DAT Question of the Day

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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Correct Answer: E. I, II, and III.

All three reagents are used for the esterification of carboxylic acids. Of these methods, III is the most general. Method I can be used if the alkyl halide is sufficiently electrophilic (methyl iodide). Method II (diazomethane) is used for making methyl esters on a small scale.

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Correct Answer: B. 2-bromo-2-methylpentane

This is an example of a free-radical halogenation. Homolysis of the bromine bond by light initiates the reaction. Chain propagation occurs whereby hydrogen abstraction yields an alkyl radical plus HBr. The alkyl radical reacts by reaction with another molecule of bromine, regenerating a bromine radical to engage in further reactions. Chain termination occurs whereby two free radicals combine. Carbon-centered radicals follow the same rules of stability as carbocations: the greater number of alkyl groups directly connected to a carbon atom, the greater the stabilization.

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