For each set of reactive intermediates, rank them in order of ...

Question

For each set of reactive intermediates, rank them in order of reactivity (1 = most reactive).

(c) Structural representations of reactive intermediates for ranking carbocation stability and reactivity.

Accepted Answer

Hi, everyone. Welcome back. Our next question says rank the following reactive intermediates. In order of increasing reactivity, we have four different molecules labeled one through four. They're all carb carb anions. So all have a negative charge on one of their carbons. Number one is a four carbon chain that ends in a carbon carbon double bond. If our cc bond is on carbons, one and two, the negative charges on carbon three, number two is a four atom chain, but instead of four carbons, it's three carbons and then a carbonel group. So on the last one. So IC O double bond to finish off the chain. If our carbon carbon is carbon one, the negative charges on carbon two molecule three is another four carbon chain with all single bonds and the negative charge on carbon two. And number four is a two carbon chain with a negative charge on one of those carbons. And then it's indicated specifically that there is a hydrogen rather than a carbon attached on the other end of that negatively charged carbon. And then our answer choices A through D are just different orders of these four molecules ranked by increasing reactivity shown by lesser than science. So we need to think about what would make a carbon more reactive. Well, always remember in chemistry that more reactive equals less stable. So if we want to put in order of increasing reactivity, that's the same thing as decreasing stability because it's often a bit easier to think in terms of what is a more stable arrangement. So we want to think about what makes a carb anion more stable recall, carbon doesn't really like to have a negative charge at all. Um When we think about the arrangements here, all of these are carbons that are sp three hybridized. So we end up in a carbon and we have a carbon with a lone pair and three bonds. So it's got three bonds. So we'll say r our prime R two prime and then a lone pair in the fourth. And so there's a negative charge since it now has five valence electrons there, although it does have a complete octet. So let's think about what would stabilize this arrangement a bit more. Well, anything that would help spread out that negative charge, it would delocalized it again, carbon doesn't like this. So if we could delocalized that negative charge a bit, that would be good. How would I do that with resonance? So de localization of the negative charge through resonance will make it more stable. Also the inductive effect, having nearby electron withdrawing elements or groups will also stabilize that negative charge. So let's look at our molecules and see what we have here. Well, right off the bat, we can see that we have one molecule that has an electron withdrawing element and that is number two. So oxygen is electron and we have two molecules, one and two that have an adjacent pi bond. So they can participate in resonance. So let's look at that. So at number one, we have our four carbons with that double bond and then a loan extra loan pair on the carbon adjacent to the double bond, there's our negative charge so that lone pair can come over and form a double bond between its carbon and the next one. And then the electrons in the double bond can come and put a negative charge on the terminal carbon there. So that creates a resonance structure. And so we have that extra, those extra electrons that extra negative charge spread out over a larger part of the molecule rather than just stuck on the one carbon. So we've got resonance structures for number one. And then number two, in an analogous way, we have that negatively charged carbon with its lone pair adjacent to a carbon that's in a carbon bond. So again, our electrons from our negatively charged carbon can come down and form a double bond to the next carbon. And the electrons from that double bond can hop up onto the oxygen oops we end up adding put the bracket in too soon have to draw the other resonance structure before I add a bracket. So then we have this resonance structure here where the carbon is no longer negatively charged. We've got a double bond between the two middle carbons, a single bond to oxygen. And now the oxygen's carrying the negative charge. Well, hopefully you can see here that this resonance structure for number two will be even more stable because now we've displaced that negative charge onto an oxygen which is way happier to have it, its electron withdrawing. So in number two, we have not only the de localization of this resonance structure, but we have an electron withdrawing element nearby. So number two will be the most stable of all. So we'll put most stable. And again, we want to go in decreasing stability. So we expect number two, the most stable to be a first in our order. So let's look for our answer choice that has number two first. A has number one. So that's incorrect. B has number two C has number three. So that's incorrect. And choice D has number four first. So that's incorrect. So on a test times of the essence, I don't have to go any further. I've eliminated everything. But choice B I don't even have to look at my other molecules, but we're just going to be thorough here and just look at the other ones really quick and double check this order here. So we've seen that number two is the most stable. Now, recall, we've got these less than signs because our original question said in order of increasing reactivity. So we're saying that number two is the least reactive, it's less than reactive than number one. So don't get thrown by that, we just keep referring to increasing reactivity means decreasing stability. So we have number two and then number one is next as it has a resonant structure, but no electron withdrawing atom, then we'll look at our two molecules that have only single bonds. Number three and number four. Well, what's the distinction there? Well, number three, our negative charge is on a secondary carbon. So secondary Carmen, how is the negative charge? And number four, our carbon, we only have two carbons and our negative charge is on the terminal carbon. There's only a hydrogen attached on the other side. So there a primary carbon has the negative charge. Well, we're often used to thinking about things like carbon anions or radicals where the more substituted carbon is more stable. This is exactly the opposite effect when we're talking about a carb anion for carbon and iron LQ groups or more specifically adjacent alkyl groups are destabilizing because they donate electron density. So don't forget that alkyl groups tend to be electron donors. They are pretty happy with partially positive charges on them. So we've taken something that already has more electrons than it wants. And we're sticking more electron density there. So the more alkyl groups attached to a carbon are negatively charged carbon, the less stable it is. So number three, which is secondary will be less stable than number one. Number four, excuse me, that's primary. So our next molecule since we're going in order of decreasing stability will be molecule four, since it's a primary carb anion and therefore more stable and finally, the least stable and therefore most reactive is molecule three which is a secondary carbonite. So again, when we look at these intermediates in our order of increasing reactivity, we have choice B being our answer two is less reactive than one is less reactive than four which is less reactive than three. See you in the next video.

For each set of reactive intermediates, rank them in order of reactivity (1 = most reactive).
(c)

Key Concepts

Reactive Intermediates

Stability of Intermediates

Reactivity Trends

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