Redox Chemistry - Organic Chemistry

Potassium permanganate is a strong oxidizing agent. It can convert secondary alcohols to ketones. It can also convert primary alcohols to carboxylic acids. 1-propanol has a hydroxy group on carbon 1, so it is primary; thus it will be converted to propanoic acid.

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.

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.

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.

In order to drive the reactant, we must first convert the alcohol group on the ethanol into a carboxylic acid. We do so by using the oxidizing agent, , a very strong oxidizing agent that is well known to oxidize primary alcohols into carboxylic acids (among other functions). Once we have our carboxylic acid, we can simply use to convert our carboxylic acid into an acid halide to attain our desired final product.

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.

is a strong oxidizing agent.

Not only can reduce secondary alcohols into ketones, but it can reduce primary alcohols and aldehydes into carboxylic acids.

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.

To solve this question we need to calculate the oxidation number of oxygen in both molecules. The formula for water is . The oxidation number of hydrogen is +1. Since there are two of them, the hydrogen atoms contribute to a charge of +2. The water molecule is neutral; therefore, the oxygen must have an oxidation number of to balance the charge. The formula for hydrogen peroxide is . Using the same logic as water, we can determine that hydrogen contributes +2. We have two oxygen atoms in this case; therefore, each oxygen atom will have an oxidation number of to give a charge of . This will balance the charges and provide a neutral hydrogen peroxide molecule.

Recall that have a negative charge suggests that an atom has extra valence electrons. A charge of suggests one extra valence electron and a charge of suggests two extra valence electrons. An oxygen typically has six valence electrons. The oxygen in water has oxidation number; therefore, it will have two extra valence electrons (eight total). On the other hand, oxygen in hydrogen peroxide will have one extra valence electron (seven total); therefore, oxygen in hydrogen peroxide has one less valence electron than oxygen in water.

Potassium bromide has a formula of . This molecule is made up of an alkali metal (potassium) and a halogen (bromine). Alkali metals have one valence electron that they readily lose to obtain octet whereas halogens have seven valence electrons and they readily gain an electron to obtain octet. Recall that losing an electron will give you a oxidation number whereas gaining an electron will give you a oxidation number. This means that alkali metals always have an oxidation number of whereas halogens always have an oxidation number of ; therefore, potassium has an oxidation number of .

The following reaction schemes show the oxidation of all substrates, indicating that substrate II is in the highest oxidation state possible, and that an oxidation of this compound will not proceed.

Remember that in sulfuric acid and water, Na2Cr2O7 will be converted to CrO3, the active oxidant species. Furthermore, the oxidation mechanism involving this species includes the key step in which a hydrogen bonded to the carbon in question is eliminated_,_ and simutaneously, a double bond from that carbon to an oxygen is installed. Thus, all substrates that feature at least one hydrogen bonded to the carbon to be oxidized can and will be oxidized in the precense of chromium trioxide.

Lastly, remember that these reactions are taking place in the prescence of water. While substrates such as compound III do not appear to be oxidizable, attack of water at the aldehyde carbon will give a dialcohol tetrathedral intermediate that can be immediately oxidized by chromium trioxide to the corresponding carboxylic acid. A similar mechanism occurs for substrate I, wherein, after the ketone oxidation state is achieved, an attack of water furnishes the same dialcohol intermediate that is oxidized to the carboxylic acid. Remember that the highest oxidation state available for organic compounds containing more than one carbon is the carboxylic acid oxidation state. Chromium trioxide will oxidize all organics to this oxidation state, unless directly-bonded hydrogens are not present in lower oxidation states, such as shown with substrate IV.

The reaction given would give a ketone. This type of reaction is called an oxidation reaction. Oxidation of a secondary alcohol as in the reaction given by (sodium dichromate) in an aqueous solution of (acetic acid) solvent yields a ketone. However, if we performed the same reaction with a primary alcohol, a carboxylic acid would have formed.

The Jones reagent can convert primary alcohol to acids and secondary alcohols to ketones. The Tollen's test only converts aldehydes to carboxylic acids. PCC can only convert primary and secondary alcohol to aldehydes and ketones, respectively. and are reducing agents.