The haloform reaction requires a carbonyl with a terminal alpha carbon. A hydrogen gets abstracted, and the enolate is formed. A halogen attacks the alpha carbon, and the ketone is reformed. This occurs three more times until the carbon has bonds to three halogens. Then the carbon leaves, forming a carbanion, and the base attacks the carbonyl carbon. An ester is formed.

During acid catalyzed hydration, a hydroxy group replaces one of the bonds in the triple bond and a double bond is formed. This is called an enol. The enol naturally turns into a ketone in a process called tautomerization. The hydroxy group can attach to either carbon across the double bond, and naming is done so that substituents have the lowest numbers. Only on 5-decyne will result in a single product, as no matter which carbon the hydroxy group bonds to, it is still on carbon 5. Thus, the only final product is 5-decone.

The other answer options will still react, but will form multiple products due to lack of symmetry.

First step: bromination across the double bond

Second step: double dehydrohalogenation and removal of terminal alkyne hydrogen

Third step: neutralization of the molecule

Fourth/fifth step: hydroboration/oxidation, followed by keto/enol tautomerization

When naming esters, the root word in the name describes the carbon chain that carries the carbonyl. Saponification describes the breaking up of an ester. This yields an alcohol and a carboxylic acid. The chain with the carbonyl is butyl, so the carboxylic acid will be butanoic acid. That means the side with the alcohol will be propanol. 

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.