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.

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 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: