Problem 84 - A Complete Resource Book in Chemistry for JEE Main

Question

\(\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CH}_{2} \stackrel{\mathrm{Cl}_{2}(\mathrm{hv})}{\longrightarrow} \mathrm{X} \stackrel{\mathrm{Nal}}{\longrightarrow} \mathrm{Y}\) The compound \(\mathrm{Y}\) in the above sequence is (a) 3-iodopropene (b) 1,2 -diodopropane (c) 1,2 -dichloro-3-iodopropane (d) 1 -chloro-2-iodopropane

Step-by-Step Solution

Answer

The compound \( \mathrm{Y} \) is 3-iodopropene (option a).

Step 1: Understanding the First Reaction

The starting compound is propene \( \mathrm{CH}_{3} \mathrm{CH}=\mathrm{CH}_{2} \). Under light (\( \mathrm{hv} \)) and chlorine \( \mathrm{Cl}_{2} \), this reaction is a free radical substitution on the allylic position (the carbon adjacent to the double bond), forming allyl chloride: \( \mathrm{CH}_{2}=\mathrm{CH}- \mathrm{CH}_{2}\mathrm{Cl} \), where chlorine is specifically added at the end carbon of the chain, resulting in \( \mathrm{X} = \mathrm{CH}_{2}=\mathrm{CH}- \mathrm{CH}_{2}\mathrm{Cl} \).

Step 2: Understanding the Second Reaction

The intermediate product \( \mathrm{X} = \mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}_{2}\mathrm{Cl} \) undergoes a nucleophilic substitution reaction with sodium iodide (\( \mathrm{NaI} \)). The iodide ion (\( \mathrm{I}^{-} \)) replaces the chlorine atom, resulting in the final compound \( \mathrm{Y} \), which is \( \mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}_{2}\mathrm{I} \). This is known as 3-iodopropene.

Key Concepts

Free Radical Substitution

Free radical substitution is a reaction that involves the replacement of an atom in a molecule with a radical. A radical is an atom or group of atoms with an unpaired electron. In organic chemistry, this process often happens under light or heat.
In the case of the exercise, propene undergoes a free radical substitution with chlorine (\(\mathrm{Cl}_{2}\)) in the presence of light (\(\mathrm{hv}\)). This light energy is enough to break the bond between the two chlorine atoms, forming two chlorine radicals. These radicals are highly reactive and seek to stabilize themselves by pairing their unpaired electrons.

Here's what happens:

• The chlorine radicals attack the hydrogen atoms at the allylic position, adjacent to the double bond, leading to the formation of hydrogen chloride (\(\mathrm{HCl}\)) and a carbon radical.
• The new carbon radical can quickly react with another chlorine radical, resulting in the addition of chlorine to the allylic position.

This results in the formation of allyl chloride (\(\mathrm{CH}_{2}=\mathrm{CH}- \mathrm{CH}_{2}\mathrm{Cl}\)), which is intermediate X in the given sequence.

Nucleophilic Substitution

Nucleophilic substitution is a type of chemical reaction where a nucleophile, which is an electron-rich species, replaces another atom or group of atoms in a molecule. This reaction is crucial for reactions in organic chemistry as it allows the formation of different compounds.
The reaction with sodium iodide (\(\mathrm{NaI}\)) is a nucleophilic substitution where the iodide ion (\(\mathrm{I}^{-}\)) serves as the nucleophile. It has a lone pair of electrons that make it highly reactive and eager to form bonds.

During the reaction:

• The iodide ion approaches the carbon atom that is bonded to chlorine in allyl chloride.
• It donates its electrons to form a new bond, displacing the chlorine atom from its position.
• The chlorine atom leaves as a chloride ion (\(\mathrm{Cl}^{-}\)).

This results in the formation of 3-iodopropene (\(\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}_{2}\mathrm{I}\)), or compound Y in the reaction sequence.

Allylic Position

The allylic position in organic molecules refers to the carbon atom adjacent to a double-bonded carbon. It is of particular interest in reactions involving double bonds because of its stability and reactivity.
In a double-bonded system like propene (\(\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CH}_{2}\)), the carbon atoms directly attached to the double bond are known as vinylic carbons. The carbon adjacent to these is considered the allylic carbon.
The allylic position is especially reactive during free radical substitutions because the carbon radical formed is stabilized by resonance with the double bond.
Here’s why this is significant:

• Substitution at the allylic position often leads to more stable intermediates due to resonance, where the charge or radical can be spread across multiple atoms.
• This stability makes the allylic position a prime target in reactions such as the one seen with chlorine in the provided exercise.

In the given exercise, the chlorine radical attacks the allylic hydrogen, showcasing the reactivity and significance of the allylic position.