydes common parent chain names, similar to those used for carboxylic acids, are used and the suffix –aldehyde is added to the end. In common names of aldehydes, carbon atoms near the carbonyl group are often designated by Greek letters. The atom adjacent to the carbonyl function is alpha, the next removed is beta and so on.
If the aldehyde moiety (-CHO) is attached to a ring the suffix –carbaldehyde is added to the name of the ring. The carbon attached to this moiety will get the #1 location number in naming the ring.
Summary of Aldehyde Nomenclature rules
1. Aldehydes take their name from their parent alkane chains. The -e is removed from the end and is replaced with -al.
2. The aldehyde funtional group is given the #1 numbering location and this number is not included in the name.
3. For the common name of aldehydes start with the common parent chain name and add the suffix -aldehyde. Substituent positions are shown with Greek letters.
4. When the -CHO functional group is attached to a ring the suffix -carbaldehyde is added, and the carbon attached to that group is C1.
Example 1
The IUPAC system names are given on top while the common name is given on the bottom in parentheses.
Aldehyde Common Names to Memorize
There are some common names that are still used and need to be memorized. Recognizing the patterns can be helpful.
Naming Ketones
The IUPAC system of nomenclature assigns a characteristic suffix of -one to ketones. A ketone carbonyl function may be located anywhere within a chain or ring, and its position is usually given by a location number. Chain numbering normally starts from the end nearest the carbonyl group. Very simple ketones, such as propanone and phenylethanone do not require a locator number, since there is only one possible site for a ketone carbonyl function. The common names for ketones are formed by naming both alkyl groups attached to the carbonyl then adding the suffix -ketone. The attached alkyl groups are arranged in the name alphabetically.
Summary of Ketone Nomenclature rules
1. Ketones take their name from their parent alkane chains. The ending -e is removed and replaced with -one.
2. The common name for ketones are simply the substituent groups listed alphabetically + ketone.
3. Some common ketones are known by their generic names. Such as the fact that propanone is commonly referred to as acetone.
Example 2
The IUPAC system names are given on top while the common name is given on the bottom in parentheses.
Ketone Common Names to Memorize
There are some common names that are still used and need to be memorized. Recognizing the patterns can be helpful.
Naming Aldehydes and Ketones in the Same Molecule
As with many molecules with two or more functional groups, one is given priority while the other is named as a substituent. Because aldehydes have a higher priority than ketones, molecules which contain both functional groups are named as aldehydes and the ketone is named as an "oxo" substituent. It is not necessary to give the aldehyde functional group a location number, however, it is usually necessary to give a location number to the ketone.
Naming Dialdehydes and Diketones
For dialdehydes the location numbers for both carbonyls are omitted because the aldehyde functional groups are expected to occupy the ends of the parent chain. The ending –dial is added to the end of the parent chain name.
Example 4
For diketones both carbonyls require a location number. The ending -dione or -dial is added to the end of the parent chain.
Naming Cyclic Ketones and Diketones
In cyclic ketones the carbonyl group is assigned location position #1, and this number is not included in the name, unless more than one carbonyl group is present. The rest of the ring is numbered to give substituents the lowest possible location numbers. Remember the prefix cyclo is included before the parent chain name to indicate that it is in a ring. As with other ketones the –e ending is replaced with the –one to indicate the presence of a ketone.
With cycloalkanes which contain two ketones both carbonyls need to be given a location numbers. Also, an –e is not removed from the end, but the suffix –dione is added.
Naming Carbonyls and Hydroxyls in the Same Molecule
When and aldehyde or ketone is present in a molecule which also contains an alcohol functional group the carbonyl is given nomenclature priority by the IUPAC system. This means that the carbonyl is given the lowest possible location number and the appropriate nomenclature suffix is included. In the case of alcohols the OH is named as a hydroxyl substituent. However, the l in hydroxyl is generally removed.
Naming Carbonyls and Alkenes in the Same Molecule
When and aldehyde or ketone is present in a molecule which also contains analkene functional group the carbonyl is given nomenclature priority by the IUPAC system. This means that the carbonyl is given the lowest possible location number and the appropriate nomenclature suffix is included. When carbonyls are included with an alkene the following order is followed:
(Location number of the alkene)-(Prefix name for the longest carbon chain minus the -ane ending)-(an -en ending to indicate the presence of an alkene)-(the location number of the carbonyl if a ketone is present)-(either an –one or and -anal ending).
Remember that the carbonyl has priority so it should get the lowest possible location number. Also, remember that cis/tran or E/Z nomenclature for the alkene needs to be included if necessary.
Aldehydes and Ketones as Fragments
• Alkanoyl is the common name of the fragment, though the older naming, acyl, is still widely used.
• Formyl is the common name of the fragment.
• Acety is the common name of the CH3-C=O- fragment.
Additional Examples of Carbonyl Nomenclature
1) Please give the IUPAC name for each compound:
Answers for Question 1
1. 3,4-dimethylhexanal
2. 5-bromo-2-pentanone
3. 2,4-hexanedione
4. cis-3-pentenal (or (Z)-3-pentenal)
5. 6-methyl-5-hepten-3-one
6. 3-hydroxy-2,4-pentanedione
7. 1,2-cyclobutanedione
8. 2-methyl-propanedial
9. 3-methyl-5-oxo-hexanal
10. cis-2,3-dihydroxycyclohexanone
11. 3-bromo-2-methylcyclopentanecarboaldehyde
12. 3-bromo-2-methylpropanal
2) Please give the structure corresponding to each name:
A) butanal
B) 2-hydroxycyclopentanone
C) 2,3-pentanedione
D) 1,3-cyclohexanedione
E) 4-hydoxy-3-methyl-2-butanone
F) (E) 3-methyl-2-hepten-4-one
G) 3-oxobutanal
H) cis-3-bromocyclohexanecarboaldehyde
I) butanedial
J) trans-2-methyl-3-hexenal
Answers to question 2: | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Aldehydes_and_Ketones/Nomenclature_of_Aldehydes_and_Ketones.txt |
• Introduction to Aldehydes and Ketones
Because of the greater electronegativity of oxygen, the carbonyl group is polar, and aldehydes and ketones have larger molecular dipole moments than do alkenes. We expect, therefore, that aldehydes and ketones will have higher boiling points than similar sized alkenes. Furthermore, the presence of oxygen with its non-bonding electron pairs makes aldehydes and ketones hydrogen-bond acceptors, and should increase their water solubility relative to hydrocarbons.
• Natural Occurrence of Aldehydes and Ketones
Aldehydes and ketones are widespread in nature and are often combined with other functional groups. Examples of naturally occurring molecules which contain a aldehyde or ketone functional group are shown in the following two figures. The compounds in the figure 1 are found chiefly in plants or microorganisms and those in the figure 2 have animal origins. Many of these molecular structures are chiral.
• Properties of Aldehydes and Ketones
This page explains what aldehydes and ketones are, and looks at the way their bonding affects their reactivity. It also considers their simple physical properties such as solubility and boiling points.
• The Carbonyl Group
A carbonyl group is a chemically organic functional group composed of a carbon atom double-bonded to an oxygen atom --> [C=O] The simplest carbonyl groups are aldehydes and ketones usually attached to another carbon compound. These structures can be found in many aromatic compounds contributing to smell and taste.
Properties of Aldehydes and Ketones
Introduction
A comparison of the properties and reactivity of aldehydes and ketones with those of the alkenes is warranted, since both have a double bond functional group. Because of the greater electronegativity of oxygen, the carbonyl group is polar, and aldehydes and ketones have larger molecular dipole moments (D) than do alkenes. The resonance structures in Figure 1 illustrate this polarity, and the relative dipole moments of formaldehyde, other aldehydes and ketones confirm the stabilizing influence that alkyl substituents have on carbocations (the larger the dipole moment the greater the polar character of the carbonyl group). We expect, therefore, that aldehydes and ketones will have higher boiling points than similar sized alkenes. Furthermore, the presence of oxygen with its non-bonding electron pairs makes aldehydes and ketones hydrogen-bond acceptors, and should increase their water solubility relative to hydrocarbons. Specific examples of these relationships are provided in the following table.
Figure 1: Resonance structures
Compound
Mol. Wt.
Boiling Point
Water
Solubility
(CH3)2C=CH2
56
-7.0 ºC
0.04 g/100
(CH3)2C=O
58
56.5 ºC
infinite
CH3CH2CH2CH=CH2
70
30.0 ºC
0.03 g/100
CH3CH2CH2CH=O
72
76.0 ºC
7 g/100
96
103.0 ºC
insoluble
98
155.6 ºC
5 g/100
The polarity of the carbonyl group also has a profound effect on its chemical reactivity, compared with the non-polar double bonds of alkenes. Thus, reversible addition of water to the carbonyl function is fast, whereas water addition to alkenes is immeasurably slow in the absence of a strong acid catalyst. Curiously, relative bond energies influence the thermodynamics of such addition reactions in the opposite sense.
The C=C of alkenes has an average bond energy of 146 kcal/mole. Since a C–C σ-bond has a bond energy of 83 kcal/mole, the π-bond energy may be estimated at 63 kcal/mole (i.e. less than the energy of the sigma bond). The C=O bond energy of a carbonyl group, on the other hand, varies with its location, as follows:
H2C=O 170 kcal/mole
RCH=O 175 kcal/mole
R2C=O 180 kcal/mole
The C–O σ-bond is found to have an average bond energy of 86 kcal/mole. Consequently, with the exception of formaldehyde, the carbonyl function of aldehydes and ketones has a π-bond energy greater than that of the sigma-bond, in contrast to the pi-sigma relationship in C=C. This suggests that addition reactions to carbonyl groups should be thermodynamically disfavored, as is the case for the addition of water. All of this is summarized in the following diagram (ΔHº values are for the addition reaction).
Although the addition of water to an alkene is exothermic and gives a stable product (an alcohol), the uncatalyzed reaction is extremely slow due to a high activation energy. The reverse reaction (dehydration of an alcohol) is even slower, and because of the kinetic barrier, both reactions are practical only in the presence of a strong acid.
In contrast, both the endothermic addition of water to a carbonyl function, and the exothermic elimination of water from the resulting geminal-diol are fast. The inherent polarity of the carbonyl group, together with its increased basicity (compared with alkenes), lowers the transition state energy for both reactions, with a resulting increase in rate. Acids and bases catalyze both the addition and elimination of water. Proof that rapid and reversible addition of water to carbonyl compounds occurs is provided by experiments using isotopically labeled water. If a carbonyl reactant composed of 16O (colored blue above) is treated with water incorporating the 18O isotope (colored red above), a rapid exchange of the oxygen isotope occurs. This can only be explained by the addition-elimination mechanism shown here.
Contributors
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry
Natural Occurrence of Aldehydes and Ketones
Aldehydes and ketones are widespread in nature and are often combined with other functional groups. Examples of naturally occurring molecules which contain a aldehyde or ketone functional group are shown in the following two figures. The compounds in the figure 1 are found chiefly in plants or microorganisms and those in the figure 2 have animal origins. Many of these molecular structures are chiral.
When chiral compounds are found in nature they are usually enantiomerically pure, although different sources may yield different enantiomers. For example, carvone is found as its levorotatory (R)-enantiomer in spearmint oil, whereas, caraway seeds contain the dextrorotatory (S)-enantiomer. In this case the change of the stereochemistry causes a drastic change in the perceived scent. Aldehydes and ketones are known for their sweet and sometimes pungent odors. The odor from vanilla extract comes from the molecule vanillin. Likewise, benzaldehyde provides a strong scent of almonds and is this author’s favorite chemical smell. Because of their pleasant fragrances aldehyde and ketone containing molecules are often found in perfumes. However, not all of the fragrances are pleasing. In particular, 2-Heptanone provides part of the sharp scent from blue cheese and (R)-Muscone is part of the musky smell from the Himalayan musk deer. Lastly, ketones show up in many important hormones such as progesterone (a female sex hormone) and testosterone (a male sex hormone). Notice how subtle differences in structure can cause drastic changes in biological activity. The ketone functionality also shows up in the anti-inflammatory steroid, Cortisone.
Figure 1. Aldehyde and ketone containing molecules isolated from plant sources.
Figure 2. Aldehyde and ketone containing molecules isolated from animal sources. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Aldehydes_and_Ketones/Properties_of_Aldehydes_and_Ketones/Introduction_to_Aldehydes_and_Ketones.txt |
This page explains what aldehydes and ketones are, and looks at the way their bonding affects their reactivity. It also considers their simple physical properties such as solubility and boiling points. Aldehydes and ketones are simple compounds which contain a carbonyl group - a carbon-oxygen double bond. They are simple in the sense that they don't have other reactive groups like -OH or -Cl attached directly to the carbon atom in the carbonyl group - as you might find, for example, in carboxylic acids containing -COOH.
Aldehydes
In aldehydes, the carbonyl group has a hydrogen atom attached to it together with either a second hydrogen atom or, more commonly, a hydrocarbon group which might be an alkyl group or one containing a benzene ring. For the purposes of this section, we shall ignore those containing benzene rings.
Notice that these all have exactly the same end to the molecule. All that differs is the complexity of the other group attached. When you are writing formulae for these, the aldehyde group (the carbonyl group with the hydrogen atom attached) is always written as -CHO - never as COH. That could easily be confused with an alcohol. Ethanal, for example, is written as CH3CHO; methanal as HCHO. The name counts the total number of carbon atoms in the longest chain - including the one in the carbonyl group. If you have side groups attached to the chain, notice that you always count from the carbon atom in the carbonyl group as being number 1.
Ketones
In ketones, the carbonyl group has two hydrocarbon groups attached. Again, these can be either alkyl groups or ones containing benzene rings. Again, we'll concentrated on those containing alkyl groups just to keep things simple. Notice that ketones never have a hydrogen atom attached to the carbonyl group.
Propanone is normally written CH3COCH3. Notice the need for numbering in the longer ketones. In pentanone, the carbonyl group could be in the middle of the chain or next to the end - giving either pentan-3-one or pentan-2-one.
Bonding and reactivity
Oxygen is far more electronegative than carbon and so has a strong tendency to pull electrons in a carbon-oxygen bond towards itself. One of the two pairs of electrons that make up a carbon-oxygen double bond is even more easily pulled towards the oxygen. That makes the carbon-oxygen double bond very highly polar.
The slightly positive carbon atom in the carbonyl group can be attacked by nucleophiles. A nucleophile is a negatively charged ion (for example, a cyanide ion, CN-), or a slightly negatively charged part of a molecule (for example, the lone pair on a nitrogen atom in ammonia, NH3).
During the reaction, the carbon-oxygen double bond gets broken. The net effect of all this is that the carbonyl group undergoes addition reactions, often followed by the loss of a water molecule. This gives a reaction known as addition-elimination or condensation. You will find examples of simple addition reactions and addition-elimination if you explore the aldehydes and ketones menu (link at the bottom of the page). Both aldehydes and ketones contain a carbonyl group. That means that their reactions are very similar in this respect.
Where aldehydes and ketones differ
An aldehyde differs from a ketone by having a hydrogen atom attached to the carbonyl group. This makes the aldehydes very easy to oxidise. For example, ethanal, CH3CHO, is very easily oxidised to either ethanoic acid, CH3COOH, or ethanoate ions, CH3COO-.
Ketones don't have that hydrogen atom and are resistant to oxidation. They are only oxidised by powerful oxidising agents which have the ability to break carbon-carbon bonds. You will find the oxidation of aldehydes and ketones discussed if you follow a link from the aldehydes and ketones menu (see the bottom of this page).
Boiling Points
Methanal is a gas (boiling point -21°C), and ethanal has a boiling point of +21°C. That means that ethanal boils at close to room temperature. The other aldehydes and the ketones are liquids, with boiling points rising as the molecules get bigger. The size of the boiling point is governed by the strengths of the intermolecular forces.
• Van der Waals dispersion forces: These attractions get stronger as the molecules get longer and have more electrons. That increases the sizes of the temporary dipoles that are set up. This is why the boiling points increase as the number of carbon atoms in the chains increases - irrespective of whether you are talking about aldehydes or ketones.
• van der Waals dipole-dipole attractions: Both aldehydes and ketones are polar molecules because of the presence of the carbon-oxygen double bond. As well as the dispersion forces, there will also be attractions between the permanent dipoles on nearby molecules. That means that the boiling points will be higher than those of similarly sized hydrocarbons - which only have dispersion forces. It is interesting to compare three similarly sized molecules. They have similar lengths, and similar (although not identical) numbers of electrons.
molecule type boiling point (°C)
CH3CH2CH3 alkane -42
CH3CHO aldehyde +21
CH3CH2OH alcohol +78
Notice that the aldehyde (with dipole-dipole attractions as well as dispersion forces) has a boiling point higher than the similarly sized alkane which only has dispersion forces. However, the aldehyde's boiling point isn't as high as the alcohol's. In the alcohol, there is hydrogen bonding as well as the other two kinds of intermolecular attraction.
Although the aldehydes and ketones are highly polar molecules, they don't have any hydrogen atoms attached directly to the oxygen, and so they can't hydrogen bond with each other.
Solubility in water
The small aldehydes and ketones are freely soluble in water but solubility falls with chain length. For example, methanal, ethanal and propanone - the common small aldehydes and ketones - are miscible with water in all proportions.The reason for the solubility is that although aldehydes and ketones can't hydrogen bond with themselves, they can hydrogen bond with water molecules. One of the slightly positive hydrogen atoms in a water molecule can be sufficiently attracted to one of the lone pairs on the oxygen atom of an aldehyde or ketone for a hydrogen bond to be formed.
There will also, of course, be dispersion forces and dipole-dipole attractions between the aldehyde or ketone and the water molecules. Forming these attractions releases energy which helps to supply the energy needed to separate the water molecules and aldehyde or ketone molecules from each other before they can mix together.
As chain lengths increase, the hydrocarbon "tails" of the molecules (all the hydrocarbon bits apart from the carbonyl group) start to get in the way. By forcing themselves between water molecules, they break the relatively strong hydrogen bonds between water molecules without replacing them by anything as good. This makes the process energetically less profitable, and so solubility decreases. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Aldehydes_and_Ketones/Properties_of_Aldehydes_and_Ketones/Properties_of_Aldehydes_and_Ketones.txt |
A carbonyl group is a chemically organic functional group composed of a carbon atom double-bonded to an oxygen atom --> [C=O] The simplest carbonyl groups are aldehydes and ketones usually attached to another carbon compound. These structures can be found in many aromatic compounds contributing to smell and taste.
Introduction
Before going into anything in depth be sure to understand that the C=O entity itself is known as the "Carbonyl group" while the members of this group are called "carbonyl compounds" --> X-C=O. The carbon and oxygen are usually sp2 hybridized and planar.
Carbonyl Group Double Bonds
The double bonds in alkenes and double bonds in carbonyl groups are VERY different in terms of reactivity. The C=C is less reactive due to C=O electronegativity attributed to the oxygen and its two lone pairs of electrons. One pair of the oxygen lone pairs are located in 2s while the other pair are in 2p orbital where its axis is directed perpendicular to the direction of the pi orbitals. The Carbonyl groups properties are directly tied to its electronic structure as well as geometric positioning. For example, the electronegativity of oxygen also polarizes the pi bond allowing the single bonded substituent connected to become electron withdrawing.
*Note: Both the pi bonds are in phase (top and botom blue ovals)
The double bond lengths of a carbonyl group is about 1.2 angstroms and the strength is about 176-179 kcal/mol). It is possible to correlate the length of a carbonyl bond with its polarity; the longer the bond meaing the lower the polarity. For example, the bond length in C=O is larger in acetaldehyde than in formaldehyde (this of course takes into account the inductive effect of CH3 in the compound).
Polarization
As discussed before, we understand that oxygen has two lone pairs of electrons hanging around. These electrons make the oxygen more electronegative than carbon. The carbon is then partially postive (electrophillic) and the oxygen partially negative (nucleophillic). The polarizability is denoted by a lowercase delta and a positive or negative superscript depending. For example, carbon would have d+ and oxygen delta^(-). The polarization of carbonyl groups also effects the boiling point of aldehydes and ketones to be higher than those of hydrocarbons in the same amount. The larger the carbonyl compound the less soluble it is in water. If the compound exceeds six carbons it then becomes insoluble.
*For more information about carbonyl solubility, look in the "outside links" section
*Amides are the most stable of the carbonyl couplings due to the high-resonance stabilization between nitrogen-carbon and carbon-oxygen.
Nucleophile Addition to a Carbonyl Group
C=O is prone to additions and nucleophillic attack because or carbon's positive charge and oxygen's negative charge. The resonance of the carbon partial positive charge allows the negative charge on the nucleophile to attack the Carbonyl group and become a part of the structure and a positive charge (usually a proton hydrogen) attacks the oxygen. Just a reminder, the nucleophile is a good acid therefore "likes protons" so it will attack the side with a positive charge.
*Remember: due to the electronegative nature of oxygen the carbon is partially positive and oxygen is partially negative
1 2 3
1. The Nucleophile (Nu) attacks the positively charged carbon and pushes one of the double bond electrons onto oxygen to give it a negative charge.
2. The Nucleophile is now a part of the carbonl structure with a negatively cahrged oxygen and a Na+ "floating" around.
3. The negatively charged oxygen attacks the proton (H+) to give the resulting product above.
Problems
1. What is the hybridization of the carbon in the C=O? the oxygen?
2. Illustrate the correct partial positive/negative or polarization of a formaldehyde.
3. Is carboxylic acid soluble in water? enone? acetaldehyde?
Answers
1. sp2;sp2
2. partial positive on the carbon and partial negative on the oxygen
3. yes; yes; yes
Contributors
• Sharleen Agvateesiri | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Aldehydes_and_Ketones/Properties_of_Aldehydes_and_Ketones/The_Carbonyl_Group.txt |
Aldehydes are typically more reactive than ketones due to the following factors.
1. Aldehydes are less hindered than ketones (a hydrogen atom is smaller than any other organic group).
2. The carbonyl carbon in aldehydes generally has more partial positive charge than in ketones due to the electron-donating nature of alkyl groups. Aldehydes only have one e- donor group while ketones have two.
• Addition-Elimination Reactions
This page looks at the reaction of aldehydes and ketones with 2,4-dinitrophenylhydrazine (Brady's reagent) as a test for the carbon-oxygen double bond. It also looks briefly at some other similar reactions which are all known as addition-elimination (or condensation) reactions.
• Addition of Alcohols to form Hemiacetals and Acetals
In this organic chemistry topic, we shall see how alcohols (R-OH) add to carbonyl groups. Carbonyl groups are characterized by a carbon-oxygen double bond. The two main functional groups that consist of this carbon-oxygen double bond are Aldehydes and Ketones.
• Addition of Secondary Amines to Form Enamines
Most aldehydes and ketones react with 2º-amines to give products known as enamines. It should be noted that, like acetal formation, these are acid-catalyzed reversible reactions in which water is lost. Consequently, enamines are easily converted back to their carbonyl precursors by acid-catalyzed hydrolysis.
• Addition of Water to form Hydrates (Gem-Diols)
It has been demonstrated that water, in the presence of an acid or a base, adds rapidly to the carbonyl function of aldehydes and ketones establishing a reversible equilibrium with a hydrate (geminal-diol or gem-diol).
• Alpha-carbon Reactions
Many aldehydes and ketones undergo substitution reactions at alpha carbons. These reactions are acid or base catalyzed, but in the case of halogenation the reaction generates an acid as one of the products, and is therefore autocatalytic. If the alpha-carbon is a chiral center, as in the second example, the products of halogenation and isotopic exchange are racemic.
• Carbonyl Group-Mechanisms of Addition
The Carbonyl Group is a polar functional group that is made up a carbon and oxygen double bonded together. There are two simple classes of the carbonyl group: Aldehydes and Ketones. Aldehydes have the carbon atom of the carbonyl group is bound to a hydrogen and ketones have the carbon atom of the carbonyl group is bound to two other carbons. Since the carbonyl group is extremely polar across the carbon-oxygen double bond, this makes it susceptible to addition reactions like the ones that occur i
• Carbonyl Group Reactions
The metal hydride reductions and organometallic additions to aldehydes and ketones, described above, both decrease the carbonyl carbon's oxidation state, and may be classified as reductions. As noted, they proceed by attack of a strong nucleophilic species at the electrophilic carbon. Other useful reductions of carbonyl compounds, either to alcohols or to hydrocarbons, may take place by different mechanisms.
• Clemmensen Reduction
The reaction of aldehydes and ketones with zinc amalgam (Zn/Hg alloy) in concentrated hydrochloric acid, which reduces the aldehyde or ketone to a hydrocarbon, is called Clemmensen reduction.
• Conjugate Addition Reactions
One of the largest and most diverse classes of reactions is composed of nucleophilic additions to a carbonyl group. Conjugation of a double bond to a carbonyl group transmits the electrophilic character of the carbonyl carbon to the beta-carbon of the double bond. These conjugated carbonyl are called enones or α, β unsaturated carbonyls.
• Cyanohydrins
Cyanohydrins have the structural formula of R2C(OH)CN. The “R” on the formula represents an alkyl, aryl, or hydrogen. In order to form a cyanohydrin, a hydrogen cyanide adds reversibly to the carbonyl group of an organic compound thus forming a hydroxyalkanenitrile adducts (commonly known and called as cyanohydrins).
• Irreversible Addition Reactions of Aldehydes and Ketones
The distinction between reversible and irreversible carbonyl addition reactions may be clarified by considering the stability of alcohols.
• Oxidation of Aldehydes and Ketones
This page looks at ways of distinguishing between aldehydes and ketones using oxidizing agents such as acidified potassium dichromate(VI) solution, Tollens' reagent, Fehling's solution and Benedict's solution.
• Reactions with Grignard Reagents
Reactions of aldehydes and ketones with Grignard reagents produce potentially quite complicated alcohols.
• Reaction with Primary Amines to form Imines
The reaction of aldehydes and ketones with ammonia or 1º-amines forms imine derivatives, also known as Schiff bases (compounds having a C=N function). Water is eliminated in the reaction, which is acid-catalyzed and reversible in the same sense as acetal formation. The pH for reactions which form imine compounds must be carefully controlled.
• Reduction of Aldehydes and Ketones
This page looks at the reduction of aldehydes and ketones by two similar reducing agents - lithium tetrahydridoaluminate(III) (also known as lithium aluminium hydride) and sodium tetrahydridoborate(III) (sodium borohydride).
• Reduction of Carbonyls to Alcohols Using Metal Hydrides
The most common sources of the hydride nucleophile are lithium aluminum hydride and sodium borohydride. In metal hydrides reductions the resulting alkoxide salts are insoluble and need to be hydrolyzed (with care) before the alcohol product can be isolated. In the sodium borohydride reduction the methanol solvent system achieves this hydrolysis automatically. In the lithium aluminum hydride reduction water is usually added in a second step.
• Reductive Amination
Aldehydes and ketones can be converted into 1°, 2° and 3° amines using reductive amination. The reaction takes place in two parts. The first step is the nucleophiic addition of the carbonyl group to form an imine. The second step is the reduction of the imine to an amine using an reducing agent.
• Reversible Addition Reactions of Aldehydes and Ketones
water adds rapidly to the carbonyl function of aldehydes and ketones. In most cases the resulting hydrate (a geminal-diol) is unstable relative to the reactants and cannot be isolated. Exceptions to this rule exist, one being formaldehyde (a gas in its pure monomeric state). Here the weaker pi-component of the carbonyl double bond, relative to other aldehydes or ketones, and the small size of the hydrogen substituents favor addition.
• Simple Addition Reactions
This page looks at the addition of hydrogen cyanide and sodium hydrogensulphite (sodium bisulphite) to aldehydes and ketones.
• The Triiodomethane (Iodoform) Reaction
This page looks at how the triiodomethane (iodoform) reaction can be used to identify the presence of a CH3CO group in aldehydes and ketones. There are two apparently quite different mixtures of reagents that can be used to do this reaction. They are, in fact, chemically equivalent.
• The Wittig Reaction
Organophosphorus ylides react with aldehydes or ketones to give substituted alkenes in a transformation called the Wittig reaction. This reaction is named for George Wittig who was awarded the Nobel prize for this work in 1979. A principal advantage of alkene synthesis by the Wittig reaction is that the location of the double bond is absolutely fixed, in contrast to the mixtures often produced by alcohol dehydration.
• Tollens’ Test
Tollens’ test, also known as silver-mirror test, is a qualitative laboratory test used to distinguish between an aldehyde and a ketone. It exploits the fact that aldehydes are readily oxidized, whereas ketones are not. Tollens’ test uses a reagent known as Tollens’ reagent, which is a colorless, basic, aqueous solution containing silver ions coordinated to ammonia
• Wolff-Kishner Reduction
Aldehydes and ketones can be converted to a hydrazine derivative by reaction with hydrazine. These "hydrazones" can be further converted to the corresponding alkane by reaction with base and heat. These two steps can be combined into one reaction called the Wolff-Kishner Reduction which represents a general method for converting aldehydes and ketones into alkanes. Typically a high boiling point solvent, such as ethylene glycol, is used to provide the high temperatures needed for this reaction.
Reactivity of Aldehydes and Ketones
This page looks at the reaction of aldehydes and ketones with 2,4-dinitrophenylhydrazine (Brady's reagent) as a test for the carbon-oxygen double bond. It also looks briefly at some other similar reactions which are all known as addition-elimination (or condensation) reactions.
The reaction with 2,4-dinitrophenylhydrazine
2,4-dinitrophenylhydrazine is often abbreviated to 2,4-DNP or 2,4-DNPH. A solution of 2,4-dinitrophenylhydrazine in a mixture of methanol and sulfuric acid is known as Brady's reagent. Although the name sounds complicated, and the structure of 2,4-dinitrophenylhydrazine looks quite complicated, it is actually very easy to work out. Start with the formula of hydrazine. That's almost all you need to remember!
Hydrazine is:
In phenylhydrazine, one of the hydrogens is replaced by a phenyl group, C6H5. This is based on a benzene ring.
In 2,4-dinitrophenylhydrazine, there are two nitro groups, NO2, attached to the phenyl group in the 2- and 4- positions. The corner with the nitrogen attached is counted as the number 1 position, and you just number clockwise around the ring.
Doing the reaction
Details vary slightly depending on the nature of the aldehyde or ketone, and the solvent that the 2,4-dinitrophenylhydrazine is dissolved in. Assuming you are using Brady's reagent (a solution of the 2,4-dinitrophenylhydrazine in methanol and sulphuric acid). Add either a few drops of the aldehyde or ketone, or possibly a solution of the aldehyde or ketone in methanol, to the Brady's reagent. A bright orange or yellow precipitate shows the presence of the carbon-oxygen double bond in an aldehyde or ketone. This is the simplest test for an aldehyde or ketone.
The overall reaction is given by the equation:
R and R' can be any combination of hydrogen or hydrocarbon groups (such as alkyl groups). If at least one of them is a hydrogen, then the original compound is an aldehyde. If both are hydrocarbon groups, then it is a ketone.
Look carefully at what has happened.
The product is known as a "2,4-dinitrophenylhydrazone". Notice that all that has changed is the ending from "-ine" to "-one". The product from the reaction with ethanal would be called ethanal 2,4-dinitrophenylhydrazone; from propanone, you would get propanone 2,4-dinitrophenylhydrazone - and so on.
The reaction is known as a condensation reaction. A condensation reaction is one in which two molecules join together with the loss of a small molecule in the process. In this case, that small molecule is water. In terms of mechanisms, this is a nucleophilic addition-elimination reaction. The 2,4-dinitrophenylhydrazine first adds across the carbon-oxygen double bond (the addition stage) to give an intermediate compound which then loses a molecule of water (the elimination stage).
Using the reaction
The reaction has two uses in testing for aldehydes and ketones.
• First, you can just use it to test for the presence of the carbon-oxygen double bond. You only get an orange or yellow precipitate from a carbon-oxygen double bond in an aldehyde or ketone.
• Secondly, you can use it to help to identify the specific aldehyde or ketone.
The precipitate is filtered and washed with, for example, methanol and then recrystallised from a suitable solvent which will vary depending on the nature of the aldehyde or ketone. For example, you can recrystallise the products from the small aldehydes and ketones from a mixture of ethanol and water.
The crystals are dissolved in the minimum quantity of hot solvent. When the solution cools, the crystals are re-precipitated and can be filtered, washed with a small amount of solvent and dried. They should then be pure.
If you then find the melting point of the crystals, you can compare it with tables of the melting points of 2,4-dinitrophenylhydrazones of all the common aldehydes and ketones to find out which one you are likely to have got.
Some other similar reactions
If you go back and look at the equations, nothing in the 2,4-dinitrophenylhydrazine changes during the reaction apart from the -NH2 group. You can get a similar reaction if the -NH2 group is attached to other things. In each case, the reaction would look like this:
In what follows, all that changes is the nature of the "X".
with hydrazine
The product is a "hydrazone". If you started from propanone, it would be propanone hydrazone.
with phenylhydrazine
The product is a "phenylhydrazone".
with hydroxylamine
The product is an "oxime" - for example, ethanal oxime.
Contributors
Jim Clark (Chemguide.co.uk) | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Aldehydes_and_Ketones/Reactivity_of_Aldehydes_and_Ketones/Addition-Elimination_Reactions.txt |
In this organic chemistry topic, we shall see how alcohols (R-OH) add to carbonyl groups. Carbonyl groups are characterized by a carbon-oxygen double bond. The two main functional groups that consist of this carbon-oxygen double bond are Aldehydes and Ketones.
Introduction
It has been demonstrated that water adds rapidly to the carbonyl function of aldehydes and ketones to form geminal-diol. In a similar reaction alcohols add reversibly to aldehydes and ketones to form hemiacetals (hemi, Greek, half). This reaction can continue by adding another alcohol to form an acetal. Hemiacetals and acetals are important functional groups because they appear in sugars.
To achieve effective hemiacetal or acetal formation, two additional features must be implemented. First, an acid catalyst must be used because alcohol is a weak nucleophile; and second, the water produced with the acetal must be removed from the reaction by a process such as a molecular sieves or a Dean-Stark trap. The latter is important, since acetal formation is reversible. Indeed, once pure hemiacetal or acetals are obtained they may be hydrolyzed back to their starting components by treatment with aqueous acid and an excess of water.
Formation of Acetals
Acetals are geminal-diether derivatives of aldehydes or ketones, formed by reaction with two equivalents (or an excess amount) of an alcohol and elimination of water. Ketone derivatives of this kind were once called ketals, but modern usage has dropped that term. It is important to note that a hemiacetal is formed as an intermediate during the formation of an acetal.
Mechanism for Hemiacetal and Acetal Formation
The mechanism shown here applies to both acetal and hemiacetal formation
1) Protonation of the carbonyl
2) Nucleophilic attack by the alcohol
3) Deprotonation to form a hemiacetal
4) Protonation of the alcohol
5) Removal of water
6) Nucleophilic attack by the alcohol
7) Deprotonation by water
Formation of Cyclic Hemiacetal and Acetals
Molecules which have an alcohol and a carbonyl can undergo an intramolecular reaction to form a cyclic hemiacetal.
Intramolecular Hemiacetal formation is common in sugar chemistry. For example, the common sugar glucose exists in the cylcic manner more than 99% of the time in a mixture of aqueous solution.
Carbonyls reacting with diol produce a cyclic acetal. A common diol used to form cyclic acetals is ethylene glycol.
Acetals as Protecting Groups
The importance of acetals as carbonyl derivatives lies chiefly in their stability and lack of reactivity in neutral to strongly basic environments. As long as they are not treated by acids, especially aqueous acid, acetals exhibit all the lack of reactivity associated with ethers in general. Among the most useful and characteristic reactions of aldehydes and ketones is their reactivity toward strongly nucleophilic (and basic) metallo-hydride, alkyl and aryl reagents. If the carbonyl functional group is converted to an acetal these powerful reagents have no effect; thus, acetals are excellent protective groups, when these irreversible addition reactions must be prevented.
In the following example we would like a Grignard reagent to react with the ester and not the ketone. This cannot be done without a protecting group because Grignard reagents react with esters and ketones. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Aldehydes_and_Ketones/Reactivity_of_Aldehydes_and_Ketones/Addition_of_Alcohols_to_form_Hemiacetals_and_Acetals.txt |
Most aldehydes and ketones react with 2º-amines to give products known as enamines. It should be noted that, like acetal formation, these are acid-catalyzed reversible reactions in which water is lost. Consequently, enamines are easily converted back to their carbonyl precursors by acid-catalyzed hydrolysis.
Mechanism
1) Nuleophilic attack
2) Proton transfer
3) Protonation of OH
4) Removal of water
5) Deprotonation
Problems
1) Please draw the products for the following reactions.
2) Please give the structure of the reactant needed to product the following product
1)
2)
Addition of Water to form Hydrates (Gem-Diols)
It has been demonstrated that water, in the presence of an acid or a base, adds rapidly to the carbonyl function of aldehydes and ketones establishing a reversible equilibrium with a hydrate (geminal-diol or gem-diol). The word germinal or gem comes from the Latin word for twin, geminus.
Reversibility of the Reaction
Isolation of gem-diols is difficult because the reaction is reversibly. Removal of the water during a reaction can cause the conversion of a gem-diol back to the corresponding carbonyl.
Factors Affecting the Gem-diol Equilibrium
In most cases the resulting gem-diol is unstable relative to the reactants and cannot be isolated. Exceptions to this rule exist, one being formaldehyde where the weaker pi-component of the carbonyl double bond, relative to other aldehydes or ketones, and the small size of the hydrogen substituents favor addition. Thus, a solution of formaldehyde in water (formalin) is almost exclusively the hydrate, or polymers of the hydrate. The addition of electron donating alkyl groups stabilized the partial positive charge on the carbonyl carbon and decreases the amount of gem-diol product at equilibrium. Because of this ketones tend to form less than 1% of the hydrate at equilibrium. Likewise, the addition of strong electron-withdrawing groups destabilizes the carbonyl and tends to form stable gem-diols. Two examples of this are chloral, and 1,2,3-indantrione. It should be noted that chloral hydrate is a sedative and has been added to alcoholic beverages to make a “Knock-out” drink also called a Mickey Finn. Also, ninhydrin is commonly used by forensic investigators to resolve finger prints.
Mechanism of Gem-diol Formation
The mechanism is catalyzed by the addition of an acid or base. Note! This may speed up the reaction but is has not effect on the equilibriums discussed above. Basic conditions speed up the reaction because hydroxide is a better nucleophilic than water. Acidic conditions speed up the reaction because the protonated carbonyl is more electrophilic.
Under Basic conditions
1) Nucleophilic attack by hydroxide
2) Protonation of the alkoxide
Under Acidic conditions
1) Protonation of the carbonyl
2) Nucleophilic attack by water
3) Deprotonation
Problems
1) Draw the expected products of the following reactions.
2) Of the following pairs of molecules which would you expect to form a larger percentage of gem-diol at equilibrium? Please explain your answer.
3) Would you expect the following molecule to form appreciable amount of gem-diol in water? Please explain your answer.
Answers
1)
2) The compound on the left would. Fluorine is more electronegative than bromine and would remove more electron density from the carbonyl carbon. This would destabilize the carbonyl allowing for more gem-diol to form.
3) Although ketones tend to not form gem-diols this compound exists almost entirely in the gem-diol form when placed in water. Ketones tend to not form gem-diols because of the stabilizing effect of the electron donating alkyl group. However, in this case the electron donating effects of alkyl group is dominated by the presence of six highly electronegative fluorines. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Aldehydes_and_Ketones/Reactivity_of_Aldehydes_and_Ketones/Addition_of_Secondary_Amines_to_Form_Enamines.txt |
Many aldehydes and ketones undergo substitution reactions at alpha carbons.
Reactions at the α-Carbon
Many aldehydes and ketones undergo substitution reactions at an alpha carbon, as shown in the following diagram (alpha-carbon atoms are colored blue). These reactions are acid or base catalyzed, but in the case of halogenation the reaction generates an acid as one of the products, and is therefore autocatalytic. If the alpha-carbon is a chiral center, as in the second example, the products of halogenation and isotopic exchange are racemic. Indeed, treatment of this ketone reactant with acid or base alone serves to racemize it. Not all carbonyl compounds exhibit these characteristics, the third ketone being an example.
Figure 1: Reactions at the $\alpha$ carbon
Two important conclusions may be drawn from these examples. First, these substitutions are limited to carbon atoms alpha to the carbonyl group. Cyclohexanone (the first ketone) has two alpha-carbons and four potential substitutions (the alpha-hydrogens). Depending on the reaction conditions, one or all four of these hydrogens may be substituted, but none of the remaining six hydrogens on the ring react. The second ketone confirms this fact, only the alpha-carbon undergoing substitution, despite the presence of many other sites. Second, the substitutions are limited to hydrogen atoms. This is demonstrated convincingly by the third ketone, which is structurally similar to the second but has no alpha-hydrogen.
Mechanism of Electrophilic α-Substitution
Kinetic studies of these reactions provide additional information. The rates of halogenation and isotope exchange are essentially the same (assuming similar catalysts and concentrations), and are identical to the rate of racemization for those reactants having chiral alpha-carbon units. At low to moderate halogen concentrations, the rate of halogen substitution is proportional (i.e. first order) to aldehyde or ketone concentration, but independent of halogen concentration. This suggests the existence of a common reaction intermediate, formed in a slow (rate-determining step) prior to the final substitution. Acid and base catalysts act to increase the rate at which the common intermediate is formed, and their concentration also influences the overall rate of substitution.
From previous knowledge and experience, we surmise that the common intermediate is an enol tautomer of the carbonyl reactant. Several facts support this proposal:
1. Compounds that do not have any α-hydrogen atoms cannot enolize and do not undergo any of the reactions described above.
2. The carbon-carbon double bond of an enol is planar, so any chirality that existed at the α carbon is lost on enolization. If chiral products are obtained from enol intermediates they will necessarily be racemic.
3. In simple aldehydes and ketones enol tautomers are present in very low concentration. Reactions that involve enol reactants will therefore be limited in rate by the enol concentration. Increasing the amounts of other reactants will have little effect on the reaction rate.
4. Enolization is catalyzed by acids and bases. These catalysts will therefore catalyze reactions proceeding via enol intermediates.
The reactions shown above, and others to be described, may be characterized as an electrophilic attack on the electron rich double bond of an enol tautomer. This resembles closely the first step in the addition of acids and other electrophiles to alkenes. Therefore, if electrophilic substitution reactions of this kind are to take place it is necessary that nucleophilic character be established at the alpha-carbon. A full description of the acid and base-catalyzed keto-enol tautomerization process (shown below) discloses that only two intermediate species satisfy this requirement. These are the enol tautomer itself and its conjugate base (common with that of the keto tautomer), usually referred to as an enolate anion.
Figure 2: Keto-enol tautomerization
Clearly, the proportion of enol tautomer present at equilibrium is a critical factor in alpha substitution reactions. In the case of simple aldehydes and ketones this is very small, as noted above. A complementary property, the acidity of carbonyl compounds is also important, since this influences the concentration of the more nucleophilic enolate anion in a reaction system. Ketones such as cyclohexanone are much more acidic than their parent hydrocarbons (by at least 25 powers of ten); nevertheless they are still very weak acids (pKa = 17 to 21) compared with water. Together with some related acidities, this is listed in the following table. Even though enol tautomers are about a million times more acidic than their keto isomers, their low concentration makes this feature relatively unimportant for many simple aldehydes and ketones.
Table 1: Acidity of α-Hydrogens in Some Activated Compounds
Compound RCH2–NO2 RCH2–COR RCH2–C≡N RCH2–SO2R
pKa 9 20 25 25
In cases where more than one activating function influences a given set of alpha-hydrogens, the enol concentration and acidity is increased. Examples of such doubly (and higher) activated carbon acids are given elsewhere.
Enols and Enolate Anions
In view of these facts it may seem surprising that alpha-substitution reactions occur at all. However, we often fail to appreciate the way in which a rapid equilibrium involving a minor reactive component may spread the consequences of its behavior throughout a much larger population. Consider, for example, a large group of hungry, active hamsters running about in a big cage. Opening onto the cage there is a small annex that can hold a maximum of three hamsters. Out of two hundred hamsters in the cage, there are an average of two hamsters in the annex at any given time. The hamsters are free to enter and exit the annex, but any hamster that does so is marked by a bright red dye. Although the hamster concentration in the annex is small relative to the whole population, it will not be long before all the hamsters are dyed red. If we substitute molecules for hamsters, their numbers will be extraordinarily large (recall the size of Avogadro's number), but the equilibrium between keto tautomers (hamsters in the cage) and enol tautomers (hamsters in the annex) is so rapid that complete turnover of all the molecules in a sample may occur in fractions of a second rather than minutes or hours. The principle is the same in both cases.
Racemization and isotope exchange are due to the rapid equilibrium between chiral keto tautomers and achiral enol tautomers, as well as statistical competition between hydrogen and its deuterium isotope. For halogenation there is also a thermodynamic driving force, resulting from increased bond energy in the products. For example, the alpha-chlorination of cyclohexane, shown above, is exothermic by over 10 kcal/mole.
The Aldol Reaction
A useful carbon-carbon bond-forming reaction known as the Aldol Reaction or the Aldol Condensation is yet another example of electrophilic substitution at the alpha carbon in enols or enolate anions. Three examples of the base-catalyzed aldol reaction are shown in the following diagram, and equivalent acid-catalyzed reactions also occur. The fundamental transformation in this reaction is a dimerization of an aldehyde (or ketone) to a beta-hydroxy aldehyde (or ketone) by alpha C–H addition of one reactant molecule to the carbonyl group of a second reactant molecule. By clicking the "Structural Analysis" button below the diagram, a display showing the nucleophilic enolic donor molecule and the electrophilic acceptor molecule together with the newly formed carbon-carbon bond will be displayed. Stepwise mechanisms for the base-catalyzed and acid-catalyzed reactions may be seen by clicking the appropriate buttons.
Figure 3: Aldol reaction
In the presence of acid or base catalysts the aldol reaction is reversible, and the beta-hydroxy carbonyl products may revert to the initial aldehyde or ketone reactants. In the absence of such catalysts these aldol products are perfectly stable and isolable compounds. Because of this reversibility, the yield of aldol products is related to their relative thermodynamic stability. In the case of aldehyde reactants (as in reactions #1 & 2 above), the aldol reaction is modestly exothermic and the yields are good. However, aldol reactions of ketones are less favorable (e.g. #3 above), and the equilibrium product concentration is small. A clever way of overcoming this disadvantage has been found. A comparatively insoluble base, Ba(OH)2, is used to catalyze the aldol reaction of acetone, and the product is removed from contact with this base by filtration and recirculation of the acetone.
Dehydration of Aldol Products
The products of aldol reactions often undergo a subsequent elimination of water, made up of an alpha-hydrogen and the beta-hydroxyl group. The product of this beta-elimination reaction is an α,β-unsaturated aldehyde or ketone, as shown in the following diagram. Acid-catalyzed conditions are more commonly used to effect this elimination (examples #1, 2 & 5), but base-catalyzed elimination also occurs, especially on heating (examples #3, 4 & 5). The additional stability provided by the conjugated carbonyl system of the product makes some ketone aldol reactions thermodynamically favorable (#4 & 5), and mixtures of stereoisomers (E & Z) are obtained from reaction #4. Reaction #5 is an interesting example of an intramolecular aldol reaction; such reactions create a new ring.
Reactions in which a larger molecule is formed from smaller components, with the elimination of a very small by-product such as water are termed Condensations. Hence the following examples are properly referred to as aldol condensations. The dehydration step of an aldol condensation is also reversible in the presence of acid and base catalysts. Consequently, on heating with aqueous solutions of strong acids or bases, many α, β-unsaturated carbonyl compounds fragment into smaller aldehyde or ketones, a process known as the retro-aldol reaction.
Figure 4: Aldol condensations
The acid-catalyzed elimination of water is not exceptional, since this was noted as a common reaction of alcohols. Nevertheless, the conditions required for the beta-elimination are found to be milder than those used for simple alcohols. The most surprising aspect of beta-elimination, however, is that it can be base-catalyzed. In earlier discussions we have noted that hydroxyl anion is a very poor leaving group. Why then should the base-catalyzed elimination of water occur in aldol products? To understand this puzzle we need to examine plausible mechanisms for beta-elimination, and these will be displayed by clicking the "Beta-Elimination Mechanism" button under the diagram.
As shown by the equations, these eliminations might proceed from either the keto or enol tautomers of the beta-hydroxy aldol product. Although the keto tautomer route is not unreasonable (recall the enhanced acidity of the alpha-hydrogens in carbonyl compounds), the enol tautomer provides a more favorable pathway for both acid and base-catalyzed elimination of the beta oxygen. Indeed, the base-catalyzed loss of hydroxide anion from the enol is a conjugated analog of the base-catalyzed decomposition of a hemiacetal.
Figure 5: Beta-elimination
Mixed Aldol Condensations
The previous examples of aldol reactions and condensations used a common reactant as both the enolic donor and the electrophilic acceptor. The product in such cases is always a dimer of the reactant carbonyl compound. Aldol condensations between different carbonyl reactants are called crossed or mixed reactions, and under certain conditions such crossed aldol condensations can be effective. Some examples are shown below, and in most cases beta-elimination of water occurs under the conditions used. The exception, reaction #3, is conducted under mild conditions with an excess of the reactive aldehyde formaldehyde serving in the role of electrophilic acceptor. The first reaction demonstrates that ketones having two sets of alpha-hydrogens may react at both sites if sufficient acceptor co-reactant is supplied. The interesting difference in regioselectivity shown in the second reaction (the reactants are in the central shaded region) illustrates some subtle differences between acid and base-catalyzed aldol reactions. The base-catalyzed reaction proceeds via an enolate anion donor species, and the kinetically favored proton removal is from the less substituted alpha-carbon. The acid-catalyzed aldol proceeds via the enol tautomer, and the more stable of the two enol tautomers is that with the more substituted double bond. Finally, reaction #4 has two reactive alpha-carbons and a reversible aldol reaction may occur at both. Only one of the two aldol products can undergo a beta-elimination of water, so the eventual isolated product comes from that reaction sequence. The aldol condensation of ketones with aryl aldehydes to form α,β-unsaturated derivatives is called the Claisen-Schmidt reaction.
Figure 6: Claisen-Schmidt reaction
The success of these mixed aldol reactions is due to two factors. First, aldehydes are more reactive acceptor electrophiles than ketones, and formaldehyde is more reactive than other aldehydes. Second, aldehydes lacking alpha-hydrogens can only function as acceptor reactants, and this reduces the number of possible products by half. Mixed aldols in which both reactants can serve as donors and acceptors generally give complex mixtures of both dimeric (homo) aldols and crossed aldols. The following abbreviated formulas illustrate the possible products in such a case, red letters representing the acceptor component and blue the donor. If all the reactions occurred at the same rate, equal quantities of the four products would be obtained. Separation and purification of the components of such a mixture would be difficult.
ACH2CHO + BCH2CHO + NaOH AA + BB + AB + BA
Directed Stereoselective Aldol Reactions
The effectiveness of the aldol reaction as a synthetic tool has been enhanced by controlling the enolization of donor compounds, and subsequent reactions with acceptor carbonyls. To see how this is done Click Here.
Irreversible Substitution Reactions
In its simplest form the aldol reaction is reversible, and normally forms the thermodynamically favored product. To fully appreciate the complex interplay of factors that underlie this important synthesis tool, we must evaluate the significance of several possible competing reaction paths.
A. The Ambident Character of Enolate Anions
Since the negative charge of an enolate anion is delocalized over the alpha-carbon and the oxygen, as shown earlier, electrophiles may bond to either atom. Reactants having two or more reactive sites are called ambident, so this term is properly applied to enolate anions. Modestly electrophilic reactants such as alkyl halides are not sufficiently reactive to combine with neutral enol tautomers, but the increased nucleophilicity of the enolate anion conjugate base permits such reactions to take place. Because alkylations are usually irreversible, their products should reflect the inherent (kinetic) reactivity of the different nucleophilic sites.
Figure 7: ambident enolates
If an alkyl halide undergoes an SN2 reaction at the carbon atom of an enolate anion the product is an alkylated aldehyde or ketone. On the other hand, if the SN2 reaction occurs at oxygen the product is an ether derivative of the enol tautomer; such compounds are stable in the absence of acid and may be isolated and characterized. These alkylations (shown above) are irreversible under the conditions normally used for SN2 reactions, so the product composition should provide a measure of the relative rates of substitution at carbon versus oxygen. It has been found that this competition is sensitive to a number of factors, including negative charge density, solvation, cation coordination and product stability.
For alkylation reactions of enolate anions to be useful, these intermediates must be generated in high concentration in the absence of other strong nucleophiles and bases. The aqueous base conditions used for the aldol condensation are not suitable because the enolate anions of simple carbonyl compounds are formed in very low concentration, and hydroxide or alkoxide bases induce competing SN2 and E2 reactions of alkyl halides. It is necessary, therefore, to achieve complete conversion of aldehyde or ketone reactants to their enolate conjugate bases by treatment with a very strong base (pKa > 25) in a non-hydroxylic solvent before any alkyl halides are added to the reaction system. Some bases having pKa's greater than 30 were described earlier, and some others that have been used for enolate anion formation are: NaH (sodium hydride, pKa > 45), NaNH2 (sodium amide, pKa = 34), and (C6H5)3CNa (trityl sodium, pKa = 32). Ether solvents like tetrahydrofuran (THF) are commonly used for enolate anion formation. With the exception of sodium hydride and sodium amide, most of these bases are soluble in THF. Certain other strong bases, such as alkyl lithium and Grignard reagents, cannot be used to make enolate anions because they rapidly and irreversibly add to carbonyl groups. Nevertheless, these very strong bases are useful in making soluble amide bases. In the preparation of lithium diisopropylamide (LDA), for example, the only other product is the gaseous alkane butane.
C4H9–Li +
butyl lithium
[(CH3)2CH]2N–H
diisopropylamine
[(CH3)2CH]2N(–) Li(+) + C4H10
LDA
O=C-C-H + LDA (–)O–C=C + [(CH3)2CH]2N–H
Because of its solubility in THF, LDA is a widely used base for enolate anion formation. In this application one equivalent of diisopropylamine is produced along with the lithium enolate, but this normally does not interfere with the enolate reactions and is easily removed from the products by washing with aqueous acid. Although the reaction of carbonyl compounds with sodium hydride is heterogeneous and slow, sodium enolates are formed with the loss of hydrogen, and no other organic compounds are produced. The following equation provides examples of electrophilic substitution at both carbon and oxygen for the enolate anion derived from cyclohexanone.
Figure 8: Enolate anion formation
A full analysis of the factors that direct substitution of enolate anions to carbon or oxygen is beyond the scope of this text. However, an outline of some significant characteristics that influence the two reactions shown above is illustrative.
Reactant
Important Factors
CH3–I
The negative charge density is greatest at the oxygen atom (greater electronegativity), and coordination with the sodium cation is stronger there. Because methyl iodide is only a modest electrophile, the SN2 transition state resembles the products more than the reactants. Since the C-alkylation product is thermodynamically more stable than the O-alkylated enol ether, this is reflected in the transition state energies.
(CH3)3Si–Cl
Trimethylsilyl chloride is a stronger electrophile than methyl iodide (note the electronegativity difference between silicon and chlorine). Relative to the methylation reaction, the SN2 transition state will resemble the reactants more than the products. Consequently, reaction at the site of greatest negative charge (oxygen) will be favored. Also, the high Si–O bond energy (over 25 kcal/mole greater than Si–C) thermodynamically favors the silyl enol ether product.
B. Alkylation Reactions of Enolate Anions
The reaction of alkyl halides with enolate anions presents the same problem of competing SN2 and E2 reaction paths that was encountered earlier in the alkyl halide chapter. Since enolate anions are very strong bases, they will usually cause elimination when reacted with 2º and 3º-halides. Halides that are incapable of elimination and/or have enhanced SN2 reactivity are the best electrophilic reactants for this purpose. Four examples of the C-alkylation of enolate anions in synthesis are displayed in the following diagram. The first two employ the versatile strong base LDA, which is the reagent of choice for most intermolecular alkylations of simple carbonyl compounds. The dichloro alkylating agent used in reaction #1 nicely illustrates the high reactivity of allylic halides and the unreactive nature of vinylic halides in SN2 reactions.
Figure 9: C-alkylation of enolate anions
The additive effect of carbonyl groups on alpha-hydrogen acidity is demonstrated by reaction #3. Here the two hydrogen atoms activated by both carbonyl groups are over 1010 times more acidic than the methyl hydrogens on the ends of the carbon chain. Indeed, they are sufficiently acidic (pKa = 9) to allow complete conversion to the enolate anion in aqueous or alcoholic solutions. As shown (in blue), the negative charge of the enolate anion is delocalized over both oxygen atoms and the central carbon. The oxygens are hydrogen bonded to solvent molecules, so the kinetically favored SN2 reaction occurs at the carbon. The monoalkylated product shown in the equation still has an acidic hydrogen on the central carbon, and another alkyl group may be attached there by repeating this sequence.
The last example (reaction #4) is an interesting case of intramolecular alkylation of an enolate anion. Since alkylation reactions are irreversible, it is possible to form small highly strained rings if the reactive sites are in close proximity. Reversible bond-forming reactions, such as the aldol reaction, cannot be used for this purpose. The use of aqueous base in this reaction is also remarkable, in view of the very low enolate anion concentration noted earlier for such systems. It is the rapid intramolecular nature of the alkylation that allows these unfavorable conditions to be used.
Figure 10: O-alkylation of enolate anion
The five-carbon chain of the dichloroketone can adopt many conformations, two of which are approximated in the preceding diagram. Although conformer II of the enolate anion could generate a stable five-membered ring by an intramolecular SN2 reaction, assuming proper orientation of the α and γ' carbon atoms, the concentration of this ideally coiled structure will be very low. In this case O-alkylation of the enolate anion, rather than C-alkylation, is preferred from stereoelectronic arguments (see Baldwin rules). On the other hand, conformations in which the α and γ-carbons are properly aligned for three-membered ring formation are much more numerous, the result being that as fast as the enolate base is formed it undergoes rapid and irreversible cyclization.
Ring closures to four, five, six and seven-membered are also possible by intramolecular enolate alkylation, as illustrated by the following example. In general, five and six-membered rings are thermodynamically most stable, whereas three-membered ring formation is favored kinetically.
Figure 11: Intramolecular enolate alkylation
Alternatives to Enolate Anions
Several enolate-like compounds and ions have been studied as alternative intermediates for synthesis. To learn more about these possibilities Click Here.
Contributors
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Aldehydes_and_Ketones/Reactivity_of_Aldehydes_and_Ketones/Alpha-carbon_Reactions.txt |
The Carbonyl Group is a polar functional group that is made up a carbon and oxygen double bonded together. There are two simple classes of the carbonyl group: Aldehydes and Ketones. Aldehydes have the carbon atom of the carbonyl group is bound to a hydrogen and ketones have the carbon atom of the carbonyl group is bound to two other carbons. Since the carbonyl group is extremely polar across the carbon-oxygen double bond, this makes it susceptible to addition reactions like the ones that occur in the pi bond of Alkenes, especially by nucleophilic and electrophilic attack.
Regions of Reactivity
There are three regions of reactivity for both aldehydes and ketones: the electron donating oxygen, the electron withdrawing carbon, and the carbon adjacent to the carbonyl group (labeled "alpha"). This module will only address the oxygen and electron withdrawing carbon areas of reactivity.
Figure 1: Regions of Reactivity
Resonance structure is defined as any of two or more possible structures of the same compound that have identical geometry but different arrangements of their paired electrons; none of the structures have physical reality or adequately account for the properties of the compound. Examination of the resonance structures of the carbonyl group clearly shows its polar nature, and highlights the areas for either electrophilic or nucleophilic attack in the addition reaction.
Figure 2: Resonance structures of the carbonyl group
Ionic Addition to Carbonyl Group
As a result of the dipole shown in the resonance structures, polar reagents such as LiAlH4 and NaBH4 (hydride reagents) or R'MgX (Grignard reagent) will reduce the carbonyl groups, and ultimately convert unsaturated aldehydes and ketones into unsaturated alcohols. Since these reagents are extremely basic, their addition reactions are irreversible.
There are, however, addition reactions with less basic nucleophiles such as water, thiols, and amines that are capable of establishing equilibria or reversible reactions. These less basic reagents can react with the carbonyl group via two pathways: nucleophilic addition-protonation and electrophilic addition-protonation.
Nucleophilic Addition-Protonation
Under neutral or basic conditions, nucleophilic attack of the electrophilic carbon takes place. As the nucleophile approaches the electrophilic carbon, two valence electrons from the nucleophile form a covalent bond to the carbon. As this occurs, the electron pair from the pie bond transfers completely over to the oxygen which produces the intermediate alkoxide ion. This alkoxide ion, with a negative charge on oxygen is susceptible to protonation from a protic solvent like water or alcohol, giving the final addition reaction.
Figure 3: Alkoxide Ion Intermediate
Electrophilic Addition-Protonation
Under acidic conditions, electrophilic attack of the carbonyl oxygen takes place. Initially, protonation of the carbonyl group at the oxygen takes place because of excess \(H^+\) all around. Once protonation has occurred, nucleophilic attack by the nucleophile finishes the addition reaction. It should be noted that electrophilic attack is extremely unlikely, however, a few carbonyl groups do become protonated initially to initiate addition through electrophilic attack. This type of reaction works best when the reagent being used is a very mildly basic nucleophile.
Figure 4: Electrophilic Addition
Problems
1. What is the best reagent to do the following conversion?
1. H2/Ni
2. Hg(OAc)2, H3O+
3. 1. BH3, THF & 2.H2O2, NaOH
4. NaBH4, THF
2. What is the Product of this reaction?
3. What is the Product of this reaction?
4. What is the Product of this reaction?
5. What is the product of this reaction?
1. D
2.
3.
4.
5. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Aldehydes_and_Ketones/Reactivity_of_Aldehydes_and_Ketones/Carbonyl_Group-Mechanisms_of_Addition.txt |
The metal hydride reductions and organometallic additions to aldehydes and ketones, described above, both decrease the carbonyl carbon's oxidation state, and may be classified as reductions. As noted, they proceed by attack of a strong nucleophilic species at the electrophilic carbon. Other useful reductions of carbonyl compounds, either to alcohols or to hydrocarbons, may take place by different mechanisms. For example, hydrogenation (Pt, Pd, Ni or Ru catalysts), reaction with diborane, and reduction by lithium, sodium or potassium in hydroxylic or amine solvents have all been reported to convert carbonyl compounds into alcohols. However, the complex metal hydrides are generally preferred for such transformations because they give cleaner products in high yield.
Aldehydes and ketones may also be reduced by hydride transfer from alkoxide salts.
The reductive conversion of a carbonyl group to a methylene group requires complete removal of the oxygen, and is called deoxygenation. In the shorthand equation shown here the [H] symbol refers to unspecified reduction conditions which effect the desired change. Three very different methods of accomplishing this transformation will be described here.
R2C=O + [H] R2CH2 + H2O
Wolff-Kishner Reduction
Reaction of an aldehyde or ketone with excess hydrazine generates a hydrazone derivative, which on heating with base gives the corresponding hydrocarbon. A high-boiling hydroxylic solvent, such as diethylene glycol, is commonly used to achieve the temperatures needed. The following diagram shows how this reduction may be used to convert cyclopentanone to cyclopentane. A second example, in which an aldehyde is similarly reduced to a methyl group, also illustrates again the use of an acetal protective group. The mechanism of this useful transformation involves tautomerization of the initially formed hydrazone to an azo isomer, and will be displayed on pressing the "Show Mechanism" button. The strongly basic conditions used in this reaction preclude its application to base sensitive compounds.
Figure 1A: Wolff-Kischner reduction of cyclopentanone
Figure 1B: Wolff-Kischner reudction of cyclopentanone
Clemmensen Reduction
This alternative reduction involves heating a carbonyl compound with finely divided, amalgamated zinc in a hydroxylic solvent (often an aqueous mixture) containing a mineral acid such as HCl. The mercury alloyed with the zinc does not participate in the reaction, it serves only to provide a clean active metal surface. The first example below shows a common application of this reduction, the conversion of a Friedel-Crafts acylation product to an alkyl side-chain. The second example illustrates the lability of functional substituents alpha to the carbonyl group. Substituents such as hydroxyl, alkoxyl & halogens are reduced first, the resulting unsubstituted aldehyde or ketone is then reduced to the parent hydrocarbon.
Figure 2A: Clemmensen reduction
Figure 2B: Clemmensen reduction Mechanism
Hydrogenolysis of Thioacetals
In contrast to the previous two procedures, this method of carbonyl deoxygenation requires two separate steps. It does, however, avoid treatment with strong base or acid. The first step is to convert the aldehyde or ketone into a thioacetal: These derivatives may be isolated and purified before continuing the reduction. The second step involves refluxing an acetone solution of the thioacetal over a reactive nickel catalyst, called Raney Nickel. All carbon-sulfur bonds undergo hydrogenolysis (the C–S bonds are broken by addition of hydrogen). In the following example, 1,2-ethanedithiol is used for preparing the thioacetal intermediate, because of the high yield this reactant usually affords. The bicyclic compound shown here has two carbonyl groups, one of which is sterically hindered (circled in orange). Consequently, a mono-thioacetal is easily prepared from the less-hindered ketone, and this is reduced without changing the remaining carbonyl function.
Figure 3: Raney Nickel hydrogenolysis of thioacetals
Oxidation
The carbon atom of a carbonyl group has a relatively high oxidation state. This is reflected in the fact that most of the reactions described thus far either cause no change in the oxidation state (e.g. acetal and imine formation) or effect a reduction (e.g. organometallic additions and deoxygenations). The most common and characteristic oxidation reaction is the conversion of aldehydes to carboxylic acids. In the shorthand equation shown here the [O] symbol refers to unspecified oxidation conditions which effect the desired change. Several different methods of accomplishing this transformation will be described here.
RCH=O + [O] RC(OH)=O
In discussing the oxidations of 1º and 2º-alcohols, we noted that Jones' reagent (aqueous chromic acid) converts aldehydes to carboxylic acids, presumably via the hydrate. Other reagents, among them aqueous potassium permanganate and dilute bromine, effect the same transformation. Even the oxygen in air will slowly oxidize aldehydes to acids or peracids, most likely by a radical mechanism. Useful tests for aldehydes, Tollens' test, Benedict's test & Fehling's test, take advantage of this ease of oxidation by using Ag(+) and Cu(2+) as oxidizing agents (oxidants).
RCH=O + 2 [Ag(+) OH(–)] RC(OH)=O + 2 Ag (metallic mirror) + H2O
When silver cation is the oxidant, as in the above equation, it is reduced to metallic silver in the course of the reaction, and this deposits as a beautiful mirror on the inner surface of the reaction vessel. The Fehling and Benedict tests use cupric cation as the oxidant. This deep blue reagent is reduced to cuprous oxide, which precipitates as a red to yellow solid. All these cation oxidations must be conducted under alkaline conditions. To avoid precipitation of the insoluble metal hydroxides, the cations must be stabilized as complexed ions. Silver is used as its ammonia complex, Ag(NH3)2(+), and cupric ions are used as citrate or tartrate complexes.
Saturated ketones are generally inert to oxidation conditions that convert aldehydes to carboxylic acids. Nevertheless, under vigorous acid-catalyzed oxidations with nitric or chromic acids ketones may undergo carbon-carbon bond cleavage at the carbonyl group. The reason for the vulnerability of the alpha-carbon bond will become apparent in the following section. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Aldehydes_and_Ketones/Reactivity_of_Aldehydes_and_Ketones/Carbonyl_Group_Reactions.txt |
The reaction of aldehydes and ketones with zinc amalgam (Zn/Hg alloy) in concentrated hydrochloric acid, which reduces the aldehyde or ketone to a hydrocarbon, is called Clemmensen reduction.
Introduction
This alternative reduction involves heating a carbonyl compound with finely divided, amalgamated zinc in a hydroxylic solvent (often an aqueous mixture) containing a mineral acid such as HCl. The mercury alloyed with the zinc does not participate in the reaction, it serves only to provide a clean active metal surface.
Possible mechanism
The mechanism of Clemmensen reduction is not fully understood; intermediacy of radicals are implicated. Clemmensen reduction is complementary to Wolff-Kishner reduction, which also converts aldehydes and ketones to hydrocarbons, in that the former is carried out in strongly acidic conditions and the latter in strongly basic conditions.
Conjugate Addition Reactions
One of the largest and most diverse classes of reactions is composed of nucleophilic additions to a carbonyl group. Conjugation of a double bond to a carbonyl group transmits the electrophilic character of the carbonyl carbon to the beta-carbon of the double bond. These conjugated carbonyl are called enones or α, β unsaturated carbonyls. A resonance description of this transmission is shown below.
From this formula it should be clear that nucleophiles may attack either at the carbonyl carbon, as for any aldehyde, ketone or carboxylic acid derivative, or at the beta-carbon. These two modes of reaction are referred to as 1,2-addition and 1,4-addition respectively. A 1,4-addition is also called a conjugate addition.
Basic reaction of 1,2 addition
Here the nucleophile adds to the carbon which is in the one position. The hydrogen adds to the oxygen which is in the two position.
Basic reaction of 1,4 addition
In 1,4 addition the Nucleophile is added to the carbon β to the carbonyl while the hydrogen is added to the carbon α to the carbonyl.
Mechanism for 1,4 addition
1) Nucleophilic attack on the carbon β to the carbonyl
2) Proton Transfer
Here we can see why this addition is called 1,4. The nucleophile bonds to the carbon in the one position and the hydrogen adds to the oxygen in the four position.
3) Tautomerization
Going from reactant to products simplified
1,2 vs. 1,4 addition
Whether 1,2 or 1,4-addition occurs depends on multiple variables but mostly it is determined by the nature of the nucleophile. During the addition of a nucleophile there is a competition between 1,2 and 1,4 addition products. If the nucleophile is a strong base, such as Grignard reagents, both the 1,2 and 1,4 reactions are irreversible and therefor are under kinetic control. Since 1,2-additions to the carbonyl group are fast, we would expect to find a predominance of 1,2-products from these reactions.
If the nucleophile is a weak base, such as alcohols or amines, then the 1,2 addition is usually reversible. This means the competition between 1,2 and 1,4 addition is under thermodynamic control. In this case 1,4-addition dominates because the stable carbonyl group is retained.
Water
Alcohols
Thiols
1o Amines
2o Amines
HBr
Cyanides
Gilman Reagents
Another important reaction exhibited by organometallic reagents is metal exchange. Organolithium reagents react with cuprous iodide to give a lithium dimethylcopper reagent, which is referred to as a Gilman reagent. Gilman reagents are a source of carbanion like nucleophiles similar to Grignard and Organo lithium reagents. However, the reactivity of organocuprate reagents is slightly different and this difference will be exploited in different situations. In the case of α, β unsaturated carbonyls organocuprate reagents allow for an 1,4 addition of an alkyl group. As we will see later Grignard and Organolithium reagents add alkyl groups 1,2 to α, β unsaturated carbonyls
Organocuprate reagents are made from the reaction of organolithium reagents and $CuI$
$2 RLi + CuI \rightarrow R_2CuLi + LiI$
This acts as a source of R:-
$2 CH_3Li + CuI \rightarrow (CH_3)_2CuLi + LiI$
Example
Nucleophiles which add 1,2 to α, β unsaturated carbonyls
Metal Hydrides
Grignard Reagents
Organolithium Reagents
Contributors
Prof. Steven Farmer (Sonoma State University)
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry
Cyanohydrins
Cyanohydrins have the structural formula of R2C(OH)CN. The “R” on the formula represents an alkyl, aryl, or hydrogen. In order to form a cyanohydrin, a hydrogen cyanide adds reversibly to the carbonyl group of an organic compound thus forming a hydroxyalkanenitrile adducts (commonly known and called as cyanohydrins).
Introduction
Cyanohydrin reactions occurs when an aldehyde or ketone gets treated by a cyanide anion (such as HCN) or a nitrile forming a cyanohydrin product. This special reaction is a nucleophilic addition, where the nucleophilic CN- attacks the electrophilic carbonyl carbon on the ketone, following a protonation by HCN, thereby the cyanide anion being regenerated. This reaction is also reversible.
Cyanohydrins are also intermediates for the Strecker amino acid synthesis. The preparation of displacements of sulfite by cyanide salts are also followed under cyanohydrins.
Mechanism of Cyanohydrin Formation
Acid-catalysed hydrolysis of silylated cyanohydrins has recently been shown to give cyanohydrins instead of ketones; thus an efficient synthesis of cyanohydrins has been found which works with even highly hindered ketones.
Acetone Cyanohydrins
Acetone cyanohydrins (ACH) have the structural formula of (CH3)2C(OH)CN and are extremely hazardous substances, since they rapidly decomposes in contact with water. In ACH, sulfuric acid is treated to give the sulfate ester of the methacrylamid. Preparations of other cyanohydrins are also used from ACH: for HACN to Michael acceptors and for the formylation of arenas. The treatment with lithium hydride affords anhydrous lithium cyanide.
Other Cyanohydrins
Other cyanohydrins, excluding acetone cyanohydrins, are: mandelonitrile and glycolonitrile.
Mandelonitrile have a structural formula of C6H5CH(OH)CN and occur in pits of some fruits. Glycolonitrile is an organic compound with the structural formula of HOCH2CN, which is the simplest cyanohydrin that is derived by formaldehydes.
Problems
Complete the following reactions for cyanohydrins:
1.)
2.)
3.) True or False: For a cyanohydrin to form, a fast additon of strong acid to cyanide salt is carried out to convert the salt into HCN.
4.) True or False: Cyanohydrin reactions are irreversible.
5.) What is the product for the overall reaction?
Answers
1.)
2.)
3.) False, slow addition
4.) False, reversible
5.)
Contributors
• Kathy Wong (UCD) | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Aldehydes_and_Ketones/Reactivity_of_Aldehydes_and_Ketones/Clemmensen_Reduction.txt |
The distinction between reversible and irreversible carbonyl addition reactions may be clarified by considering the stability of alcohols having the structure shown below in the shaded box.
If substituent Y is not a hydrogen, an alkyl group or an aryl group, there is a good chance the compound will be unstable (not isolable), and will decompose in the manner shown. Most hydrates and hemiacetals (Y = OH & OR), for example, are known to decompose spontaneously to the corresponding carbonyl compounds. Aminols (Y = NHR) are intermediates in imine formation, and also revert to their carbonyl precursors if dehydration conditions are not employed. Likewise, α-haloalcohols (Y = Cl, Br & I) cannot be isolated, since they immediately decompose with the loss of HY. In all these cases addition of H–Y to carbonyl groups is clearly reversible.
If substituent Y is a hydrogen, an alkyl group or an aryl group, the resulting alcohol is a stable compound and does not decompose with loss of hydrogen or hydrocarbons, even on heating. It follows then, that if nucleophilic reagents corresponding to H:(–), R:(–) or Ar:(–) add to aldehydes and ketones, the alcohol products of such additions will form irreversibly. Free anions of this kind would be extremely strong bases and nucleophiles, but their extraordinary reactivity would make them difficult to prepare and use. Fortunately, metal derivatives of these alkyl, aryl and hydride moieties are available, and permit their addition to carbonyl compounds.
Reduction by Complex Metal Hydrides
Addition of a hydride anion to an aldehyde or ketone would produce an alkoxide anion, which on protonation should yield the corresponding alcohol. Aldehydes would give 1º-alcohols (as shown) and ketones would give 2º-alcohols.
RCH=O + H:(–) RCH2O(–) + H3O(–) RCH2OH
Two practical sources of hydride-like reactivity are the complex metal hydrides lithium aluminum hydride (LiAlH4) and sodium borohydride (NaBH4). These are both white (or near white) solids, which are prepared from lithium or sodium hydrides by reaction with aluminum or boron halides and esters. Lithium aluminum hydride is by far the most reactive of the two compounds, reacting violently with water, alcohols and other acidic groups with the evolution of hydrogen gas. The following table summarizes some important characteristics of these useful reagents.
Reagent
Preferred Solvents
Functions Reduced
Reaction Work-up
Sodium Borohydride
NaBH4
ethanol; aqueous ethanol
15% NaOH; diglyme
avoid strong acids
aldehydes to 1º-alcohols
ketones to 2º-alcohols
inert to most other functions
1) simple neutralization
2) extraction of product
Lithium Aluminum Hydride
LiAlH4
ether; THF
avoid alcohols and amines
avoid halogenated compounds
avoid strong acids
aldehydes to 1º-alcohols
ketones to 2º-alcohols
carboxylic acids to 1º-alcohols
esters to alcohols
epoxides to alcohols
nitriles & amides to amines
halides & tosylates to alkanes
most functions react
1) careful addition of water
2) remove aluminum salts
3) extraction of product
Some examples of aldehyde and ketone reductions, using the reagents described above, are presented in the following diagram. The first three reactions illustrate that all four hydrogens of the complex metal hydrides may function as hydride anion equivalents which bond to the carbonyl carbon atom. In the LiAlH4 reduction, the resulting alkoxide salts are insoluble and need to be hydrolyzed (with care) before the alcohol product can be isolated. In the borohydride reduction the hydroxylic solvent system achieves this hydrolysis automatically. The lithium, sodium, boron and aluminum end up as soluble inorganic salts. The last reaction shows how an acetal derivative may be used to prevent reduction of a carbonyl function (in this case a ketone). Remember, with the exception of epoxides, ethers are generally unreactive with strong bases or nucleophiles. The acid catalyzed hydrolysis of the aluminum salts also effects the removal of the acetal. This equation is typical in not being balanced (i.e. it does not specify the stoichiometry of the reagent).
Figure 1: Aldehyde and ketone reductions
Reduction of α,β-unsaturated ketones by metal hydride reagents sometimes leads to a saturated alcohol, especially with sodium borohydride. This product is formed by an initial conjugate addition of hydride to the β-carbon atom, followed by ketonization of the enol product and reduction of the resulting saturated ketone (equation 1 below). If the saturated alcohol is the desired product, catalytic hydrogenation prior to (or following) the hydride reduction may be necessary. To avoid reduction of the double bond, cerium(III) chloride is added to the reaction and it is normally carried out below 0 ºC, as shown in equation 2.
1) RCH=CHCOR' + NaBH4 (aq. alcohol) ——> RCH=CHCH(OH)R' + RCH2-CH2CH(OH)R'
1,2-addition product 1,4-addition product
2) RCH=CHCOR' + NaBH4 & CeCl3 -15º ——> RCH=CHCH(OH)R'
1,2-addition product
Before leaving this topic it should be noted that diborane, B2H6, a gas that was used in ether solution to prepare alkyl boranes from alkenes, also reduces many carbonyl groups. Consequently, selective reactions with substrates having both functional groups may not be possible. In contrast to the metal hydride reagents, diborane is a relatively electrophilic reagent, as witnessed by its ability to reduce alkenes. This difference also influences the rate of reduction observed for the two aldehydes shown below. The first, 2,2-dimethylpropanal, is less electrophilic than the second, which is activated by the electron withdrawing chlorine substituents.
Figure 2: Comparative reduction rates of diborane vs metal hydride reagents
Addition of Organometallic Reagents
The two most commonly used compounds of this kind are alkyl lithium reagents and Grignard reagents. They are prepared from alkyl and aryl halides, as discussed earlier. These reagents are powerful nucleophiles and very strong bases (pKa's of saturated hydrocarbons range from 42 to 50), so they bond readily to carbonyl carbon atoms, giving alkoxide salts of lithium or magnesium. Because of their ring strain, epoxides undergo many carbonyl-like reactions, as noted previously. Reactions of this kind are among the most important synthetic methods available to chemists, because they permit simple starting compounds to be joined to form more complex structures. Examples are shown in the following diagram.
Figure 3: Organometallic addition reactions
A common pattern, shown in the shaded box at the top, is observed in all these reactions. The organometallic reagent is a source of a nucleophilic alkyl or aryl group (colored blue), which bonds to the electrophilic carbon of the carbonyl group (colored magenta). The product of this addition is a metal alkoxide salt, and the alcohol product is generated by weak acid hydrolysis of the salt. The first two examples show that water soluble magnesium or lithium salts are also formed in the hydrolysis, but these are seldom listed among the products, as in the last four reactions. Ketones react with organometallic reagents to give 3º-alcohols; most aldehydes react to produce 2º-alcohols; and formaldehyde and ethylene oxide react to form 1º-alcohols (examples #5 & 6). When a chiral center is formed from achiral reactants (examples #1, 3 & 4) the product is always a racemic mixture of enantiomers.
Two additional examples of the addition of organometallic reagents to carbonyl compounds are informative. The first demonstrates that active metal derivatives of terminal alkynes function in the same fashion as alkyl lithium and Grignard reagents. The second example again illustrates the use of acetal protective groups in reactions with powerful nucleophiles. Following acid-catalyzed hydrolysis of the acetal, the resulting 4-hydroxyaldehyde is in equilibrium with its cyclic hemiacetal.
Figure 4: Carbonyl compunds formed from organometallic addition reactions | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Aldehydes_and_Ketones/Reactivity_of_Aldehydes_and_Ketones/Irreversible_Addition_Reactions_of_Aldehydes_and_Ketones.txt |
This page looks at ways of distinguishing between aldehydes and ketones using oxidizing agents such as acidified potassium dichromate(VI) solution, Tollens' reagent, Fehling's solution and Benedict's solution.
Why do aldehydes and ketones behave differently?
You will remember that the difference between an aldehyde and a ketone is the presence of a hydrogen atom attached to the carbon-oxygen double bond in the aldehyde. Ketones don't have that hydrogen.
The presence of that hydrogen atom makes aldehydes very easy to oxidize (i.e., they are strong reducing agents). Because ketones do not have that particular hydrogen atom, they are resistant to oxidation, and only very strong oxidizing agents like potassium manganate (VII) solution (potassium permanganate solution) oxidize ketones. However, they do it in a destructive way, breaking carbon-carbon bonds. Provided you avoid using these powerful oxidizing agents, you can easily tell the difference between an aldehyde and a ketone. Aldehydes are easily oxidized by all sorts of different oxidizing agents: ketones are not.
What is formed when aldehydes are oxidized?
It depends on whether the reaction is done under acidic or alkaline conditions. Under acidic conditions, the aldehyde is oxidized to a carboxylic acid. Under alkaline conditions, this couldn't form because it would react with the alkali. A salt is formed instead.
Building equations for the oxidation reactions
If you need to work out the equations for these reactions, the only reliable way of building them is to use electron-half-equations. The half-equation for the oxidation of the aldehyde obviously varies depending on whether you are doing the reaction under acidic or alkaline conditions.
Under acidic conditions:
$RCHO + H_2O \rightarrow RCOOH + 2H^+ +2e^- \tag{1}$
Under alkaline conditions:
$RCHO + 3OH^- \rightarrow RCOO^- + 2H_2O +2e^- \tag{2}$
These half-equations are then combined with the half-equations from whatever oxidizing agent you are using. Examples are given in detail below.
Specific examples
In each of the following examples, we are assuming that you know that you have either an aldehyde or a ketone. There are lots of other things which could also give positive results. Assuming that you know it has to be one or the other, in each case, a ketone does nothing. Only an aldehyde gives a positive result.
Using acidified potassium dichromate(VI) solution
A small amount of potassium dichromate(VI) solution is acidified with dilute sulphuric acid and a few drops of the aldehyde or ketone are added. If nothing happens in the cold, the mixture is warmed gently for a couple of minutes - for example, in a beaker of hot water.
ketone No change in the orange solution.
aldehyde Orange solution turns green.
The orange dichromate(VI) ions have been reduced to green chromium(III) ions by the aldehyde. In turn the aldehyde is oxidized to the corresponding carboxylic acid. The electron-half-equation for the reduction of dichromate(VI) ions is:
$Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O \tag{3}$
Combining that with the half-equation for the oxidation of an aldehyde under acidic conditions:
$RCHO + H_2O \rightarrow RCOOH + 2H^+ +2e^- \tag{4}$
. . . gives the overall equation:
$2RCHO + Cr_2O_7^{2-} + 8H^+ \rightarrow 3RCOOH +2Cr^{3+}+ 4H_2O \tag{5}$
Using Tollens' reagent (the silver mirror test)
Tollens' reagent contains the diamminesilver(I) ion, [Ag(NH3)2]+. This is made from silver(I) nitrate solution. You add a drop of sodium hydroxide solution to give a precipitate of silver(I) oxide, and then add just enough dilute ammonia solution to redissolve the precipitate. To carry out the test, you add a few drops of the aldehyde or ketone to the freshly prepared reagent, and warm gently in a hot water bath for a few minutes.
ketone No change in the colorless solution.
aldehyde The colorless solution produces a grey precipitate of silver, or a silver mirror on the test tube.
Figure 1: Tollens' test for aldehyde: left side positive (silver mirror), right side negative. from Wikipedia
Aldehydes reduce the diamminesilver(I) ion to metallic silver. Because the solution is alkaline, the aldehyde itself is oxidized to a salt of the corresponding carboxylic acid. The electron-half-equation for the reduction of of the diamminesilver(I) ions to silver is:
$Ag(NH_3)_2^+ + e^- \rightarrow Ag + 2NH_3 \tag{6}$
Combining that with the half-equation for the oxidation of an aldehyde under alkaline conditions:
$RCHO + 3OH^- \rightarrow RCOO^- + 2H_2O +2e^- \tag{7}$
gives the overall equation:
$2Ag(NH_3)_2^+ + RCHO + 3OH^- \rightarrow 2Ag + RCOO^- + 4NH_3 +2H_2O \tag{8}$
Using Fehling's solution or Benedict's solution
Fehling's solution and Benedict's solution are variants of essentially the same thing. Both contain complexed copper(II) ions in an alkaline solution.
• Fehling's solution contains copper(II) ions complexed with tartrate ions in sodium hydroxide solution. Complexing the copper(II) ions with tartrate ions prevents precipitation of copper(II) hydroxide.
• Benedict's solution contains copper(II) ions complexed with citrate ions in sodium carbonate solution. Again, complexing the copper(II) ions prevents the formation of a precipitate - this time of copper(II) carbonate.
Both solutions are used in the same way. A few drops of the aldehyde or ketone are added to the reagent, and the mixture is warmed gently in a hot water bath for a few minutes.
ketone No change in the blue solution.
aldehyde The blue solution produces a dark red precipitate of copper(I) oxide.
Figure 2: Fehling's test. Left side negative, right side positive. from Wikipedia
Aldehydes reduce the complexed copper(II) ion to copper(I) oxide. Because the solution is alkaline, the aldehyde itself is oxidized to a salt of the corresponding carboxylic acid. The equations for these reactions are always simplified to avoid having to write in the formulae for the tartrate or citrate ions in the copper complexes. The electron-half-equations for both Fehling's solution and Benedict's solution can be written as:
$2Cu^{2+}_{complexed} + 2OH^- + 2e^- \rightarrow Cu_2O + H_2O \tag{9}$
Combining that with the half-equation for the oxidation of an aldehyde under alkaline conditions:
$RCHO + 3OH^- \rightarrow RCOO^- + 2H_2O +2e^- \tag{10}$
to give the overall equation:
$RCHO + 2Cu^{2+}_{complexed} + 5OH^- \rightarrow RCOO^- + Cu_2O + 3H_2O \tag{11}$
Contributors
Jim Clark (Chemguide.co.uk) | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Aldehydes_and_Ketones/Reactivity_of_Aldehydes_and_Ketones/Oxidation_of_Aldehydes_and_Ketones.txt |
The reaction of aldehydes and ketones with ammonia or 1º-amines forms imine derivatives, also known as Schiff bases (compounds having a C=N function). Water is eliminated in the reaction, which is acid-catalyzed and reversible in the same sense as acetal formation. The pH for reactions which form imine compounds must be carefully controlled. The rate at which these imine compounds are formed is generally greatest near a pH of 5, and drops at higher and lower pH's. At high pH there will not be enough acid to protonate the OH in the intermediate to allow for removal as H2O. At low pH most of the amine reactant will be tied up as its ammonium conjugate acid and will become non-nucleophilic.
Converting reactants to products simply
Mechanism of imine formation
1) Nucleophilic attack
2) Proton transfer
3) Protonation of OH
4) Removal of water
5) Deprotonation
Reversibility of imine forming reactions
Imines can be hydrolyzed back to the corresponding primary amine under acidic conditons.
Reactions involving other reagents of the type Y-NH2
Imines are sometimes difficult to isolate and purify due to their sensitivity to hydrolysis. Consequently, other reagents of the type Y–NH2 have been studied, and found to give stable products (R2C=N–Y) useful in characterizing the aldehydes and ketones from which they are prepared. Some of these reagents are listed in the following table, together with the structures and names of their carbonyl reaction products. Hydrazones are used as part of the Wolff-Kishner reduction and will be discussed in more detail in another module.
With the exception of unsubstituted hydrazones, these derivatives are easily prepared and are often crystalline solids - even when the parent aldehyde or ketone is a liquid. Since melting points can be determined more quickly and precisely than boiling points, derivatives such as these are useful for comparison and identification of carbonyl compounds. It should be noted that although semicarbazide has two amino groups (–NH2) only one of them is a reactive amine. The other is amide-like and is deactivated by the adjacent carbonyl group.
Problems
1)Please draw the products of the following reactions.
2) Please draw the structure of the reactant needed to produce the indicated product.
3) Please draw the products of the following reactions.
1)
2)
3)
Contributors
Prof. Steven Farmer (Sonoma State University)
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry
Reactions with Grignard Reagents
A Grignard reagent has a formula RMgX where X is a halogen, and R is an alkyl or aryl (based on a benzene ring) group. For the purposes of this page, we shall take R to be an alkyl group (e.g., CH3CH2MgBr). Grignard reagents are made by adding the halogenoalkane to small bits of magnesium in a flask containing ethoxyethane (commonly called diethyl ether or just "ether"). The flask is fitted with a reflux condenser, and the mixture is warmed over a water bath for 20 - 30 minutes.
Everything must be perfectly dry because Grignard reagents react with water. Any reactions using the Grignard reagent are carried out with the mixture produced from this reaction; you cannot separate it out in any way. Any excess Grignard reagent must be quenched before disposal.
Reactions of Grignard reagents with aldehydes and ketones
These are reactions of the carbon-oxygen double bond, and so aldehydes and ketones react in exactly the same way - all that changes are the groups that happen to be attached to the carbon-oxygen double bond. It is much easier to understand what is going on by looking closely at the general case (using "R" groups rather than specific groups) - and then slotting in the various real groups as and when you need to. The "R" groups can be either hydrogen or alkyl in any combination.
In the first stage, the Grignard reagent adds across the carbon-oxygen double bond:
Dilute acid is then added to this to hydrolyse it.
An alcohol is formed. One of the key uses of Grignard reagents is the ability to make complicated alcohols easily. What sort of alcohol you get depends on the carbonyl compound you started with - in other words, what R and R' are.
Example \(1\): Reaction with Methanal
Methanal is the simplest possible aldehyde with hydrogen as both R groups.
Assuming that you are starting with CH3CH2MgBr and using the general equation above, the alcohol you get always has the form:
Since both R groups are hydrogen atoms, the final product will be:
A primary alcohol is formed. A primary alcohol has only one alkyl group attached to the carbon atom with the -OH group on it. You could obviously get a different primary alcohol if you started from a different Grignard reagent.
Example \(2\): Reaction with Aldehyde
The next biggest aldehyde is ethanal with one of the R groups is hydrogen and the other CH3.
Again, think about how that relates to the general case. The alcohol formed is:
So this time the final product has one CH3 group and one hydrogen attached:
A secondary alcohol has two alkyl groups (the same or different) attached to the carbon with the -OH group on it. You could change the nature of the final secondary alcohol by either:
• changing the nature of the Grignard reagent - which would change the CH3CH2 group into some other alkyl group;
• changing the nature of the aldehyde - which would change the CH3 group into some other alkyl group.
Example \(3\): Reactions with Propanone
Ketones have two alkyl groups attached to the carbon-oxygen double bond. The simplest one is propanone.
This time when you replace the R groups in the general formula for the alcohol produced you get a tertiary alcohol.
A tertiary alcohol has three alkyl groups attached to the carbon with the -OH attached. The alkyl groups can be any combination of same or different. You could ring the changes on the product by:
• changing the nature of the Grignard reagent - which would change the CH3CH2 group into some other alkyl group;
• changing the nature of the ketone - which would change the CH3 groups into whatever other alkyl groups you choose to have in the original ketone. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Aldehydes_and_Ketones/Reactivity_of_Aldehydes_and_Ketones/Reaction_with_Primary_Amines_to_form_Imines.txt |
Despite the fearsome names, the structures of the two reducing agents are very simple. In each case, there are four hydrogens ("tetrahydido") around either aluminium or boron in a negative ion (shown by the "ate" ending). The "(III)" shows the oxidation state of the aluminium or boron, and is often left out because these elements only ever show the +3 oxidation state in their compounds.
The formulae of the two compounds are \(LiAlH_4\) and \(NaBH_4\). Their structures are:
In each of the negative ions, one of the bonds is a co-ordinate covalent (dative covalent) bond using the lone pair on a hydride ion (H-) to form a bond with an empty orbital on the aluminium or boron.
The reduction of an aldehyde
You get exactly the same organic product whether you use lithium tetrahydridoaluminate or sodium tetrahydridoborate. For example, with ethanal you get ethanol:
Notice that this is a simplified equation where [H] means "hydrogen from a reducing agent". In general terms, reduction of an aldehyde leads to a primary alcohol.
The reduction of a Ketone
Again the product is the same whichever of the two reducing agents you use. For example, with propanone you get propan-2-ol:
Reduction of a ketone leads to a secondary alcohol.
Using lithium tetrahydridoaluminate (lithium aluminium hydride)
Lithium tetrahydridoaluminate is much more reactive than sodium tetrahydridoborate. It reacts violently with water and alcohols, and so any reaction must exclude these common solvents. The reactions are usually carried out in solution in a carefully dried ether such as ethoxyethane (diethyl ether). The reaction happens at room temperature, and takes place in two separate stages.
In the first stage, a salt is formed containing a complex aluminium ion. The following equations show what happens if you start with a general aldehyde or ketone. R and R' can be any combination of hydrogen or alkyl groups.
The product is then treated with a dilute acid (such as dilute sulfuric acid or dilute hydrochloric acid) to release the alcohol from the complex ion.
The alcohol formed can be recovered from the mixture by fractional distillation.
Using sodium tetrahydridoborate (sodium borohydride)
Sodium tetrahydridoborate is a more gentle (and therefore safer) reagent than lithium tetrahydridoaluminate. It can be used in solution in alcohols or even solution in water - provided the solution is alkaline. Solid sodium tetrahydridoborate is added to a solution of the aldehyde or ketone in an alcohol such as methanol, ethanol or propan-2-ol. Depending on which recipe you read, it is either heated under reflux or left for some time around room temperature. This almost certainly varies depending on the nature of the aldehyde or ketone.
At the end of this time, a complex similar to the previous one is formed.
In the second stage of the reaction, water is added and the mixture is boiled to release the alcohol from the complex.
Again, the alcohol formed can be recovered from the mixture by fractional distillation.
Reduction of Carbonyls to Alcohols Using Metal Hydrides
The most common sources of the hydride Nucleophile are lithium aluminum hydride (LiAlH4) and sodium borohydride (NaBH4). Note! The hydride anion is not present during this reaction; rather, these reagents serve as a source of hydride due to the presence of a polar metal-hydrogen bond. Because aluminum is less electronegative than boron, the Al-H bond in LiAlH4 is more polar, thereby, making LiAlH4 a stronger reducing agent.
Addition of a hydride anion (H:-) to an aldehyde or ketone gives an alkoxide anion, which on protonation yields the corresponding alcohol. Aldehydes produce 1º-alcohols and ketones produce 2º-alcohols.
In metal hydrides reductions the resulting alkoxide salts are insoluble and need to be hydrolyzed (with care) before the alcohol product can be isolated. In the sodium borohydride reduction the methanol solvent system achieves this hydrolysis automatically. In the lithium aluminum hydride reduction water is usually added in a second step. The lithium, sodium, boron and aluminum end up as soluble inorganic salts at the end of either reaction. Note! LiAlH4 and NaBH4 are both capable of reducing aldehydes and ketones to the corresponding alcohol.
Mechanism
This mechanism is for a LiAlH4 reduction. The mechanism for a NaBH4 reduction is the same except methanol is the proton source used in the second step.
1) Nucleopilic attack by the hydride anion
2) The alkoxide is protonated
Properties of Hydride Sources
Two practical sources of hydride-like reactivity are the complex metal hydrides lithium aluminum hydride (LiAlH4) and sodium borohydride (NaBH4). These are both white (or near white) solids, which are prepared from lithium or sodium hydrides by reaction with aluminum or boron halides and esters. Lithium aluminum hydride is by far the most reactive of the two compounds, reacting violently with water, alcohols and other acidic groups with the evolution of hydrogen gas. The following table summarizes some important characteristics of these useful reagents.
Problems
1) Please draw the products of the following reactions:
2) Please draw the structure of the molecule which must be reacted to produce the product.
3) Deuterium oxide (D2O) is a form of water where the hydrogens have been replaced by deuteriums. For the following LiAlH4 reduction the water typically used has been replaced by deuterium oxide. Please draw the product of the reaction and place the deuterium in the proper location. Hint! Look at the mechanism of the reaction.
1)
2)
3)
Reductive Amination
Aldehydes and ketones can be converted into 1°, 2° and 3° amines using reductive amination. The reaction takes place in two parts. The first step is the nucleophiic addition of the carbonyl group to form an imine. The second step is the reduction of the imine to an amine using an reducing agent. A reducing agent commonly used for this reaction is sodium cyanoborohydride (NaBH3CN). | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Aldehydes_and_Ketones/Reactivity_of_Aldehydes_and_Ketones/Reduction_of_Aldehydes_and_Ketones.txt |
Hydration and Hemiacetal Formation
It has been demonstrated that water adds rapidly to the carbonyl function of aldehydes and ketones. In most cases the resulting hydrate (a geminal-diol) is unstable relative to the reactants and cannot be isolated. Exceptions to this rule exist, one being formaldehyde (a gas in its pure monomeric state). Here the weaker pi-component of the carbonyl double bond, relative to other aldehydes or ketones, and the small size of the hydrogen substituents favor addition. Thus, a solution of formaldehyde in water (formalin) is almost exclusively the hydrate, or polymers of the hydrate. Similar reversible additions of alcohols to aldehydes and ketones take place. The equally unstable addition products are called hemiacetals.
R2C=O + R'OH R'O–(R2)C–O–H (a hemiacetal)
Acetals are geminal-diether derivatives of aldehydes or ketones, formed by reaction with two equivalents of an alcohol and elimination of water. Ketone derivatives of this kind were once called ketals, but modern usage has dropped that term. The following equation shows the overall stoichiometric change in acetal formation, but a dashed arrow is used because this conversion does not occur on simple mixing of the reactants.
R2C=O + 2 R'OH R2C(OR')2 + H2O (an acetal)
In order to achieve effective acetal formation two additional features must be implemented. First, an acid catalyst must be used; and second, the water produced with the acetal must be removed from the reaction. The latter is important, since acetal formation is reversible. Indeed, once pure acetals are obtained they may be hydrolyzed back to their starting components by treatment with aqueous acid. The mechanism shown here applies to both acetal formation and acetal hydrolysis by the principle of microscopic reversibility.
Figure 1: Acetal Formation
Some examples of acetal formation are presented in the following diagram. As noted, p-toluenesulfonic acid (pKa = -2) is often the catalyst for such reactions. Two equivalents of the alcohol reactant are needed, but these may be provided by one equivalent of a diol (example #2). Intramolecular involvement of a gamma or delta hydroxyl group (as in examples #3 and 4) may occur, and is often more facile than the intermolecular reaction. Thiols (sulfur analogs of alcohols) give thioacetals (example #5). In this case the carbonyl functions are relatively hindered, but by using excess ethanedithiol as the solvent and the Lewis acid BF3 as catalyst a good yield of the bis-thioacetal is obtained. Thioacetals are generally more difficult to hydrolyze than are acetals.
Figure 2: Examples of acetal formation
The importance of acetals as carbonyl derivatives lies chiefly in their stability and lack of reactivity in neutral to strongly basic environments. As long as they are not treated by acids, especially aqueous acid, acetals exhibit all the lack of reactivity associated with ethers in general.
Among the most useful and characteristic reactions of aldehydes and ketones is their reactivity toward strongly nucleophilic (and basic) metallo-hydride, alkyl and aryl reagents (to be discussed shortly). If the carbonyl functional group is converted to an acetal these powerful reagents have no effect; thus, acetals are excellent protective groups, when these irreversible addition reactions must be prevented.
Formation of Imines and Related Compounds
The reaction of aldehydes and ketones with ammonia or 1º-amines forms imine derivatives, also known as Schiff bases, (compounds having a C=N function). This reaction plays an important role in the synthesis of 2º-amines. Water is eliminated in the reaction, which is acid-catalyzed and reversible in the same sense as acetal formation.
R2C=O + R'NH2 R'NH–(R2)C–O–H R2C=NR' + H2O
Imines are sometimes difficult to isolate and purify due to their sensitivity to hydrolysis. Consequently, other reagents of the type Y–NH2 have been studied, and found to give stable products (R2C=N–Y) useful in characterizing the aldehydes and ketones from which they are prepared. Some of these reagents are listed in the following table, together with the structures and names of their carbonyl reaction products. An interesting aspect of these carbonyl derivatives is that stereoisomers are possible when the R-groups of the carbonyl reactant are different. Thus, benzaldehyde forms two stereoisomeric oximes, a low-melting isomer, having the hydroxyl group cis to the aldehyde hydrogen (called syn), and a higher melting isomer in which the hydroxyl group and hydrogen are trans (the anti isomer). At room temperature or below the configuration of the double-bonded nitrogen atom is apparently fixed in one trigonal shape, unlike the rapidly interconverting pyramidal configurations of the sp3 hybridized amines.
Figure 3: Carbonyl products formed from Y-NH2 type reactants
With the exception of unsubstituted hydrazones, these derivatives are easily prepared and are often crystalline solids - even when the parent aldehyde or ketone is a liquid. Since melting points can be determined more quickly and precisely than boiling points, derivatives such as these are useful for comparison and identification of carbonyl compounds. If the aromatic ring of phenylhydrazine is substituted with nitro groups at the 2- & 4-positions, the resulting reagent and the hydrazone derivatives it gives are strongly colored, making them easy to identify. It should be noted that although semicarbazide has two amino groups (–NH2) only one of them is a reactive amine. The other is amide-like and is deactivated by the adjacent carbonyl group.
The rate at which these imine-like compounds are formed is generally greatest near a pH of 5, and drops at higher and lower pH's. This agrees with a general acid catalysis in which the conjugate acid of the carbonyl reactant combines with a free amino group, as shown in the above animation. At high pH there will be a vanishingly low concentration of the carbonyl conjugate acid, and at low pH most of the amine reactant will be tied up as its ammonium conjugate acid. With the exception of imine formation itself, most of these derivatization reactions do not require active removal of water (not shown as a product in the previous equations). The reactions are reversible, but equilibrium is not established instantaneously and the products often precipitate from solution as they are formed.
Enamine Formation
The previous reactions have all involved reagents of the type: Y–NH2, i.e. reactions with a 1º-amino group. Most aldehydes and ketones also react with 2º-amines to give products known as enamines. Two examples of these reactions are presented in the following diagram. It should be noted that, like acetal formation, these are acid-catalyzed reversible reactions in which water is lost. Consequently, enamines are easily converted back to their carbonyl precursors by acid-catalyzed hydrolysis.
Figure 4: Enamine synthesis
Cyanohydrin Formation
The last example of reversible addition is that of hydrogen cyanide (HC≡N), which adds to aldehydes and many ketone to give products called cyanohydrins.
RCH=O + H–C≡N RCH(OH)CN (a cyanohydrin)
Since hydrogen cyanide itself is an acid (pKa = 9.25), the addition is not acid-catalyzed. In fact, for best results cyanide anion, C≡N(-) must be present, which means that catalytic base must be added. Cyanohydrin formation is weakly exothermic, and is favored for aldehydes, and unhindered cyclic and methyl ketones. Two examples of such reactions are shown below.
Figure 5: Cyanohydrin formation
The cyanohydrin from benzaldehyde is named mandelonitrile. The reversibility of cyanohydrin formation is put to use by the millipede Apheloria corrugata in a remarkable defense mechanism. This arthropod releases mandelonitrile from an inner storage gland into an outer chamber, where it is enzymatically broken down into benzaldehyde and hydrogen cyanide before being sprayed at an enemy. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Aldehydes_and_Ketones/Reactivity_of_Aldehydes_and_Ketones/Reversible_Addition_Reactions_of_Aldehydes_and_Ketones.txt |
This page looks at the addition of hydrogen cyanide and sodium hydrogensulphite (sodium bisulphite) to aldehydes and ketones.
Addition of hydrogen cyanide to aldehydes and ketones
Hydrogen cyanide adds across the carbon-oxygen double bond in aldehydes and ketones to produce compounds known as hydroxynitriles. These used to be known as cyanohydrins.
Example \(1\)
For example, with ethanal (an aldehyde) you get 2-hydroxypropanenitrile:
With propanone (a ketone) you get 2-hydroxy-2-methylpropanenitrile:
The reaction isn't normally done using hydrogen cyanide itself, because this is an extremely poisonous gas. Instead, the aldehyde or ketone is mixed with a solution of sodium or potassium cyanide in water to which a little sulphuric acid has been added. The pH of the solution is adjusted to about 4 - 5, because this gives the fastest reaction. The reaction happens at room temperature.The solution will contain hydrogen cyanide (from the reaction between the sodium or potassium cyanide and the sulphuric acid), but still contains some free cyanide ions. This is important for the mechanism.
Uses of the reaction
The product molecules contain two functional groups:
• the -OH group which behaves like a simple alcohol and can be replaced by other things like chlorine, which can in turn be replaced to give, for example, an -NH2 group;
• the -CN group which is easily converted into a carboxylic acid group -COOH.
For example, starting from a hydroxynitrile made from an aldehyde, you can quite easily produce relatively complicated molecules like 2-amino acids - the amino acids which are used to construct proteins.
Addition of sodium hydrogensulphite to aldehydes and ketones
Sodium hydrogensulphite used to be known as sodium bisulphite, and you might well still come across it in organic textbooks under this name - or using the bisulfite spelling. This reaction only works well for aldehydes. In the case of ketones, one of the hydrocarbon groups attached to the carbonyl group needs to be a methyl group. Bulky groups attached to the carbonyl group get in the way of the reaction happening.
The aldehyde or ketone is shaken with a saturated solution of sodium hydrogensulphite in water. Where the product is formed, it separates as white crystals.
In the case of ethanal, the equation is:
and with propanone, the equation is:
These compounds are rarely named systematically, and are usually known as "hydrogensulphite (or bisulphite) addition compounds".
Uses of the reaction
The reaction is usually used during the purification of aldehydes (and any ketones that it works for). The addition compound can be split easily to regenerate the aldehyde or ketone by treating it with either dilute acid or dilute alkali. If you have an impure aldehyde, for example, you could shake it with a saturated solution of sodium hydrogensulphite to produce the crystals. These crystals could easily be filtered and washed to remove any other impurities. Addition of dilute acid, for example, would then regenerate the original aldehyde. It would, of course, still need to be separated from the excess acid and assorted inorganic products of the reaction.
Contributors
Template:ContribCalrk
The Triiodomethane (Iodoform) Reaction
This page looks at how the triiodomethane (iodoform) reaction can be used to identify the presence of a CH3CO group in aldehydes and ketones. There are two apparently quite different mixtures of reagents that can be used to do this reaction. They are, in fact, chemically equivalent.
Using iodine and sodium hydroxide solution
This is chemically the more obvious method. Iodine solution is added to a small amount of aldehyde or ketone, followed by just enough sodium hydroxide solution to remove the color of the iodine. If nothing happens in the cold, it may be necessary to warm the mixture very gently. A positive result is the appearance of a very pale yellow precipitate of triiodomethane (previously known as iodoform) - CHI3. Apart from its color, this can be recognised by its faintly "medical" smell. It is used as an antiseptic on the sort of sticky plasters you put on minor cuts, for example.
Using potassium iodide and sodium chlorate(I) solutions
Sodium chlorate(I) is also known as sodium hypochlorite. Potassium iodide solution is added to a small amount of aldehyde or ketone, followed by sodium chlorate(I) solution. Again, if no precipitate is formed in the cold, it may be necessary to warm the mixture very gently. The positive result is the same pale yellow precipitate as before.
The chemistry of the triiodomethane (iodoform) reaction
A positive result - the pale yellow precipitate of triiodomethane (iodoform) - is given by an aldehyde or ketone containing the grouping:
"R" can be a hydrogen atom or a hydrocarbon group (for example, an alkyl group). If "R" is hydrogen, then you have the aldehyde ethanal, CH3CHO.
• Ethanal is the only aldehyde to give the triiodomethane (iodoform) reaction.
• If "R"is a hydrocarbon group, then you have a ketone. Lots of ketones give this reaction, but those that do all have a methyl group on one side of the carbon-oxygen double bond. These are known as methyl ketones.
Equations for the triiodomethane (iodoform) reaction
We will take the reagents as being iodine and sodium hydroxide solution. The first stage involves substitution of all three hydrogens in the methyl group by iodine atoms. The presence of hydroxide ions is important for the reaction to happen - they take part in the mechanism for the reaction.
In the second stage, the bond between the C I3 and the rest of the molecule is broken to produce triiodomethane (iodoform) and the salt of an acid.
Putting all this together gives the overall equation for the reaction: | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Aldehydes_and_Ketones/Reactivity_of_Aldehydes_and_Ketones/Simple_Addition_Reactions.txt |
Organophosphorus ylides react with aldehydes or ketones to give substituted alkenes in a transformation called the Wittig reaction. This reaction is named for George Wittig who was awarded the Nobel prize for this work in 1979. A principal advantage of alkene synthesis by the Wittig reaction is that the location of the double bond is absolutely fixed, in contrast to the mixtures often produced by alcohol dehydration.
Preparation of Phosphorus Ylides
It has been noted that dipolar phosphorus compounds are stabilized by p-d bonding. This bonding stabilization extends to carbanions adjacent to phosphonium centers, and the zwitterionic conjugate bases derived from such cations are known as ylides. An ylide is defined as a compound with opposite charges on adjacent atoms both of which have complete octets. For the Wittig reaction discussed below an organophosphorus ylide, also called Wittig reagents, will be used. The ability of phosphorus to hold more than eight valence electrons allows for a resonance structure to be drawn forming a double bonded structure.
The stabilization of the carbanion provided by the phosphorus causes an increase in acidity (pKa ~35). Very strong bases, such as butyl lithium, are required for complete formation of ylides.
The ylides shown here are all strong bases. Like other strongly basic organic reagents, they are protonated by water and alcohols, and are sensitive to oxygen. Water decomposes phosphorous ylides to hydrocarbons and phosphine oxides, as shown.
Although many ylides are commercially available it is often necessary to create them synthetically. Ylides can be synthesized from an alkyl halide and a trialkyl phosphine. Typically triphenyl phosphine is used to synthesize ylides. Because a SN2 reaction is used in the ylide synthesis methyl and primary halides perform the best. Secondary halides can also be used but the yields are generally lower. This should be considered when planning out a synthesis which involves a synthesized Wittig reagent.
1) SN2 reaction
2) Deprotonation
The Wittig Reaction
The most important use of ylides in synthesis comes from their reactions with aldehydes and ketones, which are initiated in every case by a covalent bonding of the nucleophilic alpha-carbon to the electrophilic carbonyl carbon. Ylides react to give substituted alkenes in a transformation called the Wittig reaction. This reaction is named for George Wittig who was awarded the Nobel prize for this work in 1979. A principal advantage of alkene synthesis by the Wittig reaction is that the location of the double bond is absolutely fixed, in contrast to the mixtures often produced by alcohol dehydration.
Going from reactants to products simplified
Mechanism of the Wittig reaction
Following the initial carbon-carbon bond formation, two intermediates have been identified for the Wittig reaction, a dipolar charge-separated species called a betaine and a four-membered heterocyclic structure referred to as an oxaphosphatane. Cleavage of the oxaphosphatane to alkene and phosphine oxide products is exothermic and irreversible.
1) Nucleophillic attack on the carbonyl
2) Formation of a 4 membered ring
3) Formation of the alkene
Limitation of the Wittig reaction
If possible both E and Z isomer of the double bond will be formed. This should be considered when planning a synthesis involving a Wittig Reaction.
Problems
1) Please write the product of the following reactions.
2) Please indicate the starting material required to produce the product.
3) Please draw the structure of the oxaphosphetane which is made during the mechanism of the reaction given that produces product C.
4) Please draw the structure of the betaine which is made during the mechanism of the reaction given that produces product D.
5) Please give a detailed mechanism and the final product of this reaction
6) It has been shown that reacting and epoxide with triphenylphosphine forms an alkene. Please propose a mechanism for this reaction. Review the section on epoxide reactions if you need help.
Answers
1)
2)
3)
4)
5)
Nucleophillic attack on the carbonyl
Formation of a 4 membered ring
Formation of the alkene
6) Nucleophillic attack on the epoxide
Formation of a 4 membered ring
Formation of the alkene
Contributors
Prof. Steven Farmer (Sonoma State University)
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry
Tollens Test
Tollens’ test, also known as silver-mirror test, is a qualitative laboratory test used to distinguish between an aldehyde and a ketone. It exploits the fact that aldehydes are readily oxidized (see oxidation), whereas ketones are not. Tollens’ test uses a reagent known as Tollens’ reagent, which is a colorless, basic, aqueous solution containing silver ions coordinated to ammonia $[Ag(NH_3)^{2+}]$. It is prepared using a two-step procedure.
Step 1: Aqueous silver nitrate is mixed with aqueous sodium hydroxide.
\begin{align*} \ce{AgNO_3 + NaOH} &\rightarrow \ce{AgOH + NaHO_3} \[4pt] \ce{2AgOH} &\rightarrow \ce{Ag_2O + H_2O} \end{align*}
Step 2: Aqueous ammonia is added drop-wise until the precipitated silver oxide completely dissolves.
$\ce{Ag_2O + 4NH_3 + H_2O \rightarrow 2Ag(NH_3)_2^+ + 2OH^{-}} \nonumber$
Tollens’ reagent oxidizes an aldehyde into the corresponding carboxylic acid.
The reaction is accompanied by the reduction of silver ions in Tollens’ reagent into metallic silver, which, if the test is carried out in a clean glass test tube, forms a mirror on the test tube. eg:
Ketones are not oxidized by Tollens’ reagent, so the treatment of a ketone with Tollens’ reagent in a glass test tube does not result in a silver mirror (Figure $1$; right).
Wolff-Kishner Reduction
Aldehydes and ketones can be converted to a hydrazine derivative by reaction with hydrazine. These "hydrazones" can be further converted to the corresponding alkane by reaction with base and heat. These two steps can be combined into one reaction called the Wolff-Kishner Reduction which represents a general method for converting aldehydes and ketones into alkanes. Typically a high boiling point solvent, such as ethylene glycol, is used to provide the high temperatures needed for this reaction to occur. Note! Nitrogen gas is produced as part of this reaction.
Example
Mechanism of the Wolff-Kishner Reduction
1) Deprotonation of Nitrogen
2) Protonation of the Carbon
3) Deprotonation of Nitrogen
4) Protonation of Carbon
Problems
1) Please draw the products of the following reactions.
1) | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Aldehydes_and_Ketones/Reactivity_of_Aldehydes_and_Ketones/The_Wittig_Reaction.txt |
Aldehydes and ketones can be prepared using a wide variety of reactions. Although these reactions are discussed in greater detail in other sections, they are listed here as a summary and to help with planning multistep synthetic pathways. Please use the appropriate links to see more details about the reactions.
Synthesis of Aldehydes and Ketones
This page takes an introductory look at how Grignard reagents are made from halogenoalkanes (haloalkanes or alkyl halides), and introduces some of their reactions.
Introduction
A Grignard reagent has a formula $\ce{RMgX}$ where $\ce{X}$ is a halogen, and $\ce{R}$ is an alkyl or aryl (based on a benzene ring) group. For the purposes of this page, we shall take R to be an alkyl group. A typical Grignard reagent might be $\ce{CH3CH2MgBr}$. Grignard reagents are made by adding the halogenoalkane to small bits of magnesium in a flask containing ethoxyethane (commonly called diethyl ether or just "ether"). The flask is fitted with a reflux condenser, and the mixture is warmed over a water bath for 20 - 30 minutes.
Everything must be perfectly dry because Grignard reagents react with water (see below). Any reactions using the Grignard reagent are carried out with the mixture produced from this reaction. You can't separate it out in any way.
Grignard reagents and water
Grignard reagents react with water to produce alkanes. This is the reason that everything has to be very dry during the preparation above. For example:
$CH_3CH_2MgBr + H_2O \rightarrow CH_3CH_3 + Mg(OH)Br$
The inorganic product, $Mg(OH)Br$, is referred to as a "basic bromide" and is a sort of half-way stage between magnesium bromide and magnesium hydroxide.
Grignard reagents and carbon dioxide
Grignard reagents react with carbon dioxide in two stages. In the first, you get an addition of the Grignard reagent to the carbon dioxide. Dry carbon dioxide is bubbled through a solution of the Grignard reagent in ethoxyethane, made as described above. For example:
The product is then hydrolyzed (reacted with water) in the presence of a dilute acid. Typically, you would add dilute sulfuric acid or dilute hydrochloric acid to the solution formed by the reaction with the CO2. A carboxylic acid is produced with one more carbon than the original Grignard reagent. The usually quoted equation is (without the red bits):
Almost all sources quote the formation of a basic halide such as Mg(OH)Br as the other product of the reaction. That is actually misleading because these compounds react with dilute acids. What you end up with would be a mixture of ordinary hydrated magnesium ions, halide ions and sulfate or chloride ions - depending on which dilute acid you added.
Grignard reagents and carbonyl compounds
Carbonyl compounds contain the C=O double bond. The simplest ones have the form:
R and R' can be the same or different, and can be an alkyl group or hydrogen. If one (or both) of the R groups are hydrogens, the compounds are called aldehydes. For example:
If both of the R groups are alkyl groups, the compounds are called ketones. Examples include:
General Reaction between Grignards and carbonyls
The reactions between the various sorts of carbonyl compounds and Grignard reagents can look quite complicated, but in fact they all react in the same way - all that changes are the groups attached to the carbon-oxygen double bond. It is much easier to understand what is going on by looking closely at the general case (using "R" groups rather than specific groups) - and then slotting in the various real groups as and when you need to.
The reactions are essentially identical to the reaction with carbon dioxide - all that differs is the nature of the organic product. In the first stage, the Grignard reagent adds across the carbon-oxygen double bond:
Dilute acid is then added to this to hydrolyse it. (I am using the normally accepted equation ignoring the fact that the Mg(OH)Br will react further with the acid.)
An alcohol is formed. One of the key uses of Grignard reagents is the ability to make complicated alcohols easily. What sort of alcohol you get depends on the carbonyl compound you started with - in other words, what R and R' are.
Reaction between Grignard reagents and methanal
In methanal, both R groups are hydrogen. Methanal is the simplest possible aldehyde.
Assuming that you are starting with CH3CH2MgBr and using the general equation above you get always has the form:
Since both R groups are hydrogen atoms, the final product will be:
A primary alcohol is formed. A primary alcohol has only one alkyl group attached to the carbon atom with the -OH group on it. You could obviously get a different primary alcohol if you started from a different Grignard reagent.
Reaction between Grignards and other aldehydes
The next biggest aldehyde is ethanal. One of the R groups is hydrogen and the other CH3.
Again, think about how that relates to the general case. The alcohol formed is:
So this time the final product has one CH3 group and one hydrogen attached:
A secondary alcohol has two alkyl groups (the same or different) attached to the carbon with the -OH group on it. You could change the nature of the final secondary alcohol by either:
• changing the nature of the Grignard reagent - which would change the CH3CH2 group into some other alkyl group;
• changing the nature of the aldehyde - which would change the CH3 group into some other alkyl group.
The reaction between Grignard reagents and ketones
Ketones have two alkyl groups attached to the carbon-oxygen double bond. The simplest one is propanone.
This time when you replace the R groups in the general formula for the alcohol produced you get a tertiary alcohol.
A tertiary alcohol has three alkyl groups attached to the carbon with the -OH attached. The alkyl groups can be any combination of same or different. You could ring the changes on the product by
• changing the nature of the Grignard reagent - which would change the CH3CH2 group into some other alkyl group;
• changing the nature of the ketone - which would change the CH3 groups into whatever other alkyl groups you choose to have in the original ketone.
Why do Grignard reagents react with carbonyls?
The bond between the carbon atom and the magnesium is polar. Carbon is more electronegative than magnesium, and so the bonding pair of electrons is pulled towards the carbon. That leaves the carbon atom with a slight negative charge.
The carbon-oxygen double bond is also highly polar with a significant amount of positive charge on the carbon atom. The nature of this bond is described in detail elsewhere on this site. The Grignard reagent can therefore serve as a nucleophile because of the attraction between the slight negativeness of the carbon atom in the Grignard reagent and the positiveness of the carbon in the carbonyl compound. A nucleophile is a species that attacks positive (or slightly positive) centers in other molecules or ions. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Aldehydes_and_Ketones/Synthesis_of_Aldehydes_and_Ketones/Grignard_Reagents.txt |
The alkali metals (Li, Na, K etc.) and the alkaline earth metals (Mg and Ca, together with Zn) are good reducing agents, the former being stronger than the latter. These same metals reduce the carbon-halogen bonds of alkyl halides. The halogen is converted to a halide anion, and the carbon bonds to the metal which has characteristics similar to a carbanion (R:-).
Formation of Organometallic Reagents
Many organometallic reagents are commercially available, however, it is often necessary to make then. The following equations illustrate these reactions for the commonly used metals lithium and magnesium (R may be hydrogen or alkyl groups in any combination).
• An Alkyl Lithium Reagent
$\ce{R3C-X} + \ce{2Li} \rightarrow \ce{R3C-Li} + \ce{LiX}$
• A Grignard Regent
$\ce{R3C-X} + \ce{Mg} \rightarrow \ce{R3C-MgX}$
Halide reactivity in these reactions increases in the order: Cl < Br < I and Fluorides are usually not used. The alkyl magnesium halides described in the second reaction are called Grignard Reagents after the French chemist, Victor Grignard, who discovered them and received the Nobel prize in 1912 for this work. The other metals mentioned above react in a similar manner, but Grignard and Alky Lithium Reagents most widely used. Although the formulas drawn here for the alkyl lithium and Grignard reagents reflect the stoichiometry of the reactions and are widely used in the chemical literature, they do not accurately depict the structural nature of these remarkable substances. Mixtures of polymeric and other associated and complexed species are in equilibrium under the conditions normally used for their preparation.
A suitable solvent must be used. For alkyl lithium formation pentane or hexane are usually used. Diethyl ether can also be used but the subsequent alkyl lithium reagent must be used immediately after preparation due to an interaction with the solvent. Ethyl ether or THF are essential for Grignard reagent formation. Lone pair electrons from two ether molecules form a complex with the magnesium in the Grignard reagent (As pictured below). This complex helps stabilize the organometallic and increases its ability to react.
These reactions are obviously substitution reactions, but they cannot be classified as nucleophilic substitutions, as were the earlier reactions of alkyl halides. Because the functional carbon atom has been reduced, the polarity of the resulting functional group is inverted (an originally electrophilic carbon becomes nucleophilic). This change, shown below, makes alkyl lithium and Grignard reagents excellent nucleophiles and useful reactants in synthesis.
Reaction of Organometallic Reagents with Various Carbonyls
Because organometallic reagents react as their corresponding carbanion, they are excellent nucleophiles. The basic reaction involves the nucleophilic attack of the carbanionic carbon in the organometallic reagent with the electrophilic carbon in the carbonyl to form alcohols.
Both Grignard and Organolithium Reagents will perform these reactions.
Addition to formaldehyde gives 1° alcohols
Addition to aldehydes gives 2° alcohols
Addition to ketones gives 3° alcohols
Addition to carbon dioxide (CO2) forms a carboxylic acid
Example $1$:
Going from Reactants to Products Simplified
Mechanism for the Addition to Carbonyls
The mechanism for a Grignard agent is shown; the mechanism for an organometallic reagent is the same.
1) Nucleophilic attack
2) Protonation
Organometallic Reagents as Bases
These reagents are very strong bases (pKa's of saturated hydrocarbons range from 42 to 50). Although not usually done with Grignard reagents, organolithium reagents can be used as strong bases. Both Grignard reagents and organolithium reagents react with water to form the corresponding hydrocarbon. This is why so much care is needed to insure dry glassware and solvents when working with organometallic reagents.
In fact, the reactivity of Grignard reagents and organolithium reagents can be exploited to create a new method for the conversion of halogens to the corresponding hydrocarbon (illustrated below). The halogen is converted to an organometallic reagent and then subsequently reacted with water to from an alkane.
Limitation of Organometallic Reagents
As discussed above, Grignard and organolithium reagents are powerful bases. Because of this they cannot be used as nucleophiles on compounds which contain acidic hydrogens. If they are used they will act as a base and deprotonate the acidic hydrogen rather than act as a nucleophile and attack the carbonyl. A partial list of functional groups which cannot be used are: alcohols, amides, 1o amines, 2o amines, carboxylic acids, and terminal alkynes.
Problems
1) Please write the product of the following reactions.
2) Please indicate the starting material required to produce the product.
3) Please give a detailed mechanism and the final product of this reaction
4) Please show two sets of reactants which could be used to synthesize the following molecule using a Grignard reaction.
Answers
1)
2)
3)
Nucleophilic attack
Protonation
4) | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Aldehydes_and_Ketones/Synthesis_of_Aldehydes_and_Ketones/Grignard_and_Organolithium_Reagents.txt |
This page explains how aldehydes and ketones are made in the lab by the oxidation of primary and secondary alcohols.
Oxidizing alcohols to make aldehydes and ketones
The oxidizing agent used in these reactions is normally a solution of sodium or potassium dichromate(VI) acidified with dilute sulfuric acid. If oxidation occurs, the orange solution containing the dichromate(VI) ions is reduced to a green solution containing chromium(III) ions.
The net effect is that an oxygen atom from the oxidizing agent removes a hydrogen from the -OH group of the alcohol and one from the carbon to which it is attached.
[O] is often used to represent oxygen coming from an oxidising agent.
R and R' are alkyl groups or hydrogen. They could also be groups containing a benzene ring, but I'm ignoring these to keep things simple.
If at least one of these groups is a hydrogen atom, then you will get an aldehyde. If they are both alkyl groups then you get a ketone. If you now think about where they are coming from, you will get an aldehyde if your starting molecule looks like this:
In other words, if you start from a primary alcohol, you will get an aldehyde. You will get a ketone if your starting molecule looks like this:
. . . where R and R' are both alkyl groups. Secondary alcohols oxidize to give ketones.
Making aldehydes
Aldehydes are made by oxidising primary alcohols. There is, however, a problem.
The aldehyde produced can be oxidised further to a carboxylic acid by the acidified potassium dichromate(VI) solution used as the oxidising agent. In order to stop at the aldehyde, you have to prevent this from happening.
To stop the oxidation at the aldehyde, you . . .
• use an excess of the alcohol. That means that there isn't enough oxidizing agent present to carry out the second stage and oxidize the aldehyde formed to a carboxylic acid.
• distil off the aldehyde as soon as it forms. Removing the aldehyde as soon as it is formed means that it doesn't stay in the mixture to be oxidized further.
If you used ethanol as a typical primary alcohol, you would produce the aldehyde ethanal, CH3CHO. The full equation for this reaction is fairly complicated, and you need to understand about electron-half-equations in order to work it out.
In organic chemistry, simplified versions are often used which concentrate on what is happening to the organic substances. To do that, oxygen from an oxidising agent is represented as [O]. That would produce the much simpler equation:
Secondary alcohols
Secondary alcohols are oxidised to ketones. There is no further reaction which might complicate things. For example, if you heat the secondary alcohol propan-2-ol with sodium or potassium dichromate(VI) solution acidified with dilute sulphuric acid, you get propanone formed.
Playing around with the reaction conditions makes no difference whatsoever to the product. Using the simple version of the equation:
Contributors
Jim Clark (Chemguide.co.uk)
Synthesis of Aldehydes and Ketones
Aldehydes and ketones can be prepared using a wide variety of reactions. Although these reactions are discussed in greater detail in other sections, they are listed here as a summary and to help with planning multistep synthetic pathways. Please use the appropriate links to see more details about the reactions.
Hydration of an alkyne to form aldehydes
Anti-Markovnikov addition of a hydroxyl group to an alkyne forms an aldehyde. The addition of a hydroxyl group to an alkyne causes tautomerization which subsequently forms a carbonyl.
Oxidation of 2o alcohols to form ketones
Typically uses Jones reagent (CrO3 in H2SO4) but many other reagents can be used
Hydration of an alkyne to form ketones
The addition of a hydroxyl group to an alkyne causes tautomerization which subsequently forms a carbonyl. Markovnikov addition of a hydroxyl group to an alkyne forms a ketone.
Alkenes can be cleaved using ozone (O3) to form aldehydes and/or ketones
This is an example of a Ozonolysis reaction. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Aldehydes_and_Ketones/Synthesis_of_Aldehydes_and_Ketones/Preparation_of_Aldehydes_and_Ketones.txt |
The names of all alkanes end with -ane. Whether or not the carbons are linked together end-to-end in a ring (called cyclic alkanes or cycloalkanes) or whether they contain side chains and branches, the name of every carbon-hydrogen chain that lacks any double bonds or functional groups will end with the suffix -ane.
Alkanes with unbranched carbon chains are simply named by the number of carbons in the chain. The first four members of the series (in terms of number of carbon atoms) are named as follows:
1. CH4 = methane = one hydrogen-saturated carbon
2. C2H6 = ethane = two hydrogen-saturated carbons
3. C3H8 = propane = three hydrogen-saturated carbons
4. C4H10 = butane = four hydrogen-saturated carbons
Alkanes with five or more carbon atoms are named by adding the suffix -ane to the appropriate numerical multiplier, except the terminal -a is removed from the basic numerical term. Hence, C5H12 is called pentane, C6H14 is called hexane, C7H16 is called heptane and so forth.
Straight-chain alkanes are sometimes indicated by the prefix n- (for normal) to distinguish them from branched-chain alkanes having the same number of carbon atoms. Although this is not strictly necessary, the usage is still common in cases where there is an important difference in properties between the straight-chain and branched-chain isomers: e.g. n-hexane is a neurotoxin while its branched-chain isomers are not.
IUPAC nomenclature
The IUPAC nomenclature is a system on which most organic chemists have agreed to provide guidelines to allow them to learn from each others' works. Nomenclature, in other words, provides a foundation of language for organic chemistry.
Number of Hydrogen to Carbons
This equation describes the relationship between the number of hydrogen and carbon atoms in alkanes:
H = 2C + 2
where "C" and "H" are used to represent the number of carbon and hydrogen atoms present in one molecule. If C = 2, then H = 6.
Many textbooks put this in the following format:
CnH2n+2
where "Cn" and "H2n+2" represent the number of carbon and hydrogen atoms present in one molecule. If Cn = 3, then H2n+2 = 2(3) + 2 = 8. (For this formula look to the "n" for the number, the "C" and the "H" letters themselves do not change.)
Progressively longer hydrocarbon chains can be made and are named systematically, depending on the number of carbons in the longest chain.
The following table contains the systematic names for the first twenty straight chain alkanes. It will be important to familiarize yourself with these names because they will be the basis for naming many other organic molecules throughout your course of study.
Drawing Hydrocarbons
Recall that when carbon makes four bonds, it adopts the tetrahedral geometry. In the tetrahedral geometry, only two bonds can occupy a plane simultaneously. The other two bonds point in back or in front of this plane. In order to represent the tetrahedral geometry in two dimensions, solid wedges are used to represent bonds pointing out of the plane of the drawing toward the viewer, and dashed wedges are used to represent bonds pointing out of the plane of the drawing away from the viewer. Consider the following representation of the molecule methane:
Figure 1: Two dimensional representation of methane
In the above drawing, the two hydrogens connected by solid lines, as well as the carbon in the center of the molecule, exist in a plane (specifically, the plane of the computer monitor / piece of paper, etc.). The hydrogen connected by a solid wedge points out of this plane toward the viewer, and the hydrogen connected by the dashed wedge points behind this plane and away from the viewer.
In drawing hydrocarbons, it can be time-consuming to write out each atom and bond individually. In organic chemistry, hydrocarbons can be represented in a shorthand notation called a skeletal structure. In a skeletal structure, only the bonds between carbon atoms are represented. Individual carbon and hydrogen atoms are not drawn, and bonds to hydrogen are not drawn. In the case that the molecule contains just single bonds (sp3 bonds), these bonds are drawn in a "zig-zag" fashion. This is because in the tetrahedral geometry all bonds point as far away from each other as possible, and the structure is not linear. Consider the following representations of the molecule propane:
Figure 2: Full structure of propane and skeletal structure of propane
Only the bonds between carbons have been drawn, and these have been drawn in a "zig-zag" manner. Note that there is no representation of hydrogens in a skeletal structure. Since, in the absence of double or triple bonds, carbon makes four bonds total, the presence of hydrogens is implicit. Whenever an insufficient number of bonds to a carbon atom are specified in the structure, it is assumed that the rest of the bonds are made to hydrogens. For example, if the carbon atom makes only one explicit bond, there are three hydrogens implicitly attached to it. If it makes two explicit bonds, there are two hydrogens implicitly attached, etc. Note also that two lines are sufficient to represent three carbon atoms. It is the bonds only that are being drawn out, and it is understood that there are carbon atoms (with three hydrogens attached!) at the terminal ends of the structure.
Alkyl Groups
Alkanes can be described by the general formula CnH2n+2. An alkyl group is formed by removing one hydrogen from the alkane chain and is described by the formula CnH2n+1. The removal of this hydrogen results in a stem change from -ane to -yl. Take a look at the following examples.
The same concept can be applied to any of the straight chain alkane names provided in the table above.
Name Molecular Formula Condensed Structural Formula
Methane CH4 CH4
Ethane C2H6 CH3CH3
Propane C3H8 CH3CH2CH3
Butane C4H10 CH3(CH2)2CH3
Pentane C5H12 CH3(CH2)3CH3
Hexane C6H14 CH3(CH2)4CH3
Heptane C7H16 CH3(CH2)5CH3
Octane C8H18 CH3(CH2)6CH3
Nonane C9H20 CH3(CH2)7CH3
Decane C10H22 CH3(CH2)8CH3
Undecane C11H24 CH3(CH2)9CH3
Dodecane C12H26 CH3(CH2)10CH3
Tridecane C13H28 CH3(CH2)11CH3
Tetradecane C14H30 CH3(CH2)12CH3
Pentadecane C15H32 CH3(CH2)13CH3
Hexadecane C16H34 CH3(CH2)14CH3
Heptadecane C17H36 CH3(CH2)15CH3
Octadecane C18H38 CH3(CH2)16CH3
Nonadecane C19H40 CH3(CH2)17CH3
Eicosane C20H42 CH3(CH2)18CH3
Using Common Names with Branched Alkanes
Certain branched alkanes have common names that are still widely used today. These common names make use of prefixes, such as iso-, sec-, tert-, and neo-. The prefix iso-, which stands for isomer, is commonly given to 2-methyl alkanes. In other words, if there is methyl group located on the second carbon of a carbon chain, we can use the prefix iso-. The prefix will be placed in front of the alkane name that indicates the total number of carbons. Examples:
• isopentane which is the same as 2-methylbutane
• isobutane which is the same as 2-methylpropane
To assign the prefixes sec-, which stands for secondary, and tert-, for tertiary, it is important that we first learn how to classify carbon molecules. If a carbon is attached to only one other carbon, it is called a primary carbon. If a carbon is attached to two other carbons, it is called a seconday carbon. A tertiary carbon is attached to three other carbons and last, a quaternary carbon is attached to four other carbons. Examples:
• 4-sec-butylheptane (30g)
• 4-tert-butyl-5-isopropylhexane (30d); if using this example, may want to move sec/tert after iso disc
The prefix neo- refers to a substituent whose second-to-last carbon of the chain is trisubstituted (has three methyl groups attached to it). A neo-pentyl has five carbons total. Examples:
• neopentane
• neoheptane
Alkoxy Groups
Alkoxides consist of an organic group bonded to a negatively charged oxygen atom. In the general form, alkoxides are written as RO-, where R represents the organic substituent. Similar to the alkyl groups above, the concept of naming alkoxides can be applied to any of the straight chain alkanes provided in the table above.
Three Principles of Naming
1. Choose the longest, most substituted carbon chain containing a functional group.
2. A carbon bonded to a functional group must have the lowest possible carbon number. If there are no functional groups, then any substitute present must have the lowest possible number.
3. Take the alphabetical order into consideration; that is, after applying the first two rules given above, make sure that your substitutes and/or functional groups are written in alphabetical order.
Example 1
What is the name of the following molecule?
Solution
Rule #1: Choose the longest, most substituted carbon chain containing a functional group. This example does not contain any functional groups, so we only need to be concerned with choosing the longest, most substituted carbon chain. The longest carbon chain has been highlighted in red and consists of eight carbons.
Rule #2: Carbons bonded to a functional group must have the lowest possible carbon number. If there are no functional groups, then any substitute present must have the lowest possible number. Because this example does not contain any functional groups, we only need to be concerned with the two substitutes present, that is, the two methyl groups. If we begin numbering the chain from the left, the methyls would be assigned the numbers 4 and 7, respectively. If we begin numbering the chain from the right, the methyls would be assigned the numbers 2 and 5. Therefore, to satisfy the second rule, numbering begins on the right side of the carbon chain as shown below. This gives the methyl groups the lowest possible numbering.
Rule 3: In this example, there is no need to utilize the third rule. Because the two substitutes are identical, neither takes alphabetical precedence with respect to numbering the carbons. This concept will become clearer in the following examples.
Example 2
What is the name of the following molecule?
Solution
Rule #1: Choose the longest, most substituted carbon chain containing a functional group. This example contains two functional groups, bromine and chlorine. The longest carbon chain has been highlighted in red and consists of seven carbons.
Rule #2: Carbons bonded to a functional group must have the lowest possible carbon number. If there are no functional groups, then any substitute present must have the lowest possible number. In this example, numbering the chain from the left or the right would satisfy this rule. If we number the chain from the left, bromine and chlorine would be assigned the second and sixth carbon positions, respectively. If we number the chain from the right, chlorine would be assigned the second position and bromine would be assigned the sixth position. In other words, whether we choose to number from the left or right, the functional groups occupy the second and sixth positions in the chain. To select the correct numbering scheme, we need to utilize the third rule.
Rule #3: After applying the first two rules, take the alphabetical order into consideration. Alphabetically, bromine comes before chlorine. Therefore, bromine is assigned the second carbon position, and chlorine is assigned the sixth carbon position.
Example 3
What is the name of the follow molecule?
Solution
Rule #1: Choose the longest, most substituted carbon chain containing a functional group. This example contains two functional groups, bromine and chlorine, and one substitute, the methyl group. The longest carbon chain has been highlighted in red and consists of seven carbons.
Rule #2: Carbons bonded to a functional group must have the lowest possible carbon number. After taking functional groups into consideration, any substitutes present must have the lowest possible carbon number. This particular example illustrates the point of difference principle. If we number the chain from the left, bromine, the methyl group and chlorine would occupy the second, fifth and sixth positions, respectively. This concept is illustrated in the second drawing below. If we number the chain from the right, chlorine, the methyl group and bromine would occupy the second, third and sixth positions, respectively, which is illustrated in the first drawing below. The position of the methyl, therefore, becomes a point of difference. In the first drawing, the methyl occupies the third position. In the second drawing, the methyl occupies the fifth position. To satisfy the second rule, we want to choose the numbering scheme that provides the lowest possible numbering of this substitute. Therefore, the first of the two carbon chains shown below is correct.
Therefore, the first numbering scheme is the appropriate one to use.
Once you have determined the correct numbering of the carbons, it is often useful to make a list, including the functional groups, substitutes, and the name of the parent chain.
Rule #3: After applying the first two rules, take the alphabetical order into consideration. Alphabetically, bromine comes before chlorine. Therefore, bromine is assigned the second carbon position, and chlorine is assigned the sixth carbon position.
Parent chain: heptane 2-Chloro 3-Methyl 6-Bromo
6-bromo-2-chloro-3-methylheptane
Problems
What is the name of the follow molecules?
Nomenclature of Alkanes
I. The parent compound must have the longest chain of carbon atoms
Learn: methane (1 carbon atom), ethane (2 carbon atoms), propane (3 carbon atoms), butane (4 carbon atoms), pentane (5 carbon atoms), hexane (6 carbon atoms), heptane (7 carbon atoms), octane (8 carbon atoms), nonane (9 carbon atoms) and decane (10 carbon atoms).
Try to name the following compounds:
1. CH3-CH2-CH2-CH3
2. CH3-CH2-CH2-CH2-CH2-CH3
3. CH3-CH2-CH2-CH2-CH2-CH2-CH2-CH2-CH3
Try to draw structures for the following compounds:
1. propane
2. pentane
3. octane
II. The parent chain is numbered to give substituents the lowest possible numbers
Substituent names are methyl, ethyl, propyl, butyl, etc. The number showing the point of attachment to the parent chain precedes the substituent name. If you have more than one substituent with the same name, a number of attachment must be given for each substituent and the number of the substituents must be designated with di-, tri-, etc.
Try to name the following compounds...
7.
8.
9.
Try to draw structures for the following compounds...
1. 2,2-dimethylpropane
2. 3-methylheptane
3. 4,5-diethylnonane
III. Substituents are named in alphabetical order
Remember that the numbering of the parent chain must give all substituents the lowest possible numbers regardless of their names.
Try to name the following compound...
13.
IV. A large substituent is numbered to give its point of attachment
The point of attachment is number one and any other smaller groups are named as substituent groups on the larger group. This numbering is independent of the numbering of the parent chain. Try to name the following compounds...
14.
15.
Try to draw structures for the following compounds...
1. 4-(1-methylethyl)heptane
2. 5-(1,1-dimethylethyl)nonane
V. Ring compounds are designated with a cyclo- prefix and are numbered to give multiple substituents the lowest possible numbers
A single substituent does not need to be numbered. A ring can also be named as a substituent. Try to name the following compounds:
18.
19.
20.
Try to draw structures for the following compounds...
1. ethylcyclobutane
2. 1-ethyl-4-methylcyclohexane
3. cyclobutylcyclooctane
VI. Common Names
• isopropyl = 1-methylethyl
• isobutyl = 2-methylpropyl
• sec-butyl = 1-methylpropyl
• tert-butyl = 1,1-dimethylethyl
• neo-pentyl = 2,2-dimethylpropyl
• iso-pentyl = 3-methylbutyl
Try to name the following compounds using common names...
24.
25.
Try to draw structures for the following compounds...
1. 4-isopropyloctane
2. isopentylcyclohexane
Answers
1. butane
2. hexane
3. nonane
4. CH3-CH2-CH3
5. CH3-CH2-CH2-CH2-CH3
6. CH3-CH2-CH2-CH2-CH2-CH2-CH2-CH3
7. 2-methylbutane
8. 3-ethylhexane
9. 3-methyloctane (Remember that you must number the parent chain to give the substituent the lowest possible number.)
10. 4-ethyl-3-methylheptane
11. 4-(1-methylethyl)octane
12. 5-(1-methylpropyl)decane
13. 1-ethyl-2-methylcyclohexane
14. methylcyclopentane (You do not need a number since there is only one substituent and it is assumed to be on the first carbon regardless of how the ring compound is drawn.)
15. cyclopropylcyclopentane
16. neo-pentylcycloheptane (1-(2,2-dimethyl-1-propyl)cycloheptane is the proper name.)
17. isopropylcyclohexane | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkanes/Nomenclature_of_Alkanes/Nomenclature_of_Alkanes_II.txt |
Cycloalkanes are cyclic hydrocarbons, meaning that the carbons of the molecule are arranged in the form of a ring. Cycloalkanes are also saturated, meaning that all of the carbons atoms that make up the ring are single bonded to other atoms (no double or triple bonds). There are also polycyclic alkanes, which are molecules that contain two or more cycloalkanes that are joined, forming multiple rings.
Introduction
Many organic compounds found in nature or created in a laboratory contain rings of carbon atoms with distinguishing chemical properties; these compounds are known as cycloalkanes. Cycloalkanes only contain carbon-hydrogen bonds and carbon-carbon single bonds, but in cycloalkanes, the carbon atoms are joined in a ring. The smallest cycloalkane is cyclopropane.
Figure 1: The first four cycloalkanes
If you count the carbons and hydrogens, you will see that they no longer fit the general formula \(C_nH_{2n+2}\). By joining the carbon atoms in a ring,two hydrogen atoms have been lost. The general formula for a cycloalkane is \(C_nH_{2n}\). Cyclic compounds are not all flat molecules. All of the cycloalkanes, from cyclopentane upwards, exist as "puckered rings". Cyclohexane, for example, has a ring structure that looks like this:
Figure 2: This is known as the "chair" form of cyclohexane from its shape, which vaguely resembles a chair. Note: The cyclohexane molecule is constantly changing, with the atom on the left, which is currently pointing down, flipping up, and the atom on the right flipping down. During this process, another (slightly less stable) form of cyclohexane is formed known as the "boat" form. In this arrangement, both of these atoms are either pointing up or down at the same time
In addition to being saturated cyclic hydrocarbons, cycloalkanes may have multiple substituents or functional groups that further determine their unique chemical properties. The most common and useful cycloalkanes in organic chemistry are cyclopentane and cyclohexane, although other cycloalkanes varying in the number of carbons can be synthesized. Understanding cycloalkanes and their properties are crucial in that many of the biological processes that occur in most living things have cycloalkane-like structures.
Glucose (6 carbon sugar) Ribose (5 carbon sugar)
Cholesterol (polycyclic)
Although polycyclic compounds are important, they are highly complex and typically have common names accepted by IUPAC. However, the common names do not generally follow the basic IUPAC nomenclature rules. The general formula of the cycloalkanes is \(C_nH_{2n}\) where \(n\) is the number of carbons. The naming of cycloalkanes follows a simple set of rules that are built upon the same basic steps in naming alkanes. Cyclic hydrocarbons have the prefix "cyclo-".
Parent Chains
For simplicity, cycloalkane molecules can be drawn in the form of skeletal structures in which each intersection between two lines is assumed to have a carbon atom with its corresponding number of hydrogens.
same as same as
Cycloalkane Molecular Formula Basic Structure
Cyclopropane \(C_3H_6\0
Cyclobutane \(C_4H_8\)
Cyclopentane \(C_5H_{10}\)
Cyclohexane \(C_6H_{12}\)
Cycloheptane \(C_7H_{14}\)
Cyclooctane \(C_8H_{16}\)
Cyclononane \(C_9H_{18}\)
Cyclodecane \(C_{10}H_{20}\)
IUPAC Rules for Nomenclature
1. Determine the cycloalkane to use as the parent chain. The parent chain is the one with the highest number of carbon atoms. If there are two cycloalkanes, use the cycloalkane with the higher number of carbons as the parent chain.
2. If there is an alkyl straight chain that has a greater number of carbons than the cycloalkane, then the alkyl chain must be used as the primary parent chain. Cycloalkane acting as a substituent to an alkyl chain has an ending "-yl" and, therefore, must be named as a cycloalkyl.
Cycloalkane Cycloalkyl
cyclopropane cyclopropyl
cyclobutane cyclobutyl
cyclopentane cyclopentyl
cyclohexane cyclohexyl
cycloheptane cycloheptyl
cyclooctane cyclooctyl
cyclononane cyclononanyl
cyclodecane cyclodecanyl
Example 1
The longest straight chain contains 10 carbons, compared with cyclopropane, which only contains 3 carbons. Because cyclopropane is a substituent, it would be named a cyclopropyl-substituted alkane.
1. Determine any functional groups or other alkyl groups.
2. Number the carbons of the cycloalkane so that the carbons with functional groups or alkyl groups have the lowest possible number. A carbon with multiple substituents should have a lower number than a carbon with only one substituent or functional group. One way to make sure that the lowest number possible is assigned is to number the carbons so that when the numbers corresponding to the substituents are added, their sum is the lowest possible.
(1+3=4) NOT (1+5=6)
1. When naming the cycloalkane, the substituents and functional groups must be placed in alphabetical order.
(ex: 2-bromo-1-chloro-3-methylcyclopentane)
1. Indicate the carbon number with the functional group with the highest priority according to alphabetical order. A dash"-" must be placed between the numbers and the name of the substituent. After the carbon number and the dash, the name of the substituent can follow. When there is only one substituent on the parent chain, indicating the number of the carbon atoms with the substituent is not necessary.
(ex: 1-chlorocyclohexane or cholorocyclohexane is acceptable)
1. If there is more than one of the same functional group on one carbon, write the number of the carbon two, three, or four times, depending on how many of the same functional group is present on that carbon. The numbers must be separated by commas, and the name of the functional group that follows must be separated by a dash. When there are two of the same functional group, the name must have the prefix "di". When there are three of the same functional group, the name must have the prefix "tri". When there are four of the same functional group, the name must have the prefix "tetra". However, these prefixes cannot be used when determining the alphabetical priorities. There must always be commas between the numbers and the dashes that are between the numbers and the names.
Example 2
(2-bromo-1,1-dimethylcyclohexane)
Notice that "f" of fluoro alphabetically precedes the "m" of methyl. Although "di" alphabetically precedes "f", it is not used in determining the alphabetical order.
Example 3
(2-fluoro-1,1,-dimethylcyclohexane NOT 1,1-dimethyl-2-fluorocyclohexane)
8) If the substituents of the cycloalkane are related by the cis or trans configuration, then indicate the configuration by placing "cis-" or "trans-" in front of the name of the structure.
Blue=Carbon Yellow=Hydrogen Green=Chlorine
Notice that chlorine and the methyl group are both pointed in the same direction on the axis of the molecule; therefore, they are cis.
cis-1-chloro-2-methylcyclopentane
9) After all the functional groups and substituents have been mentioned with their corresponding numbers, the name of the cycloalkane can follow.
Reactivity
Cycloalkanes are very similar to the alkanes in reactivity, except for the very small ones, especially cyclopropane. Cyclopropane is significantly more reactive than what is expected because of the bond angles in the ring. Normally, when carbon forms four single bonds, the bond angles are approximately 109.5°. In cyclopropane, the bond angles are 60°.
With the electron pairs this close together, there is a significant amount of repulsion between the bonding pairs joining the carbon atoms, making the bonds easier to break.
Alcohol Substituents on Cycloalkanes
Alcohol (-OH) substituents take the highest priority for carbon atom numbering in IUPAC nomenclature. The carbon atom with the alcohol substituent must be labeled as 1. Molecules containing an alcohol group have an ending "-ol", indicating the presence of an alcohol group. If there are two alcohol groups, the molecule will have a "di-" prefix before "-ol" (diol). If there are three alcohol groups, the molecule will have a "tri-" prefix before "-ol" (triol), etc.
Example 4
The alcohol substituent is given the lowest number even though the two methyl groups are on the same carbon atom and labeling 1 on that carbon atom would give the lowest possible numbers. Numbering the location of the alcohol substituent is unnecessary because the ending "-ol" indicates the presence of one alcohol group on carbon atom number 1.
2,2-dimethylcyclohexanol NOT 1,1-dimethyl-cyclohexane-2-ol
Example 5
3-bromo-2-methylcyclopentanol NOT 1-bromo-2-methyl-cyclopentane-2-ol
Example 6
Blue=Carbon Yellow=Hydrogen Red=Oxygen
trans-cyclohexane-1,2-diol
Other Substituents on Cycloalkanes
There are many other functional groups like alcohol, which are later covered in an organic chemistry course, and they determine the ending name of a molecule. The naming of these functional groups will be explained in depth later as their chemical properties are explained.
Name Name ending
alkene -ene
alkyne -yne
alcohol -ol
ether -ether
nitrile -nitrile
amine -amine
aldehyde -al
ketone -one
carboxylic acid -oic acid
ester -oate
amide -amide
Although alkynes determine the name ending of a molecule, alkyne as a substituent on a cycloalkane is not possible because alkynes are planar and would require that the carbon that is part of the ring form 5 bonds, giving the carbon atom a negative charge.
However, a cycloalkane with a triple bond-containing substituent is possible if the triple bond is not directly attached to the ring.
Example 7
ethynylcyclooctane
Example 8
1-propylcyclohexane
Summary
1. Determine the parent chain: the parent chain contains the most carbon atoms.
2. Number the substituents of the chain so that the sum of the numbers is the lowest possible.
3. Name the substituents and place them in alphabetical order.
4. If stereochemistry of the compound is shown, indicate the orientation as part of the nomenclature.
5. Cyclic hydrocarbons have the prefix "cyclo-" and have an "-alkane" ending unless there is an alcohol substituent present. When an alcohol substituent is present, the molecule has an "-ol" ending.
Glossary
• alcohol: An oxygen and hydrogenOH hydroxyl group that is bonded to a substituted alkyl group.
• alkyl: A structure that is formed when a hydrogen atom is removed from an alkane.
• cyclic: Chemical compounds arranged in the form of a ring or a closed chain form.
• cycloalkanes: Cyclic saturated hydrocarbons with a general formula of CnH(2n). Cycloalkanes are alkanes with carbon atoms attached in the form of a closed ring.
• functional groups: An atom or groups of atoms that substitute for a hydrogen atom in an organic compound, giving the compound unique chemical properties and determining its reactivity.
• hydrocarbon: A chemical compound containing only carbon and hydrogen atoms.
• saturated: All of the atoms that make up a compound are single bonded to the other atoms, with no double or triple bonds.
• skeletal structure: A simplified structure in which each intersection between two lines is assumed to have a carbon atom with its corresponding number of hydrogens.
Problems
Name the following structures. (Note: The structures are complex for practice purposes and may not be found in nature.)
1) 2) 3) 4) 5) 6)
7)
Draw the following structures.
8) 1,1-dibromo-5-fluoro-3-butyl-7-methylcyclooctane 9) trans-1-bromo-2-chlorocyclopentane
10) 1,1-dibromo-2,3-dichloro-4-propylcyclobutane 11) 2-methyl-1-ethyl-1,3-dipropylcyclopentane 12) cycloheptane-1,3,5-triol
Name the following structures.
Blue=Carbon Yellow=Hydrogen Red=Oxygen Green=Chlorine
13) 14) 15) 16) 17)
18) 19)
Answers to Practice Problems
1) cyclodecane 2) chlorocyclopentane or 1-chlorocyclopentane 3) trans-1-chloro-2-methylcycloheptane
4) 3-cyclopropyl-6-methyldecane 5) cyclopentylcyclodecane or 1-cyclopentylcyclodecane 6) 1,3-dibromo-1-chloro-2-fluorocycloheptane
7) 1-cyclobutyl-4-isopropylcyclohexane
8) 9) 10) 11) 12)
13) cyclohexane 14) cyclohexanol 15) chlorocyclohexane 16) cyclopentylcyclohexane 17) 1-chloro-3-methylcyclobutane
18) 2,3-dimethylcyclohexanol 19) cis-1-propyl-2-methylcyclopentane
Contributors
• Pwint Zin
• Jim Clark (ChemGuide) | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkanes/Nomenclature_of_Alkanes/Nomenclature_of_Cycloalkanes.txt |
Alkanes are the simplest family of hydrocarbons - compounds containing carbon and hydrogen only with only carbon-hydrogen bonds and carbon-carbon single bonds. Alkanes are not very reactive and have little biological activity; all alkanes are colorless and odorless.
Properties of Alkanes
This is an introductory page about alkanes, such as methane, ethane, propane, butane and the remainder of the common alkanes. This page addresses their formulae and isomerism, their physical properties, and an introduction to their chemical reactivity.
Molecular Formulas
Alkanes are the simplest family of hydrocarbons - compounds containing carbon and hydrogen only. Alkanes only contain carbon-hydrogen bonds and carbon-carbon single bonds. The formula of any of the alkanes follow the general formula \(C_nH_{2n+2}\) . The first six alkanes are tabulated below:
Table 1: The first six alkanes are as follows:
name Formula
methane CH4
ethane C2H6
propane C3H8
butane C4H10
pentane C5H12
hexane C6H14
Isomerism
All of the alkanes containing 4 or more carbon atoms show structural isomerism, meaning that there are two or more different structural formulae that you can draw for each molecular formula.
Example 1: Butane or MethylPropane
C4H10 could be either of these two different molecules:
These are named butane and 2-methylpropane, respectively.
Cycloalkanes
Cycloalkanes also only contain carbon-hydrogen bonds and carbon-carbon single bonds, but the carbon atoms are joined in a ring. The smallest cycloalkane is cyclopropane.
If you count the carbons and the hydrogens, you will see that they no longer adhere to the general formula CnH2n+2. By joining the carbon atoms in a ring, two hydrogen atoms are lost. The general formula for a cycloalkane is CnH2n and these are non-planar molecules (with the exception of cyclopentane) and exist as "puckered rings".
Example 2: Cyclohexane Conformation
Cyclohexane has a ring structure that looks like this:
This is known as the "chair" form of cyclohexane because of its shape, which vaguely resembles a chair. Cyclohexane also has a boat configuration (not shown).
Physical Properties
Boiling Points
The boiling points shown are for the "straight chain" isomers in which there are more than one (Figure 1). Notice that the first four alkanes are gases at room temperature, and solids do not start to appear until about C17H36.
Figure 1:Normal boiling points of first four alkanes
The temperatures cannot be more precise than those given in this chart because each isomer has a different melting and boiling point. By the time you get 17 carbons into an alkane, there are unbelievable numbers of isomers! Cycloalkanes have boiling points that are approximately 10 - 20 K higher than the corresponding straight chain alkane.
There electronegativity difference between carbon and hydrogen (2.1 vs. 1.9) is small; therefore, there is only a slight bond polarity, meaning that the only attractions between one molecule and its neighbors will be Van der Waals dispersion forces. These forces will be very small for a molecule like methane but will increase as the size of the molecules increase. Therefore, the boiling points of the alkanes increase with the molecular size.
Regarding isomers, the more branched the chain, the lower the boiling point tends to be. Van der Waals dispersion forces are smaller for shorter molecules and only operate over very short distances between one molecule and its neighbors. It is more difficult for short, bulky molecules (with substantial amounts of branching) to lie close together (compact) compared with long, thin molecules.
Example 3
The boiling points of the three isomers of C5H12 are as follows:
• pentane (309.2 K)
• 2-methylbutane (301.0 K)
• 2,2-dimethylpropane (282.6 K)
The slightly higher boiling points for the cycloalkanes are presumably because the molecules can get closer together because the ring structure makes them tidier and less "wriggly"!
Solubility
Alkanes (both normal and cycloalkanes) are virtually insoluble in water but dissolve in organic solvents. The liquid alkanes are good solvents for many other covalent compounds. When a molecular substance dissolves in water, the following must occur:
• breaking of the intermolecular forces within the substance. In the case of the alkanes, these are the Van der Waals dispersion forces.
• breaking of the intermolecular forces in the water so that the substance can fit between the water molecules. In water, the primary intermolecular attractions are hydrogen bonds.
Breaking either of these attractions requires energy, although the amount of energy required to break the Van der Waals dispersion forces in a compound, such as methane, is relatively negligible; this is not true of the hydrogen bonds in water.
To simplify, a substance will dissolve if sufficient energy is released when the new bonds are formed between the substance and the water to make up for the energy required to break the original attractions. The only new attractions between the alkane and the water molecules are the Van der Waals forces. These forces to do not release a sufficient amount of energy to compensate for the energy required to break the hydrogen bonds in water. Therefore, the alkane does not dissolve.
Solubility in organic solvents
In most organic solvents, the primary forces of attraction between the solvent molecules are the Van der Waals forces composed of either dispersion forces or dipole-dipole attractions. Therefore, when an alkane dissolves in an organic solvent, the Van der Waals forces are broken and are replaced by new Van der Waals forces. The two processes more or less cancel each other out energetically; thus, there is no barrier to solubility.
Chemical Reactivity
Alkanes contain strong carbon-carbon single bonds and strong carbon-hydrogen bonds. The carbon-hydrogen bonds are only very slightly polar. Therefore, there is no portion of the molecule that carries any significant amount of positive or negative charge, which is required for other molecules to be attracted to it. For example, many organic reactions start because an ion or a polar molecule is attracted to a portion of an organic molecule, which carries some positive or negative charge. This attraction does not occur with alkanes because alkane molecules do not have this separation of charge. The net effect is that alkanes have a fairly restricted set of reactions, including the following:
• burn them - destroying the entire molecule;
• react them with some of the halogens, breaking the carbon-hydrogen bonds;
• crack them, breaking carbon-carbon bonds.
These reactions are covered in separate pages.
Cycloalkanes are very similar to the alkanes in reactivity, except for the very small cycloalkanes, especially cyclopropane. Cyclopropane is much more reactive than what is expected because of the bond angles in the ring. Normally, when carbon forms four single bonds, the bond angles are approximately 109.5°.
Example 4: Cyclopropane
In cyclopropane, the bond angles are 60°.
With the electron pairs this close together, there is a significant amount of repulsion between the bonding pairs joining the carbon atoms, making the bonds easier to break.
Conformational Analysis of Alkanes
A conformational analysis is a study of the energetics of different spatial arrangements of atoms relative to rotations about bonds. Conformational analyses are assisted greatly by a representation of molecules in a manner different to skeletal structures. A representation of a molecule in which the atoms and bonds are viewed along the axis about which rotation occurs is called a Newman projection (Figure 2).
Figure 2: Newman projection of ethane
In a Newman projection, the molecule is viewed along an axis containing two atoms bonded to each other and the bond between them, about which the molecule can rotate. In a Newman projection, the "substituents" of each atom composing the bond, be they hydrogens or functional groups, can then be viewed both in front of and behind the carbon-carbon bond. Specifically, one can observe the angle between a substituent on the front atom and a substituent on the back atom in the Newman projection, which is called the dihedral angle or torsion angle.
In ethane specifically, we can imagine two possible "extreme" conformations. In one case, the dihedral angle is 0° and the hydrogens on the first carbon line up with or eclipse the hydrogens on the second carbon. When the dihedral angle is 0° and the hydrogens line up perfectly, ethane has adopted the eclipsed conformation (Figure 3). The other extreme occurs when the hydrogens on the first carbon are as far away as possible from those on the second carbon; this occurs at a dihedral angle of 60° and is called the staggered conformation (Figure 3).
Figure 3: Conformations of ethane - staggered and eclipsed
The staggered conformation of ethane is a more stable, lower energy conformation than the eclipsed conformation because the eclipsed conformation involves unfavorable interactions between hydrogen atoms. Specifically, the negatively charged electrons in the bonds repel each other most when the bonds line up. Thus, ethane spends most of its time in the more stable staggered conformation.
Conformational analysis of butane
In butane, two of the substituents, one on each carbon atom being viewed, is a methyl group. Methyl groups are much larger than hydrogen atoms. Thus, when eclipsed conformations occur in butane, the interactions are especially unfavorable. There are four possible "extreme" conformations of butane: 1) Fully eclipsed, when the methyl groups eclipse each other; 2) Gauche, when the methyl groups are staggered but next to each other; 3) Eclipsed, when the methyl groups eclipse hydrogen atoms; and 4) Anti, when the methyl groups are staggered and as far away from each other as possible (Figure 4).
Figure 4: Conformations of butane: fully eclipsed, gauche, eclipsed, and anti
The fully eclipsed conformation is clearly the highest in energy and least favorable since the largest groups are interacting directly with each other. As the molecule rotates, it adopts the relatively stable gauche conformation. As it continues to rotate, it encounters less favorable eclipsed conformation in which a methyl group eclipses a hydrogen. As rotation continues, the molecule comes to the anti conformation, which is the most stable since the substituents are staggered and the methyl groups are as far away from each other as possible. An energy diagram is a graph which represents the energy of a molecule as a function of some changing parameter. An energy diagram can be made as a function of dihedral angle for butane (Figure 5).
Figure 5: Energy diagram for conformations of butane as a function of dihedral angle
Clearly, the anti and gauche conformations are significantly more stable than the eclipsed and fully eclipsed conformations. Butane spends most of its time in the anti and gauche conformations.
There is one final, very important point. At room temperature, approximately 84 kJ/mol of thermal energy is available to molecules. Thus, if the barrier to a rotation is less than 84 kJ/mol, the molecule will rotate. In ethane and butane, the barriers to rotation are significantly less than 84 kJ/mol. Therefore, even though the eclipsed conformations are unfavorable, the molecules are able to adopt them. In reality, since these conformations are not as stable, the molecules will quickly pass through them at room temperature and return a staggered conformation. Molecules are constantly converting between different staggered conformations all the time, quickly passing through eclipsed conformations in between. Thus, in alkanes, no single "true" conformation exists all the time; the molecule instead constantly converts between conformations, spending more time in those that are more stable. This constant conversion lies in stark contract to alkenes, which adopt the cis- or trans- (E- or Z-) conformations and retain them at room temperature; they do not interconvert because the barrier to rotation is too high. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkanes/Properties_of_Alkanes/Alkanes_Background.txt |
Alkanes are not very reactive when compared with other chemical species. This is because the backbone carbon atoms in alkanes have attained their octet of electrons through forming four covalent bonds (the maximum allowed number of bonds under the octet rule; which is why carbon's valence number is 4). These four bonds formed by carbon in alkanes are sigma bonds, which are more stable than other types of bond because of the greater overlap of carbon's atomic orbitals with neighboring atoms' atomic orbitals. To make alkanes react, the input of additional energy is needed; either through heat or radiation.
Gasoline is a mixture of the alkanes and unlike many chemicals, can be stored for long periods and transported without problem. It is only when ignited that it has enough energy to continue reacting. This property makes it difficult for alkanes to be converted into other types of organic molecules. (There are only a few ways to do this). Alkanes are also less dense than water, as one can observe, oil, an alkane, floats on water.
Alkanes are non-polar solvents. Since only C and H atoms are present, alkanes are nonpolar. Alkanes are immiscible in water but freely miscible in other non-polar solvents. Alkanes consisting of weak dipole dipole bonds can not break the strong hydrogen bond between water molecules hence it is not miscible in water. The same character is also shown by alkenes. Because alkanes contain only carbon and hydrogen, combustion produces compounds that contain only carbon, hydrogen, and/or oxygen. Like other hydrocarbons, combustion under most circumstances produces mainly carbon dioxide and water. However, alkanes require more heat to combust and do not release as much heat when they combust as other classes of hydrocarbons. Therefore, combustion of alkanes produces higher concentrations of organic compounds containing oxygen, such as aldehydes and ketones, when combusting at the same temperature as other hydrocarbons.
The general formula for alkanes is CNH2N+2; the simplest possible alkane is therefore methane, CH4. The next simplest is ethane, C2H6; the series continues indefinitely. Each carbon atom in an alkane has sp³ hybridization.
Alkanes are also known as paraffins, or collectively as the paraffin series. These terms are also used for alkanes whose carbon atoms form a single, unbranched chain. Branched-chain alkanes are called isoparaffins.
Methane through Butane are very flammable gases at standard temperature and pressure (STP). Pentane is an extremely flammable liquid boiling at 36 °C and boiling points and melting points steadily increase from there; octadecane is the first alkane which is solid at room temperature. Longer alkanes are waxy solids; candle wax generally has between C20 and C25 chains. As chain length increases ultimately we reach polyethylene, which consists of carbon chains of indefinite length, which is generally a hard white solid.
Reactions
Alkanes react only very poorly with ionic or other polar substances. The pKa values of all alkanes are above 50, and so they are practically inert to acids and bases. This inertness is the source of the term paraffins (Latin para + affinis, with the meaning here of "lacking affinity"). In crude oil the alkane molecules have remained chemically unchanged for millions of years.
However redox reactions of alkanes, in particular with oxygen and the halogens, are possible as the carbon atoms are in a strongly reduced condition; in the case of methane, the lowest possible oxidation state for carbon (−4) is reached. Reaction with oxygen leads to combustion without any smoke; with halogens, substitution. In addition, alkanes have been shown to interact with, and bind to, certain transition metal complexes.
Free radicals, molecules with unpaired electrons, play a large role in most reactions of alkanes, such as cracking and reformation where long-chain alkanes are converted into shorter-chain alkanes and straight-chain alkanes into branched-chain isomers.
In highly branched alkanes and cycloalkanes, the bond angles may differ significantly from the optimal value (109.5°) in order to allow the different groups sufficient space. This causes a tension in the molecule, known as steric hinderance, and can substantially increase the reactivity. The same is preferred for alkenes too. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkanes/Properties_of_Alkanes/Chemical_Properties_of_Alkanes.txt |
Cycloalkanes are cyclic hydrocarbons with the carbon atoms of the molecule are arranged in one or more rings. Cycloalkanes are also saturated, meaning that all of the carbons atoms that make up the ring are single bonded to other atoms (no double or triple bonds).
Cycloalkanes
Cycloalkanes are types of alkanes that have one or more rings of carbon atoms in their structure. The physical properties of cycloalkanes are similar to those of alkanes, but they have higher boiling points, melting points and higher densities due to the greater number of London forces that they contain.
Introduction
Cycloalkanes consist of carbon and hydrogen atoms that are saturated because of the single carbon-carbon bond (meaning that no more hydrogen atoms can be added). Cycloalkanes are also non polar and do not have intermolecular hydrogen bonding; they are usually hydrophobic (meaning they do not dissolve in water) and are less dense than water. Cycloalkanes can also be used for many different purposes. These uses are typically classified by the number of carbons in the cycloalkane ring. Many cycloalkanes are used in motor fuel, natural gas, petroleum gas, kerosene, diesel, and many other heavy oils. There are 4 general groups of cycloalkanes:
1. Small rings (cyclopropane, cyclobutane)
2. Common rings (cyclopentane, cyclohexane, cycloheptane)
3. Medium rings (from 8-12 membered)
4. Large rings (13 membered and higher)
Cycloalkanes can be substituted and named as cycloalkyl derivatives, and disubstituted cycloalkanes can be stereoisomers. In one isomer, if two substituents are placed on the same face or side of the ring, they are called cis. If the two substituents are on opposite faces, they are called trans. Substituents can also be either equitorial or axial on certain cycloalkanes, such as cyclohexane.
Figure 1: The cis and trans isomers are stereoisomers, meaning that they have identicial connectivities but a different arrangement of their atoms.
Generally, the melting point, the boiling point and the density of cycloalkanes increase as the number of carbons increases. This trend occurs because of the greater number of bonds that are in higher membered rings, thus making the bonds harder to break.
London Dispersion Forces and Cycloalkanes
Although alkanes are similar to cycloalkanes, they have higher London Dispersion forces because the ring shape allows for a greater area of contact. Ring strain also causes certain cylcoalkanes to be more reactive. London Dispersion Forces are the attractive or repulsive forces between molecules or between parts of the same molecule. For cycloalkanes, London dispersion forces refer to the repulsive forces between the molecules that cause ring strain.
Ring Strain in Cycloalkanes
Ring Strain occurs because the carbons in cycloalkanes are sp3 hybridized, which means that they do not have the expected ideal bond angle of 109.5o ; this causes an increase in the potential energy because of the desire for the carbons to be at an ideal 109.5o. An example of ring strain can be seen in the diagram of cyclopropane below in which the bond angle is 60o between the carbons.
The reason for ring strain can be seen through the tetrahedral carbon model. The C-C-C bond angles in cyclopropane (diagram above) (60o) and cyclobutane (90o) are much different than the ideal bond angle of 109.5o. This bond angle causes cyclopropane and cyclobutane to have a high ring strain. However, molecules, such as cyclohexane and cyclopentane, would have a much lower ring strain because the bond angle between the carbons is much closer to 109.5o.
Below are some examples of cycloalkanes. Ring strain can be seen more prevalently in the cyclopropane and cyclobutane models.
Below is a chart of cycloalkanes and their respective heats of combustion ( ΔHcomb). The ΔHcomb value increases as the number of carbons in the cycloalkane increases (higher membered ring), and the ΔHcomb/CH2 ratio decreases. The increase in ΔHcomb can be attributed to the greater amount of London Dispersion forces. However, the decrease in ΔHcomb/CH2can be attributed to a decrease in the ring strain.
How do cycloalkanes deal with ring strain?
Certain cycloalkanes, such as cyclohexane, deal with ring strain by forming conformers. A conformer is a stereoisomer in which molecules of the same connectivity and formula exist as different isomers, in this case, to reduce ring strain. The ring strain is reduced in conformers due to the rotations around the sigma bonds. More about cyclohexane and its conformers can be seen here.
Different Types of Strain
There are many different types of strain that occur with cycloalkanes. In addition to ring strain, there is also transannular strain, eclipsing, or torsional strain and bond angle strain.Transannular strain exists when there is steric repulsion between atoms. Eclipsing (torsional) strain exists when a cycloalkane is unable to adopt a staggered conformation around a C-C bond, and bond angle strain is the energy needed to distort the tetrahedral carbons enough to close the ring. The presence of angle strain in a molecule indicates that there are bond angles in that particular molecule that deviate from the ideal bond angles required (i.e., that molecule has conformers).
Problems
1. Which has a higher melting point?
1. cyclopentane
2. cyclopropane
3. cycloocatne
4. they all have the same melting point.
2. Why do cycloalkanes have different physical properties from normal alkanes?
3. What is bond angle strain?
4. What are London Dispersion Forces, and how do they play a part in cycloalkanes?
5. What is the "ideal bond angle" for most cycloalkanes?
Answers
1. C; as the number of carbons increases, so does the melting point. This happens because there are essentially more London dispersion forces acting upon the molecule which then makes it much harder to melt, or boil.
2. Cycloalkanes have different physical properties from normal alkanes due to the greater number of London dispersion forces that they have. Cycloalkanes also have the ability to have more steric hindrance, thus increasing their relative energy level.
3. Bond angle strain is is the energy needed to distort the tetrahedral carbons enough to close the ring
4. London Dispersion Forces are the attractive or repulsive forces between molecules, or in between parts of the same molecule. They play a role in cycloalkanes because they refer to the intramolecular forces that causes ring strain.
5. 109.5 o
Contributors
• Nicole Peiris | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkanes/Properties_of_Alkanes/Cycloalkanes/Physical_Properties_of_Cycloalkanes.txt |
Cycloalkanes are very important in components of food, pharmaceutical drugs, and much more. However, to use cycloalkanes in such applications, we must know the effects, functions, properties, and structures of cycloalkanes. Cycloalkanes are alkanes that are in the form of a ring; hence, the prefix cyclo-. Stable cycloalkanes cannot be formed with carbon chains of just any length. Recall that in alkanes, carbon adopts the sp3 tetrahedral geometry in which the angles between bonds are 109.5°. For some cycloalkanes to form, the angle between bonds must deviate from this ideal angle, an effect known as angle strain. Additionally, some hydrogen atoms may come into closer proximity with each other than is desirable (become eclipsed), an effect called torsional strain. These destabilizing effects, angle strain and torsional strain are known together as ring strain. The smaller cycloalkanes, cyclopropane and cyclobutane, have particularly high ring strains because their bond angles deviate substantially from 109.5° and their hydrogens eclipse each other. Cyclopentane is a more stable molecule with a small amount of ring strain, while cyclohexane is able to adopt the perfect geometry of a cycloalkane in which all angles are the ideal 109.5° and no hydrogens are eclipsed; it has no ring strain at all. Cycloalkanes larger than cyclohexane have ring strain and are not commonly encountered in organic chemistry.
Ring Strain and the Structures of Cycloalkanes
There are many forms of cycloalkanes, such as cyclopropane, cyclobutane, cyclopentane, cyclohexane, among others. The process of naming cycloalkanes is the same as naming alkanes but the addition of the prefix cyclo- is required. Cyclobutane is in a form of a square, which is highly unfavorable and unstable (this will be explained soon). There are different drawings for cyclobutane, but they are equivalent to each other. Cyclobutane can reduce the ring string by puckering the square cyclobutane. Cyclopentane takes the shape of a pentagon and cyclohexane is in the shape of a hexagon.
Chair Conformation of Cyclohexane - Equitorial and Axial
There are two ways to draw cyclohexane because it can be in a hexagon shape or in a different conformational form called the chair conformation and the boat conformation.
• The chair conformation drawing is more favored than the boat because of the energy, the steric hindrance, and a new strain called the transannular strain.
• The boat conformation is not the favored conformation because it is less stable and has a steric repulsion between the two H's, shown with the pink curve. This is known as the transannular strain, which means that the strain results from steric crowding of two groups across a ring. The boat is less stable than the chair by 6.9 kcal/mol. The boat conformation, however, is flexible, and when we twist one of the C-C bonds, it reduces the transannular strain.
• When we twist the C-C bond in a boat, it becomes a twisted boat.
Some Conformations of Cyclohexane Rings. (William Reusch, MSU)
Although there are multiple ways to draw cyclohexane, the most stable and major conformer is the chair because is has a lower activation barrier from the energy diagram.
Conformational Energy Profile of Cyclohexane. (William Reusch, MSU).
The transition state structure is called a half chair. This energy diagram shows that the chair conformation is lower in energy; therefore, it is more stable. The chair conformation is more stable because it does not have any steric hindrance or steric repulsion between the hydrogen bonds. By drawing cyclohexane in a chair conformation, we can see how the H's are positioned. There are two positions for the H's in the chair conformation, which are in an axial or an equitorial formation.
This is how a chair conformation looks, but you're probably wondering which H's are in the equitorial and axial form. Here are more pictures to help.
These are hydrogens in the axial form.
These hydrogens are in an equitorial form. Of these two positions of the H's, the equitorial form will be the most stable because the hydrogen atoms, or perhaps the other substituents, will not be touching each other. This is the best time to build a chair conformation in an equitorial and an axial form to demonstrate the stability of the equitorial form.
Ring Strain
Cycloalkanes tend to give off a very high and non-favorable energy, and the spatial orientation of the atoms is called the ring strain. When atoms are close together, their proximity is highly unfavorable and causes steric hindrance. The reason we do not want ring strain and steric hindrance is because heat will be released due to an increase in energy; therefore, a lot of that energy is stored in the bonds and molecules, causing the ring to be unstable and reactive. Another reason we try to avoid ring strain is because it will affect the structures and the conformational function of the smaller cycloalkanes. One way to determine the presence of ring strain is by its heat of combustion. By comparing the heat of combustion with the value measured for the straight chain molecule, we can determine the stability of the ring. There are two types of strain, which are eclipsing/torsional strain and bond angle strain. Bond angle strain causes a ring to have a poor overlap between the atoms, resulting in weak and reactive C-C bonds. An eclipsed spatial arrangement of the atoms on the cycloalkanes results in high energy.
With so many cycloalkanes, which ones have the highest ring strain and are very unlikely to stay in its current form? The figures below show cyclopropane, cyclobutane, and cyclopentane, respectively. Cyclopropane is one of the cycloalkanes that has an incredibly high and unfavorable energy, followed by cyclobutane as the next strained cycloalkane. Any ring that is small (with three to four carbons) has a significant amount of ring strain; cyclopropane and cyclobutane are in the category of small rings. A ring with five to seven carbons is considered to have minimal to zero strain, and typical examples are cyclopentane, cyclohexane, and cycloheptane. However, a ring with eight to twelve carbons is considered to have a moderate strain, and if a ring has beyond twelve carbons, it has minimal strain.
There are different types of ring strain:
• Transannular strain isdefined as the crowding of the two groups in a ring.
• Eclipsing strain, also known as torsional strain, is intramolecular strain due to the bonding interaction between two eclipsed atoms or groups.
• Bond angle strain is present when there is a poor overlap between the atoms. There must be an ideal bond angle to achieve the maximum bond strength and that will allow the overlapping of the atomic/hybrid orbitals.
Cyclohexane
Most of the time, cyclohexane adopts the fully staggered, ideal angle chair conformation. In the chair conformation, if any carbon-carbon bond were examined, it would be found to exist with its substituents in the staggered conformation and all bonds would be found to possess an angle of 109.5°.
Cyclohexane in the chair conformation. (William Reusch, MSU).
In the chair conformation, hydrogen atoms are labeled according to their location. Those hydrogens which exist above or below the plane of the molecule (shown with red bonds above) are called axial. Those hydrogens which exist in the plane of the molecule (shown with blue bonds above) are called equatorial.
Although the chair conformation is the most stable conformation that cyclohexane can adopt, there is enough thermal energy for it to also pass through less favorable conformations before returning to a different chair conformation. When it does so, the axial and equatorial substituents change places. The passage of cyclohexane from one chair conformation to another, during which the axial substituents switch places with the equatorial substituents, is called a ring flip.
Methylcyclohexane
Methylcyclohexane is cyclohexane in which one hydrogen atom is replaced with a methyl group substituent. Methylcyclohexane can adopt two basic chair conformations: one in which the methyl group is axial, and one in which it is equatorial. Methylcyclohexane strongly prefers the equatorial conformation. In the axial conformation, the methyl group comes in close proximity to the axial hydrogens, an energetically unfavorable effect known as a 1,3-diaxial interaction (Figure 3). Thus, the equatorial conformation is preferred for the methyl group. In most cases, if the cyclohexane ring contains a substituent, the substituent will prefer the equatorial conformation.
Methylcyclohexane in the chair conformation. (William Reusch, MSU).
Problems
1. Trans- 1,2-dimethylcyclopropane is more stable than cis-1,2-dimethylcyclopropane. Why? Drawing a picture of the two will help your explanation.
2. Out of all the cycloalkanes, which one has the most ring strain and which one is strain free? Explain.
3. Which of these chair conformations are the most stable and why?
3.
1. What does it mean when people say "increase in heat leads to increase in energy" and how does that statement relate to ring strains?
2. Why is that the bigger rings have lesser strains compared to smaller rings?
Answers
1. The cis isomer suffers from steric hindrance and has a larger heat of combustion.
2. Cyclopropane- ring strain. Cyclohexane chair conformation- ring strain free.
3. Top one is more stable because it is in an equitorial conformation. When assembling it with the OChem Molecular Structure Tool Kit, equitorial formation is more spread out.
4. When there is an increase in heat there will be an increase of energy released therefore there will be a lot of energy stored in the bond and molecule making it unstable.
5. Smaller rings are more compacts, which leads to steric hindrance and the angles for these smaller rings are harder to get ends to meet. Bigger rings tend to have more space and that the atoms attached to the ring won't be touching each other as much as atoms attached to the smaller ring. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkanes/Properties_of_Alkanes/Cycloalkanes/Ring_Strain_and_the_Structure_of_Cycloalkanes.txt |
The purpose of this page is to help organic chem students show how substituent groups are located on ring structures. We focus here on six-membered rings (6-rings); these are among the most common rings in organic chem (and biochem), and they suffice to raise the main issues. We will look at how to show cis and trans relationships in simple hexagon structural formulas, and we will look at structures showing the common "chair" conformation, focusing on axial vs equatorial orientations. We will also discuss the relationship between cis/trans and axial/equatorial.
Introduction
The basic approach here is to look at a series of compounds, of generally increasing complexity. They are chosen to illustrate one new feature at a time of how to draw the structures and how to see cis/trans and axial/equatorial features.
Cyclohexane
Figure 1: This basic 6-ring cycloalkane, without any special substituents (i.e., other than hydrogen), is easily shown by either of the formulas in Fig 1.
Both structures clearly imply six C in a ring, with two H at each position. Fig 1A is a simple structural formula, condensed. We say it is "condensed" because not all features are shown explicitly. In fact, in this case, not much at all is shown explicitly: the C atoms are understood (at each vertex), the H atoms are understood (enough at each C to give C its normal four bonds), and the C-H bonds are understood. About all that is explicit is the basic carbon skeleton (C-C bonds), showing that there are six C atoms in a ring. Fig 1B is very much like Fig 1A, but it adds one new feature: Fig 1B attempts to show the conformation of the molecule. In this case, we understand that cyclohexane and its simple derivatives tend to spend most of their time near this "chair" conformation. Glossary entry: Conformation.
Figure 2: If we want to show the hydrogen atoms explicitly, we can do so with structural formulas such as in Fig 2.
There is little room for confusion here, since all six C atoms are equivalent, as are all 12 H atoms.
For notes on how to draw chairs, see the section E.2. Note: How to draw chairs.
Chlorocyclohexane
This is an example of the next level of complexity, a mono-substituted cycloalkane. See Fig 3.
Figure 3
So what is new here? Not much, with the hexagon formula, Fig 3A. That type of formula shows the basic "connectivity" of the atoms -- who is connected to whom. This chemical has one Cl on the ring, and it does not matter where we show it. There is now only one H on that C, but since we are not showing H explicitly here, that is not an issue in drawing the structure. (It is an issue when you look at it and want to count H.)
With the chair formula (Fig 3B), which shows information not only about connectivity but also about conformation, there is important new information here. In a chair, there are two "types" of substituents: those pointing up or down, and called axial, and those pointing "outward", and called equatorial. I have shown the chlorine atom in an equatorial position. Why? Two reasons: it is what we would predict, and it is what is found. Why do we predict that the Cl is equatorial? Because it is bigger than H, and there is more room in the equatorial positions.
Helpful Hints...
If possible, examine a physical model of cyclohexane and chlorocyclohexane, so that you can see the axial and equatorial positions. Common ball and stick models are fine for this. It should be easy to see that the three axial H on one side can get very near each other.
If you do not have access to physical models, examining computer models can also be useful. See my RasMol page for a program and source of structures for this.
When putting substituents on chair structures, I encourage you to use the four corner positions of the chair as much as possible. It is easier to see the axial and equatorial relationship at the corners.
In Fig 3B I have shown the H atom that is on the same carbon as the Cl atom. This is perhaps not necessary, since the correct number of H atoms is understood, by counting bonds on C. But showing the H explicitly at key C atoms helps to make the structure clearer. This may be particularly important with hand-drawn structures. I often see structures where I am not sure whether a particular atom is shown axial or equatorial. But if both atoms at the position (the H as well as the Cl) are shown, then hopefully it becomes clearer which is which. I also encourage students who are not sure of their art work to annotate their drawing. Say what you mean. That allows me to distinguish whether you are unsure which direction things point or simply unsure how to draw them.)
Again, a reminder... For notes on how to draw chairs (by hand or using a drawing program), see the section E.2. Note: How to draw chairs.
D. Dichlorocyclohexanes: an introduction
The next level of complexity is a di-substituted cycloalkane, "dichlorocyclohexane". The first question we must ask is which C the two chlorine substituents are on. For now, I want to discuss 1,3- dichlorocyclohexane. This introduces another issue: are the two Cl on the same side of the ring, or on opposite sides? We call these "cis" (same side) and "trans" (opposite sides). We focus on one of these, cis-1,3-dichlorocyclohexane. And for now, we will just look at hexagon structural formulas, leaving the question of conformation for later. Let's go through this one step at a time.
1,3-Dichlorocyclohexane
Our first attempt to draw 1,3-dichlorocyclohexane might look something like Fig 4.
Figure 4
The structure in Fig 4 is indeed a dichlorocyclohexane. It is even a 1,3-dichlorocyclohexane. However, this structure provides no information about the orientation of the two Cl atoms relative to the plane of the ring. To show a specific isomer -- cis or trans -- we must somehow show how the two Cl atoms are oriented relative to the plane of the ring.
cis-1,3-Dichlorocyclohexane
Fig 5 shows two common ways to show how the substituents are oriented relative to the plane of the ring. The compound shown here is cis-1,3-dichlorocyclohexane.
Figure 5
The basic idea in both of these is that we can imagine the ring to be planar, and then show the groups above or below the plane of the ring. How we show this is different in the two parts of Fig 5. In Fig 5A, we look at the ring "edge-on". The thick line for the bottom bond is intended to convey the edge-on view (or side view); this is sometimes omitted, especially with hand-drawn structures, but be careful then that the meaning is clear. Once we understand that we are now looking at the ring edge-on, it is clear that the two Cl atoms are both above the ring, hence cis. In Fig 5B, we view the ring "face-on" (or top view), and use special bond symbols -- "stereo bonds" -- to convey up and down: the heavy wedge -- an "up wedge" -- points upward, toward you, and the dashed bond -- a "down bond" -- points downward, away from you. Again, both Cl are "up", hence cis.
Notes...
For some notes on how to draw the stereo bonds, see the section E.1. Note: How to draw stereo bonds ("up" and "down" bonds).
In discussing Fig 5, I started by saying that we imagine the ring to be planar. Emphasize that cycloalkane rings are not really planar (except for cyclopropane rings). As so often, the structural formula represents the general layout of the atoms, but not the actual molecular geometry.
Those with the Ouellette book can see examples of these two ways of showing up/down on p 81 (top) and p 80 (middle). Most organic chemistry books will show you this.
The conformation of cis-1,3-dichlorocyclohexane
Fig 6, at the right, is one way to show this. Fig 6 shows features of the compound that we have already noted -- plus one more. Let's go through these features, emphasizing what is new.
Figure 6
The structure shows cis-1,3-dichlorocyclohexane: a 6-ring; 2 Cl atoms, at positions 1 and 3; and cis, with both Cl on the same side of -- above -- the H that is on the same C.
I showed the 2 Cl atoms at corner positions, and I showed the H at the key positions explicitly. These points follow from some of the Helpful Hints discussed earlier. It is not required that you do these things, but they can make things easier for you -- and for anyone reading your structures.
Now, what is new here? The conformation. We start with the notion that the conformation of cyclohexane derivatives is based on the "chair". At each position, one substituent is axial (loosely, perpendicular to the ring), and one is equatorial (loosely, in the plane of the ring). There is more room in the equatorial positions (not easily seen with these simple drawings, but ordinary ball and stick models do help with this point). Thus we try to put the larger substituents in the equatorial positions. In this case, we put the Cl equatorial and the H axial at each position 1 and position 3.
We are now done with this compound, cis-1,3-dichlorocyclohexane. However, we have missed one very important concern -- because it is not an issue in this case. So let's look at another compound.
cis-1,2-Dichlorocyclohexane
cis-1,2-Dichlorocyclohexane is like the previous compound, except that the two chloro groups are now at 1,2, rather than at 1,3.
Fig 7 shows a simple structural formula for 1,2-dichlorocyclohexane, without orientation. This is analogous to Fig 4 above for the 1,3 isomer.
Figure 7
Figure 8
Fig 8 shows two ways to show the cis orientation in cis-1,2-dichlorocyclohexane. This Fig is analogous to Fig 5 above for the 1,3 isomer.
Figures 7 and 8 above introduce no new ideas or complications. These two figures should be straightforward.
So, what is the preferred conformation of cis-1,2-dichlorocyclohexane? This requires careful consideration; an important lesson from this exercise is to realize that we cannot propose a good conformation based simply on what we have learned so far.
The two guidelines we have so far for conformation of 6-rings are:
• The carbon ring is in a chair.
• Larger substituents are in equatorial positions.
Let's explore the difficulty here by looking at some things people might naively draw.
Figure 9
Fig 9 shows an attempt to draw a chair conformation of cis-1,2-dichlorocyclohexane. It satisfies both of the guidelines listed above. But it is wrong.
Why is Fig 9 wrong? It is the wrong chemical. The structure shown in Fig 9 is trans, not cis. Look carefully at the 1 and 2 positions. At one of them, the H is above the Cl; at the other, the Cl is above the H. Trans. Wrong chemical. The structure shown in Fig 9 is not the requested chemical.
Figure 10
Fig 10 attempts to fix the problem with Fig 9. But it is also wrong.
Why is Fig 10 wrong? After all, it seems to address the criteria presented. It contains both of the larger atoms (Cl) equatorial, and they are cis as desired. However, in Fig 10, the two axial groups on carbons # 1 and 2 (the two H that are shown) are both pointing up. This is impossible. In a valid chair, the axial groups alternate up/down as one goes around the ring; see Figure 2B, above. This follows from the tetrahedral bonding of C. Adjacent axial groups, as relevant here, must point in opposite directions; that condition is violated here. That is, Fig 10 is not a valid chair.
Those who find the above point new or surprising should check their textbook. If possible, look at models of cyclohexane and simple derivatives such as the one here. Figure 2B, above, is correct. In my experience, many students have not yet noticed this feature of chair conformations.
If you think you have an alternative that is better (or even satisfactory), please show it to me. I suspect it will turn out to be equivalent to Fig 9 or Fig 10, above. But if you think it is good, let's discuss it.
So now what? We have a contradiction. And that really is the most important point here. It is important to realize that we cannot draw a conformation for cis-1,2-dichlorocyclohexane which easily fits the criteria we have used so far: a chair, with large groups equatorial. So what do we do? Clearly, the conformation of this compound must, in some way, involve more issues than what we have considered so far.
Instructors and books will vary in how much further explanation they want to give on this matter. Therefore, how you proceed from here must take into account the preferences in your course. Here is one way to proceed.
One simple way to proceed is to re-examine the two criteria we have been using, and then state a generality about what to do in the event of a conflict. That generality is: use the chair, and then fit the groups as best you can. That is, try to put as many of the larger groups equatorial as you can, but realize that you may not get them all equatorial.
Figure 11
Fig 11 shows a plausible chair conformation of cis-1,2-dichlorocyclohexane.
What have we accomplished here? First, this is the correct compound. Convince yourself that this really is cis-1,2-dichlorocyclohexane. In particular, it is cis because at each substituted position the Cl is "above" the H; that is, both Cl are on the same side of the ring.
Second, this is a proper chair. Adjacent axial groups point in opposite directions. Only two of them are shown here: the axial Cl on the upper right C is "up", and the axial H on the lower right C is "down". Thus, these aspects satisfy the proper way to orient things, in general, on a cyclohexane chair.
How good a conformation is the one shown in Fig 11? Actually, it is quite good, in this case. The actual conformation has been measured, and it follows the basic ideas shown here.
Is this the end of the story? No, but it is about enough for now. The main purpose here was to show how one must carefully look at conformation, within the constraints of the specific isomer one is trying to draw. Some compounds cannot be easily drawn within the common "rules". cis-1,2-dichlorocyclohexane is one such example. In this case, we kept the basic chair conformation, but put one larger group in the less favored axial orientation. Measurements on many such chemicals have shown that the energetic penalty of moving a cyclohexane ring much away from the basic chair conformation is quite large -- certainly larger than the energetic penalty of putting one "somewhat large" group in an axial position. Of course, with larger groups or more groups, this might not hold.
How to draw stereo bonds ("up" and "down" bonds)
There are various ways to show these orientations. The solid (dark) "up wedge" I used is certainly common. Some people use an analogous "down wedge", which is light, to indicate a down bond; unfortunately, there is no agreement as to which way the wedge should point, and you are left relying on the lightness of the wedge to know it is "down". The "down bond" avoids this wedge ambiguity, and just uses some kind of light line. The down bond I used (e.g., in Figure 5B) is a dashed line; IUPAC encourages a series of parallel lines, something like . What I did is a variation of what is recommended by IUPAC: http://www.chem.qmul.ac.uk/iupac/stereo/intro.html.
In ISIS/Draw, the "up wedge" and "down bond" that I used, along with other variations, are available from a tool button that may be labeled with any of them, depending on most recent use. It is located directly below the tool button for ordinary C-C bonds.
In Symyx Draw, the "up wedge" and "down bond", along with other variations, are available from a tool button that may be labeled with any of them, depending on most recent use. It is located directly below the "Chain" tool button.
ChemSketch provides up and down wedges, but not the simple up and down bonds discussed above. The wedges are available from the second toolbar across the top. For an expanded discussion of using these wedges, see the section of my ChemSketch Guide on Stereochemistry: Wedge bonds.
As always, the information provided on these pages in intended to help you get started. Each program has more options for drawing bonds than discussed here. When you feel the need, look around!
How to draw chairs
Most of the structures shown on this page were drawn with the free program ISIS/Draw. I have posted a guide to help you get started with ISIS/Draw. ISIS/Draw provides a simple cyclohexane (6-ring) hexagon template on the toolbar across the top. It provides templates for various 6-ring chair structures from the Templates menu; choose Rings. There are templates for simple chairs, without substituents (e.g., Fig 1B), and for chairs showing all the substituents (e.g., Fig 2B). In either case, you can add, delete, or change things as you wish. Various kinds of stereo bonds (wedges and bars) are available by clicking the left-side tool button that is just below the regular C-C single bond button. It may have a wedge shown on it, but this will vary depending on how it has been used. To choose a type of stereo bond, click on the button and hold the mouse click; a new menu will appear to the right of the button.
The free drawing program Symyx Draw, the successor to ISIS/Draw, provides similar templates and tools. A basic chair structure is provided on the default template bar that is shown. More options are available by choosing the Rings template. See my page Symyx Draw for a general guide for getting started with this program.
The free drawing program ChemSketch provides similar templates and tools. To find the special templates for chairs, go to the Templates menu, choose Template Window, and then choose "Rings" from the drop-down menu near upper left. See my page ChemSketch for a general guide for getting started with this program.
If you want to draw chair structures by hand (and if you are going on in organic chemistry, you should)... Be careful. The precise zigs and zags, and the angles of substituents are all important. Your textbook may offer you some hints for how to draw chairs. A short item in the Journal of Chemical Education offers a nice trick, showing how the chair can be thought of as consisting of an M and a W. The article is V Dragojlovic, A method for drawing the cyclohexane ring and its substituents. J Chem Educ 78:923, 7/01. (I thank M Farooq Wahab, Chemistry, Univ Karachi, for suggesting that this article be noted here.)
Aside from drawing the basic chair, the key points in adding substituents are:
• Axial groups alternate up and down, and are shown "vertical".
• Equatorial groups are approximately horizontal, but actually somewhat distorted from that, so that the angle from the axial group is a bit more than a right angle -- reflecting the common 109 degree bond angle.
• As cautioned before, it is usually easier to draw and see what is happening at the four corners of the chair than at the two middle positions. Try to use the corners as much as possible.
Other ring sizes
The main issues raised here for showing cis and trans hold for other ring sizes. Specifically, Figure 5 (parts A and B) holds for other ring sizes. The only difference would be the geometric figure used to show the ring. Discussion of conformation is more complex, and must be considered for each ring size. The "chair" is a likely conformation for 6-rings, but not for other sizes. Discussing conformations for other ring sizes is beyond the scope of this page. 6-rings are among the most common, in both organic and biochemistry.
Contributors
>Robert Bruner (http://bbruner.org) | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkanes/Properties_of_Alkanes/Cycloalkanes/Rings%3A_cis_trans_and_axial_equatorial_relationships.txt |
Alkanes are not very reactive and have little biological activity; all alkanes are colorless and odorless.
Boiling Points
The boiling points shown are for the "straight chain" isomers of which there is more than one. The first four alkanes are gases at room temperature, and solids do not begin to appear until about \(C_{17}H_{36}\), but this is imprecise because different isomers typically have different melting and boiling points. By the time you get 17 carbons into an alkane, there are unbelievable numbers of isomers!
Cycloalkanes have boiling points that are approximately 20 K higher than the corresponding straight chain alkane.
There is not a significant electronegativity difference between carbon and hydrogen, thus, there is not any significant bond polarity. The molecules themselves also have very little polarity. A totally symmetrical molecule like methane is completely non-polar, meaning that the only attractions between one molecule and its neighbors will be Van der Waals dispersion forces. These forces will be very small for a molecule like methane but will increase as the molecules get bigger. Therefore, the boiling points of the alkanes increase with molecular size.
Where you have isomers, the more branched the chain, the lower the boiling point tends to be. Van der Waals dispersion forces are smaller for shorter molecules and only operate over very short distances between one molecule and its neighbors. It is more difficult for short, fat molecules (with lots of branching) to lie as close together as long, thin molecules.
Example \(1\): Structure dependent Boiling Points
For example, the boiling points of the three isomers of \(C_5H_{12}\) are:
• pentane: 309.2 K
• 2-methylbutane: 301.0 K
• 2,2-dimethylpropane: 282.6 K
The slightly higher boiling points for the cycloalkanes are presumably because the molecules can get closer together because the ring structure makes them tidier and less "wriggly"!
Solubility
Alkanes (both alkanes and cycloalkanes) are virtually insoluble in water, but dissolve in organic solvents. However, liquid alkanes are good solvents for many other non-ionic organic compounds.
Solubility in Water
When a molecular substance dissolves in water, the following must occur:
• break the intermolecular forces within the substance. In the case of the alkanes, these are the Van der Waals dispersion forces.
• break the intermolecular forces in the water so that the substance can fit between the water molecules. In water, the primary intermolecular attractions are hydrogen bonds.
Breaking either of these attractions requires energy, although the amount of energy to break the Van der Waals dispersion forces in something like methane is relatively negligible; this is not true of the hydrogen bonds in water.
As something of a simplification, a substance will dissolve if there is enough energy released when new bonds are made between the substance and the water to compensate for what is used in breaking the original attractions. The only new attractions between the alkane and the water molecules are Van der Waals forces. These forces do not release a sufficient amount of energy to compensate for the energy required to break the hydrogen bonds in water.; the alkane does not dissolve.
The energy only description of solvation is an oversimplification because entropic effects are also important when things dissolve.
Solubility in organic solvents
In most organic solvents, the primary forces of attraction between the solvent molecules are Van der Waals - either dispersion forces or dipole-dipole attractions. Therefore, when an alkane dissolves in an organic solvent, the Van der Waals forces are broken and are replaced by new Van der Waals forces. The two processes more or less cancel each other out energetically; thus, there is no barrier to solubility.
Straight-Chain and Branched Alkanes
Substances consisting entirely of single-bonded carbon and hydrogen atoms and lacking functional groups are called alkanes. There are three basic types of structures that classify the alkanes: (1) linear straight-chain alkanes, (2) branched alkanes, and (3) cycloalkanes.
Linear Straight-Chain Alkanes
In the straight-chain alkanes, each carbon is bound to its two neighbors and to two hydrogen atoms. (In red) Exceptions are the two terminal carbon nuclei, which are bound to only one carbon atom and three hydrogen atoms. (In blue)
CH3 - CH2 - CH2 - CH2 - CH3 CH3 - CH2 - CH3
(Pentane) (Propane)
Alkanes belong to a homologous series (homo, Greeks, same as) of organic compounds in which the members differ by a constant relative atomic mass of 14 (one carbon atom and two hydrogen atoms); one following another can be distinguished by the addition of a methylene group. The general formula for the straight-chain alkanes is H - (CH2)n - H. Methane (n=1) is the first member of the homologous series of the alkanes, followed by ethane (n=2) and so forth.
H - (CH2)1 - H H - (CH2)2 - H
CH4 H - CH2 - CH2 - H = CH3 - CH3
(Methane) (Ethane)
Branched Alkanes
Branched alkanes are derived from the straight-chain alkanes system by removing one of the hydrogen atoms from a methylene group (-CH2-) and replacing it with an alkyl group.
CH3- methyl
CH3CH2- ethyl
CH3CH2CH2- propyl
CH3CH2CH2CH2- butyl
CH3CH2CH2CH2CH2- pentyl
Straight-chained and branched alkanes follow the same general formula: CnH2n+2. The smallest branched alkane is 2-methylpropane or isobutane (Pictures shown above). 2-methylpropane has the same molecular formula as butane (C4H10) but with a different connectivity, resulting in a different structure. These 2 compounds form a pair of isomers (isos, Greeks, equal). Isomers are compounds with the same molecular formula but different structural formulae.
For higher alkane homologs (n > 4), more than two isomers are possible. For example, pentane has three possible isomers in which one is a linear straight-chain alkane and two are branched alkanes. When branched, the nomenclature can be different because of common and IUPAC names.
Problems
1. Draw linear straight-chain hexane.
2. Using the formula, calculate for the number of hydrogen in a alkane with 25 carbons.
3. How many forms of linear straight-chain can be produced with a seven-carbon alkane.
4. How many forms of branched alkane can be produced with a six-carbon alkane.
Problem Answers
1. CH3 - CH2 - CH2- CH2 - CH2 - CH3
2. 52 Hydrogens
3. Only 1 form for the straight-chain alkane.
4. There are 5 other forms of the branched alkane.
• Carl Yen | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkanes/Properties_of_Alkanes/Physical_Properties_of_Alkanes.txt |
The alkanes and cycloalkanes, with the exception of cyclopropane, are probably the least chemically reactive class of organic compounds. Alkanes contain strong carbon-carbon single bonds and strong carbon-hydrogen bonds. The carbon-hydrogen bonds are only very slightly polar; therefore, there are no portions of the molecules that carry any significant amount of positive or negative charge that can attract other molecules or ions. Alkanes can be burned, destroying the entire molecule (Alkane Heats of Combustion), alkanes can react with some of the halogens, breaking carbon-hydrogen bonds, and alkanes can crack by breaking the carbon-carbon bonds.
Reactivity of Alkanes
The combustion of carbon compounds, especially hydrocarbons, has been the most important source of heat energy for human civilizations throughout recorded history. The practical importance of this reaction cannot be denied, but the massive and uncontrolled chemical changes that take place in combustion make it difficult to deduce mechanistic paths. Using the combustion of propane as an example, we see from the following equation that every covalent bond in the reactants has been broken and an entirely new set of covalent bonds have formed in the products. No other common reaction involves such a profound and pervasive change, and the mechanism of combustion is so complex that chemists are just beginning to explore and understand some of its elementary features.
$\ce{CH3CH2CH3} + 5 \ce{O2} \rightarrow 3 \ce{CO2} + 4\ce{H2O} + \text{heat} \label{1}$
Two points concerning this reaction are important:
1. Since all the covalent bonds in the reactant molecules are broken, the quantity of heat evolved in this reaction is related to the strength of these bonds (and, of course, the strength of the bonds formed in the products). Precise heats of combustion measurements can provide useful information about the structure of molecules.
2. The stoichiometry of the reactants is important. If insufficient oxygen is supplied some of the products will consist of the less oxidized carbon monoxide $\ce{CO}$ gas.
$\ce{CH3CH2CH3} + 4 \ce{O2} \rightarrow \ce{CO2} + 2 \ce{CO} + 4\ce{H2O} + \text{heat} \label{2}$
Heat of Combustion
From the previous discussion, we might expect isomers to have identical heats of combustion. However, a few simple measurements will disabuse this belief. Thus, the heat of combustion of pentane is –782 kcal/mole, but that of its 2,2-dimethylpropane (neopentane) isomer is –777 kcal/mole. Differences such as this reflect subtle structural variations, including the greater bond energy of 1º-C–H versus 2º-C–H bonds and steric crowding of neighboring groups. In small-ring cyclic compounds ring strain can be a major contributor to thermodynamic stability and chemical reactivity. The following table lists heat of combustion data for some simple cycloalkanes and compares these with the increase per CH2 unit for long chain alkanes.
Table $1$: Heats of combustion of select hydrocarbons
Cycloalkane
(CH2)n
CH2 Units
n
ΔH25º
kcal/mole
ΔH25º
per CH2 Unit
Ring Strain
kcal/mole
Cyclopropane n = 3 468.7 156.2 27.6
Cyclobutane n = 4 614.3 153.6 26.4
Cyclopentane n = 5 741.5 148.3 6.5
Cyclohexane n = 6 882.1 147.0 0.0
Cycloheptane n = 7 1035.4 147.9 6.3
Cyclooctane n = 8 1186.0 148.2 9.6
Cyclononane n = 9 1335.0 148.3 11.7
Cyclodecane n = 10 1481 148.1 11.0
CH3(CH2)mCH3 m = large 147.0 0.0
The chief source of ring strain in smaller rings is angle strain and eclipsing strain. As noted elsewhere, cyclopropane and cyclobutane have large contributions of both strains, with angle strain being especially severe. Changes in chemical reactivity as a consequence of angle strain are dramatic in the case of cyclopropane, and are also evident for cyclobutane. Some examples are shown in the following diagram. The cyclopropane reactions are additions, many of which are initiated by electrophilic attack. The pyrolytic conversion of β-pinene to myrcene probably takes place by an initial rupture of the 1:6 bond, giving an allylic 3º-diradical, followed immediately by breaking of the 5:7 bond.
Figure $1$: Changes in chemical reactivity as a consequence of ring strain | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkanes/Reactivity_of_Alkanes/Alkane_Heats_of_Combustion.txt |
Alkanes (the most basic of all organic compounds) undergo very few reactions. One of these reactions is halogenation, or the substitution of a single hydrogen on the alkane for a single halogen to form a haloalkane. This reaction is very important in organic chemistry because it opens a gateway to further chemical reactions.
Introduction
While the reactions possible with alkanes are few, there are many reactions that involve haloalkanes. In order to better understand the mechanism (a detailed look at the step by step process through which a reaction occurs), we will closely examine the chlorination of methane. When methane (CH4) and chlorine (Cl2) are mixed together in the absence of light at room temperature nothing happens. However, if the conditions are changed, so that either the reaction is taking place at high temperatures (denoted by Δ) or there is ultra violet irradiation, a product is formed, chloromethane (CH3Cl).
Energetics
Why does this reaction occur? Is the reaction favorable? A way to answer these questions is to look at the change in enthalpy ($\Delta{H}$) that occurs when the reaction takes place.
ΔH = (Energy put into reaction) – (Energy given off from reaction)
If more energy is put into a reaction than is given off, the ΔH is positive, the reaction is endothermic and not energetically favorable. If more energy is given off in the reaction than was put in, the ΔH is negative, the reaction is said to be exothermic and is considered favorable. The figure below illustrates the difference between endothermic and exothermic reactions.
ΔH can also be calculated using bond dissociation energies (ΔH°):
$\Delta{H} = \sum \Delta{H^°} \text{ of bonds broken} - \sum \Delta{H^°} \text{ of bonds formed}$
Let’s look at our specific example of the chlorination of methane to determine if it is endothermic or exothermic:
Since, the ΔH for the chlorination of methane is negative, the reaction is exothermic. Energetically this reaction is favorable. In order to better understand this reaction we need to look at the mechanism ( a detailed step by step look at the reaction showing how it occurs) by which the reaction occurs.
Radical Chain Mechanism
The reaction proceeds through the radical chain mechanism. The radical chain mechanism is characterized by three steps: initiation, propagation and termination. Initiation requires an input of energy but after that the reaction is self-sustaining. The first propagation step uses up one of the products from initiation, and the second propagation step makes another one, thus the cycle can continue until indefinitely.
Step 1: Initiation
Initiation breaks the bond between the chlorine molecule (Cl2). For this step to occur energy must be put in, this step is not energetically favorable. After this step, the reaction can occur continuously (as long as reactants provide) without input of more energy. It is important to note that this part of the mechanism cannot occur without some external energy input, through light or heat.
Step 2: Propagation
The next two steps in the mechanism are called propagation steps. In the first propagation step, a chlorine radical combines with a hydrogen on the methane. This gives hydrochloric acid (HCl, the inorganic product of this reaction) and the methyl radical. In the second propagation step more of the chlorine starting material (Cl2) is used, one of the chlorine atoms becomes a radical and the other combines with the methyl radical.
The first propagation step is endothermic, meaning it takes in heat (requires 2 kcal/mol) and is not energetically favorable. In contrast the second propagation step is exothermic, releasing 27 kcal/mol. Since the second propagation step is so exothermic, it occurs very quickly. The second propagation step uses up a product from the first propagation step (the methyl radical) and following Le Chatelier's principle, when the product of the first step is removed the equilibrium is shifted towards it's products. This principle is what governs the unfavorable first propagation step's occurance.
Step 3: Termination
In the termination steps, all the remaining radicals combine (in all possible manners) to form more product (CH3Cl), more reactant (Cl2) and even combinations of the two methyl radicals to form a side product of ethane (CH3CH3).
Problems with the Chlorination of Methane
The chlorination of methane does not necessarily stop after one chlorination. It may actually be very hard to get a monosubstituted chloromethane. Instead di-, tri- and even tetra-chloromethanes are formed. One way to avoid this problem is to use a much higher concentration of methane in comparison to chloride. This reduces the chance of a chlorine radical running into a chloromethane and starting the mechanism over again to form a dichloromethane. Through this method of controlling product ratios one is able to have a relative amount of control over the product.
References
1. Matyjaszewski, Krzysztof, Wojciech Jakubowski, Ke Min, Wei Tang, Jinyu Huang, Wade A. Braunecker, and Nicolay V. Tsarevsky. "Diminishing Catalyst Concentration in Atom Transfer Radical Polymerization with Reducing Agents." Proceedings of the National Academy of Sciences of the United States of America 103 (2006): 15309-5314.
2. Morgan, G. T. "A State Experiment in Chemical Research." Science 72 (1930): 379-90.
3. Phillips, Francis C. "# Researches upon the Chemical Properties of Gases." Researches upon the Chemical Properties of Gases 17 (1893): 149-236.
Problems
Answers to these questions are in an attached slide
1. Write out the complete mechanism for the chlorination of methane.
2. Explain, in your own words, how the first propagation step can occur without input of energy if it is energetically unfavorable.
3. Compounds other than chlorine and methane go through halogenation with the radical chain mechanism. Write out a generalized equation for the halogenation of RH with X2including all the different steps of the mechanism.
4. Which step of the radical chain mechanism requires outside energy? What can be used as this energy?
5. Having learned how to calculate the change in enthalpy for the chlorination of methane apply your knowledge and using the table provided below calculate the change in enthalpy for the bromination of ethane.
Compound Bond Dissociation Energy (kcal/mol)
CH3CH2-H 101
CH3CH2-Br 70
H-Br 87
Br2 46
Contributors
• Kristen Kelley and Britt Farquharson | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkanes/Reactivity_of_Alkanes/Chlorination_of_Methane_and_the_Radical_Chain_Mechanism.txt |
This page deals briefly with the combustion of alkanes and cycloalkanes. In fact, there is very little difference between the two.
Complete combustion
Complete combustion (given sufficient oxygen) of any hydrocarbon produces carbon dioxide and water. It is quite important that you can write properly balanced equations for these reactions, because they often come up as a part of thermochemistry calculations. Some are easier than others. For example, with alkanes, the ones with an even number of carbon atoms are marginally harder than those with an odd number!
Example 1: Propane Combustion
For example, with propane ($\ce{C3H8}$), you can balance the carbons and hydrogens as you write the equation down. Your first draft would be:
$\ce{ C_3H_8 + O_2 \rightarrow 3CO_2 + 4H_2O} \nonumber$
Counting the oxygens leads directly to the final version:
$\ce{ C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O} \nonumber$
Example 2: Butane Combustion
With butane ($\ce{C4H10}$), you can again balance the carbons and hydrogens as you write the equation down.
$\ce{ C_4H_{10} + O_2 \rightarrow 4CO_2 + 5H_2O} \nonumber$
Counting the oxygen atoms leads to a slight problem - with 13 on the right-hand side. The simple trick is to allow yourself to have "six-and-a-half" $\ce{O2}$ molecules on the left.
$\ce{ C_4H_{10} + 6 1/2 O_2 \rightarrow 4CO_2 + 5H_2O} \nonumber$
If that offends you, double everything:
$\ce{ 2C_4H_{10} + 13 O_2 \rightarrow 8CO_2 + 10 H_2O} \nonumber$
The hydrocarbons become harder to ignite as the molecules get bigger. This is because the bigger molecules do not vaporize so easily - the reaction is much better if the oxygen and the hydrocarbon are well mixed as gases. If the liquid is not very volatile, only those molecules on the surface can react with the oxygen. Bigger molecules have greater Van der Waals attractions which makes it more difficult for them to break away from their neighbors and turn to a gas.
Provided the combustion is complete, all the hydrocarbons will burn with a blue flame. However, combustion tends to be less complete as the number of carbon atoms in the molecules rises. That means that the bigger the hydrocarbon, the more likely you are to get a yellow, smoky flame.
Incomplete combustion
Incomplete combustion (where there is not enough oxygen present) can lead to the formation of carbon or carbon monoxide. As a simple way of thinking about it, the hydrogen in the hydrocarbon gets the first chance at the oxygen, and the carbon gets whatever is left over! The presence of glowing carbon particles in a flame turns it yellow, and black carbon is often visible in the smoke. Carbon monoxide is produced as a colorless poisonous gas.
Why carbon monoxide is poisonous
Oxygen is carried around the blood by hemoglobin. Unfortunately carbon monoxide also binds to exactly the same site on the hemoglobin that oxygen does. The difference is that carbon monoxide binds irreversibly (or very strongly) - making that particular molecule of hemoglobin useless for carrying oxygen. If you breath in enough carbon monoxide you will die from a sort of internal suffocation.
Cracking Alkanes
What is cracking? Cracking is the name given to breaking up large hydrocarbon molecules into smaller and more useful bits. This is achieved by using high pressures and temperatures without a catalyst, or lower temperatures and pressures in the presence of a catalyst. The source of the large hydrocarbon molecules is often the naphtha fraction or the gas oil fraction from the fractional distillation of crude oil (petroleum). These fractions are obtained from the distillation process as liquids, but are re-vaporized before cracking. There isn't any single unique reaction happening in the cracker. The hydrocarbon molecules are broken up in a fairly random way to produce mixtures of smaller hydrocarbons, some of which have carbon-carbon double bonds. One possible reaction involving the hydrocarbon $C_{15}H_{32}$ might be:
$C_{15}H_{32} \rightarrow \underset{\text{ethene}}{2C_2H_4} + \underset{\text{propene}}{C_3H_6} + \underset{\text{octane}}{C_8H_{18}}$
Or, showing more clearly what happens to the various atoms and bonds:
This is only one way in which this particular molecule might break up. The ethene and propene are important materials for making plastics or producing other organic chemicals. The octane is one of the molecules found in petrol (gasoline).
Catalytic Cracking
Modern cracking uses zeolites as the catalyst. These are complex aluminosilicates, and are large lattices of aluminium, silicon and oxygen atoms carrying a negative charge. They are, of course, associated with positive ions such as sodium ions. You may have come across a zeolite if you know about ion exchange resins used in water softeners. The alkane is brought into contact with the catalyst at a temperature of about 500 °C and moderately low pressures.
The zeolites used in catalytic cracking are chosen to give high percentages of hydrocarbons with between 5 and 10 carbon atoms - particularly useful for petrol (gasoline). It also produces high proportions of branched alkanes and aromatic hydrocarbons like benzene. The zeolite catalyst has sites which can remove a hydrogen from an alkane together with the two electrons which bound it to the carbon. That leaves the carbon atom with a positive charge. Ions like this are called carbonium ions (or carbocations). Reorganisation of these leads to the various products of the reaction.
Thermal Cracking
In thermal cracking, high temperatures (typically in the range of 450 °C to 750 °C) and pressures (up to about 70 atmospheres) are used to break the large hydrocarbons into smaller ones. Thermal cracking gives mixtures of products containing high proportions of hydrocarbons with double bonds - alkenes. This is a gross oversimplification; tn fact, there are several versions of thermal cracking designed to produce different mixtures of products. These use completely different sets of conditions.
Thermal cracking does not go via ionic intermediates like catalytic cracking. Instead, carbon-carbon bonds are broken so that each carbon atom ends up with a single electron. In other words, free radicals are formed.
Reactions of the free radicals lead to the various products. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkanes/Reactivity_of_Alkanes/Complete_vs._Incomplete_Combustion_of_Alkanes.txt |
This page describes the reactions between alkanes and cycloalkanes with the halogens fluorine, chlorine, bromine and iodine - mainly concentrating on chlorine and bromine.
• The reaction between alkanes and fluorine: This reaction is explosive even in the cold and dark, and you tend to get carbon and hydrogen fluoride produced. It is of no particular interest. For example:
$CH_4 + 2F_2 \rightarrow C + 4HF$
• The reaction between alkanes and iodine: Iodine does not react with the alkanes to any extent - at least, under normal lab conditions.
• The reactions between alkanes and chlorine or bromine: There is no reaction in the dark.
Methane and Chlorine
In the presence of a flame, the reactions are rather like the fluorine one - producing a mixture of carbon and the hydrogen halide. The violence of the reaction drops considerably as you go from fluorine to chlorine to bromine. The interesting reactions happen in the presence of ultra-violet light (sunlight will do). These are photochemical reactions that happen at room temperature. We'll look at the reactions with chlorine, although the reactions with bromine are similar, but evolve more slowly.
Substitution reactions happen in which hydrogen atoms in the methane are replaced one at a time by chlorine atoms. You end up with a mixture of chloromethane, dichloromethane, trichloromethane and tetrachloromethane.
The original mixture of a colorless and a green gas would produce steamy fumes of hydrogen chloride and a mist of organic liquids. All of the organic products are liquid at room temperature with the exception of the chloromethane which is a gas.
If you were using bromine, you could either mix methane with bromine vapor , or bubble the methane through liquid bromine - in either case, exposed to UV light. The original mixture of gases would, of course, be red-brown rather than green. One would not choose to use these reactions as a means of preparing these organic compounds in the lab because the mixture of products would be too tedious to separate. The mechanisms for the reactions are explained on separate pages.
Larger alkanes and chlorine
You would again get a mixture of substitution products, but it is worth just looking briefly at what happens if only one of the hydrogen atoms gets substituted (monosubstitution) - just to show that things aren't always as straightforward as they seem! For example, with propane, you could get one of two isomers:
If chance was the only factor, you would expect to get three times as much of the isomer with the chlorine on the end. There are 6 hydrogens that could get replaced on the end carbon atoms compared with only 2 in the middle. In fact, you get about the same amount of each of the two isomers. If you use bromine instead of chlorine, the great majority of the product is where the bromine is attached to the center carbon atom.
Cycloalkanes
The reactions of the cycloalkanes are generally just the same as the alkanes, with the exception of the very small ones - particularly cyclopropane. In the presence of UV light, cyclopropane will undergo substitution reactions with chlorine or bromine just like a non-cyclic alkane. However, it also has the ability to react in the dark. In the absence of UV light, cyclopropane can undergo addition reactions in which the ring is broken. For example, with bromine, cyclopropane gives 1,3-dibromopropane.
This can still happen in the presence of light - but you will get substitution reactions as well. The ring is broken because cyclopropane suffers badly from ring strain. The bond angles in the ring are 60° rather than the normal value of about 109.5° when the carbon makes four single bonds. The overlap between the atomic orbitals in forming the carbon-carbon bonds is less good than it is normally, and there is considerable repulsion between the bonding pairs. The system becomes more stable if the ring is broken. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkanes/Reactivity_of_Alkanes/Halogenation_Alkanes.txt |
Chlorination of Methane by Substitution
Halogenation is the replacement of one or more hydrogen atoms in an organic compound by a halogen (fluorine, chlorine, bromine or iodine). Unlike the complex transformations of combustion, the halogenation of an alkane appears to be a simple substitution reaction in which a C-H bond is broken and a new C-X bond is formed. The chlorination of methane, shown below, provides a simple example of this reaction.
CH4 + Cl2 + energy → CH3Cl + HCl
Since only two covalent bonds are broken (C-H & Cl-Cl) and two covalent bonds are formed (C-Cl & H-Cl), this reaction seems to be an ideal case for mechanistic investigation and speculation. However, one complication is that all the hydrogen atoms of an alkane may undergo substitution, resulting in a mixture of products, as shown in the following unbalanced equation. The relative amounts of the various products depend on the proportion of the two reactants used. In the case of methane, a large excess of the hydrocarbon favors formation of methyl chloride as the chief product; whereas, an excess of chlorine favors formation of chloroform and carbon tetrachloride.
CH4 + Cl2 + energy → CH3Cl + CH2Cl2 + CHCl3 + CCl4 + HCl
Empirical Considerations
The following facts must be accomodated by any reasonable mechanism for the halogenation reaction.
1. The reactivity of the halogens decreases in the following order: F2 > Cl2 > Br2 > I2.
2. We shall confine our attention to chlorine and bromine, since fluorine is so explosively reactive it is difficult to control, and iodine is generally unreactive.
3. Chlorinations and brominations are normally exothermic.
4. Energy input in the form of heat or light is necessary to initiate these halogenations.
5. If light is used to initiate halogenation, thousands of molecules react for each photon of light absorbed.
6. Halogenation reactions may be conducted in either the gaseous or liquid phase.
7. In gas phase chlorinations the presence of oxygen (a radical trap) inhibits the reaction.
8. In liquid phase halogenations radical initiators such as peroxides facilitate the reaction.
The most plausible mechanism for halogenation is a chain reaction involving neutral intermediates such as free radicals or atoms. The weakest covalent bond in the reactants is the halogen-halogen bond (Cl-Cl = 58 kcal/mole; Br-Br = 46 kcal/mole) so the initiating step is the homolytic cleavage of this bond by heat or light, note that chlorine and bromine both absorb visible light (they are colored). A chain reaction mechanism for the chlorination of methane has been described. Bromination of alkanes occurs by a similar mechanism, but is slower and more selective because a bromine atom is a less reactive hydrogen abstraction agent than a chlorine atom, as reflected by the higher bond energy of H-Cl than H-Br.
Selectivity
When alkanes larger than ethane are halogenated, isomeric products are formed. Thus chlorination of propane gives both 1-chloropropane and 2-chloropropane as mono-chlorinated products. Four constitutionally isomeric dichlorinated products are possible, and five constitutional isomers exist for the trichlorinated propanes. Can you write structural formulas for the four dichlorinated isomers?
$CH_3CH_2CH_3 + 2Cl_2 \rightarrow \text{Four} \; C_3H_6Cl_2 \; \text{isomers} + 2 HCl$
The halogenation of propane discloses an interesting feature of these reactions. All the hydrogens in a complex alkane do not exhibit equal reactivity. For example, propane has eight hydrogens, six of them being structurally equivalent primary, and the other two being secondary. If all these hydrogen atoms were equally reactive, halogenation should give a 3:1 ratio of 1-halopropane to 2-halopropane mono-halogenated products, reflecting the primary/secondary numbers. This is not what we observe. Light-induced gas phase chlorination at 25 ºC gives 45% 1-chloropropane and 55% 2-chloropropane.
CH3-CH2-CH3 + Cl2 → 45% CH3-CH2-CH2Cl + 55% CH3-CHCl-CH3
The results of bromination ( light-induced at 25 ºC ) are even more suprising, with 2-bromopropane accounting for 97% of the mono-bromo product.
CH3-CH2-CH3 + Br2 → 3% CH3-CH2-CH2Br + 97% CH3-CHBr-CH3
These results suggest strongly that 2º-hydrogens are inherently more reactive than 1º-hydrogens, by a factor of about 3:1. Further experiments showed that 3º-hydrogens are even more reactive toward halogen atoms. Thus, light-induced chlorination of 2-methylpropane gave predominantly (65%) 2-chloro-2-methylpropane, the substitution product of the sole 3º-hydrogen, despite the presence of nine 1º-hydrogens in the molecule.
(CH3)3CH + Cl2 → 65% (CH3)3CCl + 35% (CH3)2CHCH2Cl
It should be clear from a review of the two steps that make up the free radical chain reaction for halogenation that the first step (hydrogen abstraction) is the product determining step. Once a carbon radical is formed, subsequent bonding to a halogen atom (in the second step) can only occur at the radical site. Consequently, an understanding of the preference for substitution at 2º and 3º-carbon atoms must come from an analysis of this first step.
First Step: R3CH + X· → R3C· + H-X
Second Step: R3C· + X2 → R3CX + X·
Since the H-X product is common to all possible reactions, differences in reactivity can only be attributed to differences in C-H bond dissociation energies. In our previous discussion of bond energy we assumed average values for all bonds of a given kind, but now we see that this is not strictly true. In the case of carbon-hydrogen bonds, there are significant differences, and the specific dissociation energies (energy required to break a bond homolytically) for various kinds of C-H bonds have been measured. These values are given in the following table.
R (in R–H) methyl ethyl i-propyl t-butyl phenyl benzyl allyl vinyl
Bond Dissociation Energy
(kcal/mole)
103 98 95 93 110 85 88 112
The difference in C-H bond dissociation energy reported for primary (1º), secondary (2º) and tertiary (3º) sites agrees with the halogenation observations reported above, in that we would expect weaker bonds to be broken more easily than are strong bonds. By this reasoning we would expect benzylic and allylic sites to be exceptionally reactive in free radical halogenation, as experiments have shown. The methyl group of toluene, C6H5CH3, is readily chlorinated or brominated in the presence of free radical initiators (usually peroxides), and ethylbenzene is similarly chlorinated at the benzylic location exclusively. The hydrogens bonded to the aromatic ring (referred to as phenyl hydrogens above) have relatively high bond dissociation energies and are not substituted.
C6H5CH2CH3 + Cl2 → C6H5CHClCH3 + HCl
Since carbon-carbon double bonds add chlorine and bromine rapidly in liquid phase solutions, free radical substitution reactions of alkenes by these halogens must be carried out in the gas phase, or by other halogenating reagents. One such reagent is N-bromosuccinimide (NBS), shown in the second equation below. By using NBS as a brominating agent, allylic brominations are readily achieved in the liquid phase.
The covalent bond homolyses that define the bond dissociation energies listed above may are described by the general equation:
R3C-H + energy → R3C· + H·
C-H bond dissociation energies are commonly interpreted in terms of radical stability. However, this interpretation has been questioned by Gronert. Most importantly, in terms of the selectivity of free radical reactions, it is the energies of the bonds that matter, and not why they are what they are. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkanes/Reactivity_of_Alkanes/Halogenation_of_Alkanes.txt |
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