DAT Question of the Day

Correct Answer: A. 1

LiAlH4 is a strong reducing agent of carbonyl compounds. 3-methyl-2-butanol is a secondary alcohol, therefore, reduction of a ketone is required for its preparation.

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Correct Answer: E. diastereomer.

Drawing this new molecule (right) and comparing it to the first (left), we see that the two are stereoisomers of each other, specifically, diastereomeric. They are not mirror images and non-superimposable (conditions for enantiomers). Constitutional isomerism does not apply here, as connectivity of atoms remains the same. Rotation about the bond drawn in stick form does not yield identical compounds (not rotamers).

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Correct Answer: D. 5 > 1 > 4 > 3 > 2

Mineral acids such as HBr are far more acidic than organic compounds and water; #2 is last in the ranking. From the remaining options, look for trends, and remember the factors that affect the stability of molecules: resonance, charges on atoms with appropriate electronegativities, hybridization and charge.

One way to consider the relative acidities of organic functional groups is to draw resonance structures. Remove the most acidic alpha hydrogen and see how many stable resonance structures you can draw. For instance, carboxylic acids have greater resonance stabilization because of delocalization of the negative charge over two electronegative oxygen atoms.

The alpha hydrogens of ketones (pKa = 20) are more acidic as compared to alpha-hydrogens of esters (pKa = 25); in the ester, there is also a resonance donation from the alkoxy group towards the carbonyl that competes with the stabilization of the enolate charge. This makes the ester enolate less stable than those of aldehydes and ketones so esters are even less acidic.

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Correct Answer: B.

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 charge delocalization (i.e. the more resonance
structures that can be drawn) the stronger the acid. Proton Hb is most acidic due to the charge (and lone pair of electrons) on the conjugate base being delocalized over both carbonyl groups and the benzene ring.

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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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