And the Sn2 reaction

Following these general examples, we can now analyze the opening of epoxides in more detail. The size of the ring imposes geometric constraint and the SN2 reaction demands colinearity. The approach of the nucleophile should then resemble J7. + which is different from that in displacement of a leaving group on an aliphatic chain, i. e. 19 20. 

Many radical cations derived from cyclopropane (or cyclobutane) systems undergo bond formation with nucleophiles, typically neutralizing the positive charge and generating addition products via free-radical intermediates [140, 147). In one sense, these reactions are akin to the well known nucleophilic capture of carbocations, which is the second step of nucleophilic substitution via an Sn 1 mechanism. The capture of cyclopropane radical cations has the special feature that an sp -hybridized carbon center serves as an (intramolecular) leaving group, which changes the reaction, in essence, to a second-order substitution. Whereas the SnI reaction involves two electrons and an empty p-orbital and the Sn2 reaction occurs with redistribution of four electrons, the related radical cation reaction involves three electrons. 

Transition state 11 in Section 11.2.1 was shown for the conversion of 9 to 10. The charge distribution is such that the iodide is 6-, the central carbon is 6+, and the leaving group X is 6-. If the solvent is water, the iodide ion will be solvated (surrounded by water molecules) before it collides with the sp carbon atom. This means that solvation will impede the collision of iodide with the carbon atom, and the Sn2 reaction will be slower. Water solvates both the anion and cation and the net result is that the 8 2 reaction is slower. 

Notice that the conversion of the OH into a tosylate and the Sn2 reaction are two separate synthetic steps, so we place the numbers 1 and 2 before the sets of reagents to indicate that they are separate reactions. 

FIGURE 7.65 For tertiary systems, the SnI reaction dominates the ionization produces the relatively stable tertiary carbocation and the Sn2 reaction is hindered by the three R groups that thwart approach of the ... 

It should be emphasized again that both the SnI and the Sn2 reaction show solvent effects, but that they do so for different reasons. Sn2 reactions are disfavored in protic solvents because the ground-state energy of the nucleophile is lowered by solvation. SnI reactions are favored in protic solvents because the transition-state energy leading to carbocation intermediate is lowered by solvation. 

Figure A2.3.21 Free energy profile of the SN2 reaction Cl +CH2CI— [Cl-CHg-Cl]— CICH +Cl in the gas phase, dimethyl fonnamide and in water (from [93]).

Chemists use curved arrows to show the electronic changes that occur during a chemical reaction. Fot example, the arrows describing the Sn2 reaction below show formation of a CC bond and loss of a Cl bond. 

One after the other, step through (or animate) the sequence of structures depicting the SN2 and proton transfer reactions shown above. Compare the two. From what direction does cyanide approach the hydrogen in HCl From the same side as Cl ( frontside ), or from the other side ( backside ) Does the Sn2 reaction follow a similar trajectory ... 

In contrast to S l reactions, the Sn2 type represent a one-step process. The rate-determining stage, and hence the formation of the new bond, depends on the nucleophilicity of the anion (the nucleo-philicity is essentially a function of the polarizability ). In the Sn2 reaction under discussion, Eq, (7), the nitrogen in CH3—is replaced by the atom of the mesomeric anion which possesses the greatest nucleophilicity. 

For high diazomethane concentrations, the Sn2 reaction, Eq. (7), and thus A—methylation occurs, whereas 0-methylation is favored by lower diazomethane concentrations, Eq, (6) (for an interpretation of this effect, according to Arndt, see references 33 and 42). The extent of this effect is limited by the constitution of the lactam in question. The fact that the addition of the sodium salt of saccharin to the reaction mixture leads to increased A -mcthylation for saccharin can be taken as supporting the foregoing interpretation. 

When diazomethane is slowly added to excess lactam, the anions formed can interact with unreacted lactam by means of hydrogen bonds to form ion pairs similar to those formed by acetic acid-tri-ethylamine mixtures in nonpolar solvents. The methyldiazonium ion is then involved in an ion association wdth the mono-anion of a dimeric lactam which is naturally less reactive than a free lactam anion. The velocity of the Sn2 reaction, Eq. (7), is thus decreased. However, the decomposition velocity of the methyldiazonium ion, Eq. (6a), is constant and, hence, the S l character of the reaction is increased which favors 0-methylation. It is possible that this effect is also involved in kinetic dependence investigations have shown that with higher saccharin concentrations more 0-methylsaccharin is formed. 

A mechanism that accounts for both the inversion of configuration and the second-order kinetics that are observed with nucleophilic substitution reactions was suggested in 1937 by E. D. Hughes and Christopher Ingold, who formulated what they called the SN2 reaction—short for substitution, nucleophilic, birnolecu-lar. (Birnolecular means that two molecules, nucleophile and alkyl halide, take part in the step whose kinetics are measured.)... 

What are the reasons for the reactivity differences observed in Table 11.1 Why do some reactants appear to be much more "nucleophilic" than others The answers to these questions aren t straightforward. Part of the problem is that the term micleophilicit > is imprecise. The term is usually taken to be a measure of the affinity of a nucleophile for a carbon atom in the SN2 reaction, but the reactivity of a given nucleophile can change from one reaction to the next. The exact nucleophilicity of a species in a given reaction depends on the substrate, the solvent, and even the reactant concentrations. Detailed... 

What about solvent Do solvents have the same effect in S>g 1 reactions that they have in S j2 reactions The answer is both yes and no. Yes, solvents have a large effect on S l reactions, but no, the reasons for the effects on S jl and SN2 reactions are not the same. Solvent effects in the SN2 reaction are due largely to stabilization or destabilization of the nucleophile reactant. Solvent effects in the Sjsjl reaction, however, are due largely to stabilization or destabilization of the transition state. 

The reaction of an alkyl halide or los3 late with a nucleophiJe/base results eithe in substitution or in diminution. Nucleophilic substitutions are of two types S 2 reactions and SN1 reactions, in the SN2 reaction, the entering nucleophih approaches the halide from a direction 180° away from the leaving group, result ing in an umbrella-like inversion of configuration at the carbon atom. The reaction is kinetically second-order and is strongly inhibited by increasing stork bulk of the reactants. Thus, S 2 reactions are favored for primary and secondary substrates. 

One of the most important reasons for using tosylates in S j2 reactions is stereochemical. The S]s]2 reaction of an alcohol via an alkyl halide proceeds with hvo inversions of configuration—one to make the halide from the alcohol and one to substitute the halide—and yields a product with the same stereochemistry as the starting alcohol. The SN2 reaction of an alcohol via a tosylate, however, proceeds with only one inversion and yields a product of opposite stereochemistry to the starting alcohol. Figure 17.5 shows a series of reactions on the R enantiomer of 2-octanol that illustrates these stereochemical relationships. 

LFER. Consider the Sn2 reactions of XC6H4CH2CI with I- (ki) and the SN1 reactions. with OH (fc0H)- The reaction constants are given in Table 10-2. Sketch the appearance of a plot of log ki versus log kon- What is its slope ... 

Exercise 3.6. Write a program that evaluates ex, e2, and Hl2 for the SN2 reaction ClCl-Cl- - Cl-C Cl-, neglecting the hydrogens on the carbon so that the 6 term in eq. (3.26) is not needed. Also, neglect the Uind term. Next, surround this solute system by 20 dipoles and simulate the resulting solute + solvent system with the potential U = s 1 (examine the distances between the three atoms during the simulation). 

---