Depto. Química Orgánica

Sn2 - Sn1COMPETITION 

Image Link UAM Excelencia UAM Facultad Ciencias UAM
15/07/2017
page counter
Where I am Problems and exercises Molecular panel Home
In any nucleophilic substitution, the SN1 and SN2 processes compete between them.
One will prevail over the other depending on the relative energy of the respective TS's.
What factors make the difference?
Image
SN2 Polar solvents hamper the SN2 process.
Image
SN1 (carbocation) Polar solvents favor the SN1 process.
Image
The TS doesn't develop any formal charge and won't be stabilized by a polar solvent.

On the contrary, the polar solvent will 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 - reactive intermediate - bears a net formal charge and is stabilized by solvation.
In the TS the charge is almost fully developed (it resembles closely the carbocation) and it is already stabilized by the solvent.
 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.
In the same fashion as it happened with radicals, the more substituted a carbocation, the less UNstable it is. The rationale is the same as with radicals.
ImageImage
The ratio of molecules giving reaction through SN1 or SN2 pathways - with the stereochemical implications this fact conveys (racemization or inversion, respectively) - depends on the 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 NON-protic solvents.
TertiaryVery fast in polar, protic solvents and with good leaving groups.Extremelly 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 recall our experiment!!!

(R)-2-Bromooctane rendered 80% alcohol. From it, 94% was S and only 6% R.

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

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

But,... how come the olefins?
Image