ISSUE:

German, as you are studying your page on the Hoffman elimination in amines to form less substituted alkenes, analyzing the following stages, I have some questions to ask you.

Stage 2. Treatment with aqueous silver oxide. Basic salt that forms an ammonium hydroxide, precipitating the iodide in the form of silver iodide.

At this stage I imagine that by precipitating the silver iodide the reaction is favored by Le Chatelier's principle for solutions and in this case it forms the solution, the questions are: a) why do you speak
of a basic salt?
b) How ionized is ammonium hydroxide?

Step 3. Bimolecular elimination.

Ammonium hydroxide undergoes an E2 on heating, which gives rise to the alkene. At this stage, regardless of whether an E2 is carried out, it is an endothermic reaction, to which energy must be supplied to carry it out. I imagine that the energy breaks the ionic attractions between the hydroxide and the ammonium and this would provide the possibility for the hydroxide to attack the hydrogen of the ammonium ion. What I don't understand is why you say that it attacks ammonium hydroxide if it is ionized and even more so if it is heated.

Well, I don't know if I have explained it properly, we could look at some organic chemistry book. But I mean that Ag ₂ O is a salt that has a basic character, when dissolved in water it generates hydroxide ions, which form ammonium hydroxide with ammonium iodide, by displacement of iodide, which precipitates with silver.

Ammonium hydroxide undergoes elimination by heating in the third stage of the reaction. Greetings

1. That is one of the doubts. Ag2O is not an oxide?
2. The other doubt is that the hydroxide attacks the ion because it seems to be quite ionized and more so if it is heated?

If it is silver oxide, it generates hydroxides in solution. In the elimination, the hydroxide ion removes hydrogens from the neighboring carbon to the one with the nitrogen and the nitrogen leaves as a leaving group, an E2-type mechanism.