DAT Question of the Day

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: D. 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.

A, B, and C have planes of symmetry, which makes them all achiral molecules. The only chiral molecule left is D.

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Correct Answer: A.

A: S; CH(C)2 > CH2CH(C)2 > CH2CH2 > H, here H is pointing out, so we should look from the back side.
B: R; Si > O > C > H, also we should look from the back side.
C: R; O > C=O > C(C)3 > H
D: R; O > C(C)3 > CH3 > electron pair
E: This isn’t chiral, it has a plane of symmetry!

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Correct Answer: C.

In order of increasing acidity:

ethane (pKa 42) < acetaldehyde (pKa 17) < ethanol (pKa 16) < water (pKa 15.7) < acetic acid (pKa 5) Note: You did not need the pKa values to solve this question. It is asking you to organize the functional groups from most basic to most acidic. Acetic acid would be the most acidic, and ethane being the most basic (since it lacks an O-H bond). Ethanol and aldehydes are more basic than water, so C is our answer.

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Correct Answer: C.

The reagents are indicative of the formation of an epoxide by mCPBA (m-chloroperoxybenzoic acid), followed by epoxide-opening using methylmagnesium bromide (Grignard reagent, a nucleophile). Epoxidation can occur from either side of the alkene (in this instance, epoxidation is shown from the top face). On treatment with the methyl Grignard reagent, backside attack occurs (in this instance, the nucleophile approaches from the back of the molecule), with nucleophilic addition to the least hindered epoxide carbon (steric effects).

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Correct Answer: C. SO3 / H2SO4 / heat

The sulfonation of ethylbenzene requires fuming sulfuric acid and sulfur trioxide. Reaction A would not give the product; reaction B would result in nitration rather than sulfonation; reaction D woul result in oxidation of sulfurous acid to sulfuric acids by peroxides, but without other reagents, would not react further. Note that the para isomer would be the major product for steric considerations.

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Correct Answer: A. NaI/acetone

This is a substitution reaction with inversion (SN2) of a methansulfonate (excellent leaving group) by halide, sodium iodide in this case (also known as the Finkelstein reaction. Impress your friends.). None of the other reaction conditions are suitable for this reaction.

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Correct Answer: E. HO^–

Hydroxide is the strongest base because (1) the charge is not delocalized (as in A and B) and (2) O is more electronegative than those elements further down Group 16 (S, Se).

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Correct Answer: D. 4

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 Huckel Rule — 4n+2 π electrons in the cyclic conjugated π system (n = 0, 1, 2, 3 etc.)

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