MCAT Question of the Day

Correct Answer: A. A

Due to the angle strain, deviating most from the ideal angle of 109.5°, as found in cyclohexane C, cyclobutane A (ring strain, very unstable) has the highest heat of combustion per CH2 (the most energy released).

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Correct Answer: C. separating liquids based on their different boiling points.

Using fractional distillation one can increase the temperature of a liquid mixture and as the different liquids vaporize at their respective boiling point temperatures one can collect the vapor by condensing it and collecting the desired liquid at the corresponding temperature.

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Correct Answer: B. Beta particle

Since the radioactive emission angled up toward the positive plate, it must be oppositely charged, that is, negatively charged. The only negatively charged species in the answer choices is the beta particle which is, by definition, an electron or a positron. Note that electromagnetic radiation like gamma rays (particles) and infrared radiation are not deflected in electromagnetic fields.

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Correct Answer: D. Decreased temperature

Since the reaction is exothermic (negative enthalpy), heat can be considered a product in this reaction. A decrease in temperature will result in the equilibrium shifting to the right to bring heat to the system resulting in an increase in NO production, in accordance with Le Chatelier’s principle. A change in overall pressure of the system will not affect the equilibrium because there is an equal amount of moles of gas on both sides of the equation providing an even pressure. A catalyst affects the rate of a reaction not thermodynamic properties such as equilibrium.

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Correct Answer: C. a net negative charge.

A pH above the isoelectric point means that the amino acid is in relatively basic conditions. A base will accept a proton (H⁺) from the amino acid, leaving the amino acid negatively charged.

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Correct Answer: C. The other atom involved in the hydrogen bond (not the hydrogen atom) must be covalently bonded to a hydrogen atom.

Answer choice C. is not necessary for the formation of a hydrogen bond. A classic example is that of ethers. Ethers are water soluble because the oxygen in ether, which is covalently bonded to two carbons, can hydrogen bond with the hydrogen in H2O. Note regarding answer choice C.: the 3-D shapes of many proteins and DNA is due in part to intramolecular H-bonds.

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Correct Answer: D. ²²⁶88Ra

Equation I: ²³⁸92U -> ^xyZ + 3⁴2He + 2⁰-1e^– + 3 (gamma rays)

Since the sum of the atomic numbers and mass numbers on either side of the equation must be equal:

238 = x + (3 x 4) + (2 x 0) + (3 x 0)
x = 226
92 = y + (3 x 2) + (2 x -1) + (3 x 0)
y = 88

Thus, from the answer choices, Z = Ra; note that gamma rays are a form of electromagnetic radiation and thus have no charge and no mass.

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Correct Answer: C. Only G would be positive.

G = H – TS

“The sublimation of carbon dioxide occurs quickly at room temperature” means that it is spontaneous and so G must be negative (by definition). Sublimation means:

Solid CO2 + heat -> vapor

Entropy (randomness) is clearly increasing (thus positive S) because we are moving from a structured, ordered solid to randomly moving gas particles. Heat is required so it is endothermic meaning H is positive. The question is asking about the reverse reaction so all 3 signs are reversed: G is now positive; H is now negative; S is now negative.

Going deeper: Notice that a negative S multiplied by a –T (see the Gibbs free energy equation) creates a positive term which overshadows the effect of the negative H and thus G is still positive. Also, please keep in mind that sometimes the real MCAT will provide the Gibbs free energy equation but sometimes they won’t.

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Correct Answer: C.

First, let’s number the carbons in the ring. Oxygen is attached to C1 and so N is attached to C3. So in the original molecule, we see that there are double bonds in the ring between C2=C3 and C5=C6. All of the answer choices follow the rules of drawing resonance structures except answer choice C. Let’s carefully follow the electrons for answer choice C and thus we can understand what went wrong.

Our starting material:

1) Keep in mind that oxygen is the most electronegative atom in the molecule and thus ‘wishes’ to withdraw electrons to itself.

2) The nitrogen is positively charged because it contributed its lone pair of electrons to carbon. Carbon’s double bond with the 2nd nitrogen breaks and thus nitrogen has a negative charge and an extra lone pair (note: this describes answer choice B).

3) Nitrogen’s lone pair bonds with C3 which breaks C3=C2 so now C2 has a lone pair and a formal negative charge (C2 is secondary carbanion and this describes answer choice D).

4) Remember that oxygen wants the electrons more than any other atom in this molecule. C2’s lone pair bonds with C1 as C=O breaks so now oxygen has the lone pair and the formal negative charge (this describes the following structure which is not an answer choice but to make it easy to refer to, we will call it answer choice E).

5) We just worked out why answer choice E is correct. Well, the only difference between E and C is that E correctly has the double bond C5=C6 like the starting material but answer choice C incorrectly places the double bond at C4=C5. That would only be possible if C6 had a + charge and C4 had a – charge.

6) And finally, answer choice A is like the starting material except, instead of oxygen pulling electrons from nitrogen along ‘the top’ (!!) of the molecule, oxygen pulls electrons from the C5=C6 double bond creating a C6=C1 double bond, a lone pair on oxygen and a secondary carbocation at C5 (because it lost its bond).

In case you are wondering, the current MCAT requires an understanding of resonance and Lewis structures but does NOT require specific knowledge of aromatic chemistry (i.e. EWG, EDG, addition reactions, etc.).

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Correct Answer: B. 6

Keeping in mind that a tetrapeptide (i.e. an oligopeptide or even if it were a protein) has an amino end and a carboxyl end, it means that LLVV would be different to VVLL where L = leucine and V = valine. Thus there are 6 possibilities:

1. LLVV
2. VLLV
3. VVLL
4. LVLV
5. LVVL
6. VLVL

Basic stats is only required for the 2015 MCAT. If that is what you are considering then you should describe this problem as 4 choose 2 or: 4!/(2!)(2!) = 4*3*2*1/2*2 = 24/4 = 6.

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