MCAT Question of the Day

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: B. B

In 1,3-disubstituted cyclohexanes both groups can adopt equatorial positions in the stable chair conformation if the groups are cis-. This is favored as equatorial groups have more space than in the axial positions, where torsional strain factors come into consideration.

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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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Correct Answer: A. CH3CO2H

Recall that pKa = -log Ka so their relative values, because of the negative sign in the equation, are always opposite to each other. Thus in an acid base equilibrium, AH <--> A- + H+, the largest pKa value is associated with the weakest acid. The unhalogenated carboxylic acid A is the weakest acid (acetic acid is the IUPAC name; ethanoic acid is the systematic way to name it). In the other cases, the electron withdrawing halogen substituent stabilize the carboxylate conjugate base (anion) making the parent acid more acidic.

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Correct Answer: C. Rate = k[X][Y]²

This is a common MCAT-type question. By looking at Table 1, we can see that when the concentration of X is quadrupled (factor of 4¹) while [Y] is unchanged (Exp. 1 and 3), the rate is increased by a factor of 4 = 4¹. Thus the order of the reaction with respect to X is 1 (= “first order with respect to X”).

When the concentration of Y is doubled (factor of 2¹) while [X] remains the same (Exp. 1 and 2), the rate of reaction is quadrupled (factor of 4 = 2²). Thus, the order of reaction with respect to Y is 2 (= “second order with respect to Y”).

The rate equation is Rate = [X][Y]². Since the overall rate of reaction is the sum of exponents, we can say that the reaction is (1 + 2) a third order reaction overall.

{Notice that the stoichiometric coefficients are not relevant; the order of reaction being based on data.}

Additional learning point: In Organic Chemistry, an SN2 reaction is typical of a second order reaction where the rate must be equally proportional to the concentration of both reactants; for example, first order with respect to the alkyl halide and first order with respect to the nucleophile, so overall second order.

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