Problem 88 - A Complete Resource Book in Chemistry for JEE Main

Question

\(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{Cl} \stackrel{\mathrm{Acl}_{\mathrm{K} \mathrm{OH}}}{\Delta} \longrightarrow \mathrm{A} \frac{\mathrm{SO}_{2} \mathrm{Cl}_{2}}{475 \mathrm{~K}}{\longrightarrow} \mathrm{B}\) The compound B in the above sequence is (a) 1,2 -dichloro-2-methyl propene (b) 1 -chloro-2-methyl propene (c) 1,2 -dichloro-2-methyl propene (d) 3 -chloro-2-methyl propene

Step-by-Step Solution

Answer

The compound B is (d) 3-chloro-2-methyl propene.

Step 1: Identify the Reaction Type

The first reaction involves heating tert-butyl chloride, \((\mathrm{CH}_3)_3\mathrm{C}-\mathrm{Cl}\), with alcoholic potassium hydroxide \(\mathrm{KOH}\). This is an E2 elimination reaction, where elimination of HCl occurs.

Step 2: Determine Product A

During the E2 elimination reaction, the \(\mathrm{Cl}^-\) and an adjacent \(\mathrm{H}^+\) are removed, resulting in the formation of a double bond to form 2-methylpropene \(\mathrm{A}\), which is \(\mathrm{C}\left(\mathrm{CH}_3\right)_2=\mathrm{CH}_2\).

Step 3: Understand Second Reaction Conditions

Product A, 2-methylpropene, is then reacted with \(\mathrm{SO}_{2}\mathrm{Cl}_2\) at 475 K. This condition typically involves an allylic chlorination where a chlorine atom is added at the allylic position.

Step 4: Identify the Allylic Position for Chlorination

In 2-methylpropene, the hydrogen atom at the carbon adjacent to the double bond (allylic position) is replaced by a chlorine atom.

Step 5: Predict Product B

The chlorination of the allylic hydrogen in 2-methylpropene using \(\mathrm{SO}_{2}\mathrm{Cl}_2\) results in the formation of 3-chloro-2-methylpropene.

Key Concepts

E2 Elimination Reaction

The E2 elimination reaction is a fundamental concept in organic chemistry that involves the elimination of atoms or groups from a molecule, leading to the formation of a double bond. E2 stands for bimolecular elimination, indicating that the reaction rate depends on both the substrate and the base concentration.

• This reaction typically occurs in the presence of a strong base, like alcoholic potassium hydroxide.
• For example, tert-butyl chloride in the presence of alcoholic KOH undergoes removal of hydrogen (H) and chlorine (Cl) from adjacent carbon atoms.

This leads to the formation of an alkene by creating a double bond. In the given exercise, this results in the formation of 2-methylpropene. Understanding E2 reactions is crucial for predicting the structural outcome and changes in molecular geometry during organic synthesis.

Allylic Chlorination

Allylic chlorination is a specific type of halogenation involving the replacement of a hydrogen atom by a halogen at the carbon atom next to a double bond (known as the allylic position). This process is used to modify hydrocarbons to introduce new functional groups that are more reactive.

• This reaction typically requires heat and involves reagents like sulfuryl chloride (SO₂Cl₂). It's effective in generating chlorine radicals that facilitate the substitution.
• During allylic chlorination, the allylic position is favored due to the stability of allylic radicals formed during the reaction process.

In our discussion, 2-methylpropene undergoes allylic chlorination, resulting in a chlorine atom substituting a hydrogen atom at the allylic position, forming 3-chloro-2-methylpropene. This reaction demonstrates the utility of selective transformations at specific molecular sites.

2-Methylpropene

2-Methylpropene, also known as isobutylene, is an organic compound that is part of the broader class of alkenes, which are hydrocarbons characterized by having at least one carbon-carbon double bond.

• Its molecular structure, (CH₃)₂C=CH₂, features a central double bond that makes it highly reactive, offering sites for transformations such as addition reactions or halogenations.
• Due to its double bond, it acts as a nucleophile in reactions including those involving radicals, as seen in allylic chlorination.

In the context of the discussed reactions, its role is crucial as an intermediate that undergoes further chemical modifications to achieve the final chlorinated product. This highlights the versatility and importance of 2-methylpropene in synthetic organic chemistry, especially in constructing more complex molecular structures.