(a) Rank the following carbocations from least stable to most ...

Question

(a) Rank the following carbocations from least stable to most stable. (b) Which would you expect to form first? (c) Which would you expect to react most quickly with a bromide ion (Br^⁻) ? [Carbocations, like radicals, are electron deficient.]

Four carbocation structures are shown, each marked with a positive charge, illustrating varying stability.

Accepted Answer

All right. Hi, everyone. So for this question, we have five carbo cat ions shown on the screen and the question itself has three parts. Part A says to arrange the following carbo CAD ions in decreasing order of stability. Part B says to determine which carbo CAD ion is likely to form first. And part C says which carbo CAD ion is expected to go the fastest reaction with a bromide ion. So let's begin by talking about carbo ion stability, right? And first off recall that um carbos are stabilized by hyper conjugation, right? And that essentially is a de localization of the charge of the carbo cat. So the more substituted the carbo ion, meaning the more our groups are attached to it, the more stable it's going to be right. Therefore, tertiary carbo CAD ions are the most stable, followed by secondary, followed by primary carbo CAD ions. And then the methyl carbo CAD ions are even worse than the primary ones. So with that being said, right, the reactivity of the Carbocaine itself follows a similar trend because the more stable the Carbocaine is the more reactive it is and therefore the faster the reaction proceeds. So to answer parts B and C, right, part B and C are asking about the carbo CAD ion that forms the fastest implying the most stable carbo CAD ion. And part C is also asking which carbo CAD ion undergoes the fastest reaction. So for both B and C, the answer should be the most stable Carbocaine, right? Because the more stable it is the more reactive and therefore the faster the reaction is going to proceed. So now let's look at our five carbo cat ions, right? Recall that when determining whether or not a Carbocaine, just like any other carbon is secondary tertiary, et cetera, we have to consider how many other carbons it's attached to, right. So for molecule A let's look at how many other carbons are attached to the carbon that has the positive charge. And here what you'll notice is that it has two, which means that it is a secondary carbo Kata. Now compound B on the other hand, has no carbons bound to it. So this is a methyl Carbocaine compound C is also secondary compound D is primary because it only has one other carbon attached to it and compound E therefore is tertiary. So now we can actually go ahead and answer part eight, we can answer part A by arranging these carbo catt ions from the most stable to the least stable. And we can follow our trend that we discussed before, right, our tertiary carbo catt ions are going to be the most stable, therefore, compound E wins, right? It's the most stable carbo CD ion. And it is the one that I'm going to write down first. But I need to pause here, right? Because notice how there's kind of a tie, so to speak, but not really between compounds A and C because they're both secondary but to break that tie, we have to consider the fact that compound A, the carbon that has the positive charge on compound A is in conjugation with a pi bond, right, with an all. And when I say it's in conjugation, what I mean to say is that the all keen and the positive charge are separated by one sigma bond. Now that sigma bond is really important because of resonance, which I'm going to draw out on the bottom here. So I'm going to scroll down a little bit because I'm going to actually redraw compound A and like I said before, the reason why that double bond matters is because resonance becomes possible, right. So what you'll notice is that the pi bond can go ahead and shift towards that carbo ion and have a resonance structure. So it is that resonance structure that is actually going to break the tie between compounds A and C because C has no such pi bonds and therefore resonance is not possible. So even though compounds A and C are both secondary right, compound a wince because it is further stabilized by residents. And now compound C is going to be next on the list followed by my primary carbo CAD ion compound D. And lastly my methyl carbo CAD ion compound B. So now like I said before, right, the answer to parts B and C are both going to be the most stable Carbocaine, right? Because like I said before, the more stable it is the more quickly it forms and the faster the reaction with it proceeds. So because compound E is tertiary, right? It is going to be my answer for both. Excuse me, I meant to write compound E let me fix that compound E and then compound E is going to once again be the answer to part C. So there you have it right. The first thing we did is we rearranged our carbo cat ions in order of stability. And then from there, we could deduce the answers to both B and C. So if you stuck around to the end, thank you so much for watching. And I hope you found this helpful.

(a) Rank the following carbocations from least stable to most stable. (b) Which would you expect to form first? (c) Which would you expect to react most quickly with a bromide ion (Br^⁻) ? [Carbocations, like radicals, are electron deficient.]

Key Concepts

Carbocation Stability

Reaction Mechanism

Nucleophilic Attack

Watch next

(a) Rank the following carbocations from least stable to most stable. (b) Which would you expect to form first? (c) Which would you expect to react most quickly with a bromide ion (Br⁻) ? [Carbocations, like radicals, are electron deficient.]

Key Concepts

Carbocation Stability

Reaction Mechanism

Nucleophilic Attack

Watch next

(a) Rank the following carbocations from least stable to most stable. (b) Which would you expect to form first? (c) Which would you expect to react most quickly with a bromide ion (Br^⁻) ? [Carbocations, like radicals, are electron deficient.]