Problem 214 - A Complete Resource Book in Chemistry for JEE Main

Question

When \(\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{COOH}\) is reduced with \(\mathrm{LiAlH}_{4}\) the compound obtained will be (a) \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{COOH}\) (b) \(\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}_{2} \mathrm{OH}\) (c) \(\mathrm{CH}_{3}^{2}-\mathrm{CH}_{2}-\mathrm{CH}_{2} \mathrm{OH}\) (d) \(\mathrm{CH}_{3}-\mathrm{CH}_{2}^{2}-\mathrm{CHO}\)

Step-by-Step Solution

Answer

(b) \( \mathrm{CH}_{2} = \mathrm{CH} - \mathrm{CH}_{2} \mathrm{OH} \)

Step 1: Understand the Reaction

The starting compound is an unsaturated carboxylic acid: \( \mathrm{CH}_{2} = \mathrm{CH} - \mathrm{COOH} \). The reagent \( \mathrm{LiAlH}_{4} \) is a strong reducing agent that primarily reduces carboxylic acids to alcohols.

Step 2: Identify the Product of Carboxylic Acid Reduction

When \( \mathrm{CH}_{2} = \mathrm{CH} - \mathrm{COOH} \) is treated with \( \mathrm{LiAlH}_{4} \), the carboxylic acid group \( \mathrm{-COOH} \) is reduced to a primary alcohol group \( \mathrm{-CH}_2\mathrm{OH} \). Thus, the product of the reaction is \( \mathrm{CH}_{2} = \mathrm{CH} - \mathrm{CH}_{2} \mathrm{OH} \).

Step 3: Match the Product with Options

Among the given options, (b) \( \mathrm{CH}_{2} = \mathrm{CH} - \mathrm{CH}_{2} \mathrm{OH} \) is the compound where the carboxylic acid has been reduced to a primary alcohol, while the alkene bond remains unchanged.

Key Concepts

LiAlH4

Lithium aluminium hydride, commonly written as \( \mathrm{LiAlH}_4 \), is a powerful reducing agent used frequently in organic chemistry. Its ability to donate hydride ions \( \mathrm{H}^- \) makes it particularly effective at breaking down carbon-oxygen bonds to form alcohols.
Key characteristics of \( \mathrm{LiAlH}_4 \) include:

• It is a colorless solid, often sold in powder form.
• This reagent is so potent that it is highly reactive, especially in the presence of moisture.
• It's known for reducing a variety of functional groups including carboxylic acids, esters, and aldehydes.

During reactions, \( \mathrm{LiAlH}_4 \) selectively targets the most oxidizable group, typically reducing it effectively to its alcohol counterpart. However, keep in mind that \( \mathrm{LiAlH}_4 \) does not affect carbon-carbon double or triple bonds, leaving these structures untouched in a reduction process.

Carboxylic Acid Reduction

In organic chemistry, reducing a carboxylic acid converts it into a primary alcohol. \( \mathrm{LiAlH}_4 \) is often the go-to reagent for this transformation due to its strength and reliability.
Here's how the reduction process unfolds:

• The carboxylic acid group \( \mathrm{-COOH} \) is directly converted into a \( \mathrm{-CH}_2\mathrm{OH} \) group.
• This transformation involves the removal of an oxygen atom and the addition of a hydrogen atom in its place.
• The \( \mathrm{H}^- \) ion from \( \mathrm{LiAlH}_4 \) attacks the electrophilic carbon in the carboxylic acid group, ultimately producing the alcohol.

The ability to perform this reduction without affecting other functional groups highlights the versatility of \( \mathrm{LiAlH}_4 \). It's notable that structural elements like alkenes remain unchanged during the reaction, which preserves the integrity of double bonds.

Primary Alcohol Formation

The reduction of carboxylic acids using \( \mathrm{LiAlH}_4 \) is a fantastic method for forming primary alcohols. A primary alcohol contains the \( \mathrm{-CH}_2\mathrm{OH} \) group, meaning the alcohol's hydroxyl group \( \mathrm{-OH} \) is bonded to a carbon that is itself bonded to only one other carbon atom.
Key points about primary alcohol formation:

• It's an essential step in many synthesis pathways involving organic compounds.
• The transformation is smooth and typically yields a high amount of the desired alcohol product.
• Often, the primary alcohol products are more chemically useful or biologically active than their precursor acids.

In the exercise's specific case, the double bond in the starting material, \( \mathrm{CH}_2=\mathrm{CH}-\mathrm{COOH} \), stays unchanged. This means the unsaturation remains while the carboxylic acid part transitions to a primary alcohol \( \mathrm{CH}_2=\mathrm{CH}-\mathrm{CH}_2\mathrm{OH} \), maintaining part of its original structure in the final product.