Problem 150 - A Complete Resource Book in Chemistry for JEE Main

Question

Match the following: List I (Reactants) 1\. \(\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{3} \mathrm{OH} \stackrel{\mathrm{NaBr}, \mathrm{H}_{2} \mathrm{SO}_{4}, \Delta}{\longrightarrow}\) 2\. \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{COH} \frac{\text { Conc. } \mathrm{HCl}}{\text { Roser temp }}\) 3\. \(\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH})\left(\mathrm{CH}_{2}\right)_{2} \mathrm{CH}_{3} \stackrel{\mathrm{PB}_{3}}{\longrightarrow}\) 4\. \(\mathrm{Me}_{2} \mathrm{CHCH}_{2} \mathrm{OH} \stackrel{\mathrm{soc} \mathrm{h}}{\longrightarrow}\) List II (Alkyl halides) A. \(\mathrm{CH}_{3} \mathrm{CHBr}\left(\mathrm{CH}_{2}\right)_{2} \mathrm{CH}_{3}\) B. \(\mathrm{Me}_{2} \mathrm{CHCH}_{2} \mathrm{Cl}\) C. \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCl}\) D. \(\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{3} \mathrm{Br}\) The correct matching is \(\begin{array}{lllll} & 1 & 2 & 3 & 4 \\ \text { (a) } \mathrm{C} & \mathrm{D} & \mathrm{B} & \mathrm{A}\end{array}\) (b) \(\mathrm{C}\) \(\mathrm{D} \quad \mathrm{A} \quad \mathrm{B}\) (c) D \(\mathrm{C} \quad \mathrm{A} \quad \mathrm{B}\) (d) D C \(\mathrm{A}\)

Step-by-Step Solution

Answer

The correct matching is option (c): D C A B.

Step 1: Analyzing Reaction 1

The reactant \ \( \mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{3} \mathrm{OH} \) undergoes a reaction with NaBr in the presence of H₂SO₄ and heat (\( \Delta \)). This reaction will likely replace the OH group with a Br, forming the alkyl halide \( \mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{3} \mathrm{Br} \). Thus, \( 1 \rightarrow \text{D} \).

Step 2: Analyzing Reaction 2

In this reaction, \( \left(\mathrm{CH}_{3}\right)_{3} \mathrm{COH} \) is treated with concentrated HCl at room temperature. The OH group will be replaced by Cl due to the strong acid, forming \( \left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCl} \). Thus, \( 2 \rightarrow \text{C} \).

Step 3: Analyzing Reaction 3

For the reactant \( \mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH})\left(\mathrm{CH}_{2}\right)_{2} \mathrm{CH}_{3} \), reacting with \( \mathrm{PBr}_{3} \) will replace the OH group with Br. The resulting product will be \( \mathrm{CH}_{3} \mathrm{CHBr}\left(\mathrm{CH}_{2}\right)_{2} \mathrm{CH}_{3} \). So, \( 3 \rightarrow \text{A} \).

Step 4: Analyzing Reaction 4

In the reaction \( \mathrm{Me}_{2} \mathrm{CHCH}_{2} \mathrm{OH} \) with SOCl₂, the OH group is replaced by Cl, forming \( \mathrm{Me}_{2} \mathrm{CHCH}_{2} \mathrm{Cl} \). Thus, \( 4 \rightarrow \text{B} \).

Step 5: Identifying the Correct Matching

After analyzing each reaction, we have the following matches: \( 1 \rightarrow \text{D} \), \( 2 \rightarrow \text{C} \), \( 3 \rightarrow \text{A} \), and \( 4 \rightarrow \text{B} \). Thus, the correct option corresponds to (c) \( \text{D} \text{C} \text{A} \text{B} \).

Key Concepts

Nucleophilic Substitution

In organic chemistry, nucleophilic substitution is a fundamental type of reaction where one nucleophile replaces another in a molecule. It's essential in transforming functional groups, and it underlies the conversion of alcohols to alkyl halides. When an alcohol reacts with substances like hydrogen halides, phosphorus halides, or thionyl chloride, the hydroxyl group ( OH ) is substituted by a halogen atom such as bromine ( Br ) or chlorine ( Cl ).

• **Primary Alcohols** often undergo nucleophilic substitution through an S_N2 mechanism, where the nucleophile attacks from the opposite side of the leaving group.
• **Secondary and Tertiary Alcohols** favor an S_N1 mechanism, involving a two-step process where the leaving group departs to form a carbocation, followed by the nucleophilic attack.

In the context of our exercise:

• Reaction 1 uses NaBr and H 2 SO 4 to facilitate the substitution, forming an alkyl halide by replacing the OH group with a Br atom.
• Similarly, Reaction 4 employs SOCl 2 to replace the hydroxyl group with Cl .

These examples demonstrate nucleophilic substitution's role in transforming alcohols into more reactive alkyl halides.

Reaction Mechanisms

Understanding reaction mechanisms is key to predicting the outcome of chemical reactions. Mechanisms describe step-by-step events at the molecular level that lead to the product formation. When converting alcohols to alkyl halides, different pathways can be involved, depending mainly on the structure of the alcohol. - ** S_N1 Mechanism**: Used by secondary and tertiary alcohols, involves two steps. - The OH group leaves, forming a carbocation. - Subsequently, the nucleophile attacks the C^ + . - ** S_N2 Mechanism**: Common for primary alcohols, involves a single concerted substitution step. - The nucleophile attacks the substrate, displacing the OH group from the opposite side. For instance, in Reaction 2, the tertiary alcohol under the effect of HCl follows an S_N1 mechanism as the OH leaves, forming a stable tert-butyl carbocation, quickly followed by chloride anion ( Cl^- ) attack. Reaction 3 involves PBr 3 that follows a pathway akin to S_N2, where Br replaces OH directly.

Functional Group Transformation

Functional group transformation is a pivotal concept in organic synthesis. It involves converting one type of functional group into another, thereby altering the chemical behavior and reactivity of the molecule. The alcohol to alkyl halide transformation delineated in the exercise is an exemplary functional group conversion. - **Alcohols** contain a hydroxyl group ( R - OH ) that can be transformed into an alkyl halide ( R - X , where X is a halogen) through substitution reactions.

• These halides have enhanced reactivity compared to alcohols, making them valuable intermediates in further chemical synthesis.
• For example, the transformation of (CH 3 ) 3 COH to (CH 3 ) 3 CCl under Reaction 2 is a direct conversion utilizing HCl.

This type of transformation broadens the possibilities for creating complex organic molecules, as it offers a foundation for building new chemical frameworks through subsequent reactions.