S_N2 Reaction - Factors Affecting The Rate of The Reaction

Factors Affecting The Rate of The Reaction

Four factors affect the rate of the reaction:

• Substrate. The substrate plays the most important part in determining the rate of the reaction. This is because the nucleophile attacks from the back of the substrate, thus breaking the carbon-leaving group bond and forming the carbon-nucleophile bond. Therefore, to maximise the rate of the S_N2 reaction, the back of the substrate must be as unhindered as possible. Overall, this means that methyl and primary substrates react the fastest, followed by secondary substrates. Tertiary substrates do not participate in S_N2 reactions, because of steric hindrance. Moreover, compounds like '1-chloro 1-ethene' too do not undergo nucleophillic substitution easily because the carbon to chlorine bond is said to be of partial double bond character, thus is harder to break. Another factor leading to an S_N2 reaction due to substrate involves the stability and ease by which the carbocation is formed after removing the leaving group. This means the more stable the carbocation is after removing the leaving group, the more likely it is an S_N1 reaction will occur instead of an S_N2. Among the stabilization methods to be considered are: resonance stabilization, hyper-conjugative stabilization, inductive effect stabilization, or the formation of an aromatic ring molecule (as in the case of 7-chloro cyclohept-1, 3, 5-triene, as it will form a tropolium carbocation which is aromatic).
• Nucleophile. Like the substrate, steric hindrance affects the nucleophile's strength. The methoxide anion, for example, is both a strong base and nucleophile because it is a methyl nucleophile, and is thus very much unhindered. tert-Butoxide, on the other hand, is a strong base, but a poor nucleophile, because of its three methyl groups hindering its approach to the carbon. Nucleophile strength is also affected by charge and electronegativity: nucleophilicity increases with increasing negative charge and decreasing electronegativity. For example, OH- is a better nucleophile than water, and I- is a better nucleophile than Br- (in polar protic solvents). In a polar aprotic solvent, nucleophilicity increases up a column of the periodic table as there is no hydrogen bonding between the solvent and nucleophile; in this case nucleophilicity mirrors basicity. I- would therefore be a weaker nucleophile than Br- because it is a weaker base. Verdict - A strong/anionic nucleophile always favours S_N2 manner of nucleophillic substitution.

• Solvent. The solvent affects the rate of reaction because solvents may or may not surround a nucleophile, thus hindering or not hindering its approach to the carbon atom. Polar aprotic solvents, like tetrahydrofuran, are better solvents for this reaction than polar protic solvents because polar protic solvents will be solvated by the solvent hydrogen bonding to the nucleophile and thus hindering it from attacking the carbon with the leaving group. A polar aprotic solvent with low dielectric constant or a hindered dipole end will favour S_N2 manner of nucleophillic substitution reaction. Examples-DMSO,DMF,acetone etc. In polar aprotic solvent, nucleophilicity parallels basicity.

• Leaving group. The leaving group affects the rate of reaction because the more stable it is, the more likely that it will take the two electrons of its carbon-leaving group bond with it when the nucleophile attacks the carbon. Therefore, the weaker the leaving group is as a conjugate base, and thus the stronger its corresponding acid, the better the leaving group. Examples of good leaving groups are therefore the halides (except fluoride) and tosylate, whereas HO- and H₂N- are not.