Session 184

In Blueprint MCAT full-length passage 3, we investigate the characteristic fragrance of Chanel N°5 and use our orgo knowledge to tackle some tricky questions.

We’re joined by Phil from Blueprint MCAT, formerly Next Step Test Prep. Sign up for their diagnostic and full-length exam, one that you get for free and some other goodies as well. If you want more full-length exams, you can go sign up over there as well.

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

[03:28] Passage 3

The characteristic fragrance of Chanel No. 5, one of the world’s most well-known perfumes, is due almost wholly to 2-Methylundecanal, a compound found naturally in kumquats. The compound exists in two enantiomeric forms (Figure 1), although professional performers note that the two enantiomers possess nearly identical scent profiles.

A chemistry class carried out an experiment to separate 2-Methylundecanal from 2-Methylundecanaic acid by carrying out a distillation of a liquid consisting solely of these two components. Due to the high boiling points of these compounds, the class was instructed to carry out a vacuum distillation. Students began by placing 200 mL of the liquid mixture in a round-bottom flask and adding several boiling chips. The solution was slowly heated with distillate collected in the receiving flask. Students noted that the round bottom flask still held approximately 50 ml of liquid at the end of the distillation.

Note: They give you the vacuum distillation setup, which is something that there’s a good chance you have messed around with in orgo lab back when people actually went to labs.

Check out the figures here on our YouTube channel (timestamp 01:33).

[07:15] Dealing with Pseudio Discrete Questions

Even though this passage is obviously going kind of like an organic chemistry route, orgo was like 5% of the entire exam, including the CARS section of the site. So it’s like it’s about 5% of the entire exam. This means it’s not that common.

Orgo questions on the MCAT tend to actually be much easier than orgo that you did in undergrad. 

They’re not going to make you draw mechanisms because it’s multiple choice and so you don’t have to draw anything.

[09:01] Question 10

Which of the carbons in 2-Methylundecanal below is/are chiral? [Check out the diagram YouTube channel (timestamp 08:03)]

(A) 1 only

(B) 2 only

(C) 1 and 2

(D) 2 and 3

[12:10] Thought process behind the correct answer:

Chiral means it’s attached to four different substituents.

And so if you’ll get carbon two that’s got this like methyl group sticking on the hydrogen, it’s got this weird long chain, and then it’s got the carbonyl on the other side. And so 2 is going to be a carbonyl center. 1 on the other hand, because of that double bond oxygen, it’s not connected to four different things and so it’s not going to be a chiral compound there. So the answer is B.

Bonus tip: The MCAT tends to leave in the dashes and wedges on drawings. The only reason you would ever leave in a dash or a wedge is because it matters. If it’s coming towards your way and the only reason that would matter is that it was a chiral compound. Anything that’s got a dash or a wedge attached to it is going to be a chiral compound and that makes it quite a bit easier to answer a question like this.

[13:56] Question 11

Another possible method of separating 2-Methyundecanal and 2-Methylundecanoic acid could be based on:

(A) Their differences in the rotation of plane polarized light

(B) Mass spectrometry analysis

(C) Extraction based on their differing solubilities

(D) The very different scent profiles of each molecule

[15:45] Thought Process

Most students look at the question and they think that it’s about the two figures that are drawn. But this is not asking about the enantiomers. This is asking about the aldehyde versus the carboxylic acid.

This is sort of a pseudo discrete question because there’s really nothing in the passage to help you with this other than knowing the aldehydes and carboxylic acids.

Enantiomers have differences in the way that they rotate light. And so a lot of students pick A because they’re thinking about those enantiomers. But that’s not what the question is asking about. Mass spec is going to show a difference between the two.

In (C) extraction, based on different solubilities, something that’s more hydrophilic is going to go to the aqueous layer. Something that’s more hydrophobic is going to go to the organic layer. It doesn’t even matter what molecules they asked about. Only one leads to separation and that’s extraction. Hence, this is the correct answer.

[19:11] Question 12

Boiling chips and vacuum distillation, respectively, are used in distillations to:

(A) provide nucleation sites that give the liquid a place to start forming bubbles to prevent superheating; lower the boiling points of the substances to be distilled.

(B) lower the boiling points of the substances to be distilled; to work synergistically with the vacuum system to further lower the boiling points.

(C) lower the boiling points of the substances to be distilled; provide nucleation sites that give the liquid a place to start forming bubbles to prevent superheating.

(D) provide nucleation sites that give the liquid a place to start forming bubbles to prevent superheating; speed up the distillation process by vacuuming the first distillate out of the apparatus.

[20:04] Thought Process

Vacuum distillation is very interesting thing because boiling occurs when the vapor pressure of the liquid is equal to the atmospheric pressure around it.

If you reduce the atmospheric pressure then it’s going to boil at a lower temperature which you have to worry about.

So if Phil and I had a race to boil an egg, Phil is going to win because I’m at a higher altitude and so I have a lower air pressure out here. As a result, my water is going to boil at a lower temperature like 99 degrees Celsius instead of 100 degrees Celsius. And Phil’s boiling water is hotter than my boiling water.

The boiling chips themselves just provide nucleation sites. The best example of this is Mentos. You throw those in there and they produce nucleation sites where the carbon dioxide can wax and start to accumulate into bubbles and then you get the big geyser of minty Coca Cola. So vacuum distillation lowers the boiling point.

[22:26] Possible MCAT Question

The MCAT might ask: why do you use and when do you use a vacuum distillation?

If something’s boiling point is really high, that’s fine, except that some molecules will start to break down. For example, 250 degrees Fahrenheit, and then they boil at 300 degrees Fahrenheit. And so if I tried to boil it, I’m going to break the molecules and this happens with organic molecules.

But if I can lower that boiling point down to 200, and the molecules break it to 50, now it’s going to boil off and it’s not going to break the molecules into chunks. So most of the time we see vacuum distillation with molecules that are organic in nature.

[24:42] Question 13

The liquid remaining in the round-bottom flask at the end of the procedure was most likely:

(A) A mixture consisting of equal amounts of the two components

(B) 2-Methylundecanal

(C) Water condensed from air in the lab

(D) 2-Methylundecanoic Acid

[25:33] Thought Process

The aldehyde and the carboxylic acid are going to be very similar, except for the aldehyde is going to have this double bond O at the end. The carboxylic acid is gonna have a double bond O with an OH at the end.

They key there is the OH. OH is going to build a hydrogen bond versus the aldehyde can’t. And so that the alcoholic group is going to hydrogen bond with other molecules, which makes it stickier. It’s going to stick to the other molecules and so that’s not going to boil as easily as the aldehyde, which is going to boil off a lot faster. So the one that’s left in the container is going to be the carboxylic acid because of its ability to hydrogen bond because it’s got that alcoholic group. Hence, the correct answer is D.

Links:

Check out the figures here on our YouTube channel.

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