Unit 9
Nucleophilic Substitution
SN1 & SN2
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Unit 9 Nucleophilic Substitution SN1 & SN2
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Overview
What we have now established in our preliminary studies of the alkanes, alcohols, and alkyl halides is rather significant in relationship to what we intend to accomplish in this comprehensive program of studies in organic chemistry.
We have now clearly established the importance of alcohols, their relative roles as both acids and bases, and the special position they occupy in the hierarchy of organic synthesis, especially in relationship to the aliphatics.
Alternatively, we have established the extremely limited role of the alkanes as organic precursors, and must view them from here onward in a limited and "dead-end" type of framework. Their tendency for structural and functional specificity is far too loose-ended, and what we might refer to as their "structural ambiguity" (or reaction site equivalence) is simply too prevalent to lend any sense of practicality to their use in synthesis in the chemical laboratory.
Thus, in the laboratory, alcohols are the most common starting point for the synthesis of aliphatic compounds, and the most common first step in such a synthesis is the conversion of an alcohol to an alkyl halide. Once the alky halide is produced, a number of reactions are possible, including a wide variety of both substitution and elimination reactions (see figure above).
In this section, we turn our attention to the reactions which can occur utilizing the alkyl halides as chemical precursors, with a strong emphasis on those reactions characterized by:
Nucleophilic Substitution
When methyl bromide is reacted with sodium hydroxide in a solvent that dissolves both reagents, the result is methanol (methyl alcohol) and sodium bromide. This is a simple substitution reaction, since the OH group is substituted for the halogen (Br) atom in the original compound. In this reaction, an alkyl halide has been converted into an alcohol.
The reaction as clearly heterolytic, since the departing halide ion takes with it the electron pair it has been sharing with the C atom. The OH ion brings with it the electron pair needed to bind it to the C atom. Thus, the C atom loses one pair of electrons while gaining another pair. This is a classic example of the class of reactions called nucleophilic (aliphatic) substitution.
Why is nucleophilic substitution the characteristic reaction of the alkyl halides ?
A halide ion is an extremely weak base. This is evidenced by its readiness to release (or donate) a proton to other bases, i.e. by the high acidity of the hydrogen halides. [Strong acid <--> weak conjugate base ]. In an alkyl halide, a halogen atom X is attached to a C atom. Just as X in an H-X acid readily releases a proton, so the X in an R-X alkyl halide readily releases a C atom -- again, to other bases. These bases possess an unshared pair of electrons and are seeking a relatively positive site. I.E. they are seeking a nucleus with which to share their electron pair.
Basic, electron-rich reagents which tend to attack the nucleus of carbon are called nucleophilic reagents, or simply nucleophiles. When this attack results in substitution, the reaction is called nucleophilic substitution. The compound containing the carbon atom that undergoes a particular kind of reaction is called the substrate, and is typically characterized by a leaving group. The leaving group becomes displaced from the C atom and departs form the molecule -- taking the electron pair with it.
Because the weakly basic halide ion is a good leaving group, the alkyl halides are good substrates for nucleophilic substitution. As illustrated in the table below, they react with a large number of nucleophilic reagents, both inorganic and organic, to yield a wide variety of important products. These reagents include not only negative ions like hydroxide and cyanide, but also neutral bases such as ammonia and water. Their characteristic feature is an unshared pair of electrons.
Nucleophilic substitution is the "work-horse" of organic synthesis. The synthesis of aliphatic compounds, we said, most often starts with alcohols. But as we have seen in Unit 6: Alcohols the OH group is a very poor leaving group. It is only thru the conversion of alcohols into alkyl halides - or other compounds with good leaving groups - that we open the door to nucleophilic substitution and the vast majority of higher compounds.
* Note: Other typical reactions of alkyl halides include:
1) Dehydrohalogenation / Elimination (see Unit 10: Alkenes I)
2) Preparation of a Grignard Reagent (see Unit 4: Alkanes II)
3) Reduction by Metal and Acid (see Unit 4: Alkanes II)
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Let's take a closer look at some of the nucleophiles we shall be working with. Many of the products formed are new to us. But at this point we need only see how the structure of a particular product is the natural result of the structure of a particular nucleophile. For now, we shall use alkyl halides as our examples of substrates.
Some nucleophiles are anions, like the OH- (hydroxide) ion, or another halide ion (Iodine, I-) which, while only weakly basic, does after all possess unshared electrons. Alternatively, the cyanide (CN-) ion is the strongly basic anion of the weakly basic HCN (hydrocyanic acid or Prussic acid).
Neutral molecules can also possess unshared electrons, have basic properties, and hence act as nucleophiles. E.G. Water attacks an alkyl halide to yield an alcohol. But the oxygen of water already has two hydrogens. And when it attaches itself to a C atom, it initially forms its conjugate acid, the protonated alcohol. This is easily changed into the alcohol by loss of the proton.
It should be pointed out that the existence of a lone hydrogen ion (or proton, H+) is a misnomer, as it does not actually exist in solution. In aqueous solution, H+ will immediately attach itself to a water molecule in order to form the more stable positively charged species: the hydronium ion (H3O+). See above.
Regarding other leaving groups, and thus other substrates we shall encounter, the alkyl esters of sulfonic acids (ArSO2OR), are most commonly used in place of alkyl halides. Like alkyl halides, the sulfonates are made from alcohols.
Kinetics: 2nd Order vs. 1st Order
Let us begin with a specific example: the reaction of methyl bromide with sodium hydroxide to yield methanol.
CH3Br + OH- = CH3OH + Br-
This reaction would probably be carried out in aqueous ethanol, in which both reactants are soluble. If the reaction results form collisions between an OH- ion and a CH3Br molecule, then we say that the rate depends upon the concentrations of both of these species. I.E.
Rate = k [CH3Br] [OH-]
and we find this to be so. Let us now look at the corresponding reaction between tert-butyl bromide and hydroxide ion:
In this reaction, however, we find there to be no dependence of the reaction rate on OH- concentration. The reaction rate is independent of [OH-]. Thus:
Rate = k [RBr]
To summarize our findings, we say that the methyl bromide reaction follows second-order reaction kinetics, since its rate is dependent upon the concentration of 2 substances. Alternatively, the tert-butyl bromide reaction follows first-order reaction kinetics, since its rate is dependent upon the concentration of only one substance.
In 1935, E. D. Hughes and Sir Christopher Ingold studied nucleophilic substitution reactions of alkyl halides and related compounds. Initially, the overall rate of the nucleophilic substitution was a little puzzling:
CH3X > primary > secondary < tertiary
The reaction kinetics appeared to change from second order to first order. They proposed that there were two main mechanisms at work, both of them competing with each other. The two main mechanisms are the SN1 reaction and the SN2 reaction. S stands for chemical substitution, N stands for nucleophilic, and the number represents the kinetic order of the reaction (see Unit 29: Chemical Equilibrium).
The following illustration constitutes a graph showing the relative reactivities of the different alkyl halides towards SN1 and SN2 reactions. The minimum in rate is attributed to the crossing of two opposing curves. I.E. The minimum in the curve is attributed to a shift in the reaction mechanism form SN2 to SN1.
Thus, as one passes along the series, reactivity by the SN2 mechanism decreases from CH3 to primary C atoms, and at a secondary C atom is so low that the SN1 begins to contribute significantly, rising sharply to tertiary C atoms.
In order to decipher this quandary, we need to understand the different factors influencing these 2 different reaction mechanisms: SN1 and SN2.
SN2 - Mechanism of Reaction
SN2 reactivity rates follow the trend:
CH3X > primary > secondary > tertiary
The simplest way to account for the second-order reaction kinetics is to assume that the reaction requires a collision between a hydroxide ion and a methyl bromide molecule. It is known that in its attack the OH ion stays as far away as possible from the bromine atom (i.e. it attacks the molecule from the rear). The reaction is believed to take place as follows, with a complete inversion of the molecular tetrahedral geometry taking place. This allows the distance between the (OH-) and (Br-) to be maximized.
The transition state can be pictured as structure in which carbon is partially bonded to both OH and Br. The C-OH bond is not completely formed, and the C-Br bond is not completely broken. The OH has a diminished negative charge, since it has begun to share its electrons with the C atom. Bromine has developed a partial negative charge, since it has begun to remove a pair of electrons from a C atom.
Most importantly, the OH and Br are located as far apart as possible. The three H atoms and the C atom lie within a single plane (are coplanar) with all bond angles set at 120 degrees -- due to the trigonal planar overlap of three sp2 orbitals.
The partial bonds to the leaving group and the nucleophile are formed through overlap of the remaining p orbitals: 180 degrees apart, and perpendicular to the plane of the sp2 orbitals. Thus, the C-H bonds are radial (as in the spokes of a wheel) while the C-OH and C-Br bonds lie horizontally (along the axle of the wheel). This is the mechanism that is called
SN2: Substitution Nucleophilic Bimolecular
The strongest evidence for this reaction mechanism is based on the inversion of the reaction intermediate. A reaction that yields a product whose configuration is opposite (chiral) to that of a reactant is said to proceed with inversion of configuration. I.E. The reactants and products are enantiomers. Indeed, we find this to be the case as all SN2 reactions proceed with complete stereochemical inversion.
Additional evidence for this mechanism is provided in the form of steric hindrance among alkyl groups. Direct measurement of SN2 rates for a series of substrates in DMF (dimethylformamide -- a pro-active SN2 solvent) gives results like the following, where the bottom figures represent relative rates of reaction:
These trends in reactivity match the relative rates as quoted above:
CH3X > primary > secondary > tertiary
We now need to compare the structure of the reactants to the structure of the transition state. In contrast to free radical substitution, the structure of the transition state is not intermediate between the structures of the reactants and products. Thus, we cannot simply expect that factors stabilizing the product will also stabilize the transition state.
We might begin by investigating the redistribution of charge on forming the transition state, as very often reactivity depends upon how easily the molecule accommodates that charge. Accommodation of charge, in turn, depends upon how well the substituents tend to withdraw or release electrons. What we find in the SN2 transition state, however, is that while the OH ion has brought electrons to the C atom, the halide ion has equally taken them away. Thus, there should be no net change in the electronegativity of the central C atom. Therefore, it is highly unlikely that the reactivity sequence results from the polar effects of substituent groups.
Let us next compare transition state and reactants with regard to shape and form, starting with the methyl bromide reaction. What we find as H atoms are replaced by an increasing number of methyl groups is an increasing crowding about the central carbon atom. In fact, the alkyl groups will be strategically situated on the backside of the molecule; directly opposite of the halide (Br) atom. In short, the backside of the molecule will be increasingly difficult to access for partial bonding and/or molecular tetrahedral inversion.
Thus we must conclude that differences in rates between two SN2 reactions are due chiefly to steric factors and not to polar factors. I.E. Differences in rates are related to the bulk of the substituents and not to their effect on the electron distribution (or distribution of charge). As the number of substituents attached to the carbon bearing halogens is increased, the reactivity toward SN2 substitution decreases, as experimental data has evidenced.
SN1 - Mechanism of Reaction
The reaction between tert-butyl bromide and hydroxide ion to yield tert-butyl alcohol follows first-order reaction kinetics. I.E. The reaction rate depends upon the concentration of only one reactant: tert-butyl bromide. How are we to interpret the fact that the rate is independent of [OH-] ? Our only conclusion can be that the reaction whose rate we are measuring does not involve the OH- ion.
These observations are quite consistent with the following reaction mechanism. Tert-butyl bromide slowly dissociates (step 1) into a bromide ion and a cation derived form the tert-group, called a carbocation. This carbocation then combines rapidly (step 2) with a hydroxide ion to yield tert-butyl alcohol.
The rate of the overall reaction is determined by the slow breaking of the C-Br bond to form the carbocation. Once formed, the carbocation reacts rapidly to form the product. It is step 1 that we are actually measuring. This step does not involve OH-, and thus its rate does not depend on [OH-]. A single step whose rate determines the overall rate of a stepwise reaction is called a rate-determining step (see Unit 25: Chemical Equilibrium)
It is not surprising that the (slow) rate-determining step in this reaction sequence is the one that involves the breaking of a bond, as this is a process which requires energy. We recognize this particular bond-breaking as an example of heterolysis: cleavage in which both bonding electrons go to the same fragment. This cleavage process takes even more energy than its alternative (homolysis -- the formation of 2 free radicals).
Nor is it surprising that the combining of the carbocation with a hydroxide ion is a very rapid step, since it involves only the formation of a bond, which is fundamentally an exothermic (energy-releasing) process -- and could easily be projected as a simultaneous event (no activation energy necessary).
We recognize this latter step as an acid-base reaction in the Lewis sense. We are already familiar with the OH ion as a strong base. We shall learn to recognize carbocations as extremely powerful Lewis acids.
This is the mechanism that is called:
SN1: Substitution Nucleophilic Unimolecular
Our primary evidence that alkyl halides can (and will) react by the SN1 mechanism is that the mechanism is consistent with the first-order reaction kinetics we have presented here. Thus, in general, an SN1 reaction follows first-order kinetics.
The rate of the entire reaction is determined solely by how fast the alkyl halide ionizes, and hence depends only upon the concentration of alkyl halide.
In review:
SN1 reactivity rates follow the trend:
CH3X < primary < secondary < tertiary
In the SN2 reaction, the addition of the nucleophile and the elimination of leaving group take place simultaneously. SN2 occurs where the central carbon atom is easily accessible to the nucleophile. By contrast the SN1 reaction involves two steps.
SN1 reactions tend to be important when the central carbon atom of the substrate is surrounded by bulky groups. This is due to the following two facts:
1) Bulky functional groups introduce steric interference with the SN2 reaction
2) A highly substituted carbon forms a stable carbocation.
Let us consider the nature of carbocations. Just what exactly are they, and why is their role so critical in the reaction of alkyl halides to form alcohols via the SN1 mechanism ??
SN1 Mechanism -- Carbocations
A carbocation is an ion with a positively-charged C atom. The charged carbon atom in a carbocation has only six electrons in its outer valence shell instead of the eight valence electrons that ensures maximum stability. Therefore the carbon cation is unstable and very reactive, seeking to fill its octet of valence electrons as well as regain its neutral charge.
Carbocations (or carbonium ions) are classified as primary, secondary, or tertiary depending on the number of carbon atoms bonded to the ionized carbon. Primary carbocations have one or zero carbons attached to the ionized carbon, secondary carbocations have two carbons attached to the ionized carbon, and tertiary carbocations have three carbons attached to the ionized carbon.
The carbocation has sp2 hybridization with a trigonal planar molecular geometry.
As we have seen in the methyl free radical (Unit 1: Methane), sp2 orbitals lie in a single plane, that of the carbon nucleus, and are directed towards the corners of an equilateral triangle (trigonal planar configuration, 120 degree bond angles). This part of the ion is flat, with the electron-deficient C atom and its 3 radial H atoms all coplanar.
A p orbital is situated with its two lobes lying above and below the plane of the sigma bonds. I Na carbocation, the p orbital is empty. But even in its unoccupied state, it is intimately involves in the chemistry of carbocations. This results from an overlap of the p orbital with other nearby orbitals -- overlap that is made possible by the flatness of the carbocation (as confirmed by NMR, as well as Infrared and Raman spectroscopy) .
We turn our attention now to the stereochemistry of the SN1, where we find that a typical reaction product is a biased mixture (favoring the inversion) of the the inverted compound and the racemic modification -- a condition we refer to as partial racemization. I.E. The inverted product and its enantiomer are both present in the final solution.
How do we account for this ?
In an SN2 reaction, we saw that the nucleophile attacks the substrate molecule itself, and the complete inversion observed is a direct consequence of that action. I.E. The leaving group is still attached to the C atom during the attack, which directs the attack to the backside every time.
In an SN1 reaction, the nucleophile attacks not the substrate, but the reaction intermediate (or transition state): the carbocation. In this case, the leaving group has already been detached in the rate-determining step, and thus can no longer have an influence on the spatial orientation of the attack.
Thus, the nucleophilic reagent Z attaches itself to the carbocation. But it may attach itself to either face of the planar trigonal cation, and, depending upon which face, yield one or the other of the two enantiomeric products. Together, the two enantiomers constitute the racemic modification. The racemization that accompanies these reactions is consistent with the SN1 mechanism and the formation of an intermediate carbocation as the transition state for the reaction.
If the attack on the two faces of the cation were purely random, we would expect equal amounts of the two enantiomers. But in general, the inverted product exceeds the enantiomer. I.E. The reaction proceeds with racemization plus some net inversion.
How do we accommodate even this limited net inversion ?? How do we account for the fact that the attack on the carbocation is not purely random ?? Clearly, the excess of inversion is due in some way to the leaving group, which is somehow still exerting some measure of control over the reaction's stereochemistry. *Note: In the complete absence of the leaving group, the flat cation would lose all sense of chirality.
We might speculate that before complete ionization occurs, the anion clings more or less closely to the front side of the carbocation and thus shields this side form attack. As a result, the backside is preferred -- as evidenced experimentally. This explanation supports the experimental evidence, and therefore stands. Thus, unlike an SN2 reaction which proceeds with complete inversion, an SN1 reaction proceeds with racemization.
Relative Stabilities of Carbocations
Relative stabilities of carbocations can best be measured by comparing the respective bond dissociation energies (the energy required to convert a mole of alkyl bromide molecules into carbocations and bromide ions). What we find is that the energy needed to form the various classes of carbocations decreases in the order:
CH3X > primary > secondary > tertiary
If less energy is needed to form one carbocation than another, then it can only mean that, relative to the alkyl bromide from which it was formed, the one carbocation contains less energy than the other. I.E. It is more stable. Thus, the relative stability of the carbocation increases with the number of alkyl groups bonded to the charge-bearing carbon.
CH3X < primary < secondary < tertiary
Tertiary carbocations are more stable (and form more readily) than secondary carbocations. Primary carbocations are highly unstable because, while ionized higher-order carbons are stabilized by hyperconjugation, unsubstituted (primary) carbons are not. Therefore, reactions such as the SN1 reaction normally do not occur if a primary carbocation would be formed.
[ Exception to this occurs when there is a carbon-carbon double bond next to the ionized carbon, such as that found in alkenes or aromatic compounds. E.G. Such cations as the allyl cation CH2=CH-CH2+ and benzyl cation C6H5-CH2+ are more stable than most other carbocations. Molecules which can form allyl or benzyl carbocations are especially reactive.]
Differences in stability between carbocations are much, much larger that between free radicals (say 100 kcals vs. 10-20 kcals). We thus see much larger effects on reactivity via SN1.
How do we account for these trends in relative stability in carbocations ?
I. Dispersal of charge
The relative stability of a carbocation is determined largely by how well it accommodates its electron deficiency and its corresponding positive charge.
According to the laws of electrostatics, the stability of a charged system is increased by dispersal of charge. Any factor, therefore, that tends to spread out the positive charge of the electron-deficient carbon and redistribute it over the rest of the ion will tend to stabilize a carbocation.
Consider a substituent G attached to an electron-deficient carbon in place of an H atom. Compared to hydrogen, G may either release withdraw electrons, inducing polarity in either case.
An electron-releasing substituent tends to reduce the positive charge at the electron-deficient C atom. In doing so, the substituent itself becomes somewhat more positive. The overall effect is a net dispersal of charge, which tends to stabilize the entire system -- mainly the carbocation.
An electron withdrawing substituent tends to intensify the positive charge on the electron deficient C atom, thus destabilizing the system and carbocation.
The order of stability of carbocations, we have just seen, is:
CH3X < primary < secondary < tertiary
Thus, the greater the number of alkyl groups, the more stable the carbocation.
It would therefore appear reasonable to conclude that, with regard to dispersal of charge, alkyl groups must act to release electrons here. In fact, there is increasing evidence that alkyl groups often tend to stabilize both cations and anions, indicating electron release or withdrawal on demand.
II. Rearrangement
It is noteworthy that our reactivity sequence leads directly to an identical sequence showing the relative rates of formation of carbocations. I.E. the more stable the carbocation, the faster it is formed. This is likely the most useful generalization about structure and reactivity that exists in the field of organic synthesis. Carbocations are formed form many compounds other than alkyl halides, and in reactions quite different from nucleophilic substitution. Yet in all these reactions in which carbocations are formed, carbocation stability plays the leading role in governing reactivity and mechanism orientation.
How can we account for the fact that the rate of formation of a carbocation depends upon its stability ??
In an SN1 reaction of an alkyl halide, the carbocation is formed by the heterolytic cleavage of the substrate molecule via its covalent C-H bond. In the reactant, an electron pair is shared by the C and H atoms. Except for a modest polarity, these atoms are neutral. In the products, the halogen has taken away the electron pair, and the carbon is left with only a sextet. The halide bears a full negative charge, and the carbocation bears a full positive charge, which is centered on the C atom.
In the transition state, the C-X bond must be partially broken, the halogen having begun to pull the electron pair away from the C atom. Thus, the halogen atom has begun to take on the negative charge it is to carry as a halide anion. More importantly, the C atom has begun to take on the positive charge it is to carry in the carbocation.
In 1979, Edward Arnett and Paul Schleyer compared values of activation energy of SN1 reactions with heats of ionization in super-acid solutions, and found a direct quantitative dependence of rate of formation of carbocations on carbocation stability. The more stable the carbocation, the found, the faster it is formed.
As we encounter other reactions in which carbocations are formed, we must, for each of these reactions, examine the structure of the transition state. In most ( if not all) of these reactions, we shall find that the transition state differs form the reactants chiefly in being more similar to the product. What are we seeing here is evidence for a distinct pattern of behavior. The most striking feature of this pattern is the occurrence of rearrangements within the carbocation itself.
Carbocations undergo rearrangement from less stable structures to equally stable or more stable ones with rate constants in excess of 10 ^ 9 per second. In nucleophilic substitution via SN1, for example, it is sometimes observed that the entering group, Z, becomes attached to a different carbon than the ones that originally held the leaving group, X.
In each of the above cases, we see that, in order to accommodate Z in the new position, there must be a rearrangement of H atoms in the substrate. The transformation of an n-propyl group into an isopropyl group, for example, requires the removal of one H atom from C-2 and attachment of one H atom to C-1.
This complicates synthetic pathways to many compounds. Sometimes there is even a rearrangement of the carbon skeleton.
Rearrangement lends powerful support to the particular form of the SN1 mechanism -- the intermediacy of carbocations -- by linking this mechanism to the mechanisms of those other kinds of reactions where rearrangements are observed. The correlation between rearrangement and intermediate cations is so strong that, i the absence of other information about a particular nucleophilic substitution reaction, rearrangement is generally taken as evidence that the reaction mechanism is indeed SN1.
On this basis, then, we can account for the observed products in the following way.
1) A n-propyl substrate yields the n-propyl cation. This rearranges to the isopropyl cation, which combines to give the isopropyl product.
2) The isobutyl cation rearranges to give the tert-butyl cation.
3) The 3-methyl-2-butyl cation rearranges to the 2-methyl-2-butyl cation
4) The neopentyl cation (2,2-dimethylpropane) rearranges to the tert-pentyl cation.
5) The 3,3-dimethyl-2-butyl cation rearranges to the 2,3-dimethyl-2-butyl cation
We can see in each case that rearrangement takes place in such a way that a less stable carbocation is converted into a more stable one. (e.g. a primary into a secondary, a primary into a tertiary, or a secondary into a tertiary.) Another way of saying this is that if the degree on the electron-deficient carbon can increase, then it will.
Just how does this rearrangement occur ??
Frank Whitmore pictured rearrangement as taking place this way. An H atom or alkyl group migrates with a pair of electrons form an adjacent C atom to the electron-deficient C atom. The C atom that loses the migrating group acquires a positive charge. This is referred to as a hydride shift or an alkyl shift. These are just 2 examples of the most common type of rearrangement, the 1,2-shifts. These are rearrangements in which the migrating group moves form one atom to the nearest atom.
In the case of the n-propyl cation, for example, a shift of the H atom yields the more stable isopropyl cation. Migration of a methyl group would simply form a different n-propyl cation.
In the case of the isobutyl cation, a hydride shift yields a tertiary cation, and hence is preferred over a methyl shift, which would only yield a secondary cation.
In the case of the 3,3-dimethyl-2-butyl cation, on the other hand, a methyl shift can yield a tertiary cation and is the rearrangement that takes place.
Just as the reality of carbocations has been verified, so has the reality of their rearrangement. Carbocations have actually been observed to rearrange utilizing various spectroscopic techniques, with rates of rearrangement being measured in excess of 10^9 per second. Here again, we are observing as discrete processes the individual steps proposed for the SN1 mechanism -- this time with rearrangement.
In short, we have now seen two of the actions or types of behavior which carbocations are capable of. A carbocation may either:
1) Combine with a nucleophile or rearrange to a more stable carbocation.
2) Rearrange itself to form a more stable configuration.
It is worth noting that, in rearrangement, as in every other reaction of a carbocation, the electron-deficient C atom gains a pair of electrons, this time at the expense of a neighboring C atom -- one that can better accommodate the positive charge.
SN2 vs. SN1
We now turn to the obvious questions which are surely burning inside even the brightest of bulbs in the academic box:
1) For a given substrate under a given set of conditions, which reaction will be followed ?
2) What (if anything) can be done to encourage one reaction mechanism over the other ??
In order to answer these questions, let us consider just what can happen to a molecule of substrate. It can either suffer backside attack by the nucleophile, or undergo heterolysis to form a carbocation. Whichever of these two processes goes faster determines which mechanism predominates. (Remember: Heterolysis is the first -- and rate determining -- step of the SN1 mechanism.) Once again we must turn to the matter of relative rates of competing reaction.
The role of the substrate in this competition is combination of influences by both the alkyl group and the leaving group. The nature of the leaving group is, of course, vital to the very occurrence of substitution. Whichever process is taking place (SN2 or SN1) the bond to the leaving group is being broken. The easier it is to break the bond (i.e. the better the leaving group) the faster the reaction occurs. A better leaving group thus speeds up reactions by both mechanisms -- and, as it happens, to about the same degree. Thus, the nature of the leaving group is largely inconsequential.
In contrast, the nature of the alkyl group, R, of the substrate exerts a profound effect on which mechanism is to be followed. In R, two structural factors are at work:
I. Steric hindrance -- which largely determines the ease of the backside attack.
II. Charge dispersibility -- the capacity of the R group to accommodate and redistribute a positive charge -- which largely determines the ease of heterolysis.
As we proceed along the simple alkyl series, CH3, primary, secondary, tertiary, the group R becomes more branched. The number of substituents on the C atom increases -- bulky, electron-releasing constituents. Steric hindrance increases. Backside attack becomes difficult and thus slower. Charge dispersibility increases. Heterolysis becomes easier and faster.
The result is the pattern we saw earlier. For methyl and primary substrates, SN2 is favored. For tertiary substrates, SN1 is favored. Secondary substrates show mixed tendencies.
Our remaining influence over the experimental outcome relies almost entirely on our control of the experimental conditions. For example, we have control over the concentration of the nucleophile, which directly effects the rate of the SN2 reaction. Thus, an increase in [Z] speeds up the 2nd-oreder reaction but has no effect on the 1st-order reaction. The fraction of reaction by SN2 increases. A decrease in [Z] has just the opposite effect. The fraction of reaction by SN2 decreases.
The net result is that, other experimental factors being equal, a high concentration of nucleophile favors the SN2 reaction, while a low concentration of nucleophile favors the SN1 reaction.
In a similar fashion, the rate of SN2 depends upon the nature of the nucleophile. A stronger nucleophile attacks the substrate faster. The rate of SN1 is not affected by this factor. So, other things being equal, a strong nucleophile favors the SN2 reaction mechanism, and a weak nucleophile favors the SN1 reaction mechanism.
Role of the Solvent
The last, but possibly most important experimental condition which we have under our control is the nature of the solvent. One unavoidable conclusion is that it is the nature of the substrate and the nucleophile that determine what product is formed. The major effect of the solvent is on the rate of nucleophilic substitution -- not on what the products are.
Thus we need to consider two related questions:
I. What properties of the solvent influence the rate most ?
II. How does the rate-determining step of the mechanism respond to this property of the solvent ?
It has been amply demonstrated experimentally that the rate of ionization of tert-butyl chloride increases dramatically as the polarity of the solvent increases. In the SN1 reaction, polar and non-polar solvents are similar in their interaction with the alkyl halide reactants, but differ dramatically in how they affect the intermediates characteristic of the transition state. A non-polar solvent has little effect on the intermediate species, while strongly polar solvent acts to stabilize the charge separated transition state. In summary, polar solvents increase the rate of the SN1 reaction.
Polar solvents are required in typical bimolecular substitutions (e.g. SN2) because ionic substances are not sufficiently soluble in non-polar solvents to give a high enough concentration of the nucleophile to allow the reaction to proceed at a sufficiently rapid rate. And while it has been shown that the effect of solvent polarity on the rate of the SN2 reaction is negligible, what is more important is whether or not the polar solvent is protic or aprotic.
Protic solvents carry hydrogen attached to oxygen as in a hydroxyl group, nitrogen as in an amine group, or, more generally, any molecular solvent which contains dissociable H+, such as hydrogen fluoride. The molecules of such solvents can donate an H+ (proton). Examples include water, alcohols, carboxylic acids, amines, hydrogen halides, and polyprotic acids with multiple dissociable protons such as sulfuric acid, H2SO4 and phosphoric acid, H3PO4. Conversely, aprotic solvents cannot donate hydrogen bonds.
Experiments have clearly indicated that the reaction rate by SN2 is seriously limited by polar protic solvents, and can be accelerated markedly with the use of polar aprotic solvents such as dimethyl sulfoxide. Alternatively, reaction by SN1 is favored by polar protic solvents that help to pull out the leaving group of the molecule.
Solvolysis
The most controversial aspect of nucleophilic substitution centers around the special case in which the nucleophile is the solvent: solvolysis.
R - X + :S --> R - S + :X-
There is no added strong nucleophile. And so, for many substrates, solvolysis falls into the category we have called SN1. That is, the reaction proceeds by tow or more steps, with the intermediate formation of an organic cation. It is this intermediate that lies at the center of the problem.
And what exactly is the role of the solvent ? Our choices are as follows:
1) The solvent molecules will cluster around the carbocation and the anion (and the transition state leading to their formation) and thus aid in heterolysis through the formation of ion-dipole bonds.
2) The solvent molecules will act as nucleophiles and help to eliminate the leaving groups from the substrate molecules.
Since the solvent concentration is fixed, we cannot use reaction kinetics to determine the order (and thus the mechanism) of the reaction.
It does seem clear that the solvent can give nucleophilic assistance to solvolysis. The strength of this assistance will depend on:
1) The nucleophilic power of the solvent;
2) How badly the assistance is needed;
3) How accessible (sterically) carbon is to the assisting molecule.
E.G. Water, methanol, and ethanol are strongly nucleophilic (for solvents). Acetic Acid (CH3COOH) is weaker and formic acid (HCOOH) is weaker yet. Fluoro acids and fluoro alcohols are very weak, since the highly electronegative F atom pulls electrons strongly form oxygen, lowering its basicity and nucleophilic power.
Reactivity of tertiary substrates is found to depend little upon the nucleophilic power of the solvent and chiefly upon its ionizing power. Formation of tertiary cations is relatively easy and needs little nucleophilic assistance. In any case, crowding would discourage such assistance.
Reactivity of secondary substrates is found to depend upon both the nucleophilic power and the ionizing power of the solvent. Formation of secondary cations is more difficult, and needs much nucleophilic assistance.
With most primary substrates, the reaction is probably straightforward SN2: a single step with the solvent acting as a nucleophile.
Let us focus for now on secondary alkyl substrates. What is meant by the term "nucleophilic assistance" ?
1) It differs form the typical SN2 kind of attack in that it leads to the formation, not of the product, but of an intermediate cation.
2) It differs form general solvation in that a single molecule is involved -- not a cluster of molecules. The solvent molecule attacks the substrate at the backside and, acting as a nucleophile, helps to push the leaving group out the front side.
There is formed a carbocation -- or something quite similar to a carbocation.
Clinging to its backside is the solvent molecule and to the front side, the leaving group. The geometry is similar to that of the SN2 transition state. But the difference is that this is a reaction intermediate, and thus corresponds to an energy minimum in a progress-of-reaction plot.
If the leaving group is an anion, and if the solvent is of only moderate polarity, then bonding between cation and anion may be chiefly electrostatic in nature. One would thus speak of an ion pair.
This cationic intermediate - this nucleophilically saturated carbocation - now reacts. It has open to it the wide variety of reactions, as we shall see, carbocations may undergo. In this case, it combines with the solvent molecule - with the formation of a full-fledged bond - to yield product.
If, at the time of reaction, the leaving group is still bonded to the front side (or still lurking there) reaction with solvent occurs at the backside. Alternatively, if the cation has lasted long enough for a leaving group to be exchanged for a second solvent molecule- thus forming a symmetrical reaction intermediate - reaction is equally likely on either side (front or back). Thus, solvolysis can occur with complete inversion or with inversion plus varying degrees of racemization.
Thought still widely debated, it is generally accepted that the safest description of the mechanism of solvolysis is that of a modification of SN1. There is a cationic intermediate formed which presumably has all the same capabilities of a carbocation. Dispersal of the developing positive charge provides much of the driving force for the reaction.
We shall refer to the reaction as one following the SN1 mechanism with nucleophilic assistance from the solvent. We shall call the intermediate a nucleophilically solvated carbocation (or "encumbered" carbocation). The critical factors which need to be recognized in this description are:
1) Nucleophilic attack, with its susceptibility to steric hindrance
2) Dispersal of charge, by substituents and by the solvent.
Bear in mind here that most of what we say has to do strictly with secondary alkyl substrates. The differences in stability between the various classes of carbocations are great enough that, by and large, reactions fall into three separate groups:
1) For primary substrates, single-step SN2
2) For tertiary substrates, SN1 with an intermediate that approximates our idea of a simple (solvated) carbocation
3) For secondary substrates, a two-step reaction that is SN1-like to the extent that there is a cationic intermediate, but one formed with nucleophilic assistance and still encumbered with nucleophile (solvent) and leaving group.
Nucleophilic assistance is an important factor in determining the relative reactivities among secondary substrates, and their reactivities in various solvents. But so also is the ionizing power of the solvent. And nucleophilic assistance is not so powerful a factor as the dispersal of charge that makes tertiary substrates react (without any nucleophilic assistance) more rapidly than secondary substrates.