Allylic halides have the structure H2C=CHC(X)H2

Question

b. Draw the resonance structures of the allylic cations formed by ionization of the following halides. i)

ii)

Accepted Answer

Hey everyone, let's solve this problem. It says draw the resonance forms of the olympic carbo cad ions produced as a result of ionization of the following a lily. Hey leads so a couple of things to remember about resonance forms is that they have the same connectivity, right? If they don't have the same connectivity, that would make them a totally different compound. But resonance forms have the same connectivity of atoms. The only thing that's differing is the location or the distribution of electrons. Right? And how do we know when residents is going to happen? Well, we usually see a positive or a negative charge next to a pi bond. So directly next to a carbon that's involved or an atom that's involved in a pi bond. So it could be like a nitrogen to as well. And this will create resonance I guess one or more resonance forms. Okay, so let's apply this to our compounds compound a whose name is chlor oh, metal cyclo painting, like little Pantene. Let's draw that out. And the problem mentions that the residents forms are produced as a result of ionization of this compound. So why is it becoming an ion? Well, if you learn anything about what a leaving group is, we know that chlorine is very electro negative. So it's going to be happy to leave and take these electrons from that bond with it. It will be very stable if it does that. So goodbye chlorine off you go by yourself and now let's see our resulting resonance forms. So we have our five member ring double bond that methyl group. Right? It's missing one hydrogen. So that gives our positive charge now belongs on that carbon. Okay, because it had two hydrogen and a bond to the carbon bond to the chlorine. Now it's missing that bond of the chlorine. So it only has three sigma bonds. What other resonance forms do we have? Well, we see that this positive charge is directly next to a carbon involved in a double bond. So the electrons from that double bond can jump on over and double bond with that carbon leaving our positive charge. Now on this carbon that was on the other side of the aisle, keen originally, right, because it only has one hydrogen coming off of it. So, that's three sigma bonds. So it has a plus charge. Without that. All keen. All right, That's it. For a let's move on to compound B, whose name is three chlor Oh, cyclo Pantene, three claro, cyclo Pantene panting and it's pretty similar. A five member ring with a double bond except now our chlorine is directly attached to our ring. So again, chlorine just is going to leave and stable as an ion. So that chlorine will leave have a negative charge. And then let's see what resonance forms we have for this compound. So, we have our five member ring and now that positive charge is going to be right on that carbon where the chlorine left. Okay, and again, we see a positive charge right next to the carbon that's involved in a double bond. So that double bond will swing over towards that positive charge, right? Positive or sorry, negative always goes towards the positive. And now our double bond will be with that carbon that had the positive charge before this carbon still involved in that double bond. And now the end carbon on the other side of the original double bond will now have that positive charge. Okay, so those are your resonance structures for these, a little Hey lights or rather the carbo cotton produced as a result of ionization of the olympic highlights. I hope this video is helpful and thanks for watching.

Allylic halides have the structure H2C=CHC(X)H2 b. Draw the resonance structures of the allylic cations formed by ionization of the following halides. i) ii)

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Allylic halides have the structure H2C=CHC(X)H2
b. Draw the resonance structures of the allylic cations formed by ionization of the following halides. i)
ii)