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(A)
(B)
In both of the cases when a nucleophile approaches for $\ce{S_N1}$ reaction a carbocation is formed in place of $\ce{Cl}$ atom. I know that the rate of $\ce{S_N1}$ increases with increase in carbocation stability, and in (B) the carbocation will be resonance stabilized. But at the same time won't the positive charge be delocalized as well, leading to decreased effect of positive charge at the leaving group position? Won't that negatively affect $\ce{S_N2}$ reaction?
(A) Yes, rate of $\mathrm{S_N1}$ increases with the stability of the intermediate as the energy of transition state (partially positive charged at $\alpha$-$\ce{C}$) also decreases due to stabilization by delocalization.
(B) Since the rate-determining step (RDS) $\mathrm{S_N1}$ is the formation of the carbocation, the second step (which does slow down due to delocalization, but doesn't become as slow as the RDS), and thus, the rate is dependent on how fast the carbocation forms and not how fast the nucleophile approaches the positively charged intermediate; therefore, the overall reaction will speed-up.
$\ce{1-chloropropane}$ may undergo $\mathrm{S_N2}$ because the $\alpha$-$\ce{C}$ is primary, for which $\mathrm{S_N1}$ rates are usually slow. Some $\mathrm{S_N1}$ is always observed at primary $\ce{C}$ atoms as well, but rate of $\mathrm{S_N2}$ is usually much larger.
For $\ce{3-chloroprop-1-ene}$, nucleophile can approach at both the $\alpha$ and $\gamma$ positions. Here the product is the same, but it won't be the same for $\ce{4-chloroprop-2-ene}$.
