Organic Reducing Agents - Organic Chemistry

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.

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.

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.

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.

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

All of the other listed compounds are reducing agents.

produces a trans-alkene from an alkyne whereas produces a cis-alkene. reduces an alkyne all the way down to an alkane. is a strong oxidizing agent.

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.

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 . 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.

is the only answer choice that is not an oxidizing agent. In fact, it is a reducing agent because of the lack of oxygen atoms present. This compound adds hydrogen atoms to a compound, thereby reducing it.

Alkenes can be reduced in the presence of and a metal catalyst like platinum to hydrogenate the alkene to give a saturated alkane. The reaction occurs in a heterogeneous solution rather than a homogenous solution. It occurs on the presence of a solid surface of the metal catalyst.

Note that the three carbon-carbon double bonds in the aromatic ring in the presence of the reducing agent does not get reduced because they are extremely stable due to resonance.

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.

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.