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SN2-SN1 Competition

In any nucleophilic substitution, the SN1 and SN2 processes compete with each other. One will prevail over the other depending on the relative energy of the respective TSs. What factors make the difference?

Factors affecting the energy of the TSs: steric effects slow down SN2; tertiary substrates and polar solvents favour SN1

SN2

Polar solvents hamper the SN2 process.

SN2 transition state with little charge development on the nucleophile and the leaving group

SN1 (carbocation)

Polar solvents favor the SN1 process.

Carbocation and SN1 transition state surrounded by stabilizing solvent molecules

The TS develops very little charge and is not especially stabilized by a polar solvent.

However, a polar solvent can strongly solvate ('shield') the nucleophile, thus hampering its attack.

The presence of good leaving groups favors the SN2 process.

The stronger the nucleophile, the faster the SN2 process.

Substitution at the reacting carbon heavily hinders the SN2 process.

The carbocation (a reactive intermediate) bears a net charge and is stabilized by solvation. In the TS the charge is already well developed (it closely resembles the carbocation), so it already benefits partially from this stabilization.

The presence of good leaving groups favors the SN1 process.

Nucleophile strength doesn't affect the rate of the SN1 process.

Substitution at the reacting carbon stabilizes the carbocation and favors the SN1 process.

As with radicals, the more substituted a carbocation, the less UNstable it is. The reason is the same. Note that we say LESS UNSTABLE and not MORE STABLE, to stress that carbocations are always unstable species.

Carbocation stability: primary < secondary < tertiary
Hyperconjugation in the ethyl (primary), isopropyl (secondary) and tert-butyl (tertiary) cations: neighbouring C–H sigma orbitals donate density to the empty p orbital

The proportion of molecules reacting through the SN1 or SN2 pathways, with the stereochemical consequences this entails (racemization or inversion, respectively), depends on their structure:

R-XSN1SN2
MethylNever occurs in solution.Very fast with good nucleophiles and good leaving groups.
PrimaryNever occurs in solution.Same as above but more difficult when the neighboring carbon is branched.
SecondarySlow. Rate increases with good leaving groups in protic solvents.Slow. Rate increases with a high nucleophile concentration in aprotic solvents.
TertiaryVery fast in polar, protic solvents and with good leaving groups.Extremely slow.
AllylicVery fast in polar, protic solvents and with good leaving groups.Slow.
BenzylicVery fast in polar, protic solvents and with good leaving groups.Slow.

Let's go back to our experiment.

(R)-2-bromooctane gave 80% alcohol. Of this, 94% was S and only 6% R.

We can now say that 88% of (S)-2-octanol was formed through an SN2 'inversion' process.

In turn, 6% of S isomer and 6% of R isomer were produced by an SN1 'racemization' process.

But... how were the olefins formed?

Result of the experiment: (R)-2-bromooctane gives 80% substitution (2-octanol with 94% inversion) and 20% elimination (E alkene 95%, Z 5% and traces of terminal olefin)