MCAT Question of the Day

Correct Answer: A. 1-cyclopropyl-4-isopropyl-1,3-dimethylcyclohexane

1. First, identify the (longest) parent chain or cycle -> six carbon ring = “cyclohex”

2. If several substituents are on the ring, they are listed in alphabetical order, and the first substituent is given #1 priority. The cyclohexane has the following substituents: cyclopropyl (#1), 2 methyl groups, iso-propyl (note that iso is not considered when placing in alpha order)

3. Substituents are numbered such one of them is at carbon #1 and the other locations have the lowest possible numbers, going in either a clockwise or counter-clockwise direction, giving the prefix di-, tri-, tetra-, to groups of the same kind (prefixes are not considered when alphabetizing).

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Correct Answer: B. the equilibrium mixture has more reactants than products

The chemical equilibrium constant is calculated by dividing the concentration of the products by the concentration of the reactants at equilibrium. The magnitude of the chemical equilibrium constant can shed light on the composition of the mixture at equilibrium. Since the equilibrium constant given in the question is small, the equilibrium mixture would contain more reactants than products. If the equilibrium constant were larger than 1, this would indicate that the equilibrium mixture was made up of more products than reactants. Note: the chemical equilibrium constant is not an indicator of the rate of the reaction, only the concentrations of the equilibrium mixture.

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Correct Answer: C. Diastereomers

When drawing out the structures, one can see that the basic skeletal arrangements are identical, i.e. stereoisomers (not constitutional isomers). We can dismiss conformers, since the structures cannot be interconverted exclusively by rotations about formally single bonds. The compounds are diastereomers, a class of stereoisomers that are non-superimposable, nonmirror images of one another. More specifically, the compounds are epimers: in chemistry, epimers are diastereomers that differ in configuration of only one stereogenic center.

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Correct Answer: D. Asymmetric induction

Racemization refers to the conversion of an optically active substance to a racemic form (has equal amounts of left- and right-handed enantiomers of a chiral molecule). Enantioenrichment refers to the act of enriching an already existing mixture of stereoisomers such that it contains more than 50% of one enantiomer. Chiral chromatography involves the separation of stereoisomers using a chiral stationary phase, which does not involve chemical transformations. “Optical reduction” as a technique or phrase does not exist.

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Correct Answer: C. convex.

The meniscus is the curved upper surface of liquid. This is formed because of unequal cohesion and adhesion attractions. For example, in water the adhesive forces are greater than the cohesive forces, therefore the water is able to “walk up the sides” of the surface. In the case of water, a U-shaped or concave meniscus is formed. Mercury, for example has stronger cohesive attractions between the mercury molecules than adhesive attractions between the mercury and the glass surface, therefore producing a convex meniscus, which looks like an inverted U.

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Correct Answer: C. the activation energy is lowered

A catalyst speeds up a chemical reaction by lowering the activation energy of a reaction. The catalyst is never consumed, nor does it alter the reactants or products. The total enthalpy of the reaction is also not changed from what it would have been without the catalyst. Only the rate at which the reaction occurs and the activation energy is affected. The only answer option that is correct is [C].

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Correct Answer: B. It reduces activation energy of a chemical reaction

Enzymes reduce the activation energy of chemical reactions by binding substrate(s) into energetically favorable orientations. They greatly reduce the activation energy, but do not eliminate it completely. Enzymes are completely reusable—they do not participate in the reaction as a substrate. Even when cofactors are needed for the successful action of the enzyme and though they may be tightly bound, they are not covalently bound to the enzyme. Enzymes do not add additional energy to chemical reactions as heat would, for example.

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Correct Answer: B. XeF2

Start by drawing the Lewis structure of CO2. We find that the molecular shape is linear. Here is a video to help visualize this process: CO2 Structure

Next, find which of the answer choices is also linear by drawing all of the remaining structures. We find that only XeF2 has the same linear molecular shape, even though it has a different hybridization!
Structure of H2O
Structure of XeF2
Structure of SO2
Structure of SCl2

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Correct Answer: A. 1, 4

Rotation of plane polarized light is indicative of a chiral molecule. If not, the compounds must be meso or achiral. Compounds 1 and 4 are meso compounds, which, despite having two or more stereocenters, are not chiral (superposable on its mirror image). Compounds 2 and 3 are chiral molecules.

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Correct Answer: C. Delta(Grxn) < 0 at only low temperatures

On the MCAT, numbers can be intimidating, but remember that if the answer choices have no numbers in them, we should think conceptually instead of empirically. Gibbs equation helps us determine if a reaction is spontaneous or non-spontaneous. It is given as:
G = H – TS

In this case, the enthalpy change H is a negative number (exothermic). We know that the entropy change (S) is also negative (the products are more ordered than the reactants) because the reactants are 2 moles of gas and the products are only 1 mole of gas (a synthesis of phosphorus pentachloride). Our equation now be read differently as:
G = (negative number) – temperature * (negative number)

Temperature is in Kelvin and thus always a positive number, but we must consider two possibilities regarding temperature: (1) If this positive temperature is relatively low, the second part of the equation is a small positive number. Thus, if you add a small positive number to a negative number, it will probably still be negative (G<0). (2) If the positive temperature is relatively high, the second part of the equation is a large positive number. If you add a large positive number to a negative number, it will probably become positive (G>0).

In both these cases, ambiguity is important because we don’t know the temperature or S. The only answer that appropriately describes the case we have here is C.

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