# How to know when Sn2 or Sn2' reaction will take place?

The following reaction has the product as shown, as the major product $S_N2'$(allylic nucleophilic substitution) path is followed:

However, if instead of using $\ce{H3C-O-}$, we use $\ce{H3C-OH}$ we get (both $S_N2'$ and $S_N2$ paths are followed):

It seems to me that when we use $\ce{H3C-O-}$, only $S_N2'$ reaction takes place but on using $\ce{H3C-OH}$, both $S_N2'$ and $S_N2$ are equally taking place. What is the reason?

• The second set of reactions is probably $S_{\mathrm{N}}1$. In general, it's quite difficult to tell between 2 and 2'. It's going to be fairly dependent on the specific conformation of the substrates, the nature of the solvent, and a long list of other factors. – Zhe Feb 26 '18 at 13:41

In the first case , when you're using $\ce {CH_3O^-}$, it is a very hard base, and therefore a softer nucleophile. So it prefers to attack at the softer electrophilic centre. In the compound the $\ce {C}$ centre with $\ce {Cl}$ is the hard acid centre, whereas the $\beta$ positioned carbon to it i.e. the Carbon with double bond is the soft acid centre. That's why, the major product is $\ce {S_N2' }$.
But in the next case, $\ce {CH_3OH}$ is not a very hard base , nor a very soft base. So , both the reactions are possible.