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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):enter image description here

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?

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  • $\begingroup$ 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. $\endgroup$ – Zhe Feb 26 '18 at 13:41
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The reason is the Pearson's SHAB (Soft Hard Acid Base) principle.
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.

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