Problem 86 - A Complete Resource Book in Chemistry for JEE Main

Question

Alkene \(\mathrm{R}-\mathrm{CH}=\mathrm{CH}_{2}\) reacts with \(\mathrm{B}_{2} \mathrm{H}_{6}\) in the presence of \(\mathrm{H}_{2} \mathrm{O}_{2}\) to give (a) [R]C(C)=O (b) [R]OC(C)O (c) \(\mathrm{R}-\mathrm{CH}_{2}-\mathrm{CHO}\) (d) \(\mathrm{R}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{OH}\)

Step-by-Step Solution

Answer

The correct product is option (d), \(\mathrm{R-CH_2-CH_2-OH}\).

Step 1: Identify the Reaction

The reaction given involves the alkene \(\mathrm{R-CH=CH_2}\) reacting with \(\mathrm{B_2H_6}\) followed by \(\mathrm{H_2O_2}\). This is known as the hydroboration-oxidation process, commonly used to convert alkenes into alcohols.

Step 2: Understanding Hydroboration

In the hydroboration part, \(\mathrm{B_2H_6}\) adds to the alkene. The boron atom attaches to the least substituted carbon atom of the alkene (which is anti-Markovnikov addition). This means that the \(\mathrm{B-CH}\) bond forms with the terminal carbon atom (\(\mathrm{CH_2}\) group).

Step 3: Oxidation with \(\mathrm{H_2O_2}\)

The \(\text{B-C}\) bond is then replaced by an \(\text{O-H}\) bond when the borane intermediate is treated with \(\mathrm{H_2O_2}\) and a base (often \(\mathrm{NaOH}\)). This results in the formation of an alcohol where the \(\mathrm{OH}\) group is attached to the previously less substituted carbon.

Step 4: Analyze Product Options

The product of hydroboration-oxidation should be an alcohol with the formula \(\mathrm{R-CH_2-CH_2-OH}\). Reviewing the choices: (a) is a ketone, (b) is an ester, (c) is an aldehyde, and (d) is a primary alcohol. The primary alcohol \(\mathrm{R-CH_2-CH_2-OH}\) matches our expected product.

Key Concepts

Anti-Markovnikov Addition

Understanding the concept of Anti-Markovnikov addition is crucial, especially when learning about hydroboration-oxidation reactions in organic chemistry. In a typical addition reaction, a reagent usually adds to the more substituted carbon atom of an alkene—this is known as Markovnikov's rule. However, Anti-Markovnikov addition defies this norm. Here, the reagent (in this case, boron from \(\mathrm{B}_{2}\mathrm{H}_{6}\)) attaches itself to the less substituted carbon of the alkene. This selective addition is mainly due to sterics or the electron-deficient nature of boron.

This phenomenon is crucial for predicting the final outcome of the reaction, as it determines the product’s structure. When boron attaches to the less hindered terminal carbon, it sets the stage for further transformations that ultimately yield products like primary alcohols—an essential consideration in synthetic planning.

Alkene Reactions

Alkenes are hydrocarbons that contain carbon-carbon double bonds, making them highly reactive in organic synthesis. Their reactivity stems from the \(\pi\) bond in the double bond, which is more accessible to reacting species than \(\sigma\) bonds. This allows for various reactions such as hydrogenation, halogenation, and of course, hydroboration-oxidation.

In the context of the hydroboration-oxidation reaction, the alkene undergoes addition with \(\mathrm{B}_{2}\mathrm{H}_{6}\). This initial step is unique as it generates an organoborane intermediate rather than transitioning directly to the final alcohol. Subsequent oxidation swaps the borane group with a hydroxyl group, converting the alkene into an alcohol.

• This process highlights the pivotal role of alkenes in forming more complex structures.
• Understanding alkene behavior is essential for stepwise synthesis and predicting reaction outcomes.

Primary Alcohol Formation

The ultimate goal of hydroboration-oxidation is to convert an alkene into a primary alcohol. Throughout the steps, boron first attaches to the less substituted carbon due to Anti-Markovnikov addition, which sets the path for forming the alcohol. Upon adding \(\mathrm{H}_{2} \mathrm{O}_{2}\) and a base, such as \(\mathrm{NaOH}\), an oxidation reaction takes place.

During oxidation, the boron-carbon bond is replaced with a carbon-oxygen bond. As a result, a hydroxyl group (\(\mathrm{OH}\)) takes the place of boron, finalizing the transformation into a primary alcohol with the general formula \(\mathrm{R-CH}_{2}\mathrm{-CH}_{2}\mathrm{-OH}\).

This process is highly valued in chemistry for its straightforward synthesis of primary alcohols, which are essential building blocks in pharmaceuticals and various chemical industries.

• Primary alcohols are characterized by the \(\mathrm{OH}\) group on a terminal carbon.
• They are versatile in further functional transformations, allowing for diverse chemical applications.