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MCAT Content / AAMC MCAT Practice Exam 3 Cp Solutions


Exam 3 Chemistry/Physics Section Passage 1

1) The best way to approach this question is to take what the passage says about the corresponding steps and try to visualize them. It mentions that Compound 3 undergoes an aldol condensation to cyclize, another word for forming a ring. It then goes on to say that the ring undergoes a dehydration reaction to form the two products 4a and 4b. Let’s follow the numbered carbons and determine which one corresponds to the carbon with the asterisk.

  1. C1 is directly attached to C4 and is the carbon to the right of the asterisk.
  2. C2 is indeed the carbon with the asterisk. C3 serves as the nucleophile in the aldol condensation reaction and the carbon with the asterisk is two carbons away (up and to the right) from the nucleophilic alpha carbon.
  3. C3 is the alpha carbon that serves as a nucleophile during the intramolecular cyclization. It should attach to the original five-carbon ring and is below and to the right of the carbon with the asterisk.
  4. C4 was part of the initial five-carbon ring and remains a part of the fused rings structure. The correct answer is C2 as it is the carbon represented by the asterisk.

2) The passage states that upon dehydration of the cyclized product, a thermodynamic mixture of products is formed. Recall that there are two major factors influencing which products of a reaction are preferentially formed: thermodynamics and kinetics. Thermodynamic control will favor energetically stable products even if they require a higher activation energy and more time to produce. Kinetic control will favor products that form more quickly; these products will often be less energetically stable but are formed faster due to lower activation energies.

According to Scheme A, 60% of the product is compound 4a and 40% is 4b.

  1. If both products were equally stable, thermodynamic control would produce roughly equal amounts of 4a and 4b. There is more 4a produced than 4b, so 4a should be more stable.
  2. For this to be true, the amount of 4a produced would need to be less than the amount of 4b produced. This is the opposite of what we see in Scheme A.
  3. This is true; compound 4a must be more stable than 4b because more 4a is produced (60% > 40%) and we are told that a thermodynamic mixture is produced indicating thermodynamic control. Again, thermodynamic control of a reaction will favor the energetically stable product.
  4. This is incorrect. As noted above, relative stability can be inferred because we are told in the passage that the products produced are a thermodynamic mixture. Answer choice C is the best answer.

3) To answer this question, we can look back at a modified version of the schematic we drew for the first question. We know that step 2 involves the aldol condensation followed by a dehydration reaction. If the dehydration reaction does not follow the aldol condensation, the cyclized product will be the major product.

Note that being able to follow and push electrons for common reactions is an important skill to have when test day comes along. Organic chemistry can be really difficult for many of us, but understanding the basic mechanisms and why a reaction occurs allows you to apply that knowledge more broadly. For instance: nucleophiles attack at electrophilic sites, and carbonyl carbons are electrophilic because the oxygen pulls on the carbon’s electrons. The alpha carbon of a carbonyl on the other hand, can serve as a nucleophile when it is made to carry a double bond. To form a ring from compound 2, have the alpha-carbon nucleophile on the first carbonyl attack the second electrophilic carbonyl (see the second image of the general aldol condensation).

  1. This structure looks like a neater version of what we drew above as the product of the aldol condensation reaction prior to dehydration. Looking good!
  2. If you were tempted by this answer choice, it might be worth counting how many bonds the carbon with the alcohol attached has. That carbon has five bonds, one too many! Don’t let the MCAT test writers trick you with a sneaky number of bonds.
  3. This answer choice has a few suspicious alterations. First, the carbonyl that should have been preserved during the nucleophilic attack has been reduced to an alcohol. Also, the second oxygen from the electrophilic carbonyl is missing! This answer choice is too different from the intermediate we should see.
  4. This is almost the final product, but like answer choice C, it has a reduced oxygen on the lower left corner where there should be a carbonyl. It also has a double bond where the alcohol should be located. This means that it has undergone a dehydration reaction, the opposite of what the question stem asks for. We’ll stick with the structure in answer A.

4) Before we draw out the new reaction, let’s work through this question verbally, which some of you may do on test day. Compound X has an additional methyl group attached to the “free end” of the double bond. Wherever this end is located on the original product, there should be an extra carbon/methyl group.

  1. This correctly depicts the “new” methyl group (shown in red above) at the end of the original double bond (shown in blue).
  2. The “extra” methyl group here would correspond to the other carbon on the double bond, the one noted by the green dot above.
  3. There is an extra carbon between the original ring and the methyl group on compound 1. Compound X is replacing compound 2, not compound 1. There should not be any changes to the original ring/compound 1.
  4. This answer choice is incorrect because it has the extra methyl group on the wrong side of the carbonyl, indicating that it would have been attached to the nucleophilic alpha carbon of compound 2. Compound X shows the methyl group on the beta carbon, not the nucleophilic alpha carbon. Answer choice A has the correct configuration.


Exam 3 Chemistry/Physics Section Passage 2

5) This question is meant to be relatively straightforward and highlights the importance of having chemistry and especially physics equations memorized and ready to go on test day. If you have not already started working on a master equations sheet, go ahead and add that to your “To Do” list.

Q=VC where Q is the charge stored, and both the voltage and capacitance are given at the end of the first paragraph:

Q= VC = 12V * 100mF = 12V * 100*10-3F = 1,200*10-3 C=1.2 C

U=½ CV2 where U is the energy stored, and both the voltage and capacitance are given at the end of the first paragraph

U=½ CV2 = ½ * 100mF * 12V2 = 50*10-3F *144V2=7,200*10-3 J = 7.2 J

  1. The charge in this answer choice is off by a single decimal place and the energy calculated in this answer is the result of not multiplying the ½ in the equation for energy.
  2. This is the correct pairing; see above for the math.
  3. This answer choice reverses the values for charge and energy, fails to halve the energy and is a decimal place off in each answer. Make sure to thoroughly perform all calculations and be very careful with exponents and moving decimals.
  4. This answer choice has the correct numerical values paired with the wrong measure. Answer choice B is the correct answer.

6) To answer this question, we’ll need to understand the Doppler effect and the effect of relative motion on perceived frequency. Most of us can conceptualize this by imagining an ambulance as the source of the sound/horn.

A. As you get closer to the horn/the ambulance, the frequency you hear is higher than the actual frequency, and once you’re moving away from the horn/the ambulance, the frequency you hear is lower than the true frequency. This answer choice corresponds to what we actually experience and is therefore the correct answer.

B. This is the opposite of what we see above and is incorrect. As we approach the stationary source of the sound, perceived frequency increases and is greater than the actual frequency. As we move away from the stationary source of the sound, perceived frequency drops below the true frequency.

C. The perceived frequency is equal to the actual frequency when there is no relative motion.

D. While f should be greater than f before passing the horn, it should not remain greater after passing the horn. Answer choice A is the most appropriate application of the Doppler effect.

7) P=Force*velocity=Work/time

In the absence of braking, the only force opposing motion is the one mentioned in the last sentence; there is a continuous decelerating force that is present any time the railcar is moving. This decelerating force is 1,000N.

P=Force*velocity= 1,000N * 40m/s = 40,000 W = 40kW

  1. This answer choice is too small and likely results from an arithmetic error.
  2. As shown in the calculation above, this is the correct answer. P=Force*velocity= 1,000N * 40m/s = 40,000 W = 40kW
  3. This answer is too large and is an order of magnitude off. You might have gotten this answer by also using the braking force of 14,000N instead of only using the continuous decelerating force. Recall that the question stem specifies the lack of braking.
  4. This answer choice is too large; you likely reached this answer by combining the force of the decelerating generator force, the maximum braking force and the continuous decelerating force. The question stem specifies that no braking occurs so the force should only be the continuous decelerating force. Answer B is the correct answer.

8) Recall from paragraph two that the generator braking system exerts a decelerating force that declines linearly with speed.

  1. This answer choice is super tempting because the passage mentioned that the force declines linearly with speed. However, it is the force that declines linearly, not the velocity. If force=mass*acceleration and the force decreases linearly, then in the absence of a changing mass it is the acceleration that is changing linearly. Now, if the acceleration is dropping in a linear fashion, the velocity must be dropping exponentially, not linearly. The slope of the velocity-time curve is acceleration, and a linear velocity curve would correspond to a constant deceleration, not one that keeps dropping.
  2. The railcar is slowing down when the brakes are engaged, not speeding up as shown with the increasing velocity in this curve.
  3. As in answer choice B, the railcar should be slowing down but this graph shows it speeding up.
  4. As noted in the explanation for answer choice A, it is the acceleration that is decreasing in a linear fashion, while the velocity decreases in an exponential fashion. The slope of the velocity-time graph is the acceleration and an exponential velocity-time slope corresponds to a linear acceleration-time slope; a linear velocity-time slope corresponds to a constant acceleration-time slope. This is the best answer choice and velocity-time curve.


Exam 3 Chemistry/Physics Section Discretes


  1. At first glance, this is a good answer choice because boiling points are a measure of energy required to disrupt a system. However, boiling points are a measure of the strength of INTERmolecular forces that hold molecules together. The thermodynamic stability of a compound is determined by the INTRAmolecular forces. This answer choice addresses the incorrect molecular forces.
  2. UV-vis spectroscopy is all about ground state electron excitation via the absorption of light and is related to degree of conjugation and HOMO-LUMO gaps (highest occupied molecular orbital and lowest unoccupied molecular orbital respectively). While energy is involved in the excitation of electrons, this technique will not give the thermodynamic stability of a compound as a whole. For more on UV-Vis spectroscopy, head over to
  3. Mass spectrometry is used for the identification of a compound via fragmentation and analysis of its mass-to-charge ratio. It is not appropriate for the understanding of thermodynamic stability. Here is a review of how a mass spectrometer works:
  4. Heats of combustion measures the heat produced when compounds are combusted, or burned in oxygen. If less heat is produced, the compound is more stable; when more heat is produced, the compound is less stable to begin with. This is a great way to determine the thermodynamic stability of a compound and is the best answer.

10) Because the solvent is water, whatever site we choose should not interact with water for proton exchange. If we were to label the guanine with the radioactive tritium at a site that exchanges its hydrogens for the protons in the solvent (interacts with water), the guanine would lose its radiolabel.

  1. Site I is a double bonded carbon. It is highly unlikely that a carbon at this site will interact with water to exchange its radiolabeled tritium for an unlabeled solvent proton, meaning the guanine will remain radiolabeled. This is promising and correct.
  2. Nitrogen could easily pick up a solvent proton and exchange it with its own radiolabeled tritium. This would make for a poor experiment so it is not the best site and is incorrect.
  3. Like the nitrogen at site II, the nitrogen at site III would readily lose its radiolabel. Nitrogen has a lone pair of electrons that can easily pick up a proton from the solvent.
  4. As with sites II and III, site IV is susceptible to losing its radiolabeled tritium because the nitrogen’s lone pair will readily pick up the protons from solution and could lose its radiolabel when it becomes deprotonated. Answer choice A is the most Appropriate answer.

11) The best way to answer this question is to determine the number of moles of gas particles and then use Avogadro’s number to convert the number of moles to the number of particles.

  1. This answer choice is the number of moles of gas, not the number or particles. This answer needs to be multiplied by Avogadro’s number, ~6 * 1023 molecules/mol.
  2. This answer choice results from the failure to convert degrees Celsius to Kelvin, the SI unit for temperature. Keep track of your units and convert to SI units as needed.
  3. This answer choice is incorrect because the temperature is in Celsius (wrong unit, SI unit is Kelvin) and does not account for the pressure, volume and part of Avogadro’s number.
  4. This is the best answer choice. See above for the full set of calculations.

12) When using Ka and Kb to determine whether a substance is an acid or a base, it’s important to remember that Ka*Kb=Kw so when Ka>Kb, the substance is more acidic than basic and vice versa. The table gives Ka directly, and the second column is equivalent to Kb. Note that we will need to use Ka of HCO3 but the Kw/Ka of H2CO3 because the conjugate base of H2CO3 is HCO3, but the conjugate base of HCO3 is CO32–.

Review the following content pages for a review of constants and conjugate acids and bases respectively: and

  1. a base since Ka > Kb for this ion. – If Ka were greater than Kb, the substance would be an acid, not a base.
  2. a base since Kb > Ka for this ion. – Kb of the H2CO3 (2.3 × 10–8) is indeed greater than Ka for bicarbonate (5.6 × 10–11), meaning that HCO3 is a base.
  3. an acid since Ka > Kb for this ion. – Ka (5.6 × 10–11) is less than Kb (2.3 × 10–8), not greater. Bicarbonate is a base and not an acid.
  4. an acid since Kb > Ka for this ion. – If Kb were greater than Ka, the substance would be a base, not an acid.

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