MCAT Question of the Day

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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Correct Answer: D. E

In order to answer such questions, the stability of each resulting conjugate base should be considered; the more stable this is, the stronger the parent acid.

For enolates, the greater the degree of delocalization of charge, (i.e., the more resonance structures that can be drawn), the stronger the acid. Therefore, proton C is more acidic than B. However, the carboxylate arising from loss of E is the most stable as the charge is shared between two electronegative oxygen atoms (of the CO2^– group). Proton E is the strongest acid.

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Correct Answer: D. RbCl

Ionic character increases with increasing electronegativity of the potential anion and decreasing electronegativity of the potential cation. Electronegativity increases as one moves to the top right-hand corner of the periodic table, and decreases in the opposite direction.

Since the potential anion of all four compounds in question is chloride, we only need to find the compound with the cation located furthest toward the bottom left-hand corner of the periodic table. This compound is RbCl.

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Correct Answer: C. SOCL2, then NaCN

In this reaction, there is a substitution with net retention of configuration at the chiral (stereogenic) center. An SN2 process leads to a single inversion of configuration, whereas, two consecutive SN2 processes will result in the required net retention (double inversion).

In process A, a sulfonateester leaving group is initially formed (with no change in configuration) and then cyanide displaces the leaving group with inversion; i.e., the enantiomer of the required product is formed. In process C, the alcohol is converted by addition of thionyl chloride (SOCl2) to the alkyl chloride with inversion of conversion. Displacement of the chloride by the cyanide anion (with another inversion of configuration) leads to the required product.

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Correct Answer: D. It would not be oxidized.

Carefully notice that this diol has both alcohol substituents attached to tertiary carbons. Tertiary alcohols are very, very difficult to oxidize and it certainly will not occur under mild oxidizing conditions.

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Correct Answer: C. ^aO⁰, ^bC⁰ and ^cN^-1

The normal valency of carbon is 4, oxygen is 2 and nitrogen 3. Adding electrons to make a full octet means adding two lone pairs to both oxygen and nitrogen atoms. For a -1 charge this must therefore be situated on N as it only has two bonding pairs of electrons instead of three.

If you ever get stuck during the exam, just take a moment to consider what you know for sure: (1) when oxygen is bonded twice, it has no formal charge (i.e. water); (2) when carbon is bonded 4 times, it has no formal charge (i.e. methane); (3) when nitrogen is bonded 4 times, it has a positive charge (i.e. ammonium), 3 times and it’s neutral (i.e. ammonia) and so just twice, it must be negatively charged.

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Correct Answer: A. A

An SN2 reaction is a concerted process and does not involve intermediates (ruling out profile C and D). Also, reaction transition states are higher energy than the starting material leaving profile A as the only option.

Recall that 2nd order reactions can be written of the type: rate = k[RX][Nucleophile] and so the total of the exponents = 2 (thus 2nd order overall = 1st order with respect to RX and 1st order with respect to the nucleophile).

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Correct Answer: B. B.

The polarization of a bond is determined by the relative electronegativity of each particular atom. As Cl is more electronegative than a hydrogen atom, so the dipole moment points towards the Cl atom along the H-Cl bond. As a good memory guide, from your physics review, you would know that an electric field line always points away from a positive charge.

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