DAT Question of the Day

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).

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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Correct Answer: C. 3 + 6 only

The cyclohexene product is indicative of a Diels-Alder reaction, the cycloaddition of a conjugated diene (electron rich) and dienophile (electron poor) components. For best reactivity, the diene must be able to adopt a cisoid conformation. The reaction is stereospecific with respect to both diene and dienophile, meaning a cis-dienophile gives cis substitution in the product (same applies for trans-dienophiles -> trans product substitution). If diene substituents have the same stereochemistry (i.e. both alkenes are E), both substituents will end up on the same face of the product. Mixed stereochemistry (i.e. E and Z) in the diene results in substituents on opposite faces of the product. The final product has syn-substituted diene components (diene is either E,E or Z,Z), and dienophile substituents are trans, (original dienophile has trans geometry). Hence, only answer choice [C] contains reactants that will form the desired product, with reactant 3 as the E,E diene, and reactant 6 as the trans dienophile. Answer choice [B] states reactants 1 and 6 will produce the final product in high yield, however, reactant 1 contains mixed stereochemistry (E,Z bonds) and will result in the methyl substituents on opposite faces of the product.

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Correct Answer: A. 1

Simple hydrocarbons (containing only C and H) are weakly acidic in comparison to other organic compounds. However, carbanions can be formed, with their stability depending quite heavily on the hybridization state of the orbital containing the lone pair of electrons. Alkyne 1 is most acidic as the lone pair of the conjugate base is held in a sp-orbital, rather than sp3 (for 2 and 5) or sp2 orbital (for 3 and 4). Having a greater proportion of “s character” allows for the electron density in an sp-orbital to be held closer to the positively charged nucleus, making it electrostatically more stable.

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Correct Answer: E. Addition of a nucleophile to the carbonyl is followed by loss of the leaving group.

Carbonyl compounds react with nucleophiles via an addition mechanism: the nucleophile attacks the carbonyl carbon, forming a tetrahedral intermediate. The tetrahedral intermediate itself can be an alcohol or alkoxide, depending on the pH of the reaction. The tetrahedral intermediate of an acyl compound contains a substituent attached to the central carbon that can act as a leaving group. After the tetrahedral intermediate forms, it collapses, recreating the carbonyl C=O bond and ejecting the leaving group in an elimination reaction. In nucleophilic acyl substitution, the leaving group is ejected in a separate exothermic step. The Hammond Postulate predicts that the transition state for this step resembles the tetrahedral intermediate. Therefore, in the transition state for this step minimal bond breaking has occurred, and the importance of the nature of the leaving group is diminished.

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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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