Organic Chemistry - GRE Subject Test: Chemistry

All of these methods can successfully synthesize secondary alcohols.

Lithium aluminum hydride and sodium borohydride are strong and weaker reducing agents, respectively. Both are able to reduce ketones to secondary alcohols.

Adding water and acid to an alkene (such as propene) results in Markovnikov addition of a hydroxyl group, also creating a secondary alcohol.

Grignard reagents (organometallic halides) add to the carbonyl carbon of an aldehyde, adding an alkane group and forming an alcohol product.

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.

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.

A secondary amine is an amine (nitrogen atom) that is attached to two carbon-containing groups (alkyl groups or aryl groups). The nitrogen in ephedrine is attached to two alkyl groups, making it a secondary amine.

Primary amines are generally written as . Secondary amines are generally written as . A tertiary amine will be bound to three different R-groups. Quaternary amines require a positive charge on the nitrogen atom to accommodate a fourth R-group.

Huckel's rule states that an aromatic compound must have delocalized electrons. The electrons in each double bond are delocalized for this molecule. There are nine double bonds, and thus eighteen delocalized electrons.

If 4n+2=18, then n=4.

The more stable the carbocation, the lower the activation energy for reaching that intermediate will be. The more substituted a carbocation is, the more stable it is. The carbocation bonded to three alkanes (tertiary carbocation) is the most stable, and thus the correct answer.

Secondary carbocations will require more energy than tertiary, and primary carbocations will require the most energy.

The strong sulfuric acid protonates the hydroxyl group of compound B, resulting in the loss of water as a leaving group and the generation of a carbocation intermediate. Since this carbocation carbon is attached to three other carbons, this is a tertiary carbocation. It is bound to the phenyl substituent, a methyl group, and the branched carbon chain.

The correct answer is: carbon 1 because the intermediate will then be a secondary carbocation.

This is a case of addition across a double bond where stands for any halide. The first step in this reaction is the attack of by the double bond. This will create two intermediates, the first being the halide anion (so in our case ), the second being a carbocation on our compound at one of the two carbons that formerly shared the double bond.

If the hydrogen attached the carbon 2 then we would have a positive charge on carbon 1 and vice versa. A positive charge on carbon 1 is known as a primary carbocation (a carbon attached to 1 other carbon or function group), which is rarely if ever seen due to its overwhelming instability. A positive charge on carbon 2 is known as a secondary carbocation (a carbon attached to 2 other carbons or functional groups) and is much more stable than a primary carbocation.

We would want the more thermodynamically stable intermediate for our reaction to proceed, so we would want the positive charge on carbon 2 and the hydrogen attached to carbon 1.