Problem 90 - A Complete Resource Book in Chemistry for JEE Main

Question

The compound obtained by the reaction of ethene with diborane, followed by hydrolysis with alkaline hydrogen peroxide is (a) ethanal (b) ethanol (c) triethyl boride (d) propanol

Step-by-Step Solution

Answer

The product is ethanol (option b).

Step 1: Understand the Reaction

This exercise involves the reaction of ethene with diborane (B2H6). Ethene (C2H4) and diborane react to form trialkylborane intermediates. The key is to understand that diborane adds across the double bond of ethene in a syn addition manner, forming an organoborane.

Step 2: Determine the Organo Borane Formation

When ethene reacts with diborane, trialkylborane (triethyl borane) is formed. However, the problem does not specifically ask for this intermediate but rather for the product after further reactions.

Step 3: Hydrolysis of the Organo Borane

The organoborane formed undergoes hydrolysis with alkaline hydrogen peroxide (H2O2) to convert it into an alcohol. This reaction is known as the hydroboration-oxidation of alkenes, which results in the formation of alcohols.

Step 4: Identify the Alcohol Formed

The reaction of ethene with diborane followed by hydrolysis with alkaline H2O2 gives ethanol as the final product, as a result of anti-Markovnikov addition of water.

Key Concepts

Ethene

Ethene is a fundamental organic compound in the class of hydrocarbons known as alkenes. It is composed of two carbon atoms double-bonded to each other, known by the formula \( C_2H_4 \). This simple molecule plays a pivotal role in various chemical reactions due to its reactivity of the double bond.

In most scenarios, the double bond in ethene can open up and act as a site for bonding with other atoms or groups. This makes ethene a perfect candidate for addition reactions, such as the hydroboration-oxidation process with diborane. Its structure looks like this:

• Each carbon atom is attached to two hydrogen atoms, making it a planar molecule.
• The double bond consists of one sigma bond and one pi bond.
• The pi bond provides high electron density, attracting electrophiles for addition reactions.

This reactivity is crucial for forming new bonds and compounds, like alcohols in many organic synthesis routes.

Diborane

Diborane, with the chemical formula \( B_2H_6 \), is a fascinating compound due to its unique bonding structure. It's a member of the boranes, a group of compounds made up of boron and hydrogen.

The structure of diborane is atypical because it contains several 'three-center two-electron' (3c-2e) bonds, also known as banana bonds:

• The geometry includes two boron atoms bridge-bonded by hydrogen atoms.
• This configuration permits diborane to interact with alkenes like ethene efficiently.
• The boron atoms are electron-deficient, making diborane a good electrophile.

When reacting with ethene, diborane adds across its double bonds, forming organoborane intermediates. This reaction sets the stage for the subsequent transformation to alcohols via hydroboration-oxidation, starting with the formation of boron-carbon bonds.

Anti-Markovnikov Addition

The term anti-Markovnikov addition refers to the unusual but highly useful mode of addition across double bonds. Normally, in Markovnikov addition, the more electronegative component attaches to the carbon with fewer hydrogen atoms.

In the case of the hydroboration-oxidation of alkenes such as ethene:

• Hydroboration results in the formation of an organoborane complex without a catalyst.
• The boron atom adds to the less substituted carbon atom, leading to "anti" Markovnikov addition.
• This step is stereospecific and involves the syn addition of boron to the alkene.

This pathway allows the selective conversion of alkenes into alcohols in a way that yields primary alcohols efficiently, particularly through the reaction of ethene to form ethanol.

Alcohol Formation

The final step in the process of transforming ethene into an alcohol is the hydrolysis of the organoborane intermediate. This involves the addition of hydrogen peroxide in an alkaline medium.

During this phase, boron is replaced by a hydroxyl group, culminating in the formation of an alcohol:

• The hydrogen peroxide participates in the oxidation of the organoborane.
• In basic conditions, the organoborane converts into a primary alcohol.
• For ethene, this predicts the formation of ethanol \(( C_2H_5OH )\) particularly highlighting the anti-Markovnikov outcome.

Alcohol formation through this reaction provides a reliable method to transform hydrocarbons into alcohols, expanding their functionality for subsequent chemical reactions or industrial applications.