Carbonyl Reactants - Organic Chemistry

LiAlH4 is a very strong reducing agent and it is a reactant that can reduce acyl chlorides into alcohols. There are 2 alcohols to choose from and now it needs to be known that LiAlH4 doesn't add anything but hydrogens to its substrate. (Note that in the correct answer, there is another H that is not drawn on the carbon bearing the hydroxyl group).

Lithium aluminum hydride is a reducing agent. It reduces ketones and carboxylic acids to alcohols. 2-butanone is a 4-carbon chain with a double bond between an oxygen and carbon 2. The reduction turns the double bond with oxygen into a single bond to a hydroxy group. This makes the product 2-butanol.

This is a Grignard reagent carbon-carbon bond forming reaction. This reaction is being used with an ester which goes through a ketone intermediate with requires a second attack form the organolithium to reduce to an alcohol. Esters and acyl chlorides go to an alcohol with two of the same R groups from the organolithium in a Grignard reaction. This structure fits the bill:

In the presence of a strong base and benzene, ethylene glycol will convert ketones to acetals as seen in the product shown above. The given reagents will not react with esters or alcohols. The only given compound containing a ketone is molecule IV, the correct answer. This type of reaction is useful in organic synthesis when reaction at an ester is desired, but a ketone must be protected from reacting simultaneously. The acetal formed is often called a protecting group and may be reversed by addition of acid.

This question is asking what the most acidic hydrogen is in an aldehyde. To identify the most acidic hydrogen, we'll need to consider which conjugate base will be the most stable after the loss of hydrogen. When the alpha-hydrogen is lost, the resulting carbanion will be the most stabilized due to resonance with the adjacent carbonyl group. This resonance helps to distribute the negative charge on the carbanion over a greater area, which contributes to the greater stability of this conjugate base. Abstraction of a hydrogen from any of the other position would result in a carbanion that could not participate in resonance and thus would not be as stable.

Below is the mechanism for the reaction given which is called the McMurry Reaction. It is a titanium (Ti) prompted pinacol coupling followed by deoxygenation.

Below is the mechanism for the reaction given which is called the McMurry Reaction. It is a titanium (Ti) prompted pinacol coupling followed by deoxygenation.

This reaction involves the generation of a radical by single electron transfer that couple through carbon-carbon bond formation to form the product pinacol. This reaction as shown below occurs in an aprotic solvlent:

Lithium aluminum hydride is a reducing agent. It reduces ketones and carboxylic acids to alcohols. 2-butanone is a 4-carbon chain with a double bond between an oxygen and carbon 2. The reduction turns the double bond with oxygen into a single bond to a hydroxy group. This makes the product 2-butanol.

This is a Grignard reagent carbon-carbon bond forming reaction. This reaction is being used with an ester which goes through a ketone intermediate with requires a second attack form the organolithium to reduce to an alcohol. Esters and acyl chlorides go to an alcohol with two of the same R groups from the organolithium in a Grignard reaction. This structure fits the bill:

In the presence of a strong base and benzene, ethylene glycol will convert ketones to acetals as seen in the product shown above. The given reagents will not react with esters or alcohols. The only given compound containing a ketone is molecule IV, the correct answer. This type of reaction is useful in organic synthesis when reaction at an ester is desired, but a ketone must be protected from reacting simultaneously. The acetal formed is often called a protecting group and may be reversed by addition of acid.

This question is asking what the most acidic hydrogen is in an aldehyde. To identify the most acidic hydrogen, we'll need to consider which conjugate base will be the most stable after the loss of hydrogen. When the alpha-hydrogen is lost, the resulting carbanion will be the most stabilized due to resonance with the adjacent carbonyl group. This resonance helps to distribute the negative charge on the carbanion over a greater area, which contributes to the greater stability of this conjugate base. Abstraction of a hydrogen from any of the other position would result in a carbanion that could not participate in resonance and thus would not be as stable.

LiAlH4 is a very strong reducing agent and it is a reactant that can reduce acyl chlorides into alcohols. There are 2 alcohols to choose from and now it needs to be known that LiAlH4 doesn't add anything but hydrogens to its substrate. (Note that in the correct answer, there is another H that is not drawn on the carbon bearing the hydroxyl group).

Below is the mechanism for the reaction given which is called the McMurry Reaction. It is a titanium (Ti) prompted pinacol coupling followed by deoxygenation.

Below is the mechanism for the reaction given which is called the McMurry Reaction. It is a titanium (Ti) prompted pinacol coupling followed by deoxygenation.

This reaction involves the generation of a radical by single electron transfer that couple through carbon-carbon bond formation to form the product pinacol. This reaction as shown below occurs in an aprotic solvlent:

The initially chiral carbon has an hybridization. Once treated with ,an oxidizing agent, the secondary alcohol in the compound is oxidized to a ketone. The central carbon is no longer a chiral center (a carbon with a double bond cannot be chiral), and the double bond (pi-bond) formed between carbon and oxygen gives the molecule an hybridization.

hybridization is formed with triple bonds, and an hybridization does not exist.

In acidic solution, a hydrogen will be added to the hydroxy group of the given compound. This intermediate is a good leaving group. As it leaves, a hydrogen will be abstracted and a double bond will form. The bond is across carbons two and three becasue those two are more highly substituted than carbon 1, which is primary. The naming begins at the double bond, so the name is 4-bromo-2-hexene.

Chromic acid (formed in-situ via ), the correct answer, is a powerful oxidizing agent that converts primary alcohols to carboxylic acids. PCC is also an oxidizing agent, but oxidizes primary alcohols one "rung" up the ladder to aldehydes. , and are reducing agents, which react in the opposite direction of the given transformation.

The key to answering this question is to realize that potassium permanganate is a very powerful oxidizing agent. Therefore, the primary alcohol in the reactant will be completely oxidized into a carboxylic acid. It's also important to note that because potassium permanganate is such a powerful oxidizing agent, the primary alcohol will not just stop at the aldehyde level of oxidation, but will go all the way to a carboxylic acid, which is at the highest oxidation level.