Reactions by Reactant - Organic Chemistry

First step: elimination

Second step: Addition of two bromines across the double bond

Third step: Double dehydrohalogenation to form the alkyne, and removal of alkyne hydrogen (deprotonation)

Fourth step: methylation reaction (SN2)

Fifth step: metal-ammonia reduction forms trans alkene product

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).

Whenever an alkane is reacted with excess oxygen in a high temperature reaction, the reaction is known as combustion. Combustion of an alkane always produces carbon dioxide gas and water vapor.

Therefore the correct answer is: and .

Reactions that proceed by syn addition include hydrogenation, epoxidation, and hydroboration-oxidation. Reactions that proceed by anti addition include halogenation, halohydrin addition and epoxidation followed by ring opening. Syn addition occurs because reagents must approach from only one side of the double bond.

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.

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.

Reaction of an alkene with osmium tetroxide results in syn addition of to hydroxyl groups across the double bond, resulting in a 1,2-diol. The correct answer is the option in which hydroxide groups are added to the previously double-bonded carbons on the same side of the ring. The correct answer is compound II.

This is an alkene electrophilic addition reaction. The alkene first attacks the hydrogen on the . The hydrogen always attaches to the last carbon, never the third. This form of regioselectivity is referred to as the Markovnikov rule. This occurs because it creates a secondary carbocation, rather than a primary carbocation intermediate, which is far more stable. The bromine is then free to attack the third, now positively charged carbon.

The reaction given is called a Markovnikov's addition reaction. This is a regiospecific reaction because only one orientation of the reaction occurs. It is an electrophilic addition reaction that involves the reaction of a halogenic acid such as or with an alkene. The hydrogen bonds to the carbon with the less alkyl groups bonded to the organic compound and the halogen ion bonds to the carbon with more alkyl substituents. Therefore the answer is:

When reacting with alkenes, the reagents add two syn alcohols, meaning that the alcohols face the same way in space. The alcohols can either be configured in the front (wedge) or the back (dash). Therefore, the correct answer is two of these, with the correct products being molecule 1 and molecule 2.

are the only reagents that do not elicit Markovnikov addition of the available answer choices. Instead, elicits an anti-Markovnikov addition of hydrogen and bromine to an alkene.

This is an addition reaction with being added to an alkene (ionic addition). To reach the correct answer first we need to add the to one of the carbons by using electrons from the double bond. This next step is crucial because the positive charge (carbocation) that forms needs to be in the most substituted position. However, hydride shifts can occur moving a hydrogen atom over to an adjacent carbon to form the most stable carbocation. In this case, the positive charge on the secondary carbon adjacent to the tertiary carbon will first get the positive charge and then a hydride shift will occur amongst those two carbon leading to a more stable tertiary carbocation. Finally, the will attack the carbocation from both sides, and then we will have an alkyl halide. The most stable intermediate (carbocation) produces the greatest quantity of the stable product.

This is a classic acid catalyzed alkene reaction. Because of resonance, several reactions have multiple products.

2-Methyl-2-butene will react the fastest with hydrobromic acid because when it reacts, it will form the most stable carbocation among all of the alkenes listed. This stability allows for the reaction to proceed much quicker than other compounds, therefore it will react the fastest.

The ozonolysis of alkenes forms either ketones or aldehydes. Subsequent reagents can be added to produces alcohols and/or carboxylic acids.

The reaction given is called a Markovnikov's addition reaction. This is a regiospecific reaction because only one orientation of the reaction occurs. It is an electrophilic addition reaction that involves the reaction of a halogenic acid such as or with an alkene. The hydrogen bonds to the carbon with the less alkyl groups bonded to the organic compound and the halogen ion bonds to the carbon with more alkyl substituents. Therefore the answer is:

The reaction mechanism goes through a carbocation intermediate as follows:

Therefore the answer is the following carbocation intermediate: