Problem 173 - A Complete Resource Book in Chemistry for JEE Main

Question

When \(\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}-\mathrm{CHO}\) is reduced with \(\mathrm{NaBH}_{4}\), the product formed is (a) \(\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_{2} \mathrm{OH}\) (b) \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{CHO}\) (c) \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{CH}_{3}\) (d) \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{CH}_{2} \mathrm{OH}\)

Step-by-Step Solution

Answer

The correct product is (a) \(\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_{2} \mathrm{OH}\).

Step 1: Understanding the Reactant

The given reactant is \(\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}-\mathrm{CHO}\), which is an aldehyde with a terminal double bond in an alkene structure.

Step 2: Identifying the Reagent

\(\mathrm{NaBH}_{4}\) is a reducing agent that specifically reduces carbonyl groups like aldehydes and ketones but does not typically affect carbon-carbon double bonds.

Step 3: Predicting the Reduction

Since \(\mathrm{NaBH}_{4}\) will reduce the carbonyl group (\(-\mathrm{CHO}\)) to a primary alcohol, the product will have an \(-\mathrm{CH}_2\mathrm{OH}\) group instead of \(-\mathrm{CHO}\). The alkene's double bond remains unchanged.

Step 4: Forming the Product

After reaction, \(\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}-\mathrm{CHO}\) becomes \(\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_{2}\mathrm{OH}\), since only the aldehyde group is reduced.

Step 5: Verifying the Correct Option

Comparing with the given options, the structure \(\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_{2}\mathrm{OH}\) corresponds to option (a).

Key Concepts

Sodium Borohydride (NaBH4)

Sodium Borohydride, or \(\mathrm{NaBH}_4\), is widely used as a reducing agent in organic chemistry. It plays a crucial role in the reduction of carbonyl compounds, such as aldehydes and ketones. Sodium Borohydride is a mild reducing agent, making it highly selective. This means that it will specifically target certain functional groups in a molecule for reduction, while leaving other groups untouched.

When using \(\mathrm{NaBH}_4\), the typical environment is a polar protic solvent like alcohol, which facilitates the reduction process. The presence of the borohydride ion provides the hydride, \(\mathrm{H}^-\), that acts as the reducing species. This ion transfers to the electrophilic carbon atom in the carbonyl group, effectively transforming it into an alcohol.

• **Selective Reduction:** Prefers carbonyl groups (C=O) and does not usually affect double bonds (C=C).
• **Mild Reduction:** Operates under milder conditions compared to other reducing agents, reducing specific parts of the molecule.
• **Usage:** Commonly used in synthesizing alcohols from aldehydes and ketones.

Aldehyde Reduction

Aldehyde reduction is a vital transformation in organic synthesis. This process involves the conversion of an aldehyde (an organic compound containing a carbonyl group \(-\mathrm{CHO}\)) into a primary alcohol using a reducing agent like Sodium Borohydride. Aldehydes often participate in reduction reactions due to their electrophilic carbonyl carbon.

The mechanism of aldehyde reduction using \(\mathrm{NaBH}_4\) begins with the nucleophilic attack of the hydride ion \(\mathrm{H}^-\) on the carbonyl carbon. This step results in the formation of an alkoxide ion, which is subsequently protonated by the solvent to produce the corresponding alcohol. This two-step mechanism is efficient and highly selective for the aldehyde compound. In the presence of double bonds, \(\mathrm{NaBH}_4\) will usually not interfere, thus preserving the alkene structure.

• **Mechanism:** Involves hydride transfer and subsequent protonation.
• **Specificity:** Typically targets only the carbonyl group in aldehydes and ketones.
• **Outcome:** Converts the aldehyde to a primary alcohol.

Alcohol Formation

Alcohol formation through reduction reactions is a fundamental practice in chemistry, involving the transformation of carbonyl-containing compounds into alcohols. Using Sodium Borohydride for such reductions results in the formation of primary or secondary alcohols, depending on the starting material.

For the conversion of aldehydes, the product is a primary alcohol. The reduction process does not significantly affect existing double bonds, so they remain unchanged, preserving the molecule's integrity. This selective conversion is highly beneficial for chemists aiming to synthesize alcohols without altering other parts of the structure.

• **Primary Alcohols:** Formed when aldehydes are reduced.
• **Selective Conversion:** Double bonds usually remain intact.
• **Efficiency:** Provides high yields of alcohols with minimal by-products.