Problem 37 - A Complete Resource Book in Chemistry for JEE Main

Question

Propene when heated with \(\mathrm{Cl}_{2}\) at \(500^{\circ} \mathrm{C}\) forms (a) \(\mathrm{CH}_{3}-\mathrm{CHCl}-\mathrm{CH}_{2} \mathrm{Cl}\) (b) \(\mathrm{CH}_{2} \mathrm{Cl}-\mathrm{CH}=\mathrm{CH}_{2}\) (c) \(\mathrm{CH}_{2} \mathrm{Cl}-\mathrm{CHCl}-\mathrm{CH}_{2} \mathrm{Cl}\) (d) all of these

Step-by-Step Solution

Answer

All of these products can form under the given conditions.

Step 1: Identify the Reaction Type

The reaction of propene with chlorine at a high temperature (500°C) in the absence of UV light suggests a free radical substitution reaction rather than an addition reaction because the high temperature favors substitution.

Step 2: Determine Possible Reaction Sites

In a substitution reaction, we focus on hydrogen atoms attached to carbon atoms that can be replaced by chlorine atoms. Propene, \( ext{CH}_3- ext{CH}= ext{CH}_2\), has hydrogen atoms that can be substituted by chlorine, leading to multiple possible chlorinated products.

Step 3: Analyze Potential Products

Since substitution can occur at different positions of the propene molecule, we must consider the potential products: \( ext{CH}_3- ext{CHCl}- ext{CH}_2 ext{Cl}\) where chlorine is substituted on both the first and second carbon, \( ext{CH}_2 ext{Cl}- ext{CH}= ext{CH}_2\) where chlorine is substituted on the first carbon, and \( ext{CH}_2 ext{Cl}- ext{CHCl}- ext{CH}_2 ext{Cl}\) is possible, but not typically favored due to extensive substitution under these conditions.

Step 4: Evaluate Reaction Conditions' Impact

The high-temperature condition favors the formation of multiple chlorinated isomers, meaning all these products can form in significant amounts under the given reaction conditions.

Step 5: Decide on the Set of Products

Given that under high temperature all substitution sites can react to form different products and considering typical outcomes of free radical chlorination, more than one product is likely to form. Therefore, the correct answer is that all these products (options a, b, and c) can be formed.

Key Concepts

Propene

Propene is an intriguing molecule that is widely studied in organic chemistry due to its reactivity and applications. It is a type of alkene, meaning it contains a carbon-carbon double bond, represented chemically as \( ext{C}_3 ext{H}_6\). This double bond is what makes propene particularly reactive.

The structure of propene includes three carbon atoms, with a double bond between the first two. It can be represented as \( ext{CH}_3 ext{-CH=CH}_2\). The presence of the double bond offers sites for a variety of reactions, including additions and substitutions. Because it is an unsaturated hydrocarbon, propene can undergo reactions where the double bond opens up to allow other atoms to bond with the carbon atoms.

This feature makes propene a key starting material in a multitude of chemical reactions and industrial processes. It's important in the production of polypropylene, a common plastic, and can also undergo free radical reactions like those discussed in the exercise.

Chlorination

Chlorination is a reaction that involves adding chlorine atoms to a molecule. In the context of hydrocarbons, chlorination can significantly change the compound's properties by introducing chlorine atoms in place of hydrogen.

When propene undergoes chlorination at high temperatures such as the 500°C condition mentioned in the exercise, the process is more inclined to free radical substitution rather than addition. This high temperature facilitates the formation of free radicals, which are highly reactive species and engage in substitution reactions.

The radical chlorination of propene can lead to multiple products. For a successful substitution, a chlorine atom replaces a hydrogen atom in the propene. This can happen at different spots in the molecule, leading to several potential products. For instance, chlorine can add to the middle carbon, resulting in secondary chlorides, or at either end, forming primary chlorides.

These varying substitutions are why such a range of chlorinated compounds can be produced in this reaction. The exercise highlights the products \( ext{CH}_3 ext{-CHCl-CH}_2 ext{Cl}\), \( ext{CH}_2 ext{Cl-CH=CH}_2\), and \( ext{CH}_2 ext{Cl-CHCl-CH}_2 ext{Cl}\) as possibilities.

Hydrocarbon Reactions

Hydrocarbon reactions are fundamental in organic chemistry and involve changing the structure of hydrocarbons through various chemical reactions. These reactions form the basis for creating countless organic compounds and materials.

In the realm of hydrocarbon chemistry, propene can participate in addition, substitution, and polymerization reactions. The exercise you are working on focuses on a specific type of substitution reaction, where free radicals are the main actors.

In free radical substitution, a hydrogen atom in the hydrocarbon is replaced by another atom, commonly a halogen like chlorine. Initiated by heat or light, this type of reaction is characterized by a chain mechanism with initiation, propagation, and termination steps.

Hydrocarbons like propene, with double bonds, are versatile due to their ability to react through different pathways depending on conditions like temperature and presence of catalysts. High temperatures favor radical mechanisms, as seen when propene is exposed to chlorine, opening multiple reactive pathways and leading to diverse products.

The comprehensive understanding of such reactions not only helps in predicting possible products but also in optimizing conditions for desired outcomes in industrial and laboratory settings.