Problem 17 - A Complete Resource Book in Chemistry for JEE Main

Question

The intermediate product formed during the addition of \(\mathrm{HCl}\) to propene in presence of peroxide is (a) \(\mathrm{CH}_{3} \mathrm{CHCH}_{2} \mathrm{Cl}\) (b) \(\mathrm{CH}_{3} \mathrm{CHCH}_{3}\) (c) \(\mathrm{CH}_{3} \dot{\mathrm{C}} \mathrm{H}_{2} \mathrm{CH}_{2}\) (d) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2}\)

Step-by-Step Solution

Answer

The intermediate product is \\(\mathrm{CH}_{3} \dot{\mathrm{C}}\mathrm{H}_{2}\mathrm{CH}_{2}\).

Step 1: Identify the Reaction Mechanism

The reaction described is the addition of hydrochloric acid (HCl) to propene (an alkene) in the presence of a peroxide. This is known as an anti-Markovnikov addition, which is a characteristic of the Free Radical Addition Mechanism influenced by peroxides.

Step 2: Formation of Free Radicals

In the presence of peroxides, the reaction begins with the formation of free radicals. The peroxide decomposes to form alkoxy radicals, which then generates a chlorine radical from HCl.

Step 3: Radical Addition to Propene

The chlorine radical adds to the less substituted end of the double bond in propene, forming a new carbon radical at the other end. The intermediate product is a carbon radical, specifically a secondary radical because it's more stable.

Step 4: Determine the Intermediate Structure

The addition of the chlorine radical to propene gives \(\mathrm{CH}_{3} \dot{\mathrm{C}}\mathrm{H}_{2}\mathrm{CH}_{2}\). This happens because the chlorine adds to the primary carbon (less substituted), leaving the radical on the secondary carbon (more substituted), due to the stability of the radical.

Key Concepts

Understanding Anti-Markovnikov Addition

In the world of organic chemistry, the Markovnikov rule guides us in predicting the outcome of the addition of acids to alkenes. It states that in reactions where a hydrogen halide is added to an alkene, the hydrogen atom tends to attach to the carbon with the greater number of hydrogen atoms (the more substituted carbon). However, there are certain conditions under which the opposite, known as anti-Markovnikov addition, occurs.
Anti-Markovnikov addition generally happens in the presence of peroxides, which influence the reaction mechanism to follow a different pathway. This is particularly true for additions involving hydrogen bromide (HBr), but hydrogen chloride (HCl) can also partake in similar reactions under the influence of peroxides. In this process, the halide ion attaches to the less substituted carbon atom of the double bond. This shift from the usual pattern is facilitated by free radicals and is known as a free radical addition.

Exploring Free Radical Addition

Free radical addition is a fascinating mechanism in organic chemistry that deviates from typical reaction pathways. It occurs when free radicals—unstable atoms or molecules with unpaired electrons—initiate a sequence of reactions. These free radicals can form through the decomposing effect of a peroxide.
When involved in addition reactions, the radical species add to an unsaturated substrate, such as an alkene. In the context of our exercise, the presence of peroxide leads to the cleavage of the peroxide bond to yield radicals, which then abstract a halogen radical from HCl. This chlorine radical, in turn, interacts with the alkene and continues the chain reaction.

• Initiation: Formation of radicals, often from a peroxide.
• Propagation: Radicals add to an alkene, continuing to create new radical species.
• Termination: The reaction ends when two radicals combine, stopping the chain.

The free radical mechanism is pivotal for reactions like the anti-Markovnikov addition of HCl to propene.

The Role of the Peroxide Effect

The peroxide effect, also called the Kharasch effect, is the key to understanding why certain reactions follow an anti-Markovnikov pathway. It states that the presence of a peroxide can alter the typical behavior of a reaction.
In our given exercise, peroxides decompose to create radicals, which transform the reaction mechanism from a typical ionic route to a radical one. This change in mechanism explains why the halogen ends up on the less substituted carbon atom, thereby favoring anti-Markovnikov addition.
The peroxide effect is crucial because it illustrates how catalysts and reaction conditions can dramatically alter the course of chemical transformations, enabling chemists to selectively form desired products by controlling the presence of free radicals.

Understanding Carbon Radical Formation

The formation of a carbon radical is a key intermediate step in free radical reactions like the one described in this exercise.
When a chlorine radical adds to the less substituted carbon of propene, the electron pair from the double bond is redistributed. This process leaves behind a carbon radical on the more substituted carbon atom. Carbon radicals like these are often quite unstable, though secondary radicals (those touching two other carbon atoms) are generally more stabilized due to hyperconjugation and other effects.
Recognizing the stability of these radicals is crucial for predicting reactivity and understanding why certain reaction pathways are favored. Carbon radicals are centrally important in radical chain mechanisms, where their stability can dictate the speed and selectivity of the reaction. This illustrates the significance of proper radical management in synthetic organic chemistry.