[AN.]

I don't get the difference between the two, and how do I figure out which one is the rate determining step.

Also I know that large base favors E1 mechanism, how can I tell whether a base is large base or not?

 

Thanks for the Help

[hellbraker]

SN1 mecahnisms are uni-molecular(there is only 1 molecule that is rate determining).SN1 mechanisms normally prefer tertiary or secondary substituted compounds.In SNI there is the formation of a carbocation and is therefore a two step process.Here doubling the Nucleophile has no effect on the rection,but the rate solely depends on the concentration of the substrate.
 SN2 mechanism are bi-molecular( Bimolecular reactions are reactions in which two molecules collide in the rate determing (slow) step of the reaction mechanism) and a one step process.Here if the  Organic substrate was doubled, this will results in the doubling of the rate of the reaction. In addition, doubling the concentration of the nucleophile used will double the rate as well.

In SNI the fromation of the carbocation is slow therefore,it`s the rate determiming step.
In SN2  since two reacting species are involved and is slow,it`s the rate determining step.

[Cuttlefish]

An example of each might be helpful.

S_N1 (unimolecular nucleophilic substitution):
 
   (CH₃)₃CBr ---> (CH₃)₃C⁺ + Br^-
                                                                          
This is the slow (i.e. rate determining) step.
 
   (CH₃)₃C⁺ + HO^- ---> (CH₃)₃COH
 
This is the fast step. It does not affect the rate and so the overall rate equation is
 
   Rate = k[(CH₃)₃CBr]

S_N2 (bimolecular nucleophilic substitution):

   H₃CBr + HO^- ---> H₃COH + Br^-

There is only one step in this reaction, but it involves both species (hence "bimolecular"). The rate equation is therefore

   Rate = k[H₃CBr][HO^-]
  
Also note that the tertiary alkyl bromide reacts via an S_N1 mechanism because the carbocation formed when the bromide leaves is stable, whereas the methyl bromide prefers the S_N2 mechanism because the methyl carbocation is very unstable and will not form.

[IITian]

Guidebook to Mechanism in Organic Chemistry by Peter Sykes

[ultrashogun]

The difference is that in Sn2 the leaving group leaves in the same instant in which the nucleophile attacks, while in the Sn1 reaction the leaving group leaves first, then the high energy carbocation is attacked by the nucleophile. Sn2 only has one step, so there is no question which might be the rate determining step. Sn2 has two steps and one involves the formation of a high energy molecule(carbocation) while the other involves the neutralization of the carbocation. The formation of the carbocation is obviously the slow step, so that is the rate determining step and because that step involves only the substrate only it appears in the reactions kinetic law.

[Sev]

... Sn2 has two steps and one involves the formation of a high energy molecule(carbocation) while the other involves the neutralization of the carbocation.

Typo - S_N1 has 2 steps