[chiddler]

Hello!



For both these reactions, i'm trying to understand what the ammonium chloride does. I can imagine that the NaBH4 and LiAlH4 reduce the double bond. But for #1, wouldn't acid workup be sufficient? For #2, after an alcohol is formed from LiAlH4, how does the NH4Cl remove this alcohol?

Many thanks!

Bonus question: For #1, it can be an aldehyde instead of a ketone, right?

[Honclbrif]

Its just acting as a mild proton donor.

[Dan]

It also breaks up boronate chelates of your product by forming boron-ammonia chelates. This is a significant effect - sometimes simply adding a proton source, such as acetic acid, is not sufficient to decomplex your product. I've had this problem in synthesis before.

[chiddler]

I didn't see it as a proton donor...how obvious!

Thanks very much for the help.

[fledarmus]

Echoing Dan's remark. It is surprising how clingy the reduced oxygens can be for Al or B. For many acid- and base-sensitive compounds, NH₄Cl is very useful for breaking up these adducts. I have inadvertently isolated aluminates and boronates on more than one occasion.