DAT Question of the Day

Correct Answer: C. 3

Catalytic hydrogenation reactions lead to a cis addition of hydrogen, or in this case deuterium atoms

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Correct Answer: D. 4

Aromatic compounds are named as derivatives of benzene. Substituents positions are indicated by numbers except that o- (ortho), m- (meta) and p- (para) may be used in place of 1,2-, 1,3-, and 1,4-, respectively, when only two substituents are present. If alkyl substituents are smaller than the ring (6 or fewer carbons), the compound is considered an alkyl substituted benzene. Alkyl substituents larger than the ring (7 or more carbons), are phenyl-substituted alkanes. Compounds are named by numbering the position of each so that the lowest possible numbers are used. The substituents are listed alphabetically when writing the name.

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

There are two main electronic effects that substituents can exert: RESONANCE effects include electron donating (e.g. -OMe) where p electrons are pushed toward the arene or electron withdrawing (e.g. -C=O) where p electrons are drawn away from the arene. INDUCTIVE effects include electronegativity type effects. These too can be either electron donating (e.g. -Me) where s electrons are pushed toward the arene or electron withdrawing (e.g. -CF3, -NR3) where s electrons are drawn away from the arene.

Some general pointers for recognizing the substituent effects:

• The H atom is the standard and is regarded as having no effect.
• Activating groups increase the rate (3 > 4)
• Deactivating groups decrease the rate (5 > 1)

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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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