The most acidic hydrogen of the ester gets abstracted and the enolate form of the compound is attained. The electrons from the carbon-carbon double bond attack the carbonyl carbon of the other ester. Deesterfication occurs in this same step. The final product has one ketone and one ester.

The reaction uses a Grignard reagent's nucleophilic carbon to attack the carbon in carbon dioxide. Following treatment with water the resulting molecule (a carboxylate anion, of the form ) the carboxylate oxygen will be protonated, and the result is a carboxylic acid. Thus, the resulting molecule is a benzene ring with a carboxyl group attached (this is also known as benzoic acid).

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 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.

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. 

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 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.

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.