Functional Group Reactions - GRE

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.

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 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 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.

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.

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.

N-bromosuccinimide (NBS) brominates at allylic positions. Br2 will not complete this reaction with the presence of the double bond.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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 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 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).

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.