In our discussion of chemical kinetics we described two alternative reaction profiles that are typical of nucleophilic aliphatic substitution reactions. They are reiterated in Figure 1. The profile in the left-hand panel is typical of nucleophilic aliphatic substitution reactions that proceed by a concerted, 1-step process, i.e. by an Sn2 mechansim. We are now going to look at a group of nucleophilic aliphatic substitution reactions that proceed by a non-concerted, 2-step process.
Alternative Profiles for Alternative Reaction Mechanisms
The Sn1 Mechanism
If you think of the Sn2 mechanism as the Dr. Jekyl of nucleophilic aliphatic substitution reactions, then the Sn1 mechanism is certainly the Mr. Hyde- a very different character. We will take a brief look at the differences for each of the parameters we considered during our discussion of the Sn2 mechanism.
Kinetics of the Sn1 Mechanism
When an aqueous solution of potassium hydroxide, KOH, is added to a solution of 2-chloro-2-methylpropane in ethanol, the 2-chloro-2-methylpropane is converted into 2-methyl-2-propanol as outlined in Equation 1.
Exercise 1 Suggest two procedures by which you could measure the rate of reaction 1.
In one experiment, when the initial concentrations of the 2-chloro-2-methylpropane and KOH solutions were both 0.0510 molar, the rate constant was 0.146 hr-1. In subsequent runs the initial concentration of hydroxide ion was reduced. Despite this decrease, the rate of the reaction remained 0.146 hr-1! In fact, when the concentration of hydroxide ion was reduced to 0, i.e. only water was added to the solution, the rate of the reaction remained unchanged at 0.146 hr-1!! A reaction in which the solvent acts as the nucleophile is called a solvolysis. When the solvent is water, the reaction is called hydrolysis. (The suffix lysis means "to cleave". Thus solvolysis means to cleave with the solvent and hydrolysis means to cleave with water.)
Exercise 2 If reaction 1 were a bimolecular nucleophilic aliphatic substitution, what would the rate of the reaction be when the concentration of hydroxide ion were 0?
Exercise 3 Select the best interpretation of the observation that the rate of reaction 1 does not depend on the amount of hydroxide ion added to the solution. You are measuring the rate of a reaction that produces a different product. Autoionization of water keeps the concentration of hydroxide ion constant. The hydroxide ion is not involved in the rate determining step of the reaction. Water ionizes to produce hydroxide ion in ethanol.
Exercise 4 When no KOH was added to the mixture, the solution slowly became acidic. Furthermore, the conductance of the solution increased with time. Write an equation that is consistent with this information.
Now consider the set of nucleophilic aliphatic substitution reactions represented in Scheme 1.
It All Depends. Well, Actually It Doesn't...
As the data make clear, the rate of substitution does not change significantly when the nucleophile is changed, i.e. unlike an Sn2 reaction, where the rate of substitution is affected by the reactivity of the nucleophile, the rate of this reaction is independent of the nucleophile.
Exercise 5 Draw Lewis structures of the products of the three reactions implied in Scheme 1.
Exercise 6 Select the correct order of base strengths of the ions in Scheme 1: hydroxide > carbonate > iodide hydroxide > iodide > carbonate carbonate > hydroxide > iodide carbonate > iodide > hydroxide
Exercise 7 Based on your experience with Sn2 reactions, select the correct order of nucleophilic reactivity of the ions in Scheme 1: hydroxide > carbonate > iodide hydroxide > iodide > carbonate iodide > carbonate > hydroxide carbonate > iodide > hydroxide
The lack of any correlation between the concentration or the identity of the nucleophile and the rate of substitution led chemists to postulate a 2-step process in which the first step involved ionization of the bond between the reaction center and the leaving group. For the reaction shown in Equation 1, this step may be represented as
This postulate accounts for the observation that the conductance of the solution increases with time.
The cation that is produced is called a carbocation to indicate that the carbon atom bears a formal positive charge. Note that this carbocation violates the octet rule; there are only six electrons in the valence shell of the carbon atom that bears the positive charge. It should not be a surprise then that this ion reacts with nucleophilic species very rapidly. If hydroxide ion is the nucleophile, the second step of the 2-step process looks like
In the absence of hydroxide ion, water assumes the role of the nucleophile
The intermediate oxonium ion loses a proton during the work-up of the reaction to produce the final product, 2-methyl-2-propanol.
Figure 2 identifies these species in a reaction coordinate diagram like the one in the right-hand panel of Figure 1. Here the nucleophile is hydroxide ion.
As the figure indicates, the activation energy for the first step of the reaction is much larger than that for the second step. Thus the first step is much slower. It is called the rate determining step. The rate determining step leads to the formation of the intemediate carbocation which then reacts rapidly with the nucleophile in the product determining step. Note that there are two transition states involved in this mechanism. In the first, the bond between the leaving group and the reaction center is partially broken; the carbon is acquiring a positive charge while the chlorine is becoming negative. In the second transition state the bond between the nucleophile and the reaction center is partially formed; the charge on the carbon atom is diminishing from +1 towards 0, while the charge on the hydroxide ion is changing from -1 towards 0.
Exercise 8 Draw a reaction coordinate diagram for each of the reactions presented in Scheme 1. Identify the two transition state structures in each reaction. Indicate the partial charges.
Exercise 9 Select the statement that most accurately describes why (CH3)3CS+(CH3)2 reacts with hydroxide ion, carbonate ion, and iodide ion at the same rate (See Scheme 2.). These ions are equally reactive as nucleophiles. The activation energies for the three product determining steps are the same.The rates of the product determining steps are different, but you can't measure the differences.
Exercise 10 Which nucleophile has the lowest activation energy for the product determining step of the reaction outlined in Scheme 1? iodide ion carbonate ion hydroxide ion
We'll consider one set of data to illustrate how steric factors influence the rate of nucleophilic aliphatic substitution that display unimolecular kinetics. The relevant reactions are summarized in Scheme 2
Exercise 11 Is the correlation of the reaction rate with the number of hydrogen atoms attached to the reaction center direct or indirect ?
Exercise 12 Which type of substrate reacts under the conditions described in Scheme 2? methyl 1o 2o 3o
Exercise 13 If the solvent for the reaction shown in Scheme 2 were acetone rather than formic acid, which substrate would react fastest? Which type of substrate reacts under the conditions described in Scheme 2? methyl 1o 2o 3o
In these reactions water acts as the nucleophile while formic acid, HCO2H, is the solvent. It is classified as an ionizing, non-nucleophilic solvent. It is considered to be an ionizing solvent because 1. it is protic 2. it has a high dielectric constant. Recall that the dielectric constant of a solvent is a measure of its ability to insulate a cation from the Coulombic attraction of an anion. In other words, solvents with high dielectric constants allow oppositely charged particles, cations and anions, to dissociate from each other in solution. They also promote the ionization of a polar covalent bond such as the C-Br bond in 2-bromo-2-methylpropane.
Exercise 14 Draw a picture depicting the H-bonding interaction between formic acid and 2-bromo-2-methylpropane.
Formic acid is classified as a non-nucleophilic solvent because nucleophiles such as iodide or hydroxide ion are much more reactive than the solvent. Even water is more nucleophilic than formic acid. The resonance interaction between a lone pair of electrons on the oxygen atom of the OH group and the pi system of the carbonyl group reduces the potential energy, and therefore the reactivty, of the lone pair. Scheme 3 illustrates this idea.
Resonance Once Again
Other examples of ionizing, non-nucleophilic solvents are acetic acid (CH3CO2H), trifluoroacetic acid (CF3CO2H), trifluoroethanol (CF3CH2OH), and hexafluoroisopropanol (CF3)2CHOH).
In our discussion of the Sn2 mechanism we considered a scheme for analysing the effect of changing solvents on the relative rates of bimolecular substitution reactions. That scheme required an understanding of the charge type of each reaction. The same analysis is applicable to Sn1 reactions. Consider the data in Table 1 which summarizes the change in rate of the hydrolysis of 2-chloro-2-methylpropane as the polarity of the solvent is increased by increasing the percentage of water .
Solvent Effects on the Rate of an Sn1 Reaction
| || || || || || || |
| || || || || || || |
Exercise 15 Draw the transition state for the hydrolysis of 2-chloro-2-methylpropane. Is there a build-up or a decrease in charge as the reaction proceeds from the starting materials to the transition state? build-up decrease
Exercise 16 Predict the effect of increasing the solvent polarity on the rate of the reaction shown in Scheme 1. It should increase the rate. It should decrease the rate.
Finally, let's consider the stereochemical aspects of the Sn1 mechanism.
When 2-bromooctane is refluxed in 60% water and 40% ethanol, it is converted into a mixture of 2-octanol and 2-ethoxyoctane as shown in Equation 2.
Performing this reaction with enantiomerically pure starting material provides insight into the stereochemical nature of the process. The specific rotation of (R)-2-bromooctane is -34.6 degrees. The specific rotation of (R)-2-octanol is -9.9 degrees. When reaction 2 was run using (R)-2-bromooctane, the optical rotation of the 2-octanol that was isolated was +6.5 degrees. If the reaction proceeded with complete inversion of configuration, i.e. by an Sn2 mechanism, the specific rotation of the 2-ocatnol produced should be +9.9 degrees. The fact that it is only +6.5 degrees must mean that the sample contains some of the levorotatory isomer in addition to the dextrorotatroy product. Calculating the composition of the mixture from the optical rotation is easy: +6.5 = [(X)*(+9.9) + (1-X)*(-9.9)], where X is the mole fraction of the S enantiomer and 1-X the mole fraction of the R enantiomer in the product. Doing the math shows that 83% of the 2-octanol had the S configuration while 17 % was the R enantiomer. In other words, the stereochemical configuration at the reaction center was inverted 83% of the time while being retained 17% of the time.
Exercise 17 What is the enantiomeric excess of (S)-2-octanol in the sample that has a specific rotation of +6.5 degrees? % Express your answer to the nearest whole number.
Two additional examples make it clear that unimolecular nucleophilic aliphatic substitution reactions have different stereochemical requirements than their bimolecular counterparts. Equation 3 shows the results of the hydrolysis of 1-chlorophenylethane to produce 1-phenylethanol:
The specific rotation of (R)-1-phenylethanol is +42 degrees. The rotation of the sample of 1-phenylethanol produced from optically pure (S)-1-chlorophenylethane was +7.1 degrees. Clearly this sample must contain a lot of (S)-1-phenylethanol, i.e. a lot of the stereoisomer with the same configuration as the starting material.
Exercise 18 To the nearest tenth of a percent, how much of the 1-phenylethanol has the R configuration? % How much has the S configuration? % To the nearest percent, what is the enantiomeric excess of (R)-1-phenylethanol in this sample? %
Finally, Equation 4 illustrates a unimolecular nucleophilic aliphatic substitution reaction where the product has the same configuration as the starting material, i.e. the reaction proceeds with complete retention of configuration.
Clearly the stereochemical nature of the Sn1 mechanism is more complex than that of the Sn2 process. Figure 3 summarizes the possibilities. Equation 2 exemplifies a reaction that proceeds with 83% inversion of configuration and 17% retention. In reaction 3 the 58/42 distribution of enantiomers is much closer to the 50/50 ratio in a racemic mixture. Finally, as the results of reaction 4 demonstrate, even complete retention of configuration is possible.
How do we interpret such variable behavior? In Figure 2 we postulated a 2-step process in which the leaving group departs from the substrate to produce an intermediate carbocation. Figure 4 offers a more detailed picture of the carbocationic intermediates involved in each of the processes outlined in Figure 3.
Rationalizing the Facts
The suggestion for reaction 2 is that the leaving group blocks access to the front side of the molecule, forcing the nucleophile to approach the substrate from the back side, leading to predominant inversion of configuration. When the alkyl group is changed to a phenyl ring, as in reaction 3, the carbocation is stable enough that the leaving group completely separates from the substrate; a symmetrical intermediate is formed and the product is racemic or nearly so. In reaction 4, the carboxylate anion is thought to act like an internal nucleophile, assisting the departure of the leaving group by coordinating to the back side of the C-Br bond and blocking access of the nucleophile to that side of the molecule.
Exercise 19 Figure 4 implies that the intermediate formed in reaction 3 has a long enough life time to allow the leaving group to separate from the substrate completely and for the solvent to arrange itself around the cationic center as shown before the nucleophile attacks. In other words, this intermediate has a longer life time than the one generated in reaction 2. This increased life time has been attributed to stabilization of the carbocation by the aromatic ring. Draw an orbital diagram that shows how the pi system of the aromatic ring can overlap with the p orbital of the carbocation.
Exercise 20 Draw pictures of the 1st and 2nd transition states involved in reactions 2, 3, and 4. Use dashed lines to show bonds that are partially formed or broken. Use d+/ d- symbols to depict partial charges.