MCAT Question of the Day

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: A. CH3COO^–

Carboxylic acids are considered to be among the strongest organic acids but in comparison with all types of acids, carboxylic acids are very weak. By far, the protonated form of answer choice A (acetic acid is the IUPAC name; ethanoic acid is the systematic way to name it) would be the weakest of that group (after all, we are comparing it with nitric acid, sulfuric acid and phosphoric acid).

Logically, a strong acid would create a weak conjugate base, otherwise the acid would not give up the proton readily. Thus the reverse is true: a weak acid has a strong conjugate base.

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Correct Answer: B. B.

The primary amines A and B can donate two hydrogen bonds and accept one (due to the lone pair on N). Generally, for compounds with the same functional group, the higher molecular weight compounds have a higher boiling point due to increased opportunity for intermolecular van der Waals (weak, but better than nothing) interactions. Of course, even though a periodic table is available during the exam with a click of a button, you do not need to calculate the molecular weight since you can quickly determine the number of carbons of the different N-containing structures which easily reveals B as the “heaviest.”

Helpful hint: If all other factors are the same, increased molecular weight means that it is harder to push that molecule into the vapor phase meaning the boiling point would be higher.

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Correct Answer: D. Less that 7.0 but the water is neutral.

It is not necessary to calculate anything to solve this problem but we will go through the steps.

The dissociation of water (note carefully that the ratio of the products is 1:1; also keep in mind that this is the basis of the neutrality of pure water: acid units = base units):

H2O + H2O <-> H3O⁺ + OH^–

The expression for the self-ionisation of water or the water dissociation constant:

Kw = [H3O⁺][OH^–] = [H⁺][OH^–]

At room temperature, keeping in mind that the dissociation of the ions is 1:1, we get: 10^-14 = [H⁺][OH^–] = [H⁺]²

And so [H⁺] = 10^-7 and then -log[H⁺] = 7 = pH

Doing the identical math but using Kw = 10^-13 yields a pH of 6.5 BUT because the ratio of the ions is still 1:1 (if not, you would not have done the calculation) thus it must be neutral.

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Correct Answer: D. Camphor has two stereogenic centers and exists as a pair of enantiomers.

Position 1 and 4 as labeled in the diagram below (Ref: csi.chemie.tu-darmstadt.de) are the only 2 chiral (stereogenic) carbons in camphor since they are attached to 4 different substituents, each. Notice that because of the carbonyl group being closer to position 1, and because of the methyl substituent at position 1, the molecule is NOT symmetric when looking at it from above. Because a line of symmetry cannot be drawn, thus the molecule is NOT meso, and thus it has 2 non superposable (= non superimposable) mirror images (= enantiomers).

Additional learning points: Note that the maximum number of stereoisomers is given by 2 to the power of n, where n is equal to the number of chiral carbons. So the maximum for camphor, having 2 chiral carbons, would be 2 squared which is 4. However, camphor yields only two stereoisomers because the relative configuration of both centers is constrained in the bicyclic ring system (i.e. 2 closed rings in one system).

Also, note this Mnemonic: MeSo = Mirror of Symmetry.

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Correct Answer: D. B > A > D > C

When in aqueous solution, aldehyde and ketones groups can form hydrates: this is simply the result of addition of water to the carbonyl (the partial negative charge of the oxygen in water attacking the partial positive charge of the carbonyl group).

The ability to form a hydrate (addition of water across the C=O bond) can be seen as measure of the stability of a carbonyl group.

Factors affecting this are the electronic and steric effects of substitutents attached to the electrophilic carbonyl carbon. Hydrogen bond acceptors also aid hydrate formation.

Strained cyclopropanone B is the most ready to form a hydrate as the ring bond angle is far from the ideal trigonal planar angle. This strain is relieved somewhat upon formation of a near tetrahedral hydrate.

Formaldehyde A is more reactive (unstable) than aldehyde D which in turn is more reactive than ketone C. Across this sequence more +I methyl groups are being added to the electrophilic C=O carbon atom. In addition, the less hydrophobic the carbonyl compound (and the corresponding hydrate), the better the solvation through intermolecular hydrogen bonding (i.e., A > D > C).

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Correct Answer: B. Atomic radius increases

As you move from left to right across a Period, metallic properties decrease (metals > metalloids > non-metals), atomic radius decreases (the atoms get smaller), electronegativity (the ability to attract electrons in a compound) increases, first ionization energy (the energy required to remove the outermost electron form a neutral atom) increases and electron affinity (the energy released when an electron is bonded to an atom) increases.

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Correct Answer: A. 109.5°

Tetrahedral geometry has three-dimensional placement with atoms positioned at 109.5° to each other. Trigonal planar molecules have angles of 120°, interior angles of 60° are associated with strained cyclopropanes bonding, 90° with cyclobutanes and 180° for alkynes.

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Correct Answer: B. H2SO4

A strong acid must be able to provide large numbers of hydrogen ions when dissolved in water, (H⁺(aq)). Thus a strong acid must dissociate (break into ions) completely upon dissolving in water to provide as many H⁺(aq) ions as possible.

All these compounds are water soluble, but only CH3CH2COOH and H2SO4 can provide H⁺(aq) ions. CH3CH2COOH or propanoic acid, is only partially dissociates (<1%) in water and is thus a weak acid. H2SO4 or sulfuric acid, completely (100%) dissociates into sulfate ions and hydrogen ions upon dissolving in water and is thus the strong acid of the group.

CH3CH2OH – ethanol does not dissociate, but stays as a complete molecule in solution.
NH4OH – ammonium hydroxide, a weak base which partially dissociates to give few hydroxide OH^–(aq) ions.

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Correct Answer: B. 2

The Aldol reaction combines two carbonyl compounds to form a new ?-hydroxy carbonyl compound. These products are known as aldols, from the aldehyde + alcohol, a structural motif seen in many of the products. The correct answer is the Aldol reaction of acetone. The newly formed bond is highlighted in red. Be able to recognize the ?-hydroxy carbonyl motif as an Aldol product.

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