Chemistry - Why is SN2 favored greatly over SN2' in this reaction?

Solution 1:

While pi-based resonance effects are frequently important, you must always consider inductive effects in every reaction.

Electrostatically, there is a much higher positive charge on the ipso carbon than the beta carbon, because of inductive effects. The net effect is that there is a lower energy of activation for nucleophile attack at the electrophilic center because the coloumbic forces are more favorable for attack there. The softer the nucleophile and electrophile become, that less important this effect is, but it is still quite considerable.

Without doing the QM calcs, I would rate the phosphonium electrophile as quite hard.

Solution 2:

The $\mathrm{S_N2'}$ pathway has a strict requirement (similar to $\mathrm{E2}$) for the correct alignment of the orbitals involved in the reaction, in this case, the alkene pi-system and the C-X anti-bonding orbital (in this case C-O anti-bonding orbital). That means that only particular conformation(s) of the electrophile will be reactive.

In the direct $\mathrm{S_N2}$ attack of the electrophile, while certain conformations will be more reactive than others, the C-X is always oriented correctly for reaction to occur.

I suspect that the product distribution reflects the competition between the steric hindrance of the direct $\mathrm{S_N2}$ attack and the conformational requirements of the $\mathrm{S_N2'}$. The reaction given in the question uses a relatively unhindered electrophile and a small nucleophile, which likely explains the preference for the $\mathrm{S_N2}$ product.