Grignard reagents are known for their ability to readily attack carbonyls at the point of their carbons. However, Grignard reagents do not work in the presence of protic solvents. Rather than reacting with the desired molecule, the Grignard is so unstable that it will readily accept a proton from a protic solvent. The Grignard then becomes inert and no reaction ensues with the desired molecule.

Any time we have a Grignard reagent and a primary haloalkane, we will see a substitution reaction, identical to an \(\displaystyle S_{N}2\) reaction. In this case, the Grignard can easily attack the haloalkane as the bromine leaves to create hexene.

The carbons on the epoxide compound experience a slightly positive charge. As a result, a Gringard reagent can easily attack the less substituted side of the epoxide to break the ring and to form a six membered carbon chain. \(\displaystyle H_{3}O^{+}\) is used to protonate the negatively charged oxygen atom.

\(\displaystyle LiAlH_{4}\) is a very powerful reducing agent that works to reduce almost any carbonyl compound. \(\displaystyle CH_{3}CH_{2}CONH_{2}\) is an amide and the only carbonyl compound given of the answer choices.

This problem requires that we convert our ketone group into a chlorine. However, this cannot be done directly, and requires multiple steps.

We begin by reducing the ketone with \(\displaystyle LiAlH_{4}\) to form an alcoxide. The alcoxide undergoes workup (the process whereby a negatively charged oxygen gains a proton) via \(\displaystyle H_{3}O^{+}\), depicted above as simply "\(\displaystyle H^{+}\)". We now have a secondary alcohol. From here, we can simply use the reagent \(\displaystyle SOCl_{2}\) to convert the alcohol into the desired chlorine.

First step: esterification

Second step: lithium aluminum hydride reduction

Third step: neutralization to form primary alcohol

Fourth step: SN2 reaction to form final chlorinated product

Lithium Aluminum Hydride is a potent reducing agent; it has the ability to turn esters and aldehydes into primary alcohols, and ketones into secondary alcohols. The starting material is an aldehyde, so the correct answer is thus a primary alcohol ONLY.

\(\displaystyle H_2O_2\) is not a reducing agent; peroxides (compounds with the formula R-O-O-R) are oxidizing agents. A very common peroxide is sodium peroxide \(\displaystyle (Na_2O_2)\).

All of the other listed compounds are reducing agents.

\(\displaystyle Na/NH_{3}\) produces a trans-alkene from an alkyne whereas \(\displaystyle H_{2}/Lindlar's\) \(\displaystyle Catalyst\) produces a cis-alkene. \(\displaystyle H_{2}/Pd/C\) reduces an alkyne all the way down to an alkane. \(\displaystyle MCPBA\) is a strong oxidizing agent.

This is a standard organolithium reaction.

The organolithium product can be thought of as a strong nucleophile. The carbon steals an electron from the lithium to create \(\displaystyle H_{3}C^{-}Li^{+}\). From there, the highly reactive carbo-anion is free to attack the ketone at the site of its carbon to form a tertiary alcohol on the cyclohexane.