= Solution
For Reaction 1, doubling substrate concentration doubles the rate, while doubling ethanol concentration leaves it unchanged. Thus
$$
\boxed{v_1=k_1[\mathrm A]},\qquad \boxed{\text{Reaction 1: }S_\mathrm N1}.
$$
Rate-limiting ionization of the tertiary propargylic chloride gives a carbocation stabilized by substitution and propargylic/allenylic resonance. The crowded reaction center disfavors backside attack, while neutral ethanol is a weaker nucleophile than ethoxide. Ethanol capture and deprotonation occur after the rate-determining step.
For Reaction 2, independently doubling either reactant doubles the rate, and doubling both gives a fourfold increase:
$$
\boxed{v_2=k_2[\mathrm C][\mathrm{EtO^-}]},\qquad
\boxed{\text{Reaction 2: }S_\mathrm N2}.
$$
The benzylic carbon is primary and accessible, and the strong nucleophile can displace chloride in a concerted step. The reacting bond is benzylic C–Cl, not the ring C–Cl bond. These kinetic orders support the proposed mechanisms in the stated model; the rate table alone would not identify every intermediate in a more complicated reaction network.
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