For each pair of ions, determine which ion is more stable. Use re...

Question

For each pair of ions, determine which ion is more stable. Use resonance forms to explain your answers. e.

or

f.

or

Accepted Answer

Hey everyone, Let's do this problem. It says determine which ion is more stable for each pair. Use resonance forms to explain your answers. So the question gives us a hint that we want to relate resonance forms to stability. So how can we explain stability with resonance? Well, their relationship is if there is more resonance more resonance structures, the ion will be more stable. Right? Because a charge can be de localized more over an ion. Okay, so this is what we are looking at. So we're basically seeing which ion in each pair has more resonance. So let's look at set one. So, compound A. I'm going to redraw it first. And how do we look for if residents will occur? We know that we have a pi bond pi electrons right in a double bond or a pi bond next to a charge. Right, well here I see a pipe on, right next to a charge. So these electrons can move towards that positive carbon to help give it some electrons give it some charge. So now our double bond moves to the group, the substitution group and our positive charge moves to this carbon that was in the double bond and now it is no longer in the double bond. Right. And there are no more pi bonds over here. So we can't further de localize the electrons. So that is it for module A Let's look at B. So sort of a similar structure. Six member ring methyl group. Double bond down here in a CH two plus over here. Okay, so the pi bond here in this structure is not right next to the charge anymore. It's there's a carbon in between this carbon, this positive charge carbon group. So these electrons can't move over towards that positive charge. So there are actually no further resonance forms here. So, going back to what we said about stability with resonance structures, our compound A has more resonance structures. So this one is going to be more stable. Right, does that make sense? Okay, let's apply the same thing to our next set set to compound C redraw metal group next to that ketone, an alc in over here in a negative charge at this carbon. Which means it's going to have a lone pair. Right, okay, so I see a negative charge right next to a pi bond. We know that's a formula for residents. So these electrons can swing down in between these two carbons which would move the electrons from this carbon in this pi bond onto the next carbon giving us. So now the negative charge has moved over here and we have a double bond right there and now again we have a negative charge next to a pi bond. So it just has to be one carbon away. Right, the next carbon has a pi bond. So those electrons can swing down these electrons can swing up onto that electro negative oxygen. So now we have a double bond here in our oxygen will have a negative charge with extra lone pairs. And that's as far as we can go there. Right, okay, that's that's some decent residents there. Let's see if structure D can top that. So, we have again a six member ring double on here, negative charge here. Which means there's a lone pair and similar ketone in metal. Okay, so now the electrons are going down this way, right, The lone pairs on this carbon on the next carbon, there's a pi bond attached to it so the electrons can swing down here. These electrons can go onto the next carbon giving us this structure. But now there is not another pi bond on the next carbon. Right, It's two carbons away. So we can't have any more resonance structures other than these two. So, since compound C has three resonance structures, while compound D only has two Compound C will be the more stable ion in this set. I hope this video is helpful. And thanks for watching

For each pair of ions, determine which ion is more stable. Use resonance forms to explain your answers. e. or f. or

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For each pair of ions, determine which ion is more stable. Use resonance forms to explain your answers. e.
or
f.
or