Specific Reactions and Named Reactions - Organic Chemistry

This is a Diels-Alder reaction; these reactions happen between a nucleophilic diene, shown in blue below, and an electrophilic dienophile, in green. Diels-Alder reactions install a set of bonds that connect each external carbon of the diene system to an alkene carbon in the dienophile system to create a new six-membered ring. All remaining structure of the two reactants are retained, including the six- and five-membered rings below. The red bonds are the newly installed bonds.

The electrons from one of the double bonds on the 1,3-dibutene create a new single bond. The other new single bond is created from the electrons in the double bond of the other reactant. These two new single bonds join the reactants to create a cyclic product.

The electrons from the other double bond in the 1,3-dibutene move between the carbon 2 and 3. Thus, the final product is a 6-carbon cycloalkene with a halogen substituent.

This is a standard Diels-Alder reaction. Diels-Alder reactions are driven solely by adding heat to the reagents. By looking at the reagents and the product, we can tell that this is a Diels-Alder reaction. For Diels-Alder, we need a cis-diene and an alkene as reactants. When these reactants are stimulated by heat, they form a cyclohexene product.

Diels-Alder reactions create cyclohexene rings (eliminate III, IV, and V), and starting dienophile is trans (E conformation), so product is E (Eliminate I).

This is a classic Diels-Alder reaction and it consists of a diene (cyclopentadiene) and a dienophile (ethene). The bicyclic structure forms if the electrons are moved in a circular fashion.

The Diels-Alder reaction converts a conjugated diene and a substituted alkene into a six-membered ring containing cyclohexene (a substituted cyclohexene system). In Hoffmann elimination, tetra-alkyl ammonium salts undergo elimination to form the least substituted alkene. The Wittig reaction uses phosphorus ylides, aldehydes, or ketones to form an alkene and a triphenylphosphine oxide. Lastly, Gabriel synthesis forms primary amines via the reaction of a phthalimide with an alkyl halide, followed by cleavage with hydrazine.

The Ag2O oxidizes the alcohol to form the alkoxide ion. The Ag2O is a reducing agent, and it oxidizes the alcohol because the alcohol loses a proton. Ag2O is a strong base, meaning it will accept protons readily. The given reaction is an example of Williamson ether synthesis. This reaction occurs via an SN2 pathway.

The most acidic carbon in a dicarbonyl gets abstracted to form an enolate. The reaction involves a ketone with a double bond between the alpha and beta carbons. The carbon-carbon souble bond electrons attack the beta carbon of the alkene. At this point, the electrons from the alkene move to form a double bond between the carbonyl and alpha carbon and an enolate is fromed. The negative oxygen bonds to a hydrogen, forming the enol. Through enol-ketone tautamerization, a ketone is formed.

An alkyl halide and phtalimide react under basic conditions to form a primary amine. The halogen leves and the nitrogen of the phtahlimide bonds to that carbon on the alkane. In the next step, the nitrogen leave the phthalimide. In the presence of base, the phthalimide gains two negative oxygens in the stead of the nitrogen. The final product is a primary amine.

The numbering of the molecule begins across the double bond and goes in the direction where the lowest numbers for substituents can be achieved. Thus, the methyl group will be on carbon two and the bromine on carbon 3. Substituents are ordered alphabetically. The base molecule is a six-membered ring with one double bond. This is called a cyclohexene.

The longest chain has five carbons and a double bond, so the base molecule is pentene. There is a methyl group on carbon 3. The double bond is between carbons two and three. The highest priority substituents are the ethyl group on one side and the methyl group on the other. One is up and one is down, so the molecule is E as opposed to Z.

In a free radical halogenation, a bromine group will form a bond with the most substituted carbon atom. This method leads to the most stable radical intermediate.

The reaction given is called a Fischer esterification reaction. It is a reaction between an alcohol and a carboxylic acid to form an ester. Carboxylic acids are not reactive enough to be attacked by an alcohol so the reaction requires a strong acid catalyst like hydrochloric acid for the reaction to occur. Below is the mechanism:

The compound given has an aldehyde functional group and the parent chain has to have the -CHO group. Below is how the compound given should be numbered:

Aldehydes are named by changing the ending -e from the corresponding alkane with -al. The substituents on the parent chain should be named in alphabetical order. Therefore the answer is 2-ethyl-4-methylpentanal.

We have to find the longest chain to name it as the parent chain starting at the end with the nearest the substituent. Below is how the compound given should be numbered:

The substituents on the parent chain should be named in alphabetical order and because we have 2 bromine substituents we must add the name di-bromo and add the number of the carbon atom on the parent chain in which they appear. Therefore the answer is 2,3-dibromo-4-methylhexane.

The compound given has an alcohol functional group and the parent chain has to have the hydroxyl functional group. Below is how the compound given should be numbered:

Alcohols are named by changing the ending -e from the corresponding alkane with -ol. The substituents on the parent chain should be named in alphabetical order. Therefore the answer is 3-chloro-4-methylhexanol.

The compound given has an alkene functional group and the longest hydrocarbon chain containing the double bond will be used as the parent chain. Below is how the compound given should be numbered:

We must use the E, Z system of naming alkenes that assigns priority to substituents on the based on atomic number. This compound must be labeled with a Z geometry because the substituents with the higher priority are on the same side. Because a methyl group is on the 4th carbon atom on the parent chain, we must add the name 4-methyl to the naming of this compound. Therefore the answer is Z-4-methyl-2-octene.

The compound given has an alkene functional group and the longest hydrocarbon chain containing the double bond will be used as the parent chain. Below is how the compound given should be numbered:

We must use the E, Z system of naming alkenes that assigns priority to substituents on the based on atomic number. This compound must be labeled with an E geometry because the substituents with the higher priority are on opposite sides. Because a methyl group is on the 5th carbon atom on the parent chain, we must add the name 5-methyl to the naming of this compound. Therefore the answer is E-5-methyl-2-heptene.