Blueprint MCAT Full-Length 1: Passage 6 — Organic Chemistry

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MP 190: Blueprint MCAT Full-Length 1: Passage 6 — Organic Chemistry

Session 190

Today we’re looking at organic synthesis reactions, derivatives, enzymes, and spectroscopy in passage 6 of the Blueprint MCAT full-length 1. Join us!

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

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

[02:54] Passage 6 (Questions 31-34)

Terpenes have been found to be essential building blocks of complex hormones and molecules, pigments, sterols, and even vitamins. Terpenes also play an incredibly important role by providing protection from bacteria and fungus. Terpenes have a basic structure of repeating isoprene, or (C5H8)n, units, and they are grouped according to the number of these repeating units. Monoterpenes contained 2 isoprene units; examples include menthol and pinene. Vitamin A1, a diterpene, contains 4 isoprene units. Research has revealed the Isoprene Rule, which states that adjacent isoprene units in terpenes are linked preferentially between carbon atoms located at opposite ends of the isoprene structural subunit. Head-to-head and tail-to-tail connections are rare exceptions to this rule.

Figure 1: Examples of monoterpenes and diterpenes

In human metabolism, B-carotene A, a 40-carbon terpene (Compound 1, Figure 2) is utilized as a precursor for the synthesis of vitamin A1. Retinal (Figure 2) is a key intermediate in vitamin A1 synthesis. The biosynthetic pathway by which retinal is synthesized in vivo from B-carotene has been identified by using specifically 18O-labeled O2 and 18O-labeled H2O.

Figure 2: Synthesis of vitamin A1 from B-carotene (Compound 1)

[07:30] Question 31

Which of the following reagents could be used to complete the final step of retinol synthesis shown in Figure 2?

(A) LiAIH4

(B) O3

(C) H2 with Pd

(D) KMnO4

Thought Process:

We are adding hydrogen so this is a reduction reaction. And just by looking at the picture, you can eliminate B and D because those don’t have any hydrogens.

Notice that B and D are actually oxidizing agents. And having a lot of oxygen is a good way to indicate something’s an oxidizing agent. That’s actually going to take it in the other direction going from alcohol to the aldehyde. So now, we’re looking for a reducing agent, which leaves us with A or C.

Now beyond that, it’s a little bit tricky. It does require a little bit more specific knowledge.

H2 with Pd is a very powerful reducing agent. And it’s going to turn a carbonyl or an aldehyde into alcohol. But it’s also going to add hydrogens to every single double bond and turn them into single bonds. And so that’s probably not going to work here. Because our product has 5 double bonds. And so if we used hydrogen and palladium, we’re not going to have retinol at the end of this. We’re going to have some other compound. This is something that most students should be learning in their chemistry, biochemistry, and organic chemistry kind of classes.

Hence, this would all leave us with A. A and C are both reducing agents, and they’re both going to turn this aldehyde into alcohol, but just palladium is going to go a little bit farther.

Correct Answer: A

[13:29] Don’t Give Up

Some people are saying the MCAT is really just a reasoning test and you don’t need any sort of background knowledge.

“It is a critical thinking and reasoning test but you also have to have the background knowledge.”Click To Tweet

Sometimes you have to push yourself to get enough information so you can start putting things together and making those connections. 

If you’re prepping for the MCAT right now and you feel like you’re spending so much time and your scores aren’t getting better, don’t give up and don’t think you’re not gaining ground.

The MCAT is a little bit more difficult because sometimes you have to apply data and you have to know like two or three little things in order to actually put it together.

[16:08] Question 32

Which of the following biological substances are likely derived from terpenes?

  1. Aldosterone
  2. Glucose

III. Insulin

  1. Estrogen

(A) I and II only

(B) I and III only

(C) II and III only

(D) I and IV only

Thought Process:

Glucose is obviously sugar so let’s throw that one out. Estrogen is a hormone so we can keep that. And D is the only one that has estrogen. Aldosterone is a hormone as well. So I and IV are definitely hormones. Estrogen and aldosterone are steroid hormones. Peptide hormones insulin, glucagon, ADH, adrenocorticotropic releasing hormone, TSH, and there are hundreds of them.

Correct Answer: D

[20:37] Question 33

The enzyme listed in step 1 of the retinol synthesis listed is most likely classified as a(n):

(A) transferase.

(B) lyase.

(C) isomerase.

(D) oxidoreductase.

Thought Process:

Another way to think about that is by removing a bond to oxygen. So if you remove a bond to oxygen, that’s a reduction. If you add a bond to oxygen, that’s oxidation. And so the reverse of that last step would be oxidation.

The very first step of this reaction because we have this molecule that has no oxygen on it, and then all of a sudden, it’s got a bond to oxygen in that first step. And so that’s going to be an oxidation reaction, which means this is an oxidoreductase.

With lyase, it’s going to break stuff and that’s the third reaction going on there. It’s a lyase where we cut the molecule basically in half. Isomerase is moving a carbon from one spot to another, just switching it to a different isomer of that reaction. Transferase is adding a functional group like a carboxylic acid or a phosphate or a sulfate.

Correct Answer: D

[25:24] Question 34

A researcher seeks to monitor the conversion of retinal to retinol using infrared spectroscopy. Which of the following will indicate the reaction is complete?

(A) The disappearance of peaks in the 3200-3500 cm-1 region

(B) The disappearance of a peak in the 1700-1750 cm-1 region

(C) The disappearance of a peak in the 1580-1640 cm-1 region

(D) The disappearance of a peak in the 1700-1750 cm-1 region

Thought Process:

This is definitely going to require some stuff like focusing on either losing the carbonyl, losing that aldehyde, or gaining an alcohol. And so it’s got to be one of those two.

The most obvious peak that people can identify in an IR spec is alcohol. And that’s this giant, big wide valley somewhere around like 3200 to 3500 which is what they’re going on for A. The thing as the reaction is complete, we’re not losing an alcohol, we’re gaining it.

A carbonyl has around 1700 and so after you know that, then your answer must be B.

When it comes to IRA spec, a lot of students are forced to memorize phenyl groups and sulfates and phosphates and all sorts of crazy stuff, cyano groups. But the MCAT typically doesn’t do that. The MCAT is more interested in what we see biologically.

So amines, alcohols, and carbonyls are like at the top of the list. Note that those three molecules are all found in proteins. They’re all found in DNA. And like fatty acids, they have carbonyls and alcohol but they don’t have amines.

Those three things cover a lot of biological molecules. So you want to know that an alcohol is a big wide peak around 3200 to 3500. A carbonyl is around 1700, that’s a C=O, and then amine like a CNH, that’s going to be somewhere around 3500. But it’s gonna be a really sharp peak, instead of this big wide one that we see there.

Correct Answer: B


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