A reluctant first-order substrate can be forced to ionize by a...

Question

A reluctant first-order substrate can be forced to ionize by adding some silver nitrate (one of the few soluble silver salts) to the reaction. Silver ion reacts with the halogen to form a silver halide (a highly exothermic reaction), generating the cation of the alkyl group.

Give mechanisms for the following silver-promoted rearrangements.
(b)

Give mechanisms for the following silver-promoted rearrangements.

(b) Chemical reaction showing a substrate with iodine converting to an alcohol using silver nitrate and solvents.

Accepted Answer

All right. Hello everyone. So this question says that silver ion reacts with the iodine attached to an O group to form silver iodide. And the c of the oyo group give the mechanism for the following silver promoted reaction in which we have an och iodide reacting with silver nitrate and water in the presence of propane 10 to create two distinct products. Now, what I want to point out here is that one product has a hydroxy group attached stemming from water and the other has a prop boxy group attached stemming from propane one all. So lets go ahead and get started. Now what sticks out to me when considering the text of the question is the formation of a carbo car during this mechanism recall that the formation of a carbo cion is characteristic of an SN one reaction. S and one reactions are uni molecular nucleic substitution reactions that is favored when there is a weak or neutron nucleo file reacting with a highly substituted leaving group. Also because of the formation of a planar cruel cion. If the living group is on a chiral center, a reca mixture is formed. So with this in mind, let's start with the mechanism. The first thing I want to point out here is that the leaping group, in this case, iodine is actually on a primary carbon, which normally is not suited for sn one reactions because primary carbons do not make stable carbo CS. However, the primary cul Halide is essentially forced to ionize because of that silver ion. So in the first step of the mechanism, iodine is going to go ahead and react with our silver cion. And specifically, it's going to break the bond between carbon and itself to establish a bond between iodine and silver creating silver iodide. So as a result, the formation of a carbo Canion is essentially forced. Now recall also that when a carpal cion is formed during a given reaction mechanism, there's always the possibility of rearrangement if a more stable configuration is possible because a primary carbo cion is unstable rearrangement will occur. In this case, namely, an all kill shift is going to occur to create a more stable tertiary carbo cion. And so what that means is that the ring is actually going to open to include our primary carbon here. So we go from a six carbon ring to a seven carbon ring. So after our ring expansion, we end up with a seven carbon ring and a new tertiary cruel cat, I briefly stopped the recording to go ahead and draw this intermediate just to make sure that you could see it. Well, but with this in mind, we can talk about the next step of the sn one mechanism because our carbo cion has formed and has rearranged. The next step at this point is nucleic attack followed by depot nation if necessary. But I do want to point out here that we end up with two distinct products because of this. We can see that both water and propane 10 behaved as NUC files, which produced two separate products. So here let me go ahead and include both possible nuclear files. All right. So first in blue, I'm going to go ahead and showcase the oxygen atom from water attacking this tertiary carbo cat is. All right. So here once again, I pause to go ahead and draw that ring and now we have the oxygen of water attached to this tertiary carbon. Now keep in mind that because oxygen attacked as water, right? When it is attached now to this tertiary carbon, it's going to have a positive formal charge, which is why a subsequent Depot Nation step is necessary. But before I go ahead and draw out this Depot Nation step, I want to go ahead and showcase prop pen 10 reacting as the nucleo instead in the prior step. So in red, I'm going to draw another reaction arrow showcasing the attack of our tertiary carbon by the oxygen of propane 10. So as a result, we can also have oxygen from propane 10 attached to our tertiary carbon. Though in either case, oxygen is going to end up with a positive formal charge. So the final step in both cases is going to be depot nation to yield the neutral product. Now, for both cases, I'm going to use the same molecule to detonate them. For example, earlier in blue, I used water as the Nile. So in that particular step, I'm going to go ahead and use water once more to de potentate. This gives a loan pair of electrons back to oxygen and produces a neutral hydroxy group. So going back to the other possibility I can once again or I can use another molecule of propane 10 the detonate. And so in red, we end up with a neutral proxy group attached to that tertiary card. But here we have the complete mechanism showing the formation of both products shown in the beginning of this video. So I'm going to go ahead and make this slightly smaller just so you could see it better on the screen. But with that being said, if you watch this video all the way through, thank you so very much. And I hope you found this helpful.

Key Concepts

Nucleophilic Substitution Reactions

Carbocation Stability and Rearrangement

Silver Nitrate as a Reactant

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