Problem 207 - A Complete Resource Book in Chemistry for JEE Main

Question

The main product of the following reaction is \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{CH}(\mathrm{OH}) \mathrm{CH}\left(\mathrm{CH}_{3}\right)_{2} \stackrel{\text { conc. } \mathrm{H}_{2} \mathrm{SO}_{4}}{\longrightarrow}\) ? [2010] (a) CC=CC(C)C (b) CC(C)=CCc1ccccc1 (c) CC(C)C=Cc1ccccc1 (d) CCCC(Cl)(Cl)Cl

Step-by-Step Solution

Answer

The main product is (b) CC(C)=CCc1ccccc1.

Step 1: Understanding the Reaction Type

The reaction involves treating a secondary alcohol with concentrated sulfuric acid (H₂SO₄). This is typically a dehydration reaction, where the alcohol loses a water molecule and forms an alkene.

Step 2: Identifying the Alkene Formation

The alcohol portion is \(\text{OH}\) on the second carbon (\(\text{CH}(\text{OH})\)). When dehydrating, the hydroxyl group (\(\text{OH}\)) and a hydrogen from an adjacent carbon (most likely the one with more hydrogens for stability reasons) will be removed to form the double bond.

Step 3: Determining the Most Stable Alkene

The hydrogen that is most likely to be removed is from the carbon bearing two methyl groups (a tertiary carbon). This will lead to the formation of the more stable, highly substituted alkene.

Step 4: Drawing the Expected Product

Removing a hydrogen from one of the methyl groups attached to the tertiary carbon will form a double bond between this carbon and the carbon attached to the phenyl group. The expected product will be (b) CC(C)=CCc1ccccc1, forming a more substituent-stable alkene.

Key Concepts

Alkene formation

Alkene formation is a key concept in organic chemistry, particularly when discussing the dehydration of alcohols. An alkene is a hydrocarbon containing at least one carbon-to-carbon double bond.

• These double bonds are created when alcohols lose a water molecule, a process known as dehydration.
• The formation of the double bond is a result of the removal of the hydroxyl group (-OH) from the alcohol and a hydrogen atom from an adjacent carbon atom.

When the dehydration of the alcohol occurs, the specific location of where the hydrogen is removed plays a critical role in the stability and the arrangement of the resulting alkene.

• In many cases, the hydrogen is lost from the carbon atom that is part of a more substituted carbon, leading to a more stable, substituted alkene.
• Such alkenes are favored because the increased substitution around the double bond provides greater stability due to hyperconjugation and electronic effects.

The stability of alkenes typically increases with increased substitution, making an understanding of alkene formation crucial in predicting the major product in an alcohol dehydration reaction.

Reaction mechanism

Understanding the reaction mechanism for the dehydration of alcohols is essential to grasp how alkenes form. The reaction proceeds through several stages:

• First, the alcohol is protonated by the acid leading to the formation of an oxonium ion. This step is crucial as it increases the potential of the -OH group to leave as a water molecule.
• The departure of water from the oxonium ion generates a carbocation intermediate. This intermediate is central to the next stage, as it can undergo reconfiguration to achieve greater stability if necessary.

A key step in the mechanism is the shift of hydrogen atoms to stabilize the carbocation through a mechanism known as a 1,2-hydride or alkyl shift.

• This allows the carbocation to achieve a tertiary structure if possible, thereby increasing its stability due to hyperconjugation and electronic effects associated with tertiary carbocations.
• The removal of a proton next to the carbocation leads to the formation of a double bond, resulting in the final alkene product.

Understanding each step explains the final structure of the alkene and the product distribution depending on the stability of the carbocation formed.

Sulfuric acid as a catalyst

Sulfuric acid plays a critical role as a catalyst in the dehydration of alcohols. It's important to understand how sulfuric acid is influencing this reaction:

• As a strong acid, its primary role is to protonate the -OH group of the alcohol. This converts the -OH, which is a poor leaving group, into water, a much better leaving group.
• The acidic environment thereby facilitates the formation of the carbocation by enabling the departure of the water molecule, leaving behind a positively charged carbocation.

Sulfuric acid doesn't get consumed in the reaction, which is characteristic of a catalyst as it participates in the reaction cycle but is regenerated at the end.

• It creates conditions necessary for the alcohol to lose water, promoting the conversion and rearrangement processes that eventually lead to alkene formation.
• Additionally, concentration of sulfuric acid plays a part in driving the equilibrium towards the formation of the alkene by pushing the reaction towards its products.

Understanding the function of sulfuric acid reveals why only a small amount of a catalyst is required to convert large amounts of alcohol into alkenes efficiently.