Blueprint MCAT Full-Length 1: Passage 9 – Reactions

Session 194

Passage 9 is not for the faint of heart. Follow along and listen (and watch at premed.tv) This passage dives into the heat of formation and stabilization energy.

We’re joined by Phil from Blueprint MCAT, formerly Next Step Test Prep. If you would like to follow along on YouTube, go to premed.tv.

Get your FREE copy of Blueprint MCAT’s Full-Length 1 to follow along: Go to http://medicalschoolhq.net/blueprint. In the menu, click “MCAT,” then “Free Resources.” (That’s an affiliate link, so if you end up making a purchase from Blueprint later on, I get a small commission at no extra cost to you.)

Listen to this podcast episode with the player above, or keep reading for the highlights and takeaway points.

[02:24] Passage 9 (Questions 48-52)

A prodrug must undergo chemical conversion in the body before becoming an active pharmacological agent. The metabolism can occur in the stomach before absorption takes place, in the blood, or even in the target tissue. The prodrug macetalmorade is absorbed in the duodenum and then undergoes acid-catalyzed hydrolysis of its acetal functional group to become an active metabolite.

Para substituents on aromatics can influence the relative stability of the carbocation intermediates and impact the rate of acetal hydrolysis and macetalmorade efficacy. To investigate the role of para-substitution on the stability of the macetalmorade carbocation intermediate, a chemist used isodesmic reactions to determine the relative stabilization energy (RSE) for various substituents. Isodesmic reactions are reactions in which the type of bond broken in the reactant is identical to the type of bond formed in the product. RSE provides a means of measuring the difference in the energy needed to form a cation as a result of the presence of the given substituent as shown in Figure 1. Equation 1 shows the isodesmic reactions involving macetalmorade carbocation used to calculate the RSE for a substituent (labeled X).

[04:19] Thought Process

They’re telling us that we have these reactions or we’re trying to figure out switching these different groups on the para side of our aromatic ring. It’s just a fancy way to say the opposite side of the complete 180 degree on the ring. Switching that out for different things, how stable will it make this? Will it stabilize the reaction?

We’re calculating these relative stabilization energies by switching out these different substituents. In Equation 1, this figure, they show an X. And we are pulling a hydrogen off of that version with the X and adding it to the other compound.

We’re able to do this to calculate the heat released or absorbed. And that tells us something about how stable a product is.

If it’s really stable, then it’s going to release a lot of energy. And if it’s really unstable, then we’re gonna have to put a lot of energy into it to get it there.

[05:54] Back to Passage 9

The chemist evaluated the effects of five different substituents at position X. For each compound, the heat of formation was calculated for the substituted neutral compound and cation “X” and “X+,” respectively, along with non-substituted neutral compound and cation “H” and “H+,” respectively. The change in heat of formation from “H” and “H+” and “X” and “X+” gives the energy required to deprotonate each compound, and the difference gives the RSE (Figure 1).

Figure 1 RSE based on the change in the heats of formation

The heat of formation, energy required to deprotonate, and RSE was calculated for each compound and recorded in Table 1.

Table 1 Identity of substituent X, heat of formation (J/mol), deprotonation energy (J/mol), and RSE (J/mol) for each compound

[06:36] Notes

We’re talking about the change in energy as we go “H” and “H+” and then “X” and “X+” because I’m looking back at equation one. Note that the carbon molecule on the left is losing hydrogen that’s going “X” to “X+” and the molecule on the right is going “H” and “H+.” So if we look at the the relationships between those, we can figure out.

All of this can be more or less stable. And Figure 1 helps us in conjunction with Equation 1 to understand the energy being transferred back and forth between the two forms.

In the heat of information, the energy required to deprotonate and the relative stabilization energy was calculated for each compound and recorded in Table 1. And then they give us this table. We’re probably going to have most of our questions about this Table 1 because it’s where we actually have the data up until now.

Based on the table, there are five compounds and there are different X components. Then in Equation 1, there are X on these rings. These are just the different X groups. So we have a methyl group, a hydrogen, a fluorine and an O2. But then if we look at the next three columns, those are all exactly the same. Those are probably just added there to make this passage a little bit more complex.

So we can see what happens as we go from “X” and “X+.” And if that’s going up, that means we’re going from a lower energy to a higher energy. That means it’s going to be an endothermic reaction because we’re going from a lower energy thing to a higher energy thing. And energy can’t be created or destroyed. So we must be absorbing energy from something or putting energy in. And so a positive H is going to tell it’s going to mean that we’re putting energy into this. And a negative, that’s going to mean we’re releasing energy because then we will be going in the opposite direction.

[11:00] Question 48

Which of the following cations will form most quickly from loss of a hydrogen atom?

CH3O
CH3
F
H

We have to look at the graph and see which one of these numbers is the biggest or smallest. So right off the bat, you should be able to get to 50/50 because they’re asking about forming a cation. That means we’re going to be working at the column. That’s the delta H for the X becoming an X plus so forming the cation from the natural form. So it would be the second to last column. So they’re all relatively close to each other.

The OCH3 is 181.8 and Fluorine is 191. Go for the highest and go for lowest. So the energy, the delta H, the quickest, I’m assuming the Delta is the one that requires the least. So it’s going to go faster. So it’s 181.

The ether group requires 180 joules, the fluorine is going to take 190 joules. So the one that requires the least amount of energy is going to be the fastest. So the ether, the CH3O, is the correct answer.

Correct Answer: A

[13:14] Question 49

What is the correct order of the 5 para substituents on the carbocation intermediate, if arranged from most stabilizing to least stabilizing?

F > H > CH3 > OCH3 > NO2
CH3 > NO2 > OCH3 > H > F
NO2 > OCH3 > CH3 > H > F
NO2 > CH3 > OCH3 > F > H

So we’re looking for the order. We’re trying to put these things in order of the most stable to least stable. And they’re looking for the carbocation intermediate.

The last column is where they talked about relative stabilization energy. So they’re asking us to put those in order from the most to the least stabilizing.

So A and B are out. C and D both have NO2 as the first one. And so we’re down to 50/50. The next one that we would go to negative 6.3 because that’s the next highest one. That’s OCH3. So answer choice C should be the answer without looking at anything else. We need to start with the NO2 and end of the Fluorine so C is the only viable answer.

Answer Choice: C

[15:41] Question 50

During the experiment, scientists noted that several of the reaction beakers became hot to the touch. All of the following reactions could cause this result EXCEPT:

A is the ether, B is the Fluorine, C is the methyl, and D is the NO2. Delta H for the X plus is probably what you want to look at because you’re forming it out of ingredients.

Delta H for the X plus is probably what you want to look at because you’re forming it out of ingredients.

So looking at the Delta H four X plus, all of them are hugely positive except for the NO2. And so I’m going to choose D, which is the NO2 at the end. It’s the odd one out.

Whenever you see this question, like the question saying “all of these become hot, except for one of them.” So we know three of them do one thing and one of them does something else. As we look at our table in the heat of formation for just the X, the NO2 is different. The next column in Row 2 is different. So you’re just going to pick the odd one and move on because that’s the odd one out.

Correct Answer: D

[20:23] Question 51

What effect will the addition of a fluorine substituent have on carbocation stability?

The fluorine group will be destabilizing because it is highly electronegative.
The fluorine group will be stabilizing because it is highly electronegative.
The fluorine group will be destabilizing because it has additional lone pair electrons.
The fluorine group will be stabilizing because it has additional lone pair electrons.

AB and CD are both opposites of each other so we add fluorine on the carbocation stability. And again, we know that the fluorine here was the highest for that delta H for X to X plus.

Obviously, the periodic table potentially may have to look at that to see for lone pair electrons where fluorine is. One thing we could start with is we knew that the intro to make this way more stable, and the Fluorine was on the opposite end of that spectrum.

If anything, Fluorine is going to make it less stable. And so that in and of itself, we could probably lean towards A and C because those are saying it’s destabilizing, it’s not making this more stable. B and D are both saying it’s making it more stable.

Fluorine is electronegative. It’s the most electronegative. It has a lone pair of electrons. So does the oxygen in that ether, and that was stabilizing. So it’s not destabilizing just because it has lone pairs of electrons because there’s other ones that have lone pairs of electrons and they’re not destabilizing. But the nitrogen also can’t have a lone pair with a resonance structure. So off of that Fluorine is really electronegative.

Correct Answer: A

[23:50] Question 52

The reaction to produce the “X” form of compounds 1-4 from standard-state elements will be spontaneous under what conditions?

The reactions will be spontaneous under all conditions.
The reactions will be spontaneous at high temperatures.
The reactions will be spontaneous at low temperatures.
The reactions will not be spontaneous under any conditions.

We know from the chart that delta H for X plus or actually to produce the X form, which is opposite of X plus, so we just see that the HF or just the X. So they’re negative and the NO2 is positive. And negative is more stabilizing so it’s going to release energy.

The heat of formation is releasing energy. So the products are going to be more stable than the reactants are going to be. There are two sides of things that determine whether a reaction once it becomes spontaneous. Every reaction wants to become more stable. Things want to go from less stable to more stable. The other thing is that stuff wants to break down.

If that’s a negative number, that means it’s spontaneous. If it’s a positive number, that means it’s non-spontaneous.

And so we can kind of figure this out based on the equation delta G or the Gibbs free energy is equal to delta H, the heat of formation minus the temperature times the change in entropy or disorder, the T delta S.

And so the question is asking about reactions of compounds 1-4. And the heat of formation for all those is negative. So our H is negative in all those. It doesn’t really matter what the number is because they’re not asking for a calculation here.

We have a loss and disorder as we’re building these molecules out of their their ingredients overall. And so our delta S is also going to be negative. Now, because our equation for Gibbs free energy is delta G = delta H minus T delta S, or S is negative and our H is negative. Now, because it’s H minus T delta S, the minus T delta S (because the S is negative) just becomes positive because it’s minus a negative.

So we end up with a negative number plus some temperature times another number. If we want a a like a negative delta G, which is what we’re going to have for spontaneity, we need our temperature to be a small number because we’re going to have some negative number plus temperature.

If the temperature is big, then it’s going to be a positive number. If the temperature is small, it’s gonna be a negative number.

So we want negative for a spontaneous reaction. Hence, the correct answer is C where the reaction is going to be spontaneous specifically at low temperatures.

Correct Answer: C