Sodium borohydride donates a hydride ion to a ketone or aldehyde. In order to form a ketone or aldehyde, a nucleophile must attack the carbonyl group. This is because the ketone or aldehyde has an electrophilic carbon—a nucleophile must attack it in order for any reaction to occur. A hydride ion is the only answer choice that plays the role of a nucleophile.

The correct choice, CeCl₃ and NaBH₄ in MeOH, shows reagents know as "Luche conditions," which are able to modify the reactivity of sodium borohydride to reduce the carbonyl to an alcohol without affecting alkene groups. This occurs as the cerium ion coordinates strongly to the carbonyl oxygen, which subsequently greatly enhances the electrophilicity at the carbonyl carbon. Nucleophilic attack of the hydride readily occurs, simultaneously destroying the electropilicty of the beta carbon of the alkene, such that it will not be reduced by the hydride reagent.

The incorrect answer choices would give various undesired products as detailed below:

NaBH₄ in MeOH

Use of unmodified sodium borohydride would result in a 1,4 conjugate addition reaction, saturating the alkene, with a subsequent reduction of the ketone to an alcohol.

LiAlH₄ in THF

Use of lithium aluminum hydride would give the same product as use of unmodified sodium borohydride, following the same reduction mechanism.

Pd and H₂ in hexanes

This reagent will give reduction of the alkene only.

Pd, BaSO₄, and H₂ in hexanes

This reagent combination, known as Lindlar's catalyst, will also reduce the alkene only. This reagent is typically used to selectively reduce an alkyne to an alkene.

Sodium borohydride is a reducing agent with formula . It is a reducing agent, but it is not extremely strong. It reduces ketones to alcohols, but it does not affect carboxylic acids. 3-pentanone is the only ketone of the given choices. It would be reduced to 3-pentanol.

These are all true uses of .

The reaction between magnesium and an alkyl halide in anhydrous ether results in a Grignard reagent.

An organolithium would result from the same process, but the magnesium would need to be replaced by two equivalents of lithium. Alcohols are products of reactions between a Grignard reagent and a carbonyl. 

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  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.  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.