All GRE Subject Test: Chemistry Resources
Example Questions
Example Question #1 : Reaction Mechanisms
When exposed to a good nucleophile, which molecule will most readily undergo an reaction?
reactions, also known as unimolecular nucleophilic substitution reactions, occur in two steps. Here, we are concerned with the first and second (rate-determining) steps, in which the leaving group breaks off of the molecule to form a carbocation. Alkanes that form the most stable carbocations are most likely to undergo
reactions. Tertiary carbocations are the most stable, followed by secondary. Primary and methyl carbocations are very unstable and unlikely to form at all. The tertiary alkane,
, will form a very stable tertiary carbocation compared to the other answer choices.
Example Question #1 : Reaction Types
What is created when a ketone is reacted with a phosphorus ylide?
Ester
Alkane
Aldehyde
Alkene
Alkene
The Wittig reaction involves a ketone or aldehyde reacting with a phosphorus ylide, a molecule with a negatively charged carbanion. The ketone will undergo nucleophilic addition and form a betaine. This intermediate will then form an alkene with a triphenylphosphine oxide being released. The Wittig reaction will form a mixture of both cis and trans isomers if the carbanion has two different substituents.
Wittig general reaction:
Example Question #1 : Functional Groups And Properties
Which of the following compounds would you expect to undergo a nucleophilic addition reaction?
Ethanamide
Acetic acid
Methyl ethanoate
Propanal
Propanal
When dealing with carbonyl compounds, remember that a carboxylic acid and all of its derivatives will undergo nucleophilic substitution. Aldehydes and ketones will undergo nucleophilic addition. Propanal is a three-carbon aldehyde, and will thus undergo nucleophilic addition.
Acetic acid is a carboxylic acid, methyl ethanoate is an ether, and ethanamide is an amide; each of these would undergo nucleophilic substitution.
Example Question #1 : Properties Of Hydrocarbons
Of the aromatic compounds shown above, which would be meta-directing groups for subsequent electrophilic aromatic substitution reactions?
Nitrobenzene
Anisole
Nitrobenzene and anisole
Bromobenzene
Nitrobenzene
Only nitrobenzene would be a meta-directing group for additional electrophilic aromatic substitution (EAS) reactions. Although the bromine of bromobenzene is an electron-withdrawing group, halogens are not meta-directors; therefore, additional EAS reactions with bromobenzene would result in ortho or para attached substituents.
Remember that ortho additions are adjacent to the first substituent, meta additions are two carbons displaced from the first substituent, and para additions are opposite the first substituent.
(Note that these reactions would take place much more slowly than if there was an electron-donating group attached).
Example Question #2 : Reaction Mechanisms
Which of the aromatic compounds (shown above) would undergo electrophilic aromatic substitution most quickly?
Bromobenzene
Nitrobenzene
Acetophenone
Anisole
Anisole
Electrophilic aromatic substitution occurs most rapidly when the aromatic compound has electron-donating groups attached. Due to their electron affinity, halogens are electron-withdrawing groups. Acetophenone and nitrobenzene both bear partial positive charges on the substituent directly attached to the benzene ring, which pulls electron density out of the ring as well, causing the reaction not to occur.
Anisole is the only compound with an electron-donating group, and is the correct answer. The lone pairs on the oxygen atom can be used to initiate new bonds.
Example Question #1 : Help With Alcohol Synthesis
What is the product of a hydroboration–oxidation reaction with 1-hexylcyclohexene?
2-hexylcyclohexanol
Hexylcyclohexane
Cyclohexane
3-hexylcyclohexanol
1-hexylcyclohexanol
2-hexylcyclohexanol
This reaction is an electrophilic addition reaction with an alkene. This is one of many alkene addition reactions that can add an -OH group onto your starting material. The key aspect of an hydroboration-oxidation reaction is the anti-Markovinikov addition to the double bond. The -OH group should be on the least substituted of the two carbons that originate from the double bond. In light of this information, the answer is 2-cyclohexanol.
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All GRE Subject Test: Chemistry Resources
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