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

Question

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

Step-by-Step Solution

Answer

Option (b) is the main product: \( \mathrm{CC(C)=CCc1ccccc1} \).

Step 1: Analyze the Starting Material

The given starting material is \( \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{CH}( ext{OH})\mathrm{CH}\left(\mathrm{CH}_{3}\right)_{2} \). This is a benzyl alcohol with the structure containing a phenyl ring (C6H5-), a hydroxyl group (-OH) attached to a secondary carbon, and an isopropyl group.

Step 2: Identify Reagent Conditions

Concentrated \( \mathrm{H}_{2} \mathrm{SO}_{4} \) is a strong acid and is typically used for dehydration reactions. It aids in converting alcohols to alkenes by eliminating the hydroxyl group and a hydrogen atom, resulting in a double bond formation.

Step 3: Determine the Dehydration Products

In this reaction, the secondary alcohol will lose a water molecule (\( \text{OH}^- \) from the alcohol group and \( \text{H}^+ \) from a neighboring carbon), forming an alkene. The newly formed double bond can stabilize by forming closer to the phenyl ring, due to resonance stabilization.

Step 4: Predict the Alkene's Structure

The most stable alkene is formed by eliminating the hydroxyl group and a hydrogen from the adjacent carbon in the direction that gives the more substituted alkene. The resulting product has the structure \( \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH} = \mathrm{C(CH}_{3})\mathrm{CH}_{2} \).

Step 5: Match the Product with the Given Options

Among the given options, (b) \( \mathrm{CC(C)=CCc1ccccc1} \) represents the structure where the phenyl group (benzene ring) is connected to a double bond \( -\mathrm{CH} = \mathrm{C(CH}_{3})- \). This corresponds to the main product derived from the dehydration process.

Key Concepts

Dehydration Reaction

A dehydration reaction in organic chemistry is defined by the removal of a water molecule. This type of reaction is crucial when transforming alcohols into alkenes, as demonstrated in our exercise. In the case of dehydration, you often start with an alcohol which has an -OH group that can be eliminated to form a double bond.
Dehydration occurs because of the ability of a strong acid, like concentrated sulfuric acid, to protonate the -OH group. This makes it a good leaving group as water. When water leaves, a double bond forms between the carbon atoms resulting in the formation of an alkene.

• The alcohol group and a hydrogen atom are removed.
• A carbocation intermediate is often formed during the reaction.
• Finally, a double bond forms, stabilizing the molecule."

This process transforms the alcohol into a more stable alkene product.

Alcohol to Alkene Conversion

Converting an alcohol to an alkene involves a careful selection of conditions and reagents. In our reaction exercise, the secondary alcohol is converted into an alkene through the elimination of water. This conversion is a common reaction in organic chemistry when trying to synthesize alkenes from alcohols.
During this conversion process, an important factor to bear in mind is the regioselectivity and stereochemistry of the resulting alkene. Here, we aim to form the more substituted alkene. According to Zaitsev's Rule, the more substituted alkene, which generally is more stable, tends to be the major product.

• Dehydrogenation replaces an -OH group with a carbon double bond.
• The reaction prefers a path that results in the more stable, substituted alkene.
• The process may go through a carbocation intermediate stage.

This helps explain why the double bond forms closer to the phenyl group due to its resonance stabilization capabilities.

Reagent Conditions with Conc. H2SO4

Concentrated sulfuric acid ( ext{H}_2 ext{SO}_4) is a potent reagent that facilitates dehydration reactions. Its strength as an acid means it can readily donate protons, making it highly effective in driving the dehydration of alcohols.
When using concentrated ext{H}_2 ext{SO}_4, careful laboratory conditions are necessary because of its acidity and exothermic nature. It is vital for the acid to remain concentrated to efficiently remove water molecules. The dehydration favored by conc. ext{H}_2 ext{SO}_4 follows these steps:

• Protonation of the hydroxyl group, converting it to water.
• Elimination of water, creating a carbocation.
• Formation of a double bond, resulting in the alkene.

The reagent helps steer the reaction toward producing the more thermodynamically stable alkene, as demonstrated in our exercise. This process allows for the synthesis of a targeted alkene product by careful manipulation of reaction conditions.