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: D. I-

As size increases, basicity decreases, meaning a species will be less likely to act as a base; that is, the species will be less likely to share its electrons. Iodine is the largest in the series.

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Correct Answer: D. 16

Quinine has four centers of chirality (highlighted with asterisks). Thus, in principle, quinine has 2⁴ (2 x 2 x 2 x 2 = 16) stereoisomers.

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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: C. III

The bromonium ion is formed when alkenes react with bromine. When the π cloud of the alkene (nucleophile) approaches the bromine molecule (acting as an electrophile), the σ-bond electrons of Br2 are pushed away, resulting in the departure of the bromide anion. The species formed is a high-energy intermediate along the reaction path. It will react further with some nucleophile−methanol in this case. The electronic structure of the bromonium ion has its consequences in the stereochemistry of the reaction. Nucleophilic attack takes place from the bottom face (back-side to the “departing” bromide) on the more substituted carbon. The back-side attack leads to the anti addition (Br and Nu groups end up on the opposite faces of the starting double bond), and the nucleophile adds to the more substituted carbon (Markovnikov’s rule). Structure III corresponds to the above. Structure II may form, but only in very miniscule amounts.

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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. 10

Recall that the degrees of unsaturation corresponds to the sum of pi bonds and/or rings present in a molecule. Each ring and each pi bond contributes one degree of unsaturation; triple bonds, possessing one sigma and two pi bonds, contribute two degrees of unsaturation. Gestrinone has four rings (4), three alkenes (3), one carbonyl (1) and one alkyne (2). Thus IHD = 4 + 3 + 1 + 2 = 10

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Correct Answer: C. Hydroboration-oxidation

The electronegativity of the hydrogen atom is stronger than that of boron, which means hydrogen is relatively negatively charged compared to boron. So in this case, the hydrogen will be added to the end containing more substituents and the boron part will be added to the end containing more hydrogen atoms, which will then convert to the hydroxyl group. This is the Anti-Markovnikov product.

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