Water, , has an oxygen atom with an oxidation number of  whereas hydrogen peroxide, , has an oxygen atom with an oxidation number of . A less negative oxidation number suggests that the oxygen lost an electron. Recall that oxidation is the loss of electrons whereas reduction is the gain of electrons; therefore, the oxygen atom lost an electron and was oxidized when water was converted to hydrogen peroxide.

Oxidation is the process of removing electrons from an atom. This increases the oxidation number (makes oxidation number more positive). An atom undergoes oxidation by losing electrons and donating them to another atom. Since it is facilitating reduction (gain of electrons) of another atom, an atom that is oxidized is also called a reducing agent. Oxidizing agents, on the other hand, are reduced (gain electrons) and facilitate the oxidation of other atoms (removal of electrons from other atoms). Oxidation of an atom does not depend on the Gibbs free energy; therefore, oxidation can be spontaneous or nonspontaneous.

 is the only option that will not reduce a ketone to an alcohol, simply because it is not a reducing agent like the other four—it is an oxidizing agent. The addition of oxygens to a ketone will not yield an alcohol (oxidation). The addition of hydrogen to a ketone will yield an alcohol (reduction). The other answer options are reducing agents that would facilitate this reaction.

Reduction involves gain of electrons whereas oxidation involves loss of electrons. In reduction, the oxidation number becomes more negative (due to gain of electrons) whereas in oxidation, the oxidation number becomes more positive. Sodium is an alkali metal, found on the first column of the periodic table. Every atom in this column has one valence electron; therefore, to complete its octet every alkali metal will lose an electron and will have an oxidation number of . It is very hard to oxidize and reduce these metals because of its desire to maintain octet; therefore, sodium can never be reduced or oxidized (it will always have an oxidation number of ).

Oxidation and reduction involve loss and gain of electrons, respectively. It does not involve loss or gain of protons. Recall that the identity of an atom is changed when the amount of protons change; therefore, it is very hard to change the amount of protons. Only few reactions (such as nuclear decay reactions) can change the number of protons and alter the identity of an atom.

The S_N1 mechanism involves the formation of a carbocation intermediate in the rate-determining step. The most stable carbocation will produce the fastest reaction. We can immediately eliminate any answer choices that will produce primary or secondary carbocations, since a tertiary carbocation will be much more stable. When comparing tertiary carbocations, larger and more electronegative substituents will allow for more charge stabilization.

Since the tertiary carbocation formed by the dissociation of iodide from will the be most stable, this substrate will react the fastest.

The rate of an  reaction is determined only by the concentration of the substrate. Unlike an  reaction, where the addition occurs in one step and requires the activity of the substrate and the nucleophile, an  reaction occurs in two steps and is only limited by the activity (i.e. leaving ability) of the substrate. Once the leaving group leaves the substrate, the nucleophile does not hesitate to attack the exposed carbocation.

Keep in mind that after the aldehyde is reduced into an alcohol, the molecule can undergo an intramolecular reaction, as alcohol is a good nucleophile and the halogen is a stellar leaving group.

A strong nucleophile is not required for SN1. A weak nucleophile may be used. Remember that the SN1 mechanism goes through a carbocation intermediate.

Rearrangements are possible for SN1 reactions (not SN2). A rearrangement will occur to create a more stable intermediate in the mechanism. For example, if the carbocation is secondary, a methyl shift may occur to make the carbocation intermediate tertiary.

A racemic mixture of products occurs when with the nucleophile may attack the carbocation from either the top face or bottom face. When a reaction goes through a carbocation intermediate, as in SN1, there may be a racemic mix of products.

SN1 is unimolecular, and the rate of the reaction is determined by the substrate and reaction constant.

SN1 reactions result in racemization when the nucleophile has a 50% chance of attacking the carbocation intermediate from the top face, and a 50% chance of attacking from the bottom face. SN1 reactions are favored in polar protic solvents, such as ethanol.

E2 and E1 are incorrect as they are elimination reaction mechanisms, and we are looking for a substitution mechanism. SN2 reactions result in inversion, not racemization. Additionally we know that SN2 is incorrect because SN2 is favored in polar aprotic solvents.