Problem 63 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

An alkyl halide, \(X\), of formula \(\mathrm{C}_{6} \mathrm{H}_{13} \mathrm{Cl}\) on treatment with potassium tertiary butoxide gives two isomeric alkenes \(Y\) and \(Z\left(\mathrm{C}_{6} \mathrm{H}_{12}\right)\). Both alkenes on hydrogenation give2, 3-dimethylbutane. Predict the structures of \(X, Y\) and \(Z\)

Step-by-Step Solution

Answer

The structures are: \(X\) = 2-chloro-2,3-dimethylbutane, \(Y\) = 2,3-dimethyl-2-butene, \(Z\) = 2,3-dimethyl-1-butene.

Step 1: Analyze the given compound

The given compound is an alkyl halide, specifically a chloro compound with the formula \( \mathrm{C}_{6} \mathrm{H}_{13} \mathrm{Cl} \). This indicates that the compound contains six carbon atoms, thirteen hydrogen atoms, and one chlorine atom.

Step 2: Consider the elimination reaction

Treatment with potassium tertiary butoxide is a strong base leading to an elimination reaction (E2 mechanism), resulting in the formation of alkenes. This suggests that \(X\) should be such that it can form two different alkenes \(Y\) and \(Z\) when subjected to elimination.

Step 3: Use hydrogenation result

Hydrogenation of both alkenes \(Y\) and \(Z\) yields the same saturated hydrocarbon, 2,3-dimethylbutane. This means both \(Y\) and \(Z\) are structural isomers of 2,3-dimethylbutene, which has the same carbon backbone but different double bond locations.

Step 4: Determine potential structures of alkenes

2,3-Dimethylbutane is derived from alkenes that could potentially be 2,3-dimethyl-2-butene and 2,3-dimethyl-1-butene. Both of these alkenes would give 2,3-dimethylbutane upon hydrogenation.

Step 5: Confirm alkyl halide structure

Given the alkenes, the parent structure \(X\), should be based on the positions from where the hydrogen was removed. From 2,3-dimethyl-2-butene and 2,3-dimethyl-1-butene, \(X\) should be 2-chloro-2,3-dimethylbutane because the chlorine needs to be on the 2nd carbon to facilitate E2 elimination to give the specified alkenes.

Key Concepts

Alkyl Halides

Alkyl halides are organic compounds consisting of an alkyl group bonded to a halogen atom. In our exercise, the alkyl halide is a chlorinated compound formulated as \( \mathrm{C}_{6} \mathrm{H}_{13} \mathrm{Cl} \). Here's a breakdown of what alkyl halides encompass:

• They are categorized based on the number of carbon atoms attached to the carbon bearing the halogen: primary (one carbon), secondary (two carbons), and tertiary (three carbons).
• The presence of the electronegative halogen makes the carbon-halogen bond polar, influencing its reactivity.
• These compounds often undergo substitution or elimination reactions due to their polar nature.

In the context of our problem, identifying the correct structure of the alkyl halide is crucial, as it dictates the course of the reaction when treated with bases like potassium tertiary butoxide.

Elimination Reactions

Elimination reactions involve the removal of two substituents from a molecule, forming a double bond. In organic chemistry, understanding elimination is key when dealing with alkyl halides reacting to form alkenes. Here's how it applies to our exercise:

• The reaction of the alkyl halide (\( X \)) with a strong base like potassium tertiary butoxide can lead to elimination by the E2 mechanism.
• E2 elimination involves a single concerted step where the base abstracts a proton, and the leaving group (chlorine in this case) exits, forming an alkene.
• This results in the formation of isomeric alkenes (\( Y \) and \( Z \)), meaning they have the same formula but differ in the arrangement of atoms, specifically where the double bond is located.

Such reactions are influenced by the structure of \( X \). Here, a tertiary carbon makes elimination more favorable. Understanding the E2 mechanism helps us predict the formation of the alkenes from the starting alkyl halide.

Isomerism

Isomerism in organic chemistry involves compounds with the same molecular formula but different structural arrangements. There are different types of isomerism, and in this exercise, we focus on structural isomerism and its implications:

• Structural isomers \( Y \) and \( Z \), both derived from \( X \), share the formula \( \mathrm{C}_{6} \mathrm{H}_{12} \) but have different patterns of connectivity or structure.
• In the case of the problem, these structural isomers are likely different positional isomers where the position of the double bond within the carbon chain varies.
• Upon hydrogenation, both of these alkenes give 2,3-dimethylbutane, confirming their structural similarity in terms of carbon framework, despite differences in the double bond location.

Understanding isomerism is crucial to deducing the possible structural forms of \( Y \) and \( Z \) and confirming them as isomers of 2,3-dimethylbutene, each returning to the same alkane upon hydrogenation.