11.4: The SN1 Reaction

Most nucleophilic substitutions take place by the S_N2 pathway just discussed. The reaction is favored when carried out with an unhindered substrate and a negatively charged nucleophile in a polar aprotic solvent, but is disfavored when carried out with a hindered substrate and a neutral nucleophile in a protic solvent. You might therefore expect the reaction of a tertiary substrate (hindered) with water (neutral, protic) to be among the slowest of substitution reactions. Remarkably, however, the opposite is true. The reaction of the tertiary halide 2-bromo-2-methylpropane (CH₃)₃CBr with H₂O to give the alcohol 2-methyl-2-propanol is more than 1 million times faster than the corresponding reaction of CH₃Br to give methanol.

What’s going on here? A nucleophilic substitution reaction is occurring—a hydroxyl group is replacing a halogen—yet the reactivity order seems backward. These reactions can’t be taking place by the S_N2 mechanism we’ve been discussing, so we must therefore conclude that they are occurring by an alternative substitution mechanism. This alternative mechanism is called the S_N1 reaction, for substitution, nucleophilic, unimolecular. In contrast to the S_N2 reaction of CH₃Br with OH^–, the S_N1 reaction of (CH₃)₃CBr with H₂O has a rate that depends only on the alkyl halide concentration and is independent of the H₂O concentration. In other words, the process is a first-order reaction; the concentration of the nucleophile does not appear in the rate equation. To explain this result, we need to know more about kinetic measurements.

\[ \begin{align*} \text{Reaction rate} &= \text{Rate of disappearance of alkyl halide} \\[4pt]
&= \text{Rate of appearance of product} \\[4pt]
&=k \times [\ce{RX}] \end{align*}\]

Many organic reactions occur in several steps, one of which usually has a higher-energy transition state than the others and is therefore slower. We call this step with the highest transition-state energy the rate-limiting step, or rate-determining step. No reaction can proceed faster than its rate-limiting step, which acts as a kind of traffic jam, or bottleneck.

Because the first step of the mechanism is the rate-determining step, the first transition state will be the most important. During the first step, a carbon-leaving group bond in a neutral substrate is broken to form a carbocation intermediate and a leaving group anion. The breaking bond is represented as a dashed-partial bond in the transition state. A partial positive charge is shown on the carbon to represent the formation of a carbocation, and a partial negative charge is shown on the leaving group to represent the formation of an anion.

Transition State 1 ([TS₁]^‡)

2-Steps Potential Energy Diagram

In the S_N1 reaction of (CH₃)₃CBr with H₂O, the fact that the nucleophile concentration does not appear in the first-order rate equation means that it is not involved in the rate-limiting step and must therefore be involved in some other, non-rate-limiting step. The energy diagram is shown in Figure \(\PageIndex{1}\).

Figure \(\PageIndex{1}\): An energy diagram for an S_N1 reaction. The rate-limiting step is the spontaneous dissociation of the alkyl halide to give a carbocation intermediate. Reaction of the carbocation with a nucleophile then occurs in a second, faster step.

Mechanism

Unlike what occurs in an S_N2 reaction, where the leaving group is displaced while the incoming nucleophile approaches, an S_N1 reaction takes place by loss of the leaving group before the nucleophile approaches. 2-Bromo-2-methylpropane spontaneously dissociates to the tert-butyl carbocation (CH₃)₃C⁺, plus Br^– in a slow, rate-limiting step, and the intermediate carbocation is then immediately trapped by the nucleophile water in a faster second step. Thus, water is not a reactant in the step whose rate is measured. The mechanism shown in Figure \(\PageIndex{2}\) accounts for these observations.

Figure \(\PageIndex{2}\): The mechanism of the S_N1 reaction of 2-bromo-2-methylpropane with H₂O involves three steps. Step 1 —the spontaneous, unimolecular dissociation of the alkyl bromide to yield a carbocation—is rate-limiting.

3-Steps S_N1 Reactions

A potential energy diagram for this reaction shows that each of the two positively-charged intermediates (I₁ and I₂) can be visualized as a valley in the path of the reaction, higher in energy than both the reactant and product but lower in energy than the transition states. The first, bond-breaking step is the slowest, rate-determining step - notice it has the highest activation energy and leads to the highest-energy species, I₁, the carbocation intermediate. Step 2 is rapid: a new covalent bond forms between a carbocation and a water nucleophile, and no covalent bonds are broken. The third step represents a Bronsted-Lowry proton transfer, which is rapid, with low activation energy.

Figure \(\PageIndex{5}\): An energy diagram for an S_N1 reaction. The third step represents a Bronsted-Lowry proton transfer, which is rapid, with low activation energy.

 

Stereochemistry of the Product

Because an S_N1 reaction occurs through a carbocation intermediate, its stereochemical outcome is different from that of an S_N2 reaction. Carbocations, as we’ve seen, are planar, sp²-hybridized, and achiral. Thus, if we carry out an S_N1 reaction on one enantiomer of a chiral reactant and go through an achiral carbocation intermediate, the product loses its optical activity (Section 8.12). That is, the pro-chiral intermediate carbocation can react with a nucleophile equally well from either side, leading to a racemic, 50 : 50 mixture of enantiomers (Figure \(\PageIndex{3}\)).

Figure \(\PageIndex{3}\): Stereochemistry of the S_N1 reaction. Because the reaction goes through an achiral intermediate, an enantiomerically pure reactant gives an optically inactive racemic product.

The conclusion that S_N1 reactions on enantiomerically pure substrates should give racemic products is nearly, but not exactly, what is found. In fact, few S_N1 displacements occur with complete racemization. Most give a minor (0–20%) excess of inversion. The reaction of (R)-6-chloro-2,6-dimethyloctane with H₂O, for example, leads to an alcohol product that is approximately 80% racemized and 20% inverted (80% R,S + 20% S is equivalent to 40% R + 60% S).

This lack of complete racemization in S_N1 reactions is due to the fact that ion pairs are involved. According to this explanation, the dissociation of the substrate occurs to give a structure in which the two ions are still loosely associated and in which the carbocation is effectively shielded from reaction on one side by the departing anion. If a certain amount of substitution occurs before the two ions fully diffuse apart, then a net inversion of configuration will be observed (Figure \(\PageIndex{4}\)).

Figure \(\PageIndex{4}\): Ion pairs in an S_N1 reaction. The leaving group shields one side of the carbocation intermediate from reaction with the nucleophile, thereby leading to some inversion of configuration rather than complete racemization.

Exercises

Exercise \(\PageIndex{1}\)

What product(s) would you expect from the reaction of (S)-3-chloro-3-methyloctane with acetic acid? Show the stereochemistry of both reactant and product.

Answer

Racemic 1-ethyl-1-methylhexyl acetate

Exercise \(\PageIndex{2}\)

Among the many examples of S_N1 reactions that occur with incomplete racemization, the optically pure tosylate of 2,2-dimethyl-1-phenyl-1-propanol ([α]_D = –30.3) gives the corresponding acetate ([α]_D = +5.3) when heated in acetic acid. If complete inversion had occurred, the optically pure acetate would have had [α]_D = +53.6. What percentage racemization and what percentage inversion occurred in this reaction?

Answer

90.1% racemization, 9.9% inversion

Exercise \(\PageIndex{3}\)

Assign configuration to the following substrate, and show the stereochemistry and identity of the product you would obtain by S_N1 reaction with water (reddish brown = Br):

Answer