DAT Question of the Day

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: E. 1,4-dimethylcyclopent-1-ene

The treatment of alkenes with acidic potassium permanganate under high temperatures results in oxidative cleavage. The nature of the products are dependent on alkene substitution, but all with possess a carbonyl functionality (carboxylic acid, ketone, CO2). Looking at the product, one can identify the two carbon atoms of the original alkenyl compound, and work back accordingly. Finally, standard IUPAC naming will apply.

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Correct Answer: C. The polarity of a molecule is dependent on its 3-D structure

A molecule’s dipole is an electric dipole with an inherent electric field (not be confused with a magnetic dipole which generates a magnetic field). Molecules can have dipole moments due to non-uniform distributions of positive and negative charges on the various atoms. For diatomic molecules there is only one (single or multiple) bond so the bond dipole moment is the molecular dipole moment, with typical values in the range of 0 to 11 D. For polyatomic molecules there is more than one bond, and the total molecular dipole moment may be approximated as the vector sum of all individual bond dipole moments.

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Correct Answer: D. I and III

Aromatic compounds follow four rules: (1) They are conjugated—there needs to one “p” orbital from each atom in the ring, so each atom must be either sp² or sp hybridized; 2) They are cyclic: linear systems are not aromatic; 3) They are planar: there is good overlap/interaction between p orbitals; 4) they follow the The Huckel Rule—4n+2 π electrons in the cyclic conjugated π system (n = 0, 1, 2, 3 etc.). Of the possibilities,compounds I (14 e⁻, n = 3) and III (10 e⁻, n = 2) obey the rule.

This is a little advanced and at the top of the difficulty the DAT can throw at you, you have to consider compound 3 as one molecule, not two separate rings. Aromaticity requires a planar conjugated system, so we can ignore the bridge carbon sticking out of the plane of the rest of the molecule. There cannot be any sp3 atoms IN the aromatic ring. The carbon bridge actually pushes the 10 carbon ring to be planar, without it the molecule wouldn’t be aromatic.

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Correct Answer: A. acetylene

Bond length is the distance between the centers of two covalently bonded atoms and is mostly dependent on atomic size, bond order (length: single > double > triple), and hybridization (C–H and C–C bonds shorten slightly with increased s character on carbon).

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Correct Answer: B. E2

E2 is the likely reaction to occur here:

1. Consider the leaving group: if primary, the reaction will almost certainly be SN2. If tertiary, the reaction cannot be SN2. Depending on the type of nucleophile/base, it will either proceed with concerted elimination (E2) or through carbocation formation (SN1/E1).

2. Charged nucleophiles/bases (such as –OCH3 in this example) will favor SN2/E2 pathways (i.e. eliminate SN1/E1). If a charged species is not present, the reaction is likely to be SN1/E1.

3. Polar protic solvents (like CH3OH) tend to favor elimination (E2) over substitution (SN2). Polar aprotic solvents tend to favor substitution (SN2) relative to elimination (E2).

E2 reactions occur quickest when the H-C bond and C-LG bonds involved are co-planar, or as close as possible to 180° with respect to each other. This is known as an antiperiplanar conformation. This conformation positions the sigma bonds being broken in the proper alignment to become the new pi bond.

First, determine the lowest energy chair conformation required for reaction, where the LG is axial and any other large groups remain equatorial. Once determined, two pathways must be considered. Removal of the proton that provides the more stable (tri-substituted vs. di-substituted) double bond (recall Zaitsev’s rule) is the preferred pathway.

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Correct Answer: A. I

Of the possibilities, all have the molecular formula C8H12O, but only compounds I and III can have 5 chemically unique proton signals in the H NMR spectrum. The broad IR peak at 3400 (indicative of alcohol), and sharp peak at 3300 (alkyne C-H stretch) confirms the compounds identity as I.

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