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

Short answer

The compound is hex-2-ene (option a).

Step-by-step solution

Understand the reaction with Bromine

The compound \(\mathrm{C}_{6} \mathrm{H}_{12}\) decolorizes bromine, which indicates the presence of a carbon-to-carbon double bond (alkene group) that reacts with bromine to add across the double bond. This rules out the possibility of the compound being a cyclic alkane like cyclohexane.

Analyze hydrogenation to n-hexane

Hydrogenation of the compound \(\mathrm{C}_{6} \mathrm{H}_{12}\) results in n-hexane, which is a straight-chain alkane. This suggests that the original compound must have been a straight-chain alkene as hydrogens are simply added to break the double bonds.

Consider oxidation to monobasic acids

The compound oxidizes with \(\mathrm{KMnO}_{4}\) to form two different monobasic acids of the type \(\mathrm{RCOOH}\). This implies cleavage of the carbon chain at different locations. Such a result can only be offered by an internal alkene.

Narrow down options based on oxidation products

For the oxidation to lead to two different acids, the double bond must be internal and not terminal. This rules out hex-1-ene. The internal double bond positions that would create two different oxidized products are at positions 2 and 3.

Conclude the position of the double bond

Both hex-2-ene and hex-3-ene could potentially yield two different acids. However, due to structural symmetry, hex-3-ene would yield the same acids as hex-2-ene. Thus, hex-2-ene is the correct answer, as hex-3-ene is symmetrical and would yield identical products.

Key concepts

Hydrocarbons

Hydrocarbons are organic compounds composed exclusively of carbon and hydrogen atoms. They serve as fundamental building blocks for more complex molecules. The simple structures of hydrocarbons can be divided into three main categories based on the type of carbon bond they contain:

• Saturated hydrocarbons (alkanes) have single bonds only.
• Unsaturated hydrocarbons include alkenes and alkynes, with one or more double or triple bonds.
• Aromatic hydrocarbons contain conjugated planar ring structures.

The compound \(C_{6}H_{12} \) in our problem belongs to the alkene group, characterized by at least one carbon-carbon double bond. These double bonds are reactive sites for various chemical transformations.

Alkenes

Alkenes are a family of hydrocarbons that contain at least one carbon-carbon double bond, referred to as the olefinic bond. This double bond is highly reactive, making alkenes prime candidates for various chemical reactions that convert them into different compounds.
The general formula for alkenes is \(C_{n}H_{2n}\), indicating their unsaturation compared to their alkane counterparts.

• The presence of this double bond allows alkenes to undergo addition reactions, such as with hydrogen (hydrogenation).
• They can also react with halogens like bromine (bromination), resulting in the addition of bromine atoms across the double bond.

In our exercise, the presence of a double bond is inferred from the decolorization of bromine solution, a classic test for alkenes.

Oxidation Reactions

Oxidation reactions in organic chemistry often involve adding oxygen or removing hydrogen from a molecule. In the case of alkenes, oxidative reactions usually involve the cleavage of the carbon-carbon double bond.
Potassium permanganate \(KMnO_{4}\) is a strong oxidizing agent used to transform alkenes into alcohols, ketones, or acids, depending on the reaction conditions.
In this particular reaction, the cleavage of the alkene's double bond by \(KMnO_{4}\) produces two different monobasic acids of the form \(RCOOH\). This splitting into two different products occurs because the double bond is located internally in the alkene, allowing different cleavage pathways.

Hydrogenation

Hydrogenation is a chemical reaction that involves the addition of hydrogen (H2) across the double bonds of an unsaturated hydrocarbon, converting it to a saturated one. The process generally requires a catalyst such as palladium, platinum, or nickel to proceed efficiently.
In the context of our compound \(C_{6}H_{12}\), hydrogenation yields n-hexane, which is a saturated hydrocarbon with a straight chain. This confirms that the original hydrocarbon was an alkene, as a cyclic structure would not open up to form a straight-chain alkane.

Bromination

Bromination is a specific type of addition reaction where bromine (Br2) is added to unsaturated hydrocarbons such as alkenes. The reaction is a good diagnostic test for the presence of double bonds in a molecule because the brown color of bromine disappears as it adds across these double bonds, resulting in a colorless solution.
During bromination of an alkene like \(C_{6}H_{12}\), the bromine molecules break the double bond and add to the carbon atoms, transforming the molecule into a dibromo-compound. This reaction is quick and visually straightforward, making it a useful method to identify alkenes from alkanes or other hydrocarbons.