Reduction of an alkyne with hydrogen and Lindlar's catalyst will result in a cis alkene. While  is a reducing agent, when added to an alkyne, a trans alkene is formed. Potassium permanganate is an oxidizing agent and thus will not reduce the triple bond. The Grignard reagent is used to add organic substituents onto carbonyls. Addition of one equivalent of chlorine in carbon tetrachloride solvent yields a trans alkene; addition of a second equivalent of chlorine yields a tetrachloro alkane.

Terpenes are special classes of lipids that are derived from isoprene units. An isoprene unit’s molecular formula is ; therefore, terpenes must contain a molecular formula that is derived from a empirical formula.

The molecular formula consists of seven isoprene units ( = ); therefore,  is a type of terpene.

Pentene is an alkene with the molecular formula .  () is derived from pentene units; however, terpenes do not have pentene units.

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.