There is another mechanistic pathway other than SN1 path. This is called the SN2 pathway.
The numeral 2 stands for a bimolecular reaction. Bimolecular reactions are reactions in which two molecules collide in the rate determing (slow) step of the reaction mechanism. If relative rate studeies were conducted in which the Organic substrate was doubled, this results in the doubling of the rate of the reaction. In addition, doubling the concentration of the nucleophile used will double the rate as well. We can say from these studies that the SN2 reaction is first order in respect to the organic substrate (alkyl halide in this discussion) and first order in respect to the concentration of the nucleophile. Overall the reaction is second order. If we were to constrct a Rate Law Expression based on these experimental results we would have the following expression:
R = k [R-X] [Nuc]
The reaction mechanism proposed that would account for these results has been suggested:
R-L + N- -------> [ N -****R***L] -----> N-R + L-
This mechanistic path is a one step concerted mechanism which is first order in respect to the concentration of the organic reactant, first order in respect to the concentration of nucleophile. The only step is the rate determining step and as such the reactants are the organic substrate and the nucleophile.
Methyl halides and primary halides will increase a reaction that is operating under an SN2 pathway. Tertiary halides have a very low (approaching zero) rate under SN2 conditions. This can be explained by understanding the way the incoming nucleophile approaches the organic substrate molecule. The nucleophile according to the above concerted mechanism approaches from the opposite side of the departing leaving group. It stands to reason that if the carbon where the substitution was taking place had more hydrocarbon structures attached to it then the nucleophile would find it more difficult in approaching the carbon on which substitution was occuring. We say that the nucleophile is being "sterically hindered" by the alkyl groups. Tertiary halides have the most alkyl groups attached to that carbon so this would depress the nucleophile from approaching the substrate. Methyl halides or primary halides would have much less sterical hindrance so such a pathway would be much easier to take place thus increasing the rate of the reaction. In summary:
Relative Rate under SN2 Pathway:
CH3-X > Primary halide>secondary halide>>>>Tertiary halide
In fact, you can forget teriary halides reacting under SN2 conditions. The rate of such a reaction would be approaching zero.
As the incoming nucleophile with its bonding electrons approaches the backside of the alkyl halide and begins to bond with the carbon, the leaving group is departing with the bonding electrons. The nucleophile will repel the bonding electrons of the groups attached to the carbon in which substitution is taking place. This has the effect of causing the groups to invert like an umbrella inverts in a strong gust of wind. This inversion also alters any configuration of the organic substrate so that the product will have the opposite configuration. So, for example, if the organic alkyl halide was "S" configured, its product will be "R" configured. We term this an inversion of configuration.
Since no carbocation is actually formed in this mechanistic pathway we would not expect racimization to take place. Therefore if we started with an optically active alkyl halide we would end up with an optically active product. So for example, if the alkyl halide was (S)(+32.5 degrees rotation), then the product would be (R) (-32.5 degrees rotation).
The solvent used in a substitution reaction will influence which pathway will be chosen. There are three major categories of solvent.
Polar Protic solvents are solvents in which at least one Hydrogen atom is bonded to either an Oxygen or an Nitrogen. This creates a polar molecule that will attract other molecules or ions using Hydrogen bonding. Hydrogen bonding interactions between particples are only possible when Hydrogen is bonded to an Oxygen as in water (H2O) or an alcohol (R-O-H) or is bonded to a Nitrogen atom as in an amine (R-NH2). Carboxylic Acids like formic acid, H-COOH, or acetic acid, CH3COOH are also described as Polar Protic solvents.
Polar Aprotic solvents are solvents whose molecules exhibit a molecular dipole moment but whose Hydrogen atoms are not bonded to an Oxygen or Nitrogen atom. Therefore no Hydrogen bonding interactions can take place between particles. Some examples of such solvents would include aldehydes, R-CHO, ketones, R-CO-R', Dimethyl Sulfoxide(DMSO), CH3-SO-CH3, Dimethyl Formamide(DMF), H-CO-N(CH3)2.
Non-Polar Aprotic solvents are solvents whose molecules have a zero molecular dipole and whose Hydrogen atoms are not bonded to an Oxygen or Nitrogen. Examples include all the Hydrocarbons(Alkanes, Alkenes, and Alkynes).
Studies have shown that rates of SN2 reactions are enhanced when Polar Aprotic solvents are used. These same solvents will depress the reaction rate of a nucleophilic substitution operating under SN1 conditions. Why is this so?
The Polar Aprotic solvents are not able to help stabilize any carbocation formation that would occur under SN1 conditions because there is no hydrogen bonding interaction possible with these solvents. On the other hand, Polar Aprotic solvents do not solvate or encapsulate the nucleophiles so they are free to approach the organic substrate unencumbered. This means that the rate of an SN2 reaction will be increased when Polar Aprotic solvents are used. Non-polar solvents are also unable to be attracted to the incoming nucleophile so that has a rate enhancing effect on an SN2 reaction.
Nucleophility has to do with how well the nucleophile approaches the organic molecule. If the nucleophile is a strong one then its ability to minimize any electron repulsion that inevitibly takes place will be enhanced. This has a tendency to lower the Free Energy of Activation of the Transition State which will make it easier for the product to be formed since the energy barrier is not as great. Strong nucleophiles will react more rapidly then nucleophiles that are described as poor or weak. Negative charged nucleophiles tend to be stronger than the neutral nucleophiles with the same central atom. So for example, OH- is stronger than H2O and RO- are better nucleophiles than R-O-H. Stronger nucleophiles will enhance the reaction rate of a reaction operating under SN2 conditions.
Increasing the concentration of the nucleophile will enhance the rate of an SN2 reaction since kinetically the reaction is first order in respect to the nucleophile concentration. This is in contrast to SN1 where the concentration and nucleophilicity have no effect on the rate of an SN1 reaction.
The leaving group ability (LGA) is defined as the ability of the leaving group to be displaced and remain stable. Leaving groups that are weak bases (coming from strong acids) are more stable after leaving the molecule and do not have any desire to return. However, the more basic the leaving group is (coming from weak acids), the poorer will be its ability to depart without wanting to return to the molecule. Some examples of leaving groups that have excellent LGA are the halides except for Flouride ion. Chloride, Bromide, and Iodide all come from strong acids, HCl, HBr, and HI. Strong acids produce weak conjugate bases which make fantastic leaving groups with high LGA. Flouride comes from HF which is a weak acid so its conjugate base, Flouride ion, would make a very poor leaving group. Other groups with low LGA would include acetate ion, Cyanide ion(CN -, and Nitrite ion, NO2-.
The reaction rate of a reaction occuring via SN2 path would increase with higher LGA for the leaving group. This is the same effect as was observed for an SN1 reaction.
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R. H. Logan, Instructor of Chemistry, Dallas County Community College District, North Lake College.
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