Carbocations have been postulated as short-lived intermediates in many reactions of organic molecules. The life times of these intermediates varies with their structure. In particluar, it will increase with the ability of the three substituents attached to the positively charged carbon to provide electron density to that electron deficient center. Since this electron donation involves resonance interactions, you may want to review the structural requirements for resonance before you read this discussion.
We will discuss two aspects of carbocation chemistry in this topic. First we will consider the variation in carbocation stability with structure. This discussion should enable you to evaluate the relative stabilities of various carbocations as well as to predict the relative rates of various reactions in which these cations are formed. The ability to predict relative rates rests on the assumption that the more stable the intermediate, the faster it will be formed. While this assumption is generally true, it is not without exception. The second part of the discussion will focus on molecular rearrangements, specifically a class of rearrangements called 1,2-shifts.
The relative stabilities of carbocations has been derived from the relative rates of reactions in which carbocations are postulated as intermediates, e.g. the Sn1 reactions of alkyl halides. As we saw during our consideration of the Sn1 reaction, the relative rates of reaction of RBr with formic acid increases in the order CH3Br <>3CH2Br < (CH3)2CHBr < (CH3)3CBr. This is interpreted to mean that the relative stabilities of the carbocations increase in the order CH3+ <>3CH2+ < (CH3)2C+ < (CH3)3C +. This is a trend that has been confirmed by gas phase measurements of the heterolytic C-Br bond dissociation energies of CH3Br, CH3CH2Br, (CH3)2CHBr, and (CH3)3CBr as well. Figure 1 presents the relative stabilities of these four carbocations in quantitative terms.1
A Cascade of Cations
J. Amer. Chem. Soc. 96, 7552 (1974).
In general, the relative stabilities of carbocations increase in the order methyl <>o <>o <>o. If we focus on the relative stabilities of the methyl cation and the ethyl cation, we can say that replacing one of the three hydrogen atoms in CH3+ with a methyl group produces a new carbocation that is more stable than CH3+. As you can see from Figure 1, the energy of the ethyl cation is approximately 36 kcal/mol less than that of the methyl cation. Apparently this new CH3 group is better able to provide electron density to the cationic center than a hydrogen atom is. Figure 2 shows how. We have seen this idea previously during our discussion of the structural requirements for resonance.
Lending a Helping Hand
In the methyl cation, the orbitals bonding the three hydrogen atoms to the central carbon are all perpendicular (orthogonal) to the vacant p orbital on that sp2 hybridized carbon. In the ethyl cation there are three orientations in which one of the C-H bonds of the methyl group can align in a "side-to-side" fashion with the p orbital on the cationic center. Such an orbital alignment leads to some degree of overlap between these orbitals, which, in turn, provides a means for transfer of electron density from the methyl group to the cationic center. This is a form of charge delocalization called hyperconjugation, and like all forms of charge delocalization, it results in stabilization of the system. Note that hyperconjugation is not limited to C-H bonds. It occurs with C-C bonds as well.
This idea is easily extended to carbocations in which two or three hydrogens have been replaced by methyl groups, i.e. to the isopropyl carbocation and the t-butyl carbocation. As you can see from Figure 3, there are three orientations for each of the two methyl groups in which s-p overlap is possible. For the t-butyl cation, there are 9 such stabilizing interactions possible.
Many Hands Make Light Work
Now let's consider the results of solvolysis reactions of three related systems, CH3CH2CH2Cl, CH3OCH2Cl, and CH2=CHCH2Cl, in which the structural variable is highlighted in red. These molecules may all be viewed as derivatives of CH3Cl in which one of the methyl hydrogens has been replaced by a CH3CH2 group, a CH3O group, or a CH2=CH group, respectively. The relative reactivities of these molecules under Sn1 conditions are CH3Cl <>3CH2CH2Cl <>2=CHCH2Cl <>3OCH2Cl. This means that the relative stabilities of the intemediate carbocations increase in the same order, i.e. CH3+ <>3CH2CH2+ <>2=CHCH2+ <>3OCH2+. As Figure 4 demonstrates, this order is easily rationalized by a simple extension of the logic presented in Figures 2 and 3.
To Each Accordingto Their Need...
From Each According to Their Ability
In the case of CH3OCH2Cl, the oxygen atom contributes electron density in the form of a lone pair of electrons. In CH2=CHCH2Cl the electron density is coming from a pi-bonded pair, while in the case of CH3CH2CH2Cl, the substituent is sharing a sigma-bonded pair of electrons.
In other words, the willingness of a substituent to share a pair of electrons with an adjacent carbocationic center depends upon two factors
- The proper geometry for orbital overlap. (Orthogonal orbitals have zero overlap.)
- The type of electrons being shared. (Lone pairs are more readily shared than pi-bonded pairs, which are more readily shared than sigma-bonded pairs.)
The structures in the lower row of Figure 4 represent the resonance contributorsis that result from the interactions just discussed. The structure in the lower right hand corner is highlighted in red to emphasize the fact that all the atoms have a filled valence shell. Not only are the non-bonding electrons more readily shared, but the resulting resonance contributor makes a more important contribution to the resonance hybrid. The fact that CH3OCH2Cl undergoes solvolysis 109 times faster than CH3CH2CH2Cl emphasizes the importance of this effect.
Figure 5 presents the information depicted in Figure 4 in the equivalent, but more familiar, "arrow-pushing" formalism.
It's Still the Same Old Story...
Exercise 1 Ionization of the C-Cl bond in CH3NHCH2Cl generates a carbocation, A, which is stabilized by resonance interaction with the lone pair of electrons on the nitrogen atom that is attached to the carbon adjacent to the cationic center. Draw the structure of carbocation A and the resonance contributor B.
Exercise 2 Arrange the following carbocations in order of increasing stability:
The rules of resonance theory state that the position of the nuclei must remain fixed. The only difference between one resonance structure and another is the location of the electrons. It is important to be aware of the distinction that is being made here. Figure 6 compares the hyperconjugative resonance interaction shown in Figure 5 with an alternative process in which the hydrogen atom and the pair of electrons move from one atom to the next. This change is called a molecular rearrangement, and it is the subject of the next section of this topic.
Resonance vs Rearrangement
If you imagine a molecule such as 1-bromopropane being forced to react as shown in Equation 1, it shouldn't surprise you that the carbocationic intermediate is an unstable species and that it will do whatever it can to become more stable.
One option available to this cation is to react with a nucleophile to form a neutral product. While the carbocation generated in reaction 1 certainly recombines with bomide ion, this event is not of particular interest since it does not produce any new product. The other potential nucleophile in this mixture is the solvent, but as we have seen, formic acid is used to study Sn1 reactions because it is a highly ionizing but relatively non-nucleophilic solvent. The result of using a solvent of low nucleophilic reactivity is to increase the lifetime of the carbocationic intermediate. This, in turn, provides more time for other processes to occur. In the example shown in Figure 6, one such alternative involves the migration of a hydrogen atom with a pair of electrons from one carbon atom to the adjacent, positively charged carbon atom. Such a molecular rearrangement is called a 1,2-hydride shift. It is called a 1,2-shift because the hydrogen atom moves from one atom to the adjacent atom.
Note that the 1,2-hydride shift illustrated in Figure 6 converts a 1o carbocation into a 2o carbocation. Once this more stable carbocation is formed, it can react with bromide ion to produce a new product as shown in Equation 2.
It is called a hydride shift because a hydrogen atom with a pair of electrons is a hydride ion. The animation in Figure 7 will give you a rough idea of what this process might look like.
Here I Come
Molecular rearrangements of this type frequently accompany Sn1 reactions, especially in highly ionizing, non-nucleophilic solvents such as formic acid and acetic acid. As a rule of thumb, rearrangements will occur whenever they can produce carbocations of equal or greater stability. Hydride shifts are not the only type of rearrangement seen under these conditions. Another common type of 1,2-shift is called a 1,2-methide shift.
Exercise 3 Draw the carbocation that would result from a 1,2-hydride shift of the hydrogen atom shown in red in each of the following carbocations.
Exercise 4 Which of these rearrangements would produce the most stable carbocation? A B C
The name methide refers to the H3C:- group; in other words, a methyl anion. Equation 3 describes a reaction involving a 1,2-methide shift.
In this case solvolysis of 2-bromo-3,3-dimethylbutane generates a 2o carbocation. Rearrangement of one of the three methyl groups, with a pair of electrons, from C-3 to C-2 then generates a 3o carbocation which ultimately captures a bromide ion to yield the final product, 2-bromo-2,3-dimethylbutane. Figure 8 presents a reaction profile diagram for this transformation.
Over Hill Over Dale...
While 1,2-shifts of other alkyl groups can occur, 1,2-methide shifts are the most common. Another group that undergoes frequent rearrangement is a phenyl ring.
Exercise 5 Draw the carbocation that would result from a 1,2-methide shift in each of the following carbocations.
Exercise 6 Which of these rearrangements would produce the most stable carbocation? A B C
Exercise 7 Which of these rearrangements would produce a less stable carbocation? A B C
When 3-methyl-3-phenyl-2-butanol is treated with aqueous HBr, the reaction outlined in Equation 4 occurs.
It seems likely that the formation of 3-methyl-3-phenyl-2-bromobutane involves simple replacement of the OH group of the starting material with a bromide ion. The interesting question concerns the origin of the isomeric 2-methyl-3-phenyl-2-bromobutane. Before we suggest an answer to this question, let's consider the role of aqueous HBr in reaction 4. There are three points to be noted:
- the sovent is water, which means that it supports the formation of ions
- HBr is a strong acid, which means that a significant percentage of the starting material will be protonated
- HBr is a source of the nucleophilic bromide ions that are incorporated into the product
With these points in mind, Figure 7 suggests a plausible pathway by which the 2-methyl-3-phenyl-2-bromobutane may be formed.
Move Over Neighbor
The process is initiated by an acid-base reaction in which the HBr protonates the the alcohol. This converts the OH group to a better leaving group, water. Ionization of the C-O bond generates the 2o carbocation shown in the upper right hand corner of the figure. When this ion captures a bromide ion, a molecule of 3-methyl-3-phenyl-2-bromobutane is formed. However, capture of a bromide ion competes with the intramolecular donation of electron density from the adjacent phenyl ring as shown by the red arrow in the lower left hand corner of the figure. This change produces a "bridged" ion in which the aromatic ring is simultaneously bonded to two carbon atoms. The 1,2-shift of the phenyl ring is then completed by the electronic movement indicated by the blue arrow. This process generates a 3o carbocation, which ultimately traps a bromide ion to produce the second product, 2-methyl-3-phenyl-2-bromobutane.
The details of the scheme outlined in Figure 7 were vigorously debated for over a quarter century. Many brilliant chemists devised many brilliant experiments to try to answer such questions as "What is the timing of this rearrangement? Does the phenyl group assist in the ionization of the C-O bond or does it rearrange after the carbocation is fully formed?" "What is the structure of the bridged ion?" Admittedly the picture presented in Figure 7 is an oversimplification. It purpose is not to summarize all the data generated by the attempts to answer these questions, but rather to illustrate the application of basic principles to the rationalization of experimental fact.
Exercise 8 What is the approximate equilibrium constant for the reaction? Exercise 9 Draw the resonance structure of the bridged ion shown in Figure 7, but with the positive charge in the para position of the aromatic ring.