`RBr` when treated with `AgCN` in a highly polar solvent gives `RNC` whereas when it is treated with `NaCN` it gives `RCN`. Explain.

Answer

As `[CN^(-)]` is an ambident nuicleophile which ahave two nucleophile which have two nucleophilic sites and can attack from either side. In a highly polar solvent, `AgCN` promotes the formation of carbocation `R^(+)`, precipitation of `AgBr`.
`R-BR+Ag+[CN^(-)]overset(underset(dot)overset(ɵ)(C) -= overset(dot)(N) to C = underset(dot)overset(dot)(N^(ɵ))(to) R + CN - + Agdot(Br) darr overset("fast")(to) R-N + -= C -`
In teh absence of such promotion by `Ag^(+)` with `Na+[CN]^(-)`, te resulting `S_(N)2` reaction is found to proceed with preferential attack on the atom in nucleophile which is more polarisable i.e C.
`NC^(-)+R-Brrarrunderset("Transition State")([NC^(delta-)...R...Br^(delta-)])rarrN-=C-R +Br^(-)`