Re-Cap of Orgo II

Hey everyone,

First and foremost, if you have not yet done Full-Length #1, that is your official homework for MSCT II! Do NOT forget to complete this test! When you take Full-Length #1, do so in Internet Explorer. There have occasionally been glitches reported with Chrome, Safari and Firefox. Also, don’t forget to turn off spell-check, grammar-check, and auto correct if you plan to complete the writing samples.

Let’s review your required homework for Biology II:

• Biology Review Notes Chapters 6, 7, 9, 10 & 13 (Musculoskeletal System, Digestion, Cardiovascular System, Immune System, Nervous System)
  • Comment: Yeah, it is a lot, but it’s Bio! You love Bio!
    At least, I hope you do because that’s what you’re choosing to become an expert on for the rest of your life!
• Skeletal & Immune System Workshop & Quiz
  • Comment: the portion of this workshop that reviews the skeletal system goes into a lot of depth, so I’ll let you in on a secret. The key information you really need to know about the skeletal system is on slide 8 (bone turnover). Also, the immunology component of this workshop more than makes up for the first half of the workshop. The animation on slide 22 is particularly useful. Make sure you’re comfortable with the immune system and don’t hesitate to ask me questions.

For those of you who would like some additional Biology review before class, I recommend the following:

• Biology Foundation Review Unit 2

To reinforce what we have covered in Organic Chemistry II, complete the following topical tests:

• Oxygen Containing Compounds Test 1
  • Comment: Both of the passages on this exam are testing concepts we do not review until Organic Chemistry III, thus I would recommend setting this topical test aside until then.
• Molecular Spectroscopy Test 1
  • Comment: spectroscopy is not likely to be the focus of an entire passage or question set on the MCAT, but this exam provides a neat little review of the basic concepts discussed in class and in the online workshop.
• Hydrocarbons Test 1
  • Comment: this topical test was assigned as a review assignment for Organic Chemistry I, but since I didn’t comment on this test earlier I’ll share some comments now. Question #2 is particularly nice as it tests your understanding of how kinetics and thermodynamics may not always see eye to eye. The other questions are nothing special, but the MCAT has played around with these concepts in the BS section before so make sure you can follow along with the logic on this question. Passage 2 is about free radicals and it’s no big deal. A few discrete questions are slightly out of scope in that the AAMC will no longer ask you about alkene addition reactions without a passage that describes various alkene reactions. Still, it’s good to be clear on Markovnikov v. anti-Markovnikov.
• Molecular Structure of Organic Compounds Test 1

Helpful Hints:

• Much of the reactivity discussed in class boils down to nucleophilicity and electrophilicity. Whenever presented with an organic reaction you’ve never seen before, always identify the nucleophile (which will have a concentration of electron density, in the form of a double bond, lone pair or negative charge) and the electrophile (which will be electron deficient).
• Carboxylic acids and the carboxylic acid derivatives can undergo nucleophilic acyl substitution because they have leaving groups. Aldehydes and ketones can undergo nucleophilic addition, but not substitution, because they do not have leaving groups. As long as you understand how these reactions look under acidic and basic conditions you’re in good shape.
• We are almost to the first full length practice test (the “midterm”)! Start gearing up for it by scheduling at least 5 hours of uninterrupted time between now and the MSCT II class. Don’t be afraid and don’t procrastinate – it’s important that you take the first full length now, even though you know and feel you aren’t fully prepared, so that you can assess the progress you’ve made so far and be more informed about what you need to continue to work on.
• When you take your practice test, be sure to replicate Test Day procedures and protocol. This means that you should start your practice test at the same time that your actual test will start. Take your practice test in a quiet but not isolated area, such as the Kaplan computer lab or your school’s library. Follow testing regulations regarding food, drink, calculators and cell phones. None of these are allowed into the testing room. Allow yourself the 10-minute breaks but don’t take more than the allotted time. Finally don’t forget about your computer functionalities: highlighting text and crossing out wrong answer choices. Disable autocorrect, autocapitalization and spell check when writing your writing samples. The whole point of the practice tests is to help you get into to Test Day mode and get used to Test Day procedures. Do everything that you will do on Test Day and nothing that you won’t.

• A table of C-NMR values has been attached to this email, but we can reduce this down to a few key numbers (last two are most important). Do not misconstrue this (like standing waves) to mean that C-NMR is a common or high-yield topic; it’s not!
  • Regular sp³ (0-40); if ewg is present, then slightly higher
  • Regular sp (65-90); if ewg is present, then slightly higher
  • regular sp² (100-150)
  • benzene (110-175)
  • carboxylic acids & esters (160-185)
  • aldehydes & ketones (190-220)

I’ll see you all at Biology II!

Re: Explanation for 119

I understand all of the explanation except the initial determination of peaks. Why are there 3 peaks? I thought there would be 5 peaks. What happens to the hydrogens attached to C1 and C4?
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Sorry that I didn't respond to this earlier -- I didn't realize that the question had been put as a comment on the previous one.

The hydrogens on C1 are part of the two methyl groups bound to C-2. While, in nomenclature, we would distinguish this carbon from the carbon in the methyl group on C-2, you can see that they're actually equivalent.

Please check out my unparalleled MS Paint skills that show the three groups of protons by color:

Question #119 on AAMC Practice Test #9

This is asking what the H NMR of the Compound II would look like (I tried to recreate Compound II but it didn't show up well).


The explanation provided was rather brief. Can you please provide a bit more detail? Thanks.

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This question asks us what the proton NMR of 2-bromo-2,3-dimethylbutane would look like. Just to walk through the nomenclature real quick, our longest carbon chain is four carbons. Carbon "2" would be the one that has the bromine on it, to give it the lowest possible number. This carbon also has a methyl group attached. Carbon "3" has just a methyl group.

With proton NMR, there are four things to consider:

(1) groups of equivalent hydrogens (corresponds to the number of peaks)

(2) number of hydrogens within each of these groups (corresponds to the "integration," or height of the peaks)

(3) electron-withdrawing groups and hybridization of the carbons (corresponds to "left-shifting" or "de-shielding" of the protons)

(4) splitting of peaks (corresponds to how many hydrogens are connected to the adjacent carbon).

For this question, since it's only asking us how many peaks there are, and what the splitting is, we can focus only on #1 and #4 above.

First step: how many groups of equivalent hydrogens are there? To answer this, I usually think about how I would describe the hydrogens in the molecule. Here, there are three types:

-Those that are in the methyl groups connected to carbon 2

-Those that are in the methyl groups connected to carbon 3

-The one hydrogen by itself bound to carbon 3

Automatically, this means there are three peaks. Now, let's talk about splitting.

For the first group (methyl group hydrogens on carbon 2), there will be no splitting. This is because there are no hydrogens on carbon 2. Remember that splitting will correspond to n+1, where n is the number of hydrogens on the adjacent carbon. For this peak, we end up with a singlet.

For the second group (methyl groups on carbon 3), there will be splitting caused by the lone hydrogen on carbon 3. This will split this peak into n+1 = 1+1 = 2 "subpeaks," or a doublet.

Finally, for the last carbon (the one by itself on carbon 3), it will be split by the six carbons found on the methyl groups attached to carbon 3. This will split the peak into n+1 = 6+1 = 7 "subpeaks," or a septet.

Thus, we end up with a singlet, a doublet, and a septet.

Question #50, Physical Sciences, Kaplan Full Length #4

Question #50, Physical Sciences, Kaplan Full Length #4

The question asks which factor would NOT increase the acceleration of a ball going down an inclined plane.

The correct answer is the mass of the ball. However, I didn't even read that answer because the first answer choice was the angle of the incline.

I thought that increasing the angle of incline would decrease the acceleration and decreasing the angle would increase the acceleration.

Can you please explain?

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Be careful with questions like this. Our intuition, as humans, is not particularly good when it comes to changing angles and thinking about the effects. Since it mentions acceleration in the question stem, think... force!

Three steps:

1. Make a free-body diagram.

2. Write Newton's Second Law.

3. Plug and chug (or, in this case, think about the variables they mention in the answers).

For step 1, let's think about the forces we have:

-Gravity. It will point straight down.

-The normal force, pointing 90 degrees from the incline.

-Friction, if it exists. It will point up the plane, if the object is going down the plane.

Since we don't have the forces acting in two directions only (like x/y axes), we'll have to break the gravity up into components. The force of gravity acting parallel to the plane (down the plane) will be equal to mg sinθ; the force of gravity acting perpendicular to the plane (opposite the normal force) will be equal to mg cosθ. If you don't remember why these components are what they are, make sure to check out the section on inclined planes in the Review Notes or Lesson Book.

So, let's move onto Newton's Second Law. The dimension we're concerned with here is everything parallel to the plane. This would be F = ma, where the net force is equal to mg sinθ - f (which is the force of gravity parallel to the plane, minus the friction. So mg sinθ - f = ma.

Remember that friction can also be expressed as the coefficient of friction, μ, times the normal force. We also pointed out that the normal force is equal and opposite to the component of gravity pointing perpendicular to the plane (mg cosθ). So, friction = μN = μmg cosθ.

Putting that into our equation, we have mg sinθ - μmg cosθ = ma. We can cancel out m, and have just g sinθ - μg cosθ = a. This is perfect for the question -- let's see how each variable will affect acceleration.

Starting with A (increasing the angle of inclination), what will that affect? It will affect θ throughout. As θ increases from 0 to 90 degrees, sinθ also increases (from 0 to 1); conversely, cosθ decreases (from 1 to 0). So acceleration must be increasing -- gsinθ is increasing, and we're subtracting a smaller and smaller number with μg cosθ. So increasing the angle increases the acceleration. Also, think intuitively -- the acceleration when θ=0 should be zero (it's flat on a table), and when θ=90, acceleration will be equal to g, since the ball will be in freefall then.

Now for B and C. Decreasing the friction will decrease μ, so that will increase acceleration (we'll be subtracting a smaller number). We didn't talk about air resistance, but removing that drag force should also logically increase the acceleration.

Finally, for D, note that mass actually cancelled out of our equation. Whether we increase the mass or decrease it, the acceleration shouldn't change. In reality, this scenario is very common. Mass often cancels out in kinematics equations, since most forces are dependent on the mass (gravity, for example) and we use the equation F = ma. We can often divide both sides by m once we've gotten it set up.

Question #44, Physical Sciences, Kaplan Full Length #4

Question #44, Physical Sciences, Kaplan Full Length #4

When looking at the passage and the 5 reactions given, it seems the ratio of moles of thiosulfate to oxygen would be 2:1, however that is not the case. And the explanation goes through a conversion.

I don't understand. Can you please explain?

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This question is probably easier to answer using the method you mention here -- looking at the 5 reactions they give you. The conversion they work with is tedious and long.

However, you have to be careful when you use these reactions. Reaction 2 is different from the rest -- because it has 2 Mn(OH)2 in it, and the rest only work with 1 Mn(OH)2. So, to make the reactions all equivalent, let's just multiply Reactions 1, 3, 4 and 5 by 2. This way, we can imagine that we start with 1 mol O2, and make 2 mol MnO(OH)2. Those 2 mol Mn(OH)2 make two mol Mn(SO4)2 (Reaction 3), which go on to make 2 mol I2 (Reaction 4), which finally are titrated with 4 mol S2O3(2-) (Reaction 5). Or, just jumping from Reaction 2 to Reaction 5, we have 1 mol O2 that we start with and, now that we've adjusted, we get out 4 mol thiosulfate. This gives us that 4:1 ratio they have as the answer.

This is similar to a mistake many people make in biology, actually -- in the area of cell respiration. Remember how 1 glucose makes 2 pyruvate? And then the reactions from then are are written with just 1 pyruvate? It's the same idea. We have to be consistent with the Reactions here.

Question #42, Physical Sciences, Kaplan Full Length #4

Question #42, Physical Sciences, Kaplan Full Length #4

The question asks for which half reaction would the electrode potential increase as the pH of the solution increases.

The explanation states that as the pH is increasing, the H+ is decreasing and OH- is increasing. However, the correct answer has OH- in the reactants rather than the products. I would think it would be the other way around. Can you please further explain?

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Exactly as you stated, when the pH goes up, that means [H+] is decreasing, and [OH-] is increasing (it is getting more basic). The question itself says: "For which of the following half-reactions should the electrode potential be expected to increase as the pH of the solution increases?"

What is this "electrode potential"? Think of it as a measure of how much the reaction wants to occur, similar to Gibbs free energy. Indeed, this is confirmed mathematically by the equation ΔG = -nFE, where E (the emf of the cell) is based off of the standard reduction potentials of the cathode and anode (essentially the same thing as electrode potentials). Note that if ΔG is negative (spontaneous), then E will be positive because of the sign in the equation. Therefore, a positive E will represent a more favorable reaction.

By that logic, what this question is really asking is when having additional OH- around will be "good" for the reaction - more favorable.

The reduction potentials in the passage are not written as equilibrium reactions (with double-headed arrows), but these reactions must be reversible. A Galvanic cell and an electrolytic cell with the same electrodes will be running reverse reactions (that is, if Mn is being oxidized in a Galvanic cell, it's being reduced in the electrolytic cell). Therefore, these reactions are reversible.

That (reversibility) should make you think about Le Châtelier's Principle. If having OH- is favorable, that must mean that OH- is a reactant -- it will push the reaction forward, encouraging it to occur. The only answer that shows that is D.

For answer choice C, having additional OH- around would actually encourage this reaction to reverse, and thus makes it (as written) less favorable.

Question #29, Physical Sciences, Kaplan Full Length #4

Question #29, Physical Sciences, Kaplan Full Length #4

Per the explanation of this question and the correct answer, I understand that Pipe A opens to air at ground level where the air is moving slower, therefore exerting more pressure. I understand that Pipe B opens above ground where the air is moving quicker and exerting less pressure. I don't quite understand how those 2 situations affect the underground tunnel and the direction of the air flow.

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For this question, they give us Bernoulli's equation right in the question stem. Therefore, let's consider what variables we're concerned about in the equation.

Bernoulli's equation says P + 1/2ρv² + ρgy = constant. The constants 1/2 and g, and the variable ρ aren't really something we're considering here (the density of the air, it tells us, should be assumed to be constant). So what we're actually concerned about is P, v, and y. Let's consider what these represent here. P is the pressure of the fluid, v is the velocity of the fluid, and y is the height (altitude) of the fluid. In the question, we are told that both the velocity of air and the altitude of the air are higher at Pipe B. If this is true, the only way for the sum (P + v + y) to stay constant is if the pressure at Pipe B is much lower. This should also make some intuitive sense to you because pressure goes down with increasing altitude. So, P(A) > P(B).

How does this affect the air in the tunnel? Well, remember that air (and any fluid, for that matter) will follow a pressure gradient, going from the area of high pressure to the area of low pressure. We just identified Pipe A's opening as that point of high pressure and Pipe B's opening as the point of low pressure. Thus, the air will flow down Pipe A, across the tunnel (a left-to-right direction), and up and out of Pipe B.

This is the same phenomenon as what is occurring when you have two windows open in different rooms of your apartment, and may sometimes experience an "air tunnel". The air is rushing because it's following a pressure gradient in the building.