Reactions by Product - Organic Chemistry

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.

Unlike E2 reactions, in which hydrogen abstraction occurs simultaneously with the dissociation of the leaving group (limiting the configuration of the reaction's product), E1 reactions occur in two distinct steps. The slow rate-determining step that must first occur is the dissociation of the leaving group. Leaving behind a carbocation intermediate, it is often necessary to consider possible carbocation rearrangements that would stabilize the positive charge.

In this case, no such rearrangement is favorable as their are no locations of greater stability available.

However, what must be considered is that the intermediate is free to orient itself in its most stable conformation prior to the formation of the double bond in the second step. As a result, the E product (the larger substituents are on oriented opposite one another with respect to the double bond) is yielded primarily.

To carry out this reaction, we need to create a radical as an intermediate, which is an unpaired electron. We do so by introducing , UV light, and heat to the 1-methyl cyclohexane. The light and the heat react with the to break the bond and create two radical bromine atoms. One of the radical bromine atoms removes a hydrogen from the carbon on the 1-methyl cyclohexane that is most substituted, and a radical carbon is formed. Finally, the second radical bromine reacts with the radical carbon to form the final product.

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.

Two sets of reagents are required to convert butanone into 2-butene. First, we use to reduce the butanone into a 2-butanol. Second, we use heat and acid to dehydrate the butanol and yield the final desired product.

1. ; 2. Heat/ may seem like an acceptable answer choice. However, note that the Grignard reagent converts the butanone into a tertiary alcohol, rather than a secondary alcohol as needed.

2-butone is a carbonyl compound that can readily be reduced by into a secondary alcohol, 2-butanol. When 2-butanol is heated in acid, we get dehydration, which leads to 2-butene.

This is an addition reaction with 3 products. The unknown reactant reacts with and gives those three products. Addition reactions begin with double bonded compounds and so these electrons are used to react with some reagent . One needs to work backwards to figure out how something was formed and in this case, there are mechanistic pathways, and one of the pathways involves a hydride shift. These 3 products often exist in different concentrations after the reaction.

Metallic sodium in liquid ammonia creates solvated electrons which can convert an alkyne to an E alkene. The same will not happen when sodium is combined with water, where sodium reacts violently to create sodium hydroxide and hydrogen gas. Lindlar's catalyst is a poisoned catalyst used to form alkenes from alkynes, bud results in a Z conformation. Without the poisoned catalyst, an alkane will be formed.

Below is the mechanism for the reaction given to form the alkene:

The reason this is the major product is because on tertiary alcohols are best dehydrated based on the E1 mechanism below:

The reason this is the major product is because tertiary alcohols are best dehydrated based on the E1 mechanism below:

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.

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.

In the first step of the reaction, a chlorine is abstracted and a double bond is formed. In the ozonolysis step, the molecule is broken at the double bond and each carbon at that bond gets double bonded to an oxygen. The presence of zinc keeps it from becoming a carboxylic acid.

Working backwards, you must remove the oxygens from the final product and redraw the molecule at the double bond. This intermediate is 1,3-dimethylcyclopentene. Next, the chlorine must be added to one of the carbons on the double bond. The only possible choice is 2-chloro-1,3-dimethylcyclopentane.

In order to convert a secondary alcohol into a ketone, we must employ as a reagent.

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.