Why do different products form when alkali halide reacts with zinc in polar protic and aprotic solvents?

Let's take an alkali halide such as $$\ce{CH3CH2Cl}$$ the two reactions in polar protic and aprotic solvents are as follows:

\begin{align} \ce{CH3CH2Cl &->[Zn, CH3OH] CH3CH3}\tag{R1} \\ \ce{2CH3CH2Cl &->[Zn, dry ether] CH3CH2CH2CH3}\tag{R2} \end{align}

What is the reason that we obtain different products in different solvents? How does the mechanism of reaction differ? Also, please correct me if I'm wrong regarding any of the following reactions, they were mentioned in my class notes.

• $\ce{CH3\color{blue}{C}H2ZnCl}$ is potentially proton-basic at the blue carbon atom. Jul 16 '21 at 12:38
• Think of the zinc acting like magnesium, forming Grignard (they're not exactly same, of course, but a fair comparison). In the first case you have a proton source, which then will form alkane. In the second, it will undergo SN2.
– TRC
Jul 16 '21 at 14:17