Dear Andrew, as you can see from the attached schematic formula drawing of cannabidiol (taken from the respective Wikipedia entry), the molecule has a variety of differently bonded hydrogen atoms. None of them will be replaced by chlorine upon treatment with lithium chloride (LiCl). Lithium chloride is a neutral salt which is unreactive with respect to organic C–H bonds. To give you a more specific answer to your interesting technical question it would help if you could specify in more detail what your goal is.

Good luck with your work!

I was inquiring to a solution of of LiCl, or LiOH, or CH3NaO in water and combined cannabidiol/ethanol mixture, then apply the electrical current (electrolysis) should cause a replacement by a lithium Ion or add. I guess the electrical current would break LiCl into Li+ and Cl- ions the Li+ mates with a hydrogen and the Cl- replace the hydrogen. Basically I want to find the easiest way to break the C-H bond and replace the hydrogen(s) of the CBD molecule. I know diatomic halogen gassed will replace at least one hydrogen on the CBD.

LiCl + COHCHC = LiH + COHCClC Reaction type: double replacement

https://www.webqc.org/balance.php

Balanced equation:

2 LiCl + COHCHC = 2 LiH + COClCClC Reaction type: double replacement

so why not? I say lets bring in the sodium methoxide and add some methoxy groups...I have a saying when it comes to offers of free food, "why not"

Here's what we're gunna dew. dehydrated ethanol, cannabidiol, sodium methoxide, ammonia, HCL and add H2O2 and Au if its possible to gold plate it. watcher eyes! maybe zap it with 240 volts blow powder all over or melt it put on ice bath etc.

Dear Andrew, thank you for your and response and further explanations. Using "dehydrated ethanol, cannabidiol, sodium methoxide, ammonia, HCL and add H2O2 and Au if its possible to gold plate it" sounds like a wonderful idea taken directly from the laboratory of an alchemist. Don't forget to apply cell phone radiation to double the reaction rate! 😎😂 However, what looks good on paper does not always work in practice. For example, you will never be able to form lithium hydride (LiH) from lithium chloride and solution. In earnest, it would be helpful if you could tell us what your actual goal is. First of all, it might be useful for you to read the following very instructive review article about the derivative chemistry of cannabidiol:

An Overview on Medicinal Chemistry of Synthetic and Natural Derivatives of Cannabidiol

Article An Overview on Medicinal Chemistry of Synthetic and Natural ...

Fortunately this article has been posted by the authors as public full text on RG. Thus you can freely download it as pdf file. On page 7 of this review article you will find a paragraph on halogenated cannabidiol derivatives. It is stated here that (citation): "The first reported halogenations occurred at the 3′ and/or 5′ positions by chlorine, bromine or iodine substitution, allowing the preparation of 3′-ClCBD, 3′,5′-diCl-CBD, 3′-Br-CBD, 3′,5′-diBr-CBD, 3′-I-CBD, and3′,5′-diI-CBD." I hope this helps.

Good luck with your work and best wishes!

Thank you prof. Frank Edwlmann, Think thermonuclear, Sir. We are dealing with ALL FINE POWDERS HERE Professor. I understand the halogenation reaction H is removed from cbd and h mates with Cl forming hydrogen chloride. the other Cl atom replaces the H on CBD. Okay lets say I add two moles LiOH anhydrous powder to CBD powder and dehydrated ethanol C2H5OH powder. Let me ask you this professor, would one amp ignight cbd powder disolved in dehydrated ethanol powder. Please share an example of when you dissolved powder in powder Pfrofessor. I suggest a tesla coil or a lightning rod for excitment, static:

190 LiOH + 53 C21H30O2C2H5OH = 29 H2O2 + 95 C2H5OHLi2 + 49 C21H30O4 what do you make of it..... Huh......Huh......:)