GRE Subject Test: Chemistry : Other Carbonyl Chemistry

Study concepts, example questions & explanations for GRE Subject Test: Chemistry

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Example Questions

Example Question #15 : Organic Chemistry

When butanoic acid undergoes the Hell-Volhard-Zelinsky (HVZ) reaction, the final product is __________.

Possible Answers:

None of the other answers

butanoyl bromide

2-bromobutanoic acid

heptanoic acid

3-bromobutanoic acid

Correct answer:

2-bromobutanoic acid


In the HVZ reaction of a carboxylic acid, a bromine is added to the alpha carbon. Phosphorus catalyzes the reaction and allows for the formation of an acyl halide. Acyl halides readily undergo enol-ketone tautomerization. The enol form uses electrons from the carbon-carbon double bond to bond to the bromine. In water, the two bromine molecules convert back to a carboxylic acid with one bromine on the alpha carbon.

Example Question #22 : Carbonyl Reactants

Which of the following statements is false?

Possible Answers:

A nucleophile attacks an epoxide to yield an alcohol

Treatment of an unsymmetrical epoxide with methanol and acid will result in nucleophilic attack at the more hindered carbon

The catalyzed ring-opening of an epoxide in aqueous acid will yield a cis glycol

Treatment of an unhindered epoxide with a Grignard reagent will result in a nucleophilic attack at the less hindered carbon

The base-promoted ring-opening of an epoxide using aqueous NaOH will yield a trans glycol

Correct answer:



The statement "The catalyzed ring opening of an epoxide in aqueous acid will yield a cis glycol" is incorrect. These reaction conditions will yield a trans glycol. In fact, regardless of conditions, the opening of an epoxide will always yield a trans glycol (the two alcohol groups are on opposite sides).

Example Question #23 : Carbonyl Reactants

All of the following electrophilic substrates can theoretically undergo substitution reactions, however, at different rates. Rank them from most to least reactive in the presence of a nucleophile.


Possible Answers:

I > III > II > IV

II > III > I > IV

III > I > II > IV

I > IV > III > II

II > IV > III > I

Correct answer:

II > III > I > IV


As the first step in a substitution reaction involves a nucleophilic attack at an electrophilic carbonyl carbon, we must consider the varying reactivity of the electrophilic carbonyl center. Resonance diagrams, as well as an understanding of electronegativity, will help us understand the degree to which this effect is observed in a substrate. 

Resonance diagrams for all four substrates show how electrons contained in the leaving group's heteroatom may be shared throughout the carbonyl system, effectively placing a partial negative charge on the electrophilic carbon. To determine which is the most electrophilic, we must identify the resonance diagram below that contributes the least to the overall molecule. This molecule will be least stable and most reactive.

Note: Remember, resonance diagrams show possible electron distributions, and a molecule exists as a weighted average of these possibilities, favoring the more stable ones. 

Compound II is the most electrophilic substrate, as the lone pair on the central oxygen molecule must be shared between two carbonyls. The resonance forms below each contribute very little to the overall molecule. This is not the case in any other pictured substrate.

Now compare compounds I and III. Resonance for these molecules is essentially identical, with a nitrogen atom in compound I and an oxygen atom in compound III. We may conclude that the resonance form of compound III contributes less to the true existence of the molecule, as oxygen is more electronegative. The sharing of electrons will be less favorable in the resonance form of compound III than the resonance form of compound I. 

For compound IV, both resonance structures are equally stable, and the molecule will exist as an average of both structures, placing a fair amount of electron density at the carbonyl carbon, drastically reducing the electrophilicity of the central carbon.



If this above explanation is confusing to you, you may also compare how good the leaving groups are. Acetate, the leaving group of compound II, is a stable ion and will readily leave in a substitution reaction. Methoxide is the next best leaving group, from compound III, followed by the negatively charged ethanamine leaving group from compound I. As  will be a terrible leaving group, a substitution reaction with carboxylate substrates, such as compound IV, will never occur.

The compounds, in order of reactivity, are II > III > I > IV.

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