DAT Question of the Day

Correct Answer: C. III

The carbon-chlorine bonds are polar due to the difference in electronegativities of C (2.55) and Cl (3.16) atoms. These individual dipole moments cancel each other out only in an anti-periplanar arrangement (compound III).

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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: 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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Correct Answer: E. ultraviolet light

Ultraviolet–visible spectroscopy (UV-Vis or UV/Vis) refers to absorption/reflectance spectroscopy UV-Vis spectral region. The absorption or reflectance in the visible range directly affects the perceived color of the chemicals involved. In this region of the electromagnetic spectrum, molecules undergo electronic transitions. Molecules containing π-electrons or non-bonding electrons (n-electrons) can absorb the energy in the form of ultraviolet or visible light to excite these electrons to higher anti-bonding molecular orbitals. The more easily excited the electrons (i.e. lower energy gap between HOMO and the LUMO), the longer the wavelength of light it can absorb.

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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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