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\. \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CBr}+\mathrm{alc} . \mathrm{KOH}\) 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) \(\mathrm{C} \quad \mathrm{B} \mathrm{D}\) (b) \(\mathrm{C} \mathrm{A} \mathrm{D}\) (c) \(\mathrm{C}\) D A (d) \(\mathrm{B} \mathrm{A} \mathrm{C}\)

Step-by-Step Solution

Answer

The correct matching is (a) C, B, D.

Step 1: Understand the Reactants and Reagents

We have three reactants with alcoholic KOH as a reagent. Alcoholic KOH is used for elimination reactions, specifically dehydrohalogenation, where the hydrogen and the halide are removed from the reactant to form an alkene.

Step 2: Analyze Reactant 1

The first reactant is 1-bromobutane: \( \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{Br} \). This undergoes dehydrohalogenation to form but-1-ene (C). This is because the hydrogen is removed from the primary carbon adjacent to the bromine.

Step 3: Analyze Reactant 2

The second reactant is 2-bromobutane: \( \text{CH}_3\text{CH}_2\text{CH(Br)}\text{CH}_3 \). The possible alkenes are cis- and trans-but-2-ene. Trans-but-2-ene (B) is generally more stable than cis-but-2-ene, making it the major product.

Step 4: Analyze Reactant 3

The third reactant is 2-bromo-2-methylpropane: \( (\text{CH}_3)_3\text{CBr} \). This forms 2-methylprop-1-ene (D) upon elimination since the structure leads to a highly substituted and thus more stable alkene.

Step 5: Match Each Reactant with Its Product

Based on the earlier analysis, the correct matches are: 1) 1-bromobutane with but-1-ene (C), 2) 2-bromobutane with trans-but-2-ene (B), 3) 2-bromo-2-methylpropane with 2-methylprop-1-ene (D). Thus, the answer is option (a): C, B, D.

Key Concepts

Dehydrohalogenation

Dehydrohalogenation is a fundamental chemical reaction that involves the removal of a hydrogen atom (dehydrogenation) and a halogen atom (halogenation) from an organic compound. It is a type of elimination reaction crucial for forming alkenes from alkyl halides. The process requires the presence of a reagent like alcoholic potassium hydroxide (alc. KOH), which facilitates the removal of the hydrogen and halogen atoms.
The reaction proceeds by breaking the C-H and C-X (X = halogen) bonds. This produces a double bond between two carbon atoms, resulting in the formation of an alkene.

• This reaction typically requires heat and often involves a strong base.
• It is favored by higher temperatures, leading to a higher yield of alkenes.

Dehydrohalogenation is key in organic chemistry because it allows chemists to create a variety of alkene structures, essential in producing various biochemicals and pharmaceuticals.

Alkenes

Alkenes are hydrocarbons that contain at least one carbon-carbon double bond (C=C). This bond gives them unique properties, distinguishing them from alkanes, which only have single bonds. The presence of a double bond in alkenes introduces a planar structure and results in restricted rotation around the double bond, influencing the molecule's chemical behavior.
The simplest alkene is ethene (ethylene), which is used as a key building block in the chemical industry. Alkenes are represented with the suffix "-ene," indicating the presence of the double bond.

• Alkenes are unsaturated, which means they have fewer hydrogen atoms than alkanes.
• The presence of the double bond makes them reactive, allowing them to participate in a variety of chemical reactions, such as polymerization and reactions with halogens.

The versatility and reactivity of alkenes make them central in organic chemistry and industrial applications.

E2 Mechanism

The E2 mechanism (Bimolecular Elimination) is one of the pathways by which dehydrohalogenation can occur. This mechanism involves the simultaneous removal of a hydrogen and a halogen from neighboring carbon atoms, resulting in the formation of an alkene.
The E2 mechanism is characterized by:

• A single, concerted step where bonds break and form simultaneously.
• The involvement of a strong base, such as alcoholic KOH, which deprotonates a hydrogen while the leaving group (the halogen) exits.
• No carbocation intermediate, making the reaction faster and more controlled.

Because the E2 mechanism relies on a good leaving group and a strong base, it is sensitive to the structure of the alkyl halide and the conditions of the reaction. The choice of base and substrate can greatly influence the reaction's outcome, making it a valuable tool for synthesizing alkenes with specific desired configurations.

Stability of Alkenes

The stability of alkenes is an essential factor in determining the outcome of elimination reactions, such as those involving the E2 mechanism. The stability of an alkene depends largely on its structure, particularly the number of substituents attached to the carbon atoms involved in the double bond.
Key points regarding alkene stability include:

• More substituted alkenes (those with more alkyl groups attached to the carbons of the double bond) are generally more stable due to hyperconjugation and the electron-donating effects of alkyl groups.
• Trans-configured alkenes are usually more stable than their cis counterparts because the larger groups are on opposite sides of the double bond, reducing steric hindrance.
• Conjugation with double bonds or aromatic rings also contributes to increased stability.

Understanding alkene stability is critical for predicting the major products of elimination reactions, as seen in reactions where trans alkenes are typically favored. This knowledge helps chemists design reactions to synthesize alkenes with desired geometries and levels of substitution.