Problem 117 - A Complete Resource Book in Chemistry for JEE Main

Question

A hydrocarbon \(\mathrm{C}_{6} \mathrm{H}_{12}\) decolourizes bromine solution and yields n-hexane on hydrogenation. On oxidation with \(\mathrm{KMnO}_{4}\) it forms two different monobasic acids of the type \(\mathrm{RCOOH}\). The compound is (a) hex-2-ene (b) hex-3-ene (c) cyclohexene (d) hex-1-ene

Step-by-Step Solution

Answer

The compound is (b) hex-3-ene.

Step 1: Analyze the Given Information

We are given a hydrocarbon with the molecular formula \(C_6H_{12}\). The compound decolourizes bromine solution, suggesting it contains a carbon-carbon double bond (unsaturation). On hydrogenation, it forms n-hexane, and oxidative cleavage with \(KMnO_4\) results in two different monobasic acids.

Step 2: Evaluate Unsaturation Type

Since bromine decolourizes and the compound becomes fully saturated (n-hexane) upon hydrogenation, the hydrocarbon must contain a single double bond. The molecular formula \(C_6H_{12}\) supports this as one degree of unsaturation (one double bond) is present.

Step 3: Determine Structure Based on Oxidation Products

The oxidation of the compound by \(KMnO_4\) yielding two different monobasic acids indicates cleavage of the double bond. If the double bond is internal, different acids can result from cleavage at different positions along the chain.

Step 4: Compare Molecular Structures

Option (a) hex-2-ene and option (b) hex-3-ene are internal alkenes. Both can produce two different monobasic acids upon oxidation; however, hex-3-ene will give more distinctly different acids compared to hex-2-ene due to the position of the break in the carbon chain.

Step 5: Deduce the Correct Compound

Considering the above analysis, option (c) cyclohexene forms only one type of acid upon oxidation (adipic acid). Hence, the possible structures are hex-2-ene or hex-3-ene. Among these, hex-3-ene is more likely due to how oxidation typically yields distinctly different acids from cleaving more centrally located double bonds.

Key Concepts

Alkenes

Alkenes are a fascinating group of hydrocarbons. They are characterized by the presence of at least one carbon-carbon double bond. This double bond is the source of their reactive nature. In the formula for an alkene, you can see that it follows the pattern \(C_nH_{2n}\). In our example, \(C_6H_{12}\), it clearly indicates that it's an alkene.

These compounds are distinct from alkanes, which contain only single carbon-carbon bonds. The presence of a double bond not only affects the chemical reactivity but also the shape of the molecule. This pi bond in the double bond locks the carbon atoms in a fixed position, preventing rotation and leading to geometric isomerism. This is why alkenes can exist in forms such as cis or trans-isomers.

• They are typically unsaturated hydrocarbons.
• Can engage in addition reactions easily.
• The double bond is a site of high electron density making alkenes reactive.

Alkenes are versatile, playing key roles in both synthetic and natural chemical processes.

Oxidative Cleavage

Oxidative cleavage is a powerful reaction used to break carbon-carbon double bonds. When employed on alkenes, it cleaves the double bond entirely, introducing oxygen at the positions where the carbons were initially bonded. This reaction is typically carried out using a strong oxidizing agent, like potassium permanganate (KMnO4). In this reaction, each carbon atom bonded by the double bond is oxidized, often yielding carboxylic acids (RCOOH) as products.

This transformation is significant as it can be used to predict the position and structure of the double bond within the original alkene. If a hydrocarbon, such as hex-3-ene, undergoes oxidative cleavage, it will form two different acids depending on where the double bond is split. The ability to study oxidative cleavage helps scientists determine ancient structures of unknown alkenes by interpreting the resulting oxidation products.

• Utilizes strong oxidizing agents like KMnO4.
• Produces cleaved products like aldehydes or acids.
• Essential for understanding alkene structures.

Bromine Decolourization

The decolourization of bromine, often in a bromine-water or bromine-carbon tetrachloride solution, is a classic test for unsaturation in hydrocarbons. In our exercise, the hydrocarbon \(C_6H_{12}\) helps demonstrate this reaction. When a bromine solution loses its characteristic brown-red color after coming into contact with an alkene, it indicates the presence of a double bond.

Here's how it works: the pi bond of the alkene interacts with the bromine molecules, leading to an addition reaction. This results in bromine atoms attaching across the former double bond, forming a dibromo-compound.

• Indicates unsaturation due to double bonds.
• Involves addition reactions forming dibromo products.
• Useful for distinguishing alkenes from alkanes.

This ability to turn from brown to colorless serves as a simple yet effective method of identifying the presence of alkenes.

Unsaturation

Unsaturation in molecules refers to the presence of double or triple bonds between carbon atoms. These unsaturations create points of reactivity, allowing a variety of chemical reactions to occur. In our hydrocarbon, \(C_6H_{12}\), the presence of one double bond contributes one degree of unsaturation.

The notion of unsaturation is essential for understanding the chemical properties and behaviors of hydrocarbons. It alters molecular geometry by limiting rotation and enabling geometric isomerism, such as cis-trans isomerism in alkenes. Importantly, it imparts chemical reactivity, allowing compounds to participate in reactions such as addition reactions, where reactants "add" across the double bond, or oxidative reactions that can cleave these bonds.

• Indicates reactive double or triple bonds.
• Crucial for chemical reactivity.
• Effects molecular geometry and isomerism.

Understanding unsaturation helps in predicting the reactivity and the types of reactions alkenes can undergo.

KMnO4 Reaction

The KMnO4 reaction is a distinguishing reaction for alkenes, due to its strong oxidizing nature. When an alkene reacts with potassium permanganate, also known for its deep purple color, this oxidizing agent causes the alkene to cleave, initially forming glycol (with syn addition of hydroxyl groups), and depending on conditions, further oxidation to acids may occur.

In a basic medium, this reaction turns into a dihydroxylation process. However, under strong oxidative conditions, as is utilized in the exercise with \(C_6H_{12}\), it breaks the alkene at the double bond, turning each carbon into a carboxylic acid (RCOOH). The reaction’s tell-tale sign is the color change from purple to brown or colorless due to the reduction of MnO4 to MnO2 or other manganese compounds, confirming unsaturation.

• Turns alkenes into glycols or acids.
• Color shift indicates reaction occurrence.
• Utilized to cleave double bonds in hydrocarbons.

This reaction serves as a powerful tool for the analysis and transformation of unsaturated compounds.

Hydrogenation

Hydrogenation is a widely used chemical reaction that converts unsaturated hydrocarbons into saturated ones. This process involves the addition of hydrogen () across the double bonds in alkenes, converting them into single bonds. For example, treating an alkene like \(C_6H_{12}\) with hydrogen gas in the presence of a catalyst such as palladium, platinum, or nickel results in the saturated hydrocarbon n-hexane.

The reaction is crucial in various industries, notably in the food industry to convert vegetable oils, which are unsaturated, into semi-solid forms such as margarine. In the exercise, hydrogenation confirms that \(C_6H_{12}\) is initially unsaturated and contains a double bond, as the end product is a fully saturated alkane.

• Requires catalysts like Pt, Pd, or Ni.
• Transforms unsaturated to saturated molecules.
• Widely used in industrial applications.

This ability to saturate compounds significantly alters their physical and chemical properties, making hydrogenation an essential process in synthetic chemistry.