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.  

Lithium aluminum hydride is a reducing agent. It reduces ketones and carboxylic acids to alcohols. 2-butanone is a 4-carbon chain with a double bond between an oxygen and carbon 2. The reduction turns the double bond with oxygen into a single bond to a hydroxy group. This makes the product 2-butanol.

The reaction of a primary amine with a ketone or aldehyde produces an imine by nucleophilic addition. Below is the mechanism for the reaction given:

PCC is considered a weak oxidizing agent. The reaction of a primary alcohol with PCC would only yield an aldehyde, while reaction with a secondary alcohol will yield a ketone. PCC will not be used to generate carboxylic acids.

A stronger oxidation, like or , is required to oxidize up to the carboxylic acid. Treatment of an ester with a base or treatment of carbon dioxide with a Grignard reagent are other ways of making carboxylic acids.

After recognizing the reagents given, we know we need to begin with an alcohol. The alcohol choices differ only by how substituted they are.

For a carboxylic acid to form from a reaction with sodium dichromate and sulfuric acid, a primary alcohol needs to be available. Therefore, 1-pentanol is the correct answer.

The malonic ester is deprotonated at the most acidic hydrogen, the one on the carbon between the two oxygens. The electrons from that bond to the hydrogen form a carbon-carbon double bond while electrons from the oxygen-carbon double bond go to the oxygen atom. This is the enolate form. The chlorine leaves the 3-chloroheptane and the electrons from the carbon-carbon double bond bond to the carbon that the chlorine left. At the same time, the carbonyl is reformed. After deesterfication, 3-ethylheptanoic acid is formed.

In the HVZ reaction of a carboxylic acid, a bromine is added to the alpha carbon. Phosphorus catalyzes the reaction and allows for the formation of an acyl halide. Acyl halides readily undergo enol-ketone tautomerization. The enol form uses electrons from the carbon-carbon double bond to bond to the bromine. In water, the two bromine molecules convert back to a carboxylic acid with one bromine on the alpha carbon.

The statement "The catalyzed ring opening of an epoxide in aqueous acid will yield a cis glycol" is incorrect. These reaction conditions will yield a trans glycol. In fact, regardless of conditions, the opening of an epoxide will always yield a trans glycol (the two alcohol groups are on opposite sides).

As the first step in a substitution reaction involves a nucleophilic attack at an electrophilic carbonyl carbon, we must consider the varying reactivity of the electrophilic carbonyl center. Resonance diagrams, as well as an understanding of electronegativity, will help us understand the degree to which this effect is observed in a substrate. 

Resonance diagrams for all four substrates show how electrons contained in the leaving group's heteroatom may be shared throughout the carbonyl system, effectively placing a partial negative charge on the electrophilic carbon. To determine which is the most electrophilic, we must identify the resonance diagram below that contributes the least to the overall molecule. This molecule will be least stable and most reactive.

Note: Remember, resonance diagrams show possible electron distributions, and a molecule exists as a weighted average of these possibilities, favoring the more stable ones. 

Compound II is the most electrophilic substrate, as the lone pair on the central oxygen molecule must be shared between two carbonyls. The resonance forms below each contribute very little to the overall molecule. This is not the case in any other pictured substrate.

Now compare compounds I and III. Resonance for these molecules is essentially identical, with a nitrogen atom in compound I and an oxygen atom in compound III. We may conclude that the resonance form of compound III contributes less to the true existence of the molecule, as oxygen is more electronegative. The sharing of electrons will be less favorable in the resonance form of compound III than the resonance form of compound I. 

For compound IV, both resonance structures are equally stable, and the molecule will exist as an average of both structures, placing a fair amount of electron density at the carbonyl carbon, drastically reducing the electrophilicity of the central carbon.

If this above explanation is confusing to you, you may also compare how good the leaving groups are. Acetate, the leaving group of compound II, is a stable ion and will readily leave in a substitution reaction. Methoxide is the next best leaving group, from compound III, followed by the negatively charged ethanamine leaving group from compound I. As  will be a terrible leaving group, a substitution reaction with carboxylate substrates, such as compound IV, will never occur.

The compounds, in order of reactivity, are II > III > I > IV.