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-l-ene The correct matching is \(\begin{array}{lll}1 & 2 & 3\end{array}\) 123 (a) \(\mathrm{C} \mathrm{B} \mathrm{D}\) (b) \(\mathrm{C} \mathrm{A} \mathrm{D}\) (c) \(\mathrm{C} \mathrm{D} \mathrm{A}\) (d) \(\mathrm{B} \mathrm{A} \mathrm{C}\)

Step-by-Step Solution

Answer

Option (a): C B D.

Step 1: Understand the Reaction Type

We are given organic compounds that react with alcoholic KOH, which is a reagent known for dehydrohalogenation, a type of elimination reaction. This will typically form alkenes from haloalkanes.

Step 2: Identify the Structure of Reactants

The given reactants are: 1. 1-bromobutane, 2. 2-bromobutane, and 3. t-butyl bromide. Each of these molecules features a bromine atom that will be eliminated along with a hydrogen atom (from a beta carbon) to form a double bond.

Step 3: Predict the Products

1. For 1-bromobutane, removal of H and Br will create but-1-ene, as the double bond will form at the terminal position. 2. For 2-bromobutane, elimination can occur in two ways, potentially forming cis-but-2-ene or trans-but-2-ene, with trans being more stable. 3. For t-butyl bromide, elimination results in the formation of 2-methylprop-1-ene due to its branched structure.

Step 4: Match with List II

Now that we know the products: 1. 1-bromobutane results in but-1-ene, which is option C. 2. 2-bromobutane predominantly forms the more stable trans-but-2-ene, which is option B. 3. t-butyl bromide produces 2-methylprop-1-ene, which is option D. Thus, the matching is 1C, 2B, 3D.

Step 5: Final Answer

Based on our matching, the correct answer is option (a): 1C 2B 3D.

Key Concepts

Alcoholic KOH

Alcoholic potassium hydroxide, also known simply as alcoholic KOH, is a key reagent in organic chemistry. It's notably famous for its role in elimination reactions. When dissolved in alcohol, KOH behaves as a strong base.
This means it has a high tendency to remove hydrogen ions (H+). When used in reactions, its primary function is to promote dehydrohalogenation. In many cases involving haloalkanes (alkyl halides), alcoholic KOH facilitates the removal of a halogen atom (like bromine) and a hydrogen atom from adjacent carbon atoms.
Here's a simple breakdown on how it works:

• Dissolve KOH in alcohol to form alcoholic KOH.
• It acts as a nucleophile because of its ability to donate an electron pair.
• The reaction typically results in the formation of an unsaturated compound, such as an alkene.

As a strong base, its reactivity influences the formation of the desired alkene structure.

Dehydrohalogenation

Dehydrohalogenation is a specific type of elimination reaction involving the removal of a hydrogen halide (HX) from a haloalkane. The prefix 'dehydro-' indicates the removal of hydrogen, while '-halogenation' signifies the removal of a halogen.
In the context of organic reactions, dehydrohalogenation usually leads to the creation of alkenes - hydrocarbons with carbon-carbon double bonds. Here's a look at the process:

• A beta-hydrogen (hydrogen on the carbon adjacent to the one bonded to the halogen) is often removed.
• The halogen atom (bromine, chlorine, etc.) is also simultaneously removed.
• This dual removal results in the formation of a double bond between the two involved carbon atoms.

In the case of using alcoholic KOH, the solution's strong basicity aids in the easy removal of these atoms, especially in haloalkanes.

Elimination Reaction

An elimination reaction is fundamental in organic chemistry. It refers to the process where two substituents are removed from a molecule, resulting in the formation of a double or triple bond.
In simpler terms, think of it as the chemistry equivalent of decluttering a molecule. The most characteristic examples are dehydrohalogenation reactions, where a halogen and a hydrogen are removed to form a desired product.

• Elimination reactions result in the decrease in the number of single bonds in a molecule.
• They can occur via various mechanisms like E1, E2, etc.
• Commonly leads to the production of alkenes from alkyl halides or alcohols.

Elimination reactions are also remarkable because they can lead to different stereochemical configurations, such as cis and trans isomers, depending on the involved substituents and conditions.

Alkene Formation

Alkene formation is the highlight of many chemical transformations. Alkenes are hydrocarbons characterized by the presence of a carbon-carbon double bond.
The dehydrohalogenation of haloalkanes under controlled conditions often results in alkene formation. With understanding the conditions like using alcoholic KOH, one can predictably generate specific alkenes.

• The double bond in alkenes increases molecule stability, hence why trans-isomers (with substituents opposite each other) are often more stable than cis-isomers.
• The number and type of substituents along with the reaction conditions affect which alkene is formed.
• Alkenes serve as crucial intermediates and starting points in numerous synthetic pathways.

In the example reactions, varying the structure of the haloalkane precursor results in different alkene configurations and structural isomers.