The correct answer is that the aldehyde is reduced to an alkane. In viewing the final product, we see that acetaldehyde would be reduced to ethane. The reaction of any aldehyde or ketone with hydrazine and the subsequent addition of base and heat will result in that aldehyde or ketone being reduced to an alkane, and is referred to as the Wolf-Kishner reaction. The Wolf-Kishner reagent is a commonly tested reducing agent.

This is an example of a Grignard reagent reaction. Because we are adding three carbons to our chain, the Grignard reagent we need must have three carbons on it. We can therefore rule out methyl grignard and ethyl grignard.

N-propyl is the straight-chained 3-carbon alkane, while isopropyl is branched. Looking at our final product, we can see the carbon chain we have added is straight-chained, and thus N-propyl Grignard is the best option. Because Grignard reagents are relatively basic, we must add an hydronium ion workup to protonate our alcohol.  

The reaction of a Grignard reagent with oxirane (a type of epoxide) in addition to the work-up with dilute acid will yield a primary alcohol solely because there is a work-up with dilute acid. This shows that there is an excess of hydrogen, which will yield to a primary alcohol versus a secondary or tertiary alcohol.

Grignard reagents (often formed in-situ due to their highly non-specific reactivity) act as reducing agents by forming carbanions, which are strong bases and nucleophiles. Nitriles may react with Grignard reagents to form imines. The reaction proceeds in an analogous fashion to a standard nucleophilic carbonyl addition, converting the triple bond to a double bond by forming an intermediate in which nitrogen carries a negative charge. A protic solvent, such as ethanol, is then added to neutralize the intermediate. Thus, the correct answer is the only compound in which an imine is formed, which is found in compound I.

Step 1 converts the carboxylic acid into an ester.

Step 2 adds two equivalents of Grignard reagent: one to turn the molecule into acetone, and a second one to turn it into tert-butoxide. There is no way to stop the reaction after the first addition of Grignard reagent.

Step 3 will neutralize the base, leaving only t-butyl alcohol (I).

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.