Rates of Intramolecular Reactions - Online Tutor, Practice Problems & Exam Prep

Intramolecular reactions occur within the same molecule, leading to faster rates compared to intermolecular reactions due to higher collision frequency (z) and orientation factor (ρ). The rate constant (k) is influenced by activation energy (E_a), temperature (T), and the frequency factor (A = z × ρ). For example, forming 5 or 6-membered rings enhances reaction rates significantly. Increasing temperature by 10 Kelvin can double the reaction rate, demonstrating the importance of these factors in chemical kinetics.

Rates of Intramolecular Reactions Concept 1

Hey, everyone. So here we're going to take a look at the rates of intramolecular reactions. Remember, intramolecular means it happens within the same molecule. This is different from intermolecular reactions where it's two different molecules or two different compounds reacting with one another. Here we're going to recall, the rate constant is defined as so rate constant is our lowercase k here.

N equals a times, we're going to say the inverse of the natural log, 2 negative e⁄ea/rt. A here, well, this is our frequency factor. A, if we break it down even further, it equals z⁢⁣⁢⁣ρ. Z here represents our collision frequency. How often are molecules colliding with each other?

And we're going to say, in terms of the collisions, the number of times they hit each other that's z. And then we're going to say ρ here, this represents our orientation factor. Remember, for two different molecules to successfully combine together, they have to hit each other with enough force and in the right orientation. So this involves the ρ, the orientation factor. Now, here we'd say that A which is our frequency factor which is a product of these two variables, this is the number of collisions per second with proper orientation.

And then here we're going to say that the inverse of the natural log over, to the negative ea over rt, this deals with the collisions with sufficient energy. So, you have to hit each other a number of times. You have to hit each other with the proper orientation, but you also have to have enough force to have molecules connect together. These variables in this portion, we're not going to spend much time on it. We already know that ea is activation energy, r is our gas constant, and t is temperature in Kelvin.

Now, intramolecular reactions, we're going to say are faster, than analogous intermolecular reactions due to higher z and ρ. Right? So again, it's happening all within the same molecule, therefore, the possibilities of colliding are much greater because it's all on the same chain. And then here, also because it's on the same chain, there's a greater likelihood of having the proper orientation. This results in a higher z and ρ.

Right? So keep this in mind when we're talking about intramolecular reactions and we're comparing it to our rate constant, which some of you might remember from general chemistry.

Rates of Intramolecular Reactions Concept 2

In this feeder, we're going to compare intramolecular reactions to intermolecular reactions. Here we recall that intramolecular reactions that form 5 and 6 membered rings are faster than intermolecular reactions. We're going to say the presence of both reacting groups within the same molecule increases the probability of collision which, remember, is variable z. We're going to say molecules with reacting groups locked in the proper orientation, so ρ, react at very high rates. Now, here, we're going to take a look at some reactions.

These reactions and the relative rates of reaction connected to them are specific here. If I were to change the length of the chain, if I had changed what x is, then the relative rate values would change. Okay? So these numbers are unique to these reactions. Now here, notice we have x.

In this case, x is specifically this. X could be another group in another example that would change my relative rates. Alright. So for the first one, we have two different molecules interacting with each other. We know that this would come here kicking this carbonyl group up, the pi bond up to oxygen, but then coming back down kicking out the halogen.

I'm skipping that middle step to show how we make this with the x jettisoned. Now the contributing factor here, there is none. And we're going to say the relative rate because there's no contributing factor is just 1. By now, let's look and see how this number changes. So if we take a look here, we have now it all happening on the same chain so we're dealing with intramolecular reactions now.

Same thing effect would happen. This eventually gets kicked out closing in to give us this. We created a 5-membered ring which is faster than an intermolecular reaction so we know that our relative rate should be greater than 1. Now, here, we would say that the way this is set up is that this reacting oxygen and this halogen that's or this group that's leaving, this x group that's leaving, they're oriented on the same side with each other which is good. Because it's on the same chain, we say the contributing factor is z, our collision frequency factor here.

And we're going to say that collisions are greater because all happening on the same molecule. There is no pi bond here so there is free rotation. So these two groups are not locked in the position that we show here. But because it's happening on the same molecule we see that the rate will jump up. The rate now becomes 2.2×105.

So that's a huge difference in the relative rate. Now for this one, it's all on the same chain so c is a factor. And because we have a pi bond here, there's no free rotation around that double bond so these groups are locked in the position that they're in. So that means that my orientation factor would also be a contributing factor here. This comes in here, hits here, and kicks this out.

So we're on the same chain and we're locked in place. That's going to help to increase my rate even more. So here our number is 1.0×107. Again, these relative rates are the calculated rates for these specific examples. If I change what my x is, if I change the length of the chains, if I make a 6 member ring instead of a 5 member ring, these relative rates would be different.

And then finally, here in this one, we have still, we have z involved since it's on the same chain. We have ρ involved because of our locked positions. But then look, we also included temperature. We've increased the temperature by 10 Kelvin. So temperature would be a factor as well.

So this will come in here and create a 5 membered ring. Now, the rule of thumb is if you change the temperature by just 10 Kelvin, that typically doubles the rate of reaction. So we would double this number here. So it become, just multiply this by 22.0×107. So we can see as we add more and more contributing factors that our relative rate starts to skyrocket, go up much higher.

So, for the first one was intermolecular, that's why it's just a relative rate of 1, but then once things start happening on the same chain through intramolecular processes, we can see that the rate starts jumping up high. We had just our collision here and then we had collision with orientation and then collision, orientation, and temperature. The trifecta helps to give us a really high relative rate at the end. So keep this in mind how these factors together can help to drastically increase your relative rate.

Rates of Intramolecular Reactions Example 1

In this example, it asks which one of these molecules will form a lactone faster in an acidic solution. So remember, a lactone is just a cyclic ester. In all of these cases, we have a carboxylic acid and an alcohol. Remember, alcohol and carboxylic acid together help to make an ester.

A simple way to look at this is knowing that we'd have a condensation reaction when the carboxylic acid reacts with the OH group of the alcohol. In terms of simplicity, we'd say we lose water. This would be gone. And what happens next is this carbonyl carbon would form a connection to this oxygen over here. As a result, our ring would be a 5-membered ring.

The structure looks something like this: since it's composed of 1-2-3-4-5 elements. Thus, we made a five-member lactone ring. For this one, we'd lose water again.

What's left behind connects to make a ring. So counting it out as 1, 2, 3, 4, 5, 6, 7, and this one here as 1, 2, 3, 4, 5, 6. Remember, we said that when you're making a 5-membered or a 6-membered ring, they have faster rates than typical intermolecular reactions. Five-membered and six-membered rings have less angle strain involved with them, so they're the preferred sizes for rings.

Considering this, the seven-membered option is out. Between the 5 and 6, the 6-membered is better. Thinking back to concepts such as chair formations commonly seen in six-membered rings, a six-membered ring would be considered better than a five-membered ring. Therefore, we'd say option C is the best because it makes a six-membered ring, which is preferable to the five-membered ring from option A.

So, the final answer here would be C. C will form a lactone faster within an acidic solution.

Which one of the following compounds cyclizes the fastest in a basic solution?

• A. 4-bromobutan-1-ol
• B. (Z)-4-bromobut-2-en-1-ol
• C. (E)-4-bromobut-2-en-1-ol

Which of the following compounds will undergo elimination the fastest in a basic solution?

• A.
• B.
• C.