Functional Group Reactions - GRE Subject Test: Chemistry
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What is the product of the reaction between 2-butanone and lithium aluminum hydride?
What is the product of the reaction between 2-butanone and lithium aluminum hydride?
Lithium aluminum hydride is a reducing agent. It reduces ketones and carboxylic acids to alcohols. 2-butanone is a 4-carbon chain with a double bond between an oxygen and carbon 2. The reduction turns the double bond with oxygen into a single bond to a hydroxy group. This makes the product 2-butanol.
Lithium aluminum hydride is a reducing agent. It reduces ketones and carboxylic acids to alcohols. 2-butanone is a 4-carbon chain with a double bond between an oxygen and carbon 2. The reduction turns the double bond with oxygen into a single bond to a hydroxy group. This makes the product 2-butanol.
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Which of the following can reduce an alkene to an alkane?
Which of the following can reduce an alkene to an alkane?
Neither lithium aluminum hydride, nor sodium borohydride will reduce C–C double bonds.
H2/Raney nickel and H2/Pd can each (individually) reduce an alkene to an alkane. Since both H2/Raney nickel and H2/Pd can reduce the alkene, the answer is both of those reagents. This is a catalytic hydrogenation reaction, and H2/Raney nickel not only reduces C–C double bonds, but also carbonyl compounds.
Neither lithium aluminum hydride, nor sodium borohydride will reduce C–C double bonds.
H2/Raney nickel and H2/Pd can each (individually) reduce an alkene to an alkane. Since both H2/Raney nickel and H2/Pd can reduce the alkene, the answer is both of those reagents. This is a catalytic hydrogenation reaction, and H2/Raney nickel not only reduces C–C double bonds, but also carbonyl compounds.
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Identify the major organic product expected from the acid-catalyzed dehydration of 2-methyl-2-pentanol.
Identify the major organic product expected from the acid-catalyzed dehydration of 2-methyl-2-pentanol.
The initial compound is a five-carbon alkane chain with methyl and hydroxy groups on the second carbon. Dehydration involves the hydrogenation of the hydroxy group. That group then leaves, and a double bond is formed. Zaitsev's rule states that double bonds are more stable on more highly substituted carbons. The double bond forms across carbons two and three.
The initial compound is a five-carbon alkane chain with methyl and hydroxy groups on the second carbon. Dehydration involves the hydrogenation of the hydroxy group. That group then leaves, and a double bond is formed. Zaitsev's rule states that double bonds are more stable on more highly substituted carbons. The double bond forms across carbons two and three.
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What is the IUPAC name of the given molecule?
What is the IUPAC name of the given molecule?
The longest carbon chain that can be formed is eight carbons. The base molecule is octane.
Using IUPAC rules, substituents should have the lowest possible numbers; thus, we start counting carbons from the right side rather than the left. If you count from the correct side, there are two methyl groups on carbon 3 and one on carbon 5. Thus, the name of the moleculue is 3,3,5-trimethyloctane.
The longest carbon chain that can be formed is eight carbons. The base molecule is octane.
Using IUPAC rules, substituents should have the lowest possible numbers; thus, we start counting carbons from the right side rather than the left. If you count from the correct side, there are two methyl groups on carbon 3 and one on carbon 5. Thus, the name of the moleculue is 3,3,5-trimethyloctane.
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How could you brominate the compound?
How could you brominate the compound?
The given molecule is an alkane. The only way to brominate an alkane is with bromine gas and UV light. The energy from the light serves to creat two radical bromines. These radicals are capable of bonding with alkanes. If the given compound were an alkene, either hydrobromic acid or bromine and peroxides would work.
The given molecule is an alkane. The only way to brominate an alkane is with bromine gas and UV light. The energy from the light serves to creat two radical bromines. These radicals are capable of bonding with alkanes. If the given compound were an alkene, either hydrobromic acid or bromine and peroxides would work.
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The stereochemical pathway for the hydrogenation of an alkene with a metal catalyst, such as platinum, occurs via __________.
The stereochemical pathway for the hydrogenation of an alkene with a metal catalyst, such as platinum, occurs via __________.
Hydrogenation of an alkene with a metal catalyst, such as platinum, occurs via syn addition.
It is important to note the three main types of reactions for alkenes. The first type of reaction is a 2-step mechanism in which the electrophile attacks the carbocation nucleophile. This can yield syn or anti products. The second type of reaction is a 2-step mechanism that forms a bridged carbocation as the intermediate. This can yield only anti products. The third and last type of reaction is a 1-step addition. This can only yield syn products.
An example of the third type of reaction is the addition of a hydrogen with palladium, platinum, or nickel as demonstrated in the picture.
Hydrogenation of an alkene with a metal catalyst, such as platinum, occurs via syn addition.
It is important to note the three main types of reactions for alkenes. The first type of reaction is a 2-step mechanism in which the electrophile attacks the carbocation nucleophile. This can yield syn or anti products. The second type of reaction is a 2-step mechanism that forms a bridged carbocation as the intermediate. This can yield only anti products. The third and last type of reaction is a 1-step addition. This can only yield syn products.
An example of the third type of reaction is the addition of a hydrogen with palladium, platinum, or nickel as demonstrated in the picture.
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What is(are) the product(s) in the Pd-catalyzed hydrogenation if 1,2-dimethylcyclopentene?
What is(are) the product(s) in the Pd-catalyzed hydrogenation if 1,2-dimethylcyclopentene?
The product for this hydrogenation is _cis-_1,2-dimethylcyclopentane.
It is important to note the three main types of reactions for alkenes. The first type of reaction is a 2-step mechanism in which the electrophile attacks the carbocation nucleophile. This can yield syn or anti products. The second type of reaction is a 2-step mechanism that forms a bridged carbocation as the intermediate. This can yield only anti products. The third and last type of reaction is a 1-step addition. This can only yield syn products.
The cis product alone forms because the reagents used were hydrogen and a metal catalyst palladium (other common metal catalysts are platinum and nickel). This type of reagent with an alkene will always be a 1-step addition that yields solely syn products. Cis-1,2-dimethylcyclopentane is the only answer that solely indicates syn products.
The product for this hydrogenation is _cis-_1,2-dimethylcyclopentane.
It is important to note the three main types of reactions for alkenes. The first type of reaction is a 2-step mechanism in which the electrophile attacks the carbocation nucleophile. This can yield syn or anti products. The second type of reaction is a 2-step mechanism that forms a bridged carbocation as the intermediate. This can yield only anti products. The third and last type of reaction is a 1-step addition. This can only yield syn products.
The cis product alone forms because the reagents used were hydrogen and a metal catalyst palladium (other common metal catalysts are platinum and nickel). This type of reagent with an alkene will always be a 1-step addition that yields solely syn products. Cis-1,2-dimethylcyclopentane is the only answer that solely indicates syn products.
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What reagent will complete this reaction?
What reagent will complete this reaction?
N-bromosuccinimide (NBS) brominates at allylic positions. Br2 will not complete this reaction with the presence of the double bond.
N-bromosuccinimide (NBS) brominates at allylic positions. Br2 will not complete this reaction with the presence of the double bond.
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What is the IUPAC name of the given diene?
What is the IUPAC name of the given diene?
You must begin counting the carbons so that the first functional substituent has the lowest possible number. In this case, C1 is connected to C2 by the double bond, meaning we start counting from the left.
The longest carbon chain is seven carbons so the parent molecule is heptane. With this numbering, there are methyl groups on carbons 3 and 6 and a chlorine on carbon 5.
Substituents are named in alphabetical order and two double bonds result in a diene. Thus, the correct answer is 5-chloro-3,6-dimethyl-1,5-heptadiene.
You must begin counting the carbons so that the first functional substituent has the lowest possible number. In this case, C1 is connected to C2 by the double bond, meaning we start counting from the left.
The longest carbon chain is seven carbons so the parent molecule is heptane. With this numbering, there are methyl groups on carbons 3 and 6 and a chlorine on carbon 5.
Substituents are named in alphabetical order and two double bonds result in a diene. Thus, the correct answer is 5-chloro-3,6-dimethyl-1,5-heptadiene.
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Starting with an alkyne, synthesis of a cis alkene is driven upon addition of which of the following reagents?
Starting with an alkyne, synthesis of a cis alkene is driven upon addition of which of the following reagents?
Reduction of an alkyne with hydrogen and Lindlar's catalyst will result in a cis alkene. While 
is a reducing agent, when added to an alkyne, a trans alkene is formed. Potassium permanganate is an oxidizing agent and thus will not reduce the triple bond. The Grignard reagent is used to add organic substituents onto carbonyls. Addition of one equivalent of chlorine in carbon tetrachloride solvent yields a trans alkene; addition of a second equivalent of chlorine yields a tetrachloro alkane.
Reduction of an alkyne with hydrogen and Lindlar's catalyst will result in a cis alkene. While
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What is the best reagent for abstracting a hydrogen from ethyne?
What is the best reagent for abstracting a hydrogen from ethyne?
The triple bond in ethyne makes the hydrogens slightly more acidic than those found in ethane. A very strong base, such as the conjugate base of ammonia, would be able to abstract that hydrogen. The abstraction turns the base into ammonia. It also creates a carbanion that can be used for chain extension and alkyne synthesis.
The triple bond in ethyne makes the hydrogens slightly more acidic than those found in ethane. A very strong base, such as the conjugate base of ammonia, would be able to abstract that hydrogen. The abstraction turns the base into ammonia. It also creates a carbanion that can be used for chain extension and alkyne synthesis.
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What is the product of the compound when it reacts with two equivalents of base?
What is the product of the compound when it reacts with two equivalents of base?
For each equivalent of base, a pi bond is formed between the carbons initially bound to the bromine atoms. For each bond formed, a bromine leaving group leaves the hydrocarbon. One equivalent of base abstracts a hydrogen. The electrons from the bond to the hydrogen create a pi bond. This occurs twice, and a triple bond is formed. The result is a 5-carbon chain with a triple bond between the second and third carbons. This molecule is 2-pentyne.
For each equivalent of base, a pi bond is formed between the carbons initially bound to the bromine atoms. For each bond formed, a bromine leaving group leaves the hydrocarbon. One equivalent of base abstracts a hydrogen. The electrons from the bond to the hydrogen create a pi bond. This occurs twice, and a triple bond is formed. The result is a 5-carbon chain with a triple bond between the second and third carbons. This molecule is 2-pentyne.
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Which of the following molecule is a type of terpene?
Which of the following molecule is a type of terpene?
Terpenes are special classes of lipids that are derived from isoprene units. An isoprene unit’s molecular formula is 
; therefore, terpenes must contain a molecular formula that is derived from a 
empirical formula.
The molecular formula 
consists of seven isoprene units (
= 
); therefore, 
is a type of terpene.
Pentene is an alkene with the molecular formula 
. 
(
) is derived from pentene units; however, terpenes do not have pentene units.
Terpenes are special classes of lipids that are derived from isoprene units. An isoprene unit’s molecular formula is
The molecular formula
Pentene is an alkene with the molecular formula
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Acetaldehyde 
undergoes a Wolf-Kishner reaction, which is the addition of hydrazine 
with subsequent addition of a base and heat. In this reaction, the aldehyde is __________, resulting in a(n) __________ product.
Acetaldehyde
The correct answer is that the aldehyde is reduced to an alkane. In viewing the final product, we see that acetaldehyde would be reduced to ethane. The reaction of any aldehyde or ketone with hydrazine and the subsequent addition of base and heat will result in that aldehyde or ketone being reduced to an alkane, and is referred to as the Wolf-Kishner reaction. The Wolf-Kishner reagent is a commonly tested reducing agent.
The correct answer is that the aldehyde is reduced to an alkane. In viewing the final product, we see that acetaldehyde would be reduced to ethane. The reaction of any aldehyde or ketone with hydrazine and the subsequent addition of base and heat will result in that aldehyde or ketone being reduced to an alkane, and is referred to as the Wolf-Kishner reaction. The Wolf-Kishner reagent is a commonly tested reducing agent.
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What reagent(s) will successfully complete the synthesis reaction shown above?
What reagent(s) will successfully complete the synthesis reaction shown above?
This is an example of a Grignard reagent reaction. Because we are adding three carbons to our chain, the Grignard reagent we need must have three carbons on it. We can therefore rule out methyl grignard and ethyl grignard.
N-propyl is the straight-chained 3-carbon alkane, while isopropyl is branched. Looking at our final product, we can see the carbon chain we have added is straight-chained, and thus N-propyl Grignard is the best option. Because Grignard reagents are relatively basic, we must add an hydronium ion workup to protonate our alcohol.
This is an example of a Grignard reagent reaction. Because we are adding three carbons to our chain, the Grignard reagent we need must have three carbons on it. We can therefore rule out methyl grignard and ethyl grignard.
N-propyl is the straight-chained 3-carbon alkane, while isopropyl is branched. Looking at our final product, we can see the carbon chain we have added is straight-chained, and thus N-propyl Grignard is the best option. Because Grignard reagents are relatively basic, we must add an hydronium ion workup to protonate our alcohol.
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Which of the following reactions would NOT produce a carboxylic acid?
Which of the following reactions would NOT produce a carboxylic acid?
PCC is considered a weak oxidizing agent. The reaction of a primary alcohol with PCC would only yield an aldehyde, while reaction with a secondary alcohol will yield a ketone. PCC will not be used to generate carboxylic acids.
A stronger oxidation, like 
or 
, is required to oxidize up to the carboxylic acid. Treatment of an ester with a base or treatment of carbon dioxide with a Grignard reagent are other ways of making carboxylic acids.
PCC is considered a weak oxidizing agent. The reaction of a primary alcohol with PCC would only yield an aldehyde, while reaction with a secondary alcohol will yield a ketone. PCC will not be used to generate carboxylic acids.
A stronger oxidation, like
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Which one of the following compounds can produce a carboxylic acid only when reacted with sodium dichromate and sulfuric acid?
Which one of the following compounds can produce a carboxylic acid only when reacted with sodium dichromate and sulfuric acid?
After recognizing the reagents given, we know we need to begin with an alcohol. The alcohol choices differ only by how substituted they are.
For a carboxylic acid to form from a reaction with sodium dichromate and sulfuric acid, a primary alcohol needs to be available. Therefore, 1-pentanol is the correct answer.
After recognizing the reagents given, we know we need to begin with an alcohol. The alcohol choices differ only by how substituted they are.
For a carboxylic acid to form from a reaction with sodium dichromate and sulfuric acid, a primary alcohol needs to be available. Therefore, 1-pentanol is the correct answer.
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When 3-chloroheptane undergoes malonic ester synthesis, the final product is __________.
When 3-chloroheptane undergoes malonic ester synthesis, the final product is __________.
The malonic ester is deprotonated at the most acidic hydrogen, the one on the carbon between the two oxygens. The electrons from that bond to the hydrogen form a carbon-carbon double bond while electrons from the oxygen-carbon double bond go to the oxygen atom. This is the enolate form. The chlorine leaves the 3-chloroheptane and the electrons from the carbon-carbon double bond bond to the carbon that the chlorine left. At the same time, the carbonyl is reformed. After deesterfication, 3-ethylheptanoic acid is formed.
The malonic ester is deprotonated at the most acidic hydrogen, the one on the carbon between the two oxygens. The electrons from that bond to the hydrogen form a carbon-carbon double bond while electrons from the oxygen-carbon double bond go to the oxygen atom. This is the enolate form. The chlorine leaves the 3-chloroheptane and the electrons from the carbon-carbon double bond bond to the carbon that the chlorine left. At the same time, the carbonyl is reformed. After deesterfication, 3-ethylheptanoic acid is formed.
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Which of the following would be a major product of the given reaction?
Which of the following would be a major product of the given reaction?
The reaction of a primary amine with a ketone or aldehyde produces an imine by nucleophilic addition. Below is the mechanism for the reaction given:
The reaction of a primary amine with a ketone or aldehyde produces an imine by nucleophilic addition. Below is the mechanism for the reaction given:
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Which of the following statements is false?
Which of the following statements is false?
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).
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).
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