Question

The following conversion can be accomplished by: $$ \mathrm{CH}_{3} \mathrm{CH}=\mathrm{CHCOOH} \longrightarrow \mathrm{CH}_{3} \mathrm{CH}=\mathrm{CHCH}_{2} \mathrm{OH} $$ (A) \(\mathrm{NaBH}_{4}\) (B) \(\mathrm{Al}\left(\mathrm{Me}_{2} \mathrm{CHO}\right)_{3}, \mathrm{Me}_{2} \mathrm{CHOH}\) (C) \(\mathrm{H}_{2}, \mathrm{Ni}\) (D) \(\mathrm{LiAlH}_{4}\)

Short answer

The correct reducing agent for the given reaction is (B) \(\mathrm{Al}(\mathrm{Me}_{2} \mathrm{CHO})_{3}, \mathrm{Me}_{2}\mathrm{CHOH}\), as it can efficiently reduce the alkyne to an alkene and carboxylic acid to an alcohol group.

Step-by-step solution

Identify the reactant and product

The reactant molecule given is a 3-carbon chain that contains an alkyne and a carboxylic acid group. The product is the same 3-carbon chain, but there is an alkene instead of the alkyne, and the carboxylic acid group has been replaced by an alcohol group. Reactant: \(\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CHCOOH}\) Product: \(\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CHCH}_{2} \mathrm{OH}\)

Analyze each of the given reducing agents

Here, we look at each of the options provided and determine their efficiency and relevance in the given reaction. (A) \(\mathrm{NaBH}_{4}\) - Sodium borohydride is a selective reducing agent that reduces aldehydes and ketones to alcohols. It is not effective in reducing alkynes or carboxylic acids. (B) \(\mathrm{Al}(\mathrm{Me}_{2} \mathrm{CHO})_{3}, \mathrm{Me}_{2}\mathrm{CHOH}\) - Aluminium isopropoxide, an organometallic reagent, can be used in reduction reactions. In the presence of an alcohol (dimethylcarbinol), it can reduce carboxylic acid to an alcohol group and can also reduce an alkyne to an alkene. This seems to be a suitable reagent for the given reaction. (C) \(\mathrm{H}_{2}, \mathrm{Ni}\) - Hydrogen gas in the presence of a catalyst like Nickel is used for hydrogenation, which can reduce alkynes to alkanes (instead of alkene). Also, it doesn't have the ability to reduce carboxylic acid to alcohol. (D) \(\mathrm{LiAlH}_{4}\) - Lithium aluminum hydride is a strong reducing agent that can reduce carboxylic acid to an alcohol group. However, it is not able to reduce the alkyne to an alkene in this case.

Choosing the correct reducing agent

From the analysis above, we can see that Option (B) \(\mathrm{Al}(\mathrm{Me}_{2} \mathrm{CHO})_{3}, \mathrm{Me}_{2}\mathrm{CHOH}\) can efficiently reduce the alkyne to an alkene and carboxylic acid to an alcohol group in the given reaction. Therefore, the correct choice is (B).

Key concepts

Carboxylic Acid Reduction

Carboxylic acids are distinctive in organic chemistry due to their functional carboxyl group ( -COOH"). This group makes them highly reactive especially in reduction processes. Reduction involves the addition of hydrogen or removal of oxygen, resulting in a decrease in the oxidation state. For carboxylic acids, this process can be used to convert them into primary alcohols.

Why Reduce Carboxylic Acids?

• Reduction changes the solubility and reactivity of the molecule, significantly altering its chemical behavior.
• This transformation is critical in the synthesis of various pharmaceuticals, fragrances, and polymers.

LiAlH4 (Lithium Aluminum hydride) is often used for reducing carboxylic acids. It is very effective but might not be the best choice if you're also looking to alter other functionalities within the same molecule.

Aluminum isopropoxide can also reduce carboxylic acids under the right conditions, which is the case in our exercise. By reducing the carboxylic acid, you achieve a fundamental change in the molecule, making it more versatile for further chemical transformations.

Alkene Formation

Alkenes, characterized by the carbon-carbon double bond, are important intermediates in organic chemistry. Their formation typically requires a reduction step, especially when starting from alkynes (triple carbon-carbon bonds).

The Importance of Alkenes:

• Alkenes act as key building blocks in the synthesis of complex molecules.
• Their double bond facilitates additional reactions, making them versatile intermediates.

Reducing an alkyne to an alkene requires the partial reduction of the triple bond without completely hydrogenating it to an alkane. The use of specific reagents and conditions ensures the selective hydrogenation to alkenes. In the context of the exercise, aluminum isopropoxide functions in this precise manner, selectively reducing the alkyne present in the starting material to an alkene.

Reducing Agents in Organic Chemistry

Reducing agents are chemicals that donate electrons in reactions, playing a pivotal role in altering the state of functional groups in organic molecules. They are crucial for transformations that involve the reduction of various functional groups.

Types and Applications:

• NaBH4 (Sodium borohydride): Selective for the reduction of ketones and aldehydes.
• LiAlH4 (Lithium Aluminum Hydride): Remarkably strong and versatile, capable of reducing esters and carboxylic acids to alcohols.
• Aluminum isopropoxide: Specially acts under the presence of alcohols to facilitate the reduction of carboxylic acids and alkynes.

Choosing the right reducing agent depends on the target functionality and the desired transformation. Each agent has its strengths and limitations, and understanding these intricacies allows chemists to perform precise and efficient reactions. In our exercise, aluminum isopropoxide efficiently handles the dual task of converting both the carboxylic acid and alkyne groups.