Keep in mind the following principles: Cyclization is favored when a five/six-member ring may be formed. Addition at an aldehyde is favored relative to the same reaction at a ketone. 

As a result, abstraction of a hydrogen bound to carbon 6 (an alpha-carbon) is favored since the resulting carbanion may attack the aldehyde (carbon 1) to form a six-member ring, resulting in compound II. Compound I results from abstracting a hydrogen from carbon 2, generating a carbanion which may then attack the ketone. Based on the latter of the above principles, this is a minor product.

First step: Friedel-Crafts acylation of benzene

Second step: Formation of enolate

Third step: aldol addition (enolate attacks carbonyl carbon in benzaldehyde)

Fourth step: neutralization of anion and dehydration forming alkene

This is a classic esterification reaction. Esterfication occurs when a carboxylic acid and an alcohol are reacted together. Only one answer choice is an ester. 

First step: esterification

Second step: reduction

Third step: neutralization

Fourth step: oxidation to aldehyde

Fifth step: alkene metathesis

The Wittig reaction involves the reaction of a phosphonium ylide (generated by treating a phosphonium salt with a strong base) with a ketone or aldehyde.

The reaction proceeds through a phosphaoxetane (4-membered ring containing both phosphorus and oxygen) intermediate to generate a new compound containing a carbon-carbon double bond, plus a phosphine oxide byproduct. It does not form trans double bonds exclusively; sometimes, a mixture of cis and trans isomers are obtained, and sometimes the cis isomer is the predominant product. 

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.