DAT Question of the Day

Correct Answer: C. has a lower frequency than the analogous C=O in a non-conjugated system

All carbonyl compounds absorb in the IR region of 1760–1665 cm^–1 owing to the stretching vibration of the C=O bond. This carbonyl band is distinct with a characteristic high intensity, and therefore is quite useful for diagnostic purposes because few other functional groups absorb in this region. Conjugation with a double bond or benzene ring lowers the stretching frequency, usually by 30 to 40 cm^–1. The stretching frequency of the conjugated double bond is also lowered and may be enhanced in intensity. An example of this phenomenon is provided:

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Correct Answer: A. 1,1-dibromopropane

An approximated 1H NMR spectrum of 1,1-dibromopropane (CH3CH2CHBr2) is shown below:

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Correct Answer: B. The two compounds can be separated by ordinary physical chemical separation methods

Cis and trans-1,2-dibromocyclopropane (pictured below) share a diastereomeric relationship; that is, the molecules are stereoisomers that are not enantiomers. Diastereomerism occurs when two or more stereoisomers (i.e. same connectivities) of a compound have different configurations at one or more (but not all) of the equivalent stereocenters and are not mirror images of each other. Accordingly, diastereomers are distinct molecules with different chemical properties, and can be separated by ordinary physical means. Recall that enantiomers are mirror images of each other and are identical in all physical aspects except for their direction of rotation of plane polarized light.

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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: 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. 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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