# What is role of copper powder, calcium chloride and cuprous chloride in the SN1 reaction of hydrochloric acid with propargylic alcohol?

For the synthesis of a 2H-cromene from a phenol I need to prepare 3-chloro-2-methyl-but-1-yne from the corresponding alcohol 2-methylbut-3-yn-2-ol. This reaction could be seen as a simple $$S_\text N1$$ substitution.

Plausible mechanism:

However it requires a catalytic amount of copper powder, and almost half equivalent of $$\ce{CaCl2}$$ and $$\ce{CuCl}$$, in addition to concentrated $$\ce{HCl}$$ at 0°C temperature. The authors say that these additives are necessary to avoid the formation of a series of subproducts but they don't talk about mechanistic explanation about the role of these additives.

Possible subproducts:

In my opinion, the formation of subproducts derived from the eneyne, obtained from dehydration, is suppressed using low temperature.

So the additives act probably on the carbocation. We may suppose the formation of a copper acetilyde that makes the resonance structure of carbocation less important. This is corroborated by the use of concentrated HCl during the work-up. But I don't understand how it is possible to form a copper acetilyde in acid media (usually it requires the presence of a base that deprotonate the alkyne).

The autors says that the formation of subproducts is more evident for propargylic alcohol that have a low solubilty in water (this is not the case of 2-methylbut-3-yn-2-ol).Regarding the presence of $$\ce{CaCl2}$$, I know that this salt is used in the distillation of some halocompounds because it interacts in some way with alcohol. So maybe $$\ce{CaCl2}$$ helps the solubilization of alcohol.

Reference:

G. F. Hennion, A. P. Boisselle, J. Org. Chem. 1961, 26, 725–727.

My take as to the likely role for a catalytic amount of copper powder, $$\ce{CaCl2, HCl},$$ and $$\ce{CuCl}$$ is as follows:

$$\ce{ Cu + Cu(II) <=> 2 Cu(I) }$$

where the $$\ce{Cl-}$$ presence forms a soluble complex to keep the reaction reversible:

$$\ce{ CuCl + Cl- <=> CuCl2- }$$

The first equation is an example of a working so-called redox couple with an associated presence of solvated electrons $$\ce{(e-)}$$ in the alcohol medium. Further, in the presence of H+ (from the $$\ce{HCl}$$), we have:

$$\ce{ H+ + e- <=> H^.}$$

The formed hydrogen radical can further attack an organic complex resulting in either hydrogen or water with the introduction of an associated organic radical:

$$\ce{ RH + H^. <=> ^.R + H2}$$

Or:

$$\ce{ ROH + ^.H <=> ^.R + H2O}$$

At low temperatures, the above reaction is moved to the right. A minor formation reaction is also possible:

$$\ce{ R^. + ^.R -> R2 (g)}$$

where the above reaction could account for any reported associated formation of a metal acetylide.

Lastly, I suspect, presence of $$\ce{CaCl2}$$ adds $$\ce{Cl-}$$ and also removes water and further engages with the alcohol creating an addition compound.

I trust this helps.