Editing SN2 reaction (section)

==Factors affecting the rate of the reaction==
The four factors that affect the rate of the reaction, in the order of decreasing importance, are:<ref>{{March6th}}</ref><ref>{{Cite journal|last1=Hamlin|first1=Trevor A.|last2=Swart|first2=Marcel|last3=Bickelhaupt|first3=F. Matthias|date=2018|title=Nucleophilic Substitution (SN2): Dependence on Nucleophile, Leaving Group, Central Atom, Substituents, and Solvent|journal=ChemPhysChem|language=en|volume=19|issue=11|pages=1315–1330|doi=10.1002/cphc.201701363|issn=1439-7641|pmc=6001448|pmid=29542853}}</ref>

===Substrate===
The substrate plays the most important part in determining the rate of the reaction. For S<sub>N</sub>2 reaction to occur more quickly, the nucleophile must easily access the sigma antibonding orbital between the central carbon and leaving group.

S<sub>N</sub>2 occurs more quickly with substrates that are more [[Steric effects|sterically accessible]] at the central carbon, i.e. those that do not have as much sterically hindering substituents nearby. Methyl and primary substrates react the fastest, followed by secondary substrates. Tertiary substrates do not react via the S<sub>N</sub>2 pathway, as the greater steric hindrance between the nucleophile and nearby groups of the substrate will leave the S<sub>N</sub>1 reaction to occur first.
[[File:Steric effects on SN2 reactivity.svg|center|750px]]

Substrates with adjacent pi C=C systems can favor both S<sub>N</sub>1 and S<sub>N</sub>2 reactions. In S<sub>N</sub>1, allylic and benzylic carbocations are stabilized by delocalizing the positive charge. In S<sub>N</sub>2, however, the [[Conjugated system|conjugation]] between the reaction centre and the adjacent pi system stabilizes the transition state. Because they destabilize the positive charge in the carbocation intermediate, electron-withdrawing groups favor the S<sub>N</sub>2 reaction. Electron-donating groups favor leaving-group displacement and are more likely to react via the S<sub>N</sub>1 pathway.<ref name="Clayden-2012" />
[[File:Benzylic chloride nucleophilic substitution.svg|center|650px]]

===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. [[Potassium tert-butoxide|''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<sup>−</sup> is a better nucleophile than water, and I<sup>−</sup> is a better nucleophile than Br<sup>−</sup> (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<sup>−</sup> would therefore be a weaker nucleophile than Br<sup>−</sup> because it is a weaker base. Verdict - A strong/anionic nucleophile always favours S<sub>N</sub>2 manner of nucleophillic substitution.

===Leaving group===
Good leaving groups on the substrate lead to faster S<sub>N</sub>2 reactions. A good leaving group must be able to stabilize the [[electron density]] that comes from breaking its bond with the carbon center. This leaving group ability trend corresponds well to the [[Acid dissociation constant|p''K''<sub>a</sub>]] of the leaving group's conjugate acid (p''K''<sub>aH</sub>); the lower its p''K''<sub>aH</sub> value, the faster the leaving group is displaced.

Leaving groups that are neutral, such as [[water]], [[alcohols]] ({{chem2|R\sOH}}), and [[amines]] ({{chem2|R\sNH2}}), are good examples because of their positive charge when bonded to the carbon center prior to nucleophilic attack. Halides ([[Chloride|{{chem2|Cl-}}]], [[Bromide|{{chem2|Br-}}]], and [[Iodide|{{chem2|I-}}]], with the exception of [[Fluoride|{{chem2|F-}}]]), serve as good anionic leaving groups because electronegativity stabilizes additional electron density; the fluoride exception is due to its strong bond to carbon.

Leaving group reactivity of alcohols can be increased with [[sulfonates]], such as [[Tosyl group|tosylate]] ({{chem2|-OTs}}), [[triflate]] ({{chem2|-OTf}}), and [[mesylate]] ({{chem2|-OMs}}). Poor leaving groups include [[hydroxide]] ({{chem2|-OH}}), [[alkoxides]] ({{chem2|-OR}}), and [[Azanide|amides]] ({{chem2|-NR2}}).
[[File:Alcohol to tosylate.svg|center|450px]]

The [[Finkelstein reaction]] is one S<sub>N</sub>2 reaction in which the leaving group can also act as a nucleophile. In this reaction, the substrate has a halogen atom exchanged with another halogen. As the negative charge is more-or-less stabilized on both halides, the reaction occurs at equilibrium.
[[File:Finkelstein reaction example.svg|center|400px]]

===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.<ref>{{Cite journal|last1=Hamlin|first1=Trevor A.|last2=van Beek|first2=Bas|last3=Wolters|first3=Lando P.|last4=Bickelhaupt|first4=F. Matthias|date=2018|title=Nucleophilic Substitution in Solution: Activation Strain Analysis of Weak and Strong Solvent Effects|journal=Chemistry – A European Journal|language=en|volume=24|issue=22|pages=5927–5938|doi=10.1002/chem.201706075|issn=1521-3765|pmc=5947303|pmid=29457865}}</ref> [[Polar aprotic solvents]], like [[tetrahydrofuran]], are better solvents for this reaction than polar [[protic solvent]]s because polar protic solvents will [[hydrogen bond]] to the nucleophile, 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<sub>N</sub>2 manner of nucleophilic substitution reaction. Examples: [[dimethylsulfoxide]], [[dimethylformamide]], [[acetone]], etc. In parallel, solvation also has a significant impact on the intrinsic strength of the nucleophile, in which strong interactions between solvent and the nucleophile, found for polar [[protic solvent]]s, furnish a weaker nucleophile. In contrast, polar aprotic solvents can only weakly interact with the nucleophile, and thus, are to a lesser extent able to reduce the strength of the nucleophile.<ref>{{cite journal |last1=Hansen |first1=Thomas |last2=Roozee |first2=Jasper C. |last3=Bickelhaupt |first3=F. Matthias |last4=Hamlin |first4=Trevor A. |title=How Solvation Influences the S N 2 versus E2 Competition |journal=The Journal of Organic Chemistry |date=4 February 2022 |volume=87 |issue=3 |pages=1805–1813 |doi=10.1021/acs.joc.1c02354|pmid=34932346 |pmc=8822482 }}</ref><ref>{{cite journal |last1=Vermeeren |first1=Pascal |last2=Hansen |first2=Thomas |last3=Jansen |first3=Paul |last4=Swart |first4=Marcel |last5=Hamlin |first5=Trevor A. |last6=Bickelhaupt |first6=F. Matthias |title=A Unified Framework for Understanding Nucleophilicity and Protophilicity in the S N 2/E2 Competition |journal=Chemistry – A European Journal |date=December 2020 |volume=26 |issue=67 |pages=15538–15548 |doi=10.1002/chem.202003831|pmid=32866336 |pmc=7756690 }}</ref>