Any time we have a Grignard reagent and a primary haloalkane, we will see a substitution reaction, identical to an reaction. In this case, the Grignard can easily attack the haloalkane as the bromine leaves to create hexene.

Any time we have a Grignard reagent and a primary haloalkane, we will see a substitution reaction, identical to an reaction. In this case, the Grignard can easily attack the haloalkane as the bromine leaves to create hexene.

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

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.

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. is used to protonate the negatively charged oxygen atom.

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.

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

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. is used to protonate the negatively charged oxygen atom.

is a very powerful reducing agent that works to reduce almost any carbonyl compound. 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 to form an alcoxide. The alcoxide undergoes workup (the process whereby a negatively charged oxygen gains a proton) via , depicted above as simply "". We now have a secondary alcohol. From here, we can simply use the reagent to convert the alcohol into the desired chlorine.