Of the choices given, all can be made from some type of aldehyde reduction except choice II. Choices I, III, and IV each have the terminal (or primary) alcohol that is characteristic of a former aldehyde. In contrast, choice II has a secondary alcohol, characteristic of a former ketone. In other words, if choice II was oxidized then the product would be a ketone, not an aldehyde.

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.

Treatment of a carboxylic acid with acid results in decarboxylation, and the evolution of , especially if the resulting compound contains a benzylic or allylic carbon, as is the case here.

Radical stability increases as carbon substitution increases. In addition, radicals in conjugation with double bonds via resonance are more stable than the corresponding non-conjugated radical.

In this case radicals B, C, and D are all tertiary radicals, but only radical C has additional stabilization from resonance. Radical C is therefore the most stable. Radical E is the least substituted of the five radicals, and is the least stable.

The strong sulfuric acid protonates the hydroxyl group of compound B, resulting in the loss of water as a leaving group and the generation of a carbocation intermediate. Since this carbocation carbon is attached to three other carbons, this is a tertiary carbocation. It is bound to the phenyl substituent, a methyl group, and the branched carbon chain.

A balanced equation of the combustion of pentane indicates that one mole of pentane reacts with eight moles of oxygen gas to produce five moles of carbon dioxide and six moles of water.

Because there are nine moles of reactant and eleven moles of product, entropy increases in this reaction. Exothermic reactions that increase entropy are favorable at all temperatures, as seen in the Gibb's free energy equation.

In our scenario, H is negative and S is positive.

The heat of combustion for cycloalkanes can be quantified based on two factors: the number of carbons in the ring, and the amount of strain in the ring. Cyclohexane does in fact have a larger molar heat of combustion than cyclobutane. This is because there are more carbons in the ring; however, ring stability will determine the heat of combustion per  group in the ring. Even though cyclohexane has more carbons than cyclobutane, the heat of combustion per  group in cyclobutane will be greater compared to cyclohexane due to cyclobutane's ring strain. 

The answer is 1,3,5-hexatriene. Chemicals that are more stable will give off less heat when they are reduced or hydrogenated. Although these two compounds contain the same number of pi bonds to be reduced, benzene is aromatic, and therefore is much more stable than the conjugated non-cyclic hexatriene. We would find that the heat of hydrogenation for hexatriene would be noticeably greater than that of benzene.

A combustion reaction of any hydrocarbon yields the same products: carbon dioxide and water. The heat of combustion for the reaction shows how much energy is released as the hydrocarbon is converted to those products.

Since the products are the same for each alkene, any difference in heat of combustion will arise from differences in energy of the starting alkenes. Since the heat of combustion is negative, that means the reactants are higher in energy than the products, creating an exothermic reaction. Furthermore, the more negative the heat of combustion, the higher in energy the reactants are, and the less stable they are as well. Since compound D has the largest heat of combustion, it is highest in energy and therefore the least stable. By similar reasoning, B is the next highest in energy, followed by C, and then finally A, the most stable compound.

The three steps of a free radical chlorination reaction are, in order, initiation, propagation, and termination.

Free radicals are produced in the initiation and propagation steps. The termination steps combine any two free radicals formed in the reaction to produce a compound that has no unpaired electrons (free radicals).