Stereochemistry of Sn2 reaction

Sn2 reactions proceed with inversion at the electrophilic carbon. This suggests that the nucleophile attacks from the backside of carbon, i.e., the side of carbon furthest away from the leaving group. 

Backside attack may be favored for electrostatic reasons. Examine electrostatic potential maps fox bromide + methyl bromide frontside attack and bromide + methyl bromide backside attack, transition states involving frontside and backside attack of Br (the nucleophile) onto CHsBr, respectively. Which atoms in the transition states are most electron-rich Which trajectory better minimizes electrostatic repulsion  

Backside attack may be favored in order to facilitate transfer of nonbonding electrons from the nucleophile into the electrophile s lowest-unoccupied molecular orbital (LUMO). Efficient electron transfer requires maximal overlap of the LUMO and the donor orbital (usually a nonbonded electron pair on the nucleophile). Examine the LUMO of methyl bromide. How would a nucleophile have to approach in order to obtain the best overlap Is your answer more consistent with preferential backside or frontside attack  

Electron transfer into the LUMO might also cause bonding changes. What are the CBr bonding characteristics of the LUMO in methyl bromide Is it bonding (one surface extends over the bond) or antibonding (two surfaces meet in middle of the bond) How would electron transfer from a nucleophile affect the CBr bond length  

Apply the same analysis to trimethyloxonium ion, MesO a cationic electrophile. Examine the LUMO. What are the best sites for nucleophile-LUMO overlap How would electron transfer affect CO bond lengths  

What is the structure of the transition state in an Sn2 reaction In particular, what is the spatial arrangement of the nucleophile in relation to the leaving group as reactants pass through the transition state on their way to products  

Which of these two opposite stereochemical possibilities operates was determined in experiments with optically active alkyl halides. In one such experiment, Hughes and Ingold determined that the reaction of 2-bromooctane with hydroxide ion gave 2-octanol having a configuration opposite that of the starting alkyl halide. 

Although the alkyl halide and alcohol given in this example have opposite configurations when they have opposite signs of rotation, it cannot be assumed that this will be true for all alkyl halide/alcohol pairs. (See Section 7.5) 

For a change of pace, try doing Problem 8.4 with molecu lar models instead of making structural drawings. 

The first example of a stereo-electronic effect in this text concerned anti elimination in E2 reactions of alkyl halides (Section 5.16). 

Nucleophilic substitution had occurred with inversion of configuration, consistent with the following transition state  

Exercise 8-6 What inference as to reaction mechanism might you make from the observation that the rate of hydrolysis of a certain alkyl chloride in aqueous 2-propa-none is retarded by having a moderate concentration of lithium chloride in the solution  

The stereochemical consequences of front- and back-side displacements are different. With cyclic compounds, the two types of displacement lead to different products. For example, an SN2 reaction between cis-3-methylcyclopentyl chloride and hydroxide ion would give the cis alcohol by front-side approach but the trans alcohol by back-side approach. The actual product is the trans alcohol, from which we know that reaction occurs by backside displacement  

Exercise 8-7 Equations 8-3 through 8-5 show how Kenyon and Phillips established that inversion of configuration accompanies what we now recognize to be SN2 substitutions. For each reaction, we indicate whether R—0 or O—H is broken by an appropriately placed vertical line. Explain how the sequence of steps shows that inversion occurs in the SN2 reaction of Equation 8-5. The symbols (+) or (—) designate for each compound the sign of the rotation a of the plane of polarized light that it produces. 

Exercise 8-8 Explain how, in the presence of bromide ion, either enantiomer of 2-bromobutane racemizes (Section 5-1B) in 2-propanone solution at a rate that is first order in Breand first order in 2-bromobutane. 

Exercise 8-9 When either of the enantiomers of 1-deuterio-1-bromobutane is heated with bromide ion in 2-propanone, it undergoes an SN2 reaction that results in a slow loss of its optical activity. If radioactive bromide ion (Br e) is present in the solution, radioactive 1-deuterio-1-bromobutane is formed by the same SN2 mechanism in accord with the following equation  

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The stereochemistry of Sn2 reactions has been investigated. It has been found that both syn " (the nucleophile enters on the side from which the leaving group departs) and anti ° reactions can take place, depending on the nature of X and though the syn pathway predominates in most cases. 

In papers on the Walden inversion, Ingold s group (Harvey et al., 1960) established that electrostatic forces do not contravene the normal stereochemistry of SN2 reactions. Even in the favorable case of negative nucleophile (azide) and positive substrate (sulfonium salt), exclusive inversion occurs. Obviously, the quantal forces of orbital symmetry and energy obliterate the coulombic factor here. 

The stereochemistry of Sn2 reactions has been investigated. It has been found... 

Early synthetic observations illustrated the great preference for an anti-orientation of the elimination groups . In this conformation the electron pair released from the beta carbon hydrogen bond enters the C -octet on the side remote from the leaving group. Repulsion energy between the electron pairs is thus minimised in the transition state and by analogy to the known stereochemistry of Sn2 reactions, the Cg-H electrons are most favourably disposed... 

The stereochemistry of Sn2 reactions is directly related to key features of the mechanism that we learned earlier ... 

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