During $\ce{S_N1}$, there is an equilibrium step where the leaving group leaves, in this case, a halide. This is the mechanism for an alkyl bromide substitution.
What $\ce{Ag+}$ allows is the precipitation $\ce{Br-}$ anions produced during the equilibrium step and therefore shift the equilibrium more to the right (the side with the carbocation) according to Le Chatelier's principle. This will promote $\ce{S_N1}$ reactions.
During $\ce{S_N1}$, there is an equilibrium step where the leaving group leaves, in this case, a halide. This is the mechanism for an alkyl bromide substitution.
What $\ce{Ag+}$ allows is the precipitation $\ce{Br-}$ anions produced during the equilibrium step and therefore shift the equilibrium more to the right (the side with the carbocation) according to Le Chatelier's principle. This will promote $\ce{S_N1}$ reactions.
In solution, Sn1 and Sn2 reactions occur pretty much simultaneously I believe. In general, for primary carbocations, Sn2 will predominate, about 1000 times the rate of sn1. Adding Ag+ may increase the rate of Sn1 but I would say not too much. Again, there are many factors to this: solvent, ksp of Ag-Hal, any other functional groups, etc. However, I think Sn2 would still be predominant for primary halides even with Ag+ added. I could be wrong though.More
During $\ce{S_N1}$, there is an equilibrium step where the leaving group leaves, in this case, a halide. This is the mechanism for an alkyl bromide substitution.
What $\ce{Ag+}$ allows is the precipitation $\ce{Br-}$ anions produced during the equilibrium step and therefore shift the equilibrium more to the right (the side with the carbocation) according to Le Chatelier's principle. This will promote $\ce{S_N1}$ reactions.
During $\ce{S_N1}$, there is an equilibrium step where the leaving group leaves, in this case, a halide. This is the mechanism for an alkyl bromide substitution.
What $\ce{Ag+}$ allows is the precipitation $\ce{Br-}$ anions produced during the equilibrium step and therefore shift the equilibrium more to the right (the side with the carbocation) according to Le Chatelier's principle. This will promote $\ce{S_N1}$ reactions.
More
VOTE
VOTE
VOTE