DAT Question of the Day

Correct Answer: E. 1, because it is in a cis conformation

An important criterion for diene reactivity is for the diene to maintain a cis conformation. Cyclopentadiene (1) has the double bonds permanently cis oriented due to its cyclical nature, thus is the most reactive of the two. Re-drawing diene (2) into a cis conformation shows the steric hindrance experienced by the terminal methyl groups; diene 2 is unable to maintain the geometry necessary for a Diels-Alder reaction.

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Correct Answer: E. cyclopentanone

Beta-keto carboxylic acids undergo decarboxylation through ring formations. The carbonyl bond of the ketone will react with the alcohol group of the carboxylic acid to form a ring and release the carboxylic acid as CO2. The remaining enol will tautomerize into a carbonyl group again, leaving cyclopentanone. Notice a cyclopentaldehyde can’t even be a possible answer, aldehydes cannot exist in cyclo compounds.

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Correct Answer: A. Change the water solubility of an unknown compound

Adding acid or base to a mixture of unknown compounds is designed to change the water solubility. The loss or gain of protons will make the compound become ionic in nature, and thus soluble in water. On the other hand, compounds that are already ionic may become neutrally charged, resulting in lower solubility in water.

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Correct Answer: B. 15

Dynemicin A has two aromatic systems, three carbonyl substituents (two ketones, one carboxylic acid), two alkenes and two alkynes. Total = 15. Recall that alkynes possess two pi bonds in addition to one sigma bond, therefore C is not correct.

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Correct Answer: B. PCC

The conversion of 2,2-dimethylbutan-1-ol to 2,2-dimethylbutanal is an oxidation (alcohol -> aldehyde). Lithium aluminum hydride (E) is a strong reducing agent. Potassium permanganate (C) is used in the oxidation of alkenes to diols, and is generally not strong enough to further oxidize aliphatic alcohols. Ozone (D) is used in the oxidative cleavage of alkenes into the corresponding aldehydes or ketones. Chromic acid (A) is a harsh oxidant, oxidizing primary alcohols up to the corresponding carboxylic acid, and secondary alcohols up to the corresponding ketone. PCC (B, pyridinium chlorochromate) is a milder chromium-based oxidant used in the selective oxidation of primary alcohols to aldehydes.

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Correct Answer: B. E2

E2 is the likely reaction to occur here:

1. Consider the leaving group: if primary, the reaction will almost certainly be SN2. If tertiary, the reaction cannot be SN2. Depending on the type of nucleophile/base, it will either proceed with concerted elimination (E2) or through carbocation formation (SN1/E1).

2. Charged nucleophiles/bases (such as –OCH3 in this example) will favor SN2/E2 pathways (i.e. eliminate SN1/E1). If a charged species is not present, the reaction is likely to be SN1/E1.

3. Polar protic solvents (like CH3OH) tend to favor elimination (E2) over substitution (SN2). Polar aprotic solvents tend to favor substitution (SN2) relative to elimination (E2).

First, determine the lowest energy chair conformation required for reaction, where the LG is axial and any other large groups remain equatorial. Once determined, two pathways must be considered. Removal of the proton that provides the more stable (tri-substituted vs. di-substituted) double bond (recall Zaitsev’s rule) is the preferred pathway.

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Correct Answer: A. I

Of the possibilities, all have the molecular formula C8H12O, but only compounds I and III can have 5 chemically unique proton signals in the H NMR spectrum. The broad IR peak at 3400 (indicative of alcohol), and sharp peak at 3300 (alkyne C-H stretch) confirms the compounds identity as I.

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Correct Answer: C. The polarity of a molecule is dependent on its 3-D structure

A molecule’s dipole is an electric dipole with an inherent electric field (not be confused with a magnetic dipole which generates a magnetic field). Molecules can have dipole moments due to non-uniform distributions of positive and negative charges on the various atoms. For diatomic molecules there is only one (single or multiple) bond so the bond dipole moment is the molecular dipole moment, with typical values in the range of 0 to 11 D. For polyatomic molecules there is more than one bond, and the total molecular dipole moment may be approximated as the vector sum of all individual bond dipole moments.

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Correct Answer: C. 3 + 6 only

The cyclohexene product is indicative of a Diels-Alder reaction, the cycloaddition of a conjugated diene (electron rich) and dienophile (electron poor) components. For best reactivity, the diene must be able to adopt a cisoid conformation. The reaction is stereospecific with respect to both diene and dienophile, meaning a cis-dienophile gives cis substitution in the product (same applies for trans-dienophiles -> trans product substitution). If diene substituents have the same stereochemistry (i.e. both alkenes are E), both substituents will end up on the same face of the product. Mixed stereochemistry (i.e. E and Z) in the diene results in substituents on opposite faces of the product. The final product has syn-substituted diene components (diene is either E,E or Z,Z), and dienophile substituents are trans, (original dienophile has trans geometry). Hence, only answer choice [C] contains reactants that will form the desired product, with reactant 3 as the E,E diene, and reactant 6 as the trans dienophile. Answer choice [B] states reactants 1 and 6 will produce the final product in high yield, however, reactant 1 contains mixed stereochemistry (E,Z bonds) and will result in the methyl substituents on opposite faces of the product.

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Correct Answer: A. 1

Simple hydrocarbons (containing only C and H) are weakly acidic in comparison to other organic compounds. However, carbanions can be formed, with their stability depending quite heavily on the hybridization state of the orbital containing the lone pair of electrons. Alkyne 1 is most acidic as the lone pair of the conjugate base is held in a sp-orbital, rather than sp3 (for 2 and 5) or sp2 orbital (for 3 and 4). Having a greater proportion of “s character” allows for the electron density in an sp-orbital to be held closer to the positively charged nucleus, making it electrostatically more stable.

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