Effect of Substrate

S_N1 Substrate Effects

The rate determining step (RDS) in S_N1 mechanism is dissociation of leaving group to form carbocation.  Thus 3^o halides and tosylates react fast by this mechanism since they form relatively stable 3^o carbocation intermediates.  Imagine what would happen if methyl bromide (CH₃Br) were to react by this mechanism.  Do you think the cation formed would be very stable?

• 3^o > 2^o > 1^o > methyl
• Nucleophile not important (it aint in RDS)
• Rate=k[R-LG]
• no stereochemistry
• Favors polar solvents

S_N2 Substrate Effects

S_N2 mechanism occurs in one step in which the nucleophile attacks the alkyl halide from the backside of the C-LG bond.  When a nucleophile attacks a methyl or primary halide it does not encounter much steric hinderance (i.e. the Nu can easily attack the carbon without bumping into too many other groups).  On the contrary if a nucleophile attempted tp attack a tertiary halide it would encounter steric hinderance.

On the contrary if a nucleophile attempted to attack a tertiary halide it would encounter severe hinderance and be pushed away.

• methyl > 1^o > 2^o > 3^o (Sterics are important here, see also β-effect)
• Nucleophile sterics and electronics important.
• Rate=k[Nu][R-LG]
• Stereochemical inversion (Walden Inversion) resulting from backside attack
• Favors polar aprotic solvents

How to determine which mechanism is occuring?

Its easy for methy, primary (1^o) and tertiary (3^o) haildes and tosylates. Methyl and 1^o halides and tosylates will undoubtly follow the S_N2 pathway since they can not form stable carbocations required for SN1 and they have least steric hinderance (favors S_N2). Tertiary (3^o) will follow S_N1 pathway since they can form stable cations and have the most steric congestion which limits nucleophilic attack in S_N2 like reactions. Secondary halides and tosylates are difficult to predict (in practice they can be problematic as well). Bare in mind that elimination reactions (E1/E2) compete with substitution and often times elimination products form as well.