Problem 151 - A Complete Resource Book in Chemistry for JEE Main

Question

Match the following: List I (Reactants) 1\. \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Br}+\) alc. \(\mathrm{KOH}\) 2\. \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}(\mathrm{Br}) \mathrm{CH}_{3}+\) alc. \(\mathrm{KOH}\) 3\. \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CBr}+\) alc. \(\mathrm{KOH}\) List II (Products) A. cis-but-2-ene B. trans-but-2-ene C. but-1-ene D. 2-methylprop-1-ene The correct matching is \(\begin{array}{lll}1 & 2 & 3\end{array}\) \(\begin{array}{lll}1 & 2 & 3\end{array}\) (a) C B D (b) \(\mathrm{C} \mathrm{A} \mathrm{D}\) (c) \(\mathrm{C} \mathrm{D} \quad \mathrm{A}\) (d) B A C

Step-by-Step Solution

Answer

The correct matching is option (a) CBD.

Step 1: Identify the Reaction Type

This exercise involves elimination reactions, specifically dehydrohalogenation, where alcoholic KOH is used to eliminate hydrogen and a halogen atom (like Br) from a molecule, forming an alkene.

Step 2: Analyze Reaction 1

For reaction 1, the reactant is \(\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{CH}_{2}\mathrm{CH}_{2}\mathrm{Br}\) with alcoholic KOH. This leads to the elimination of HBr resulting in the formation of a terminal alkene due to beta-elimination, forming but-1-ene (option C).

Step 3: Analyze Reaction 2

For reaction 2, the reactant is \(\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{CH} (\mathrm{Br})\mathrm{CH}_{3}\) with alcoholic KOH. The elimination yields butene. Due to the chance for cis-trans isomerism in the structure, this can form trans-but-2-ene, which is more stable (option B).

Step 4: Analyze Reaction 3

For reaction 3, the reactant is \((\mathrm{CH}_{3})_{3}\mathrm{CBr}\) with alcoholic KOH. Elimination in such a tertiary structure forms 2-methylprop-1-ene due to the major elimination path, yielding the more substituted and stable alkene (option D).

Step 5: Match with List II

Based on the analysis: - Reaction 1 forms but-1-ene (C) - Reaction 2 forms trans-but-2-ene (B) - Reaction 3 forms 2-methylprop-1-ene (D) Thus, the correct matching sequence is option (a) CBD.

Key Concepts

Elimination Reactions

In organic chemistry, elimination reactions are a type of reaction where two substituents are removed from a molecule, resulting in the formation of a multiple bond, typically a double bond, thus creating an alkene.
Elimination reactions are often categorized based on the number of steps involved in the reaction mechanism:

• E1 reaction: A two-step process where the leaving group departs first, forming a carbocation intermediate, followed by deprotonation to form the double bond.
• E2 reaction: A one-step process involving a concerted removal of a proton and a leaving group, leading directly to the formation of the double bond.

In the given exercise, dehydrohalogenation is the specific type of elimination reaction involved.
This involves the removal of a hydrogen atom and a halogen (like bromine), facilitated by alcoholic KOH, to form an alkene.

Alkenes

Alkenes are hydrocarbons that contain at least one carbon-carbon double bond. They are unsaturated, meaning they have fewer hydrogen atoms than their corresponding alkanes.
This double bond is a region of high electron density, which makes alkenes more reactive than alkanes.

• Structure: Alkenes have a planar configuration around the double bond, with bond angles of approximately 120 degrees.
• Nomenclature: The name of an alkene is derived from the longest chain containing the double bond and ending in "-ene." Position numbers indicate the location of the double bond.

In the exercise, the formation of alkenes like but-1-ene and trans-but-2-ene comes from the elimination reactions.
These compounds exhibit different stabilities based on the substitution pattern on the double bond.

Stereochemistry

Stereochemistry refers to the spatial arrangement of atoms in molecules, which can affect physical and chemical properties. In organic reactions, the details of how reactants convert to products include considering the three-dimensional orientations.

• Cis and Trans Isomerism: This is common in alkenes, where the two substituents attached to each carbon in the double bond can vary in spatial arrangement. "Cis" indicates substituents on the same side, while "trans" means they are on opposite sides. In reaction 2, trans-but-2-ene is more stable due to reduced steric hindrance compared to its cis counterpart.
• Stability: In alkenes, trans isomers are usually more stable than cis isomers due to lower steric strain.

Understanding stereochemistry is crucial for predicting product distribution and stability in chemical reactions.
Stereoisomers may have different chemical behaviors, which can significantly affect their roles in chemical and biological processes.