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In \(\mathrm{S}_{\mathrm{N}}{ }^{2}\) reactions, the correct order of reactivity for the following compounds: \(\mathrm{CH}_{3} \mathrm{Cl}, \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Cl}\), \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCl}\) and \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCl}\) is: \([2014]\) (a) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Cl}>\mathrm{CH}_{3} \mathrm{Cl}>\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCl}>\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCl}\) (b) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCl}>\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Cl}>\mathrm{CH}_{3} \mathrm{Cl}>\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCl}\) (c) \(\mathrm{CH}_{3} \mathrm{Cl}>\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCl}>\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Cl}>\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCl}\) (d) \(\mathrm{CH}_{3} \mathrm{Cl}>\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Cl}>\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCl}>\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCl}\)

Short Answer

Expert verified
Option (d): \(\text{CH}_3\text{Cl} > \text{CH}_3\text{CH}_2\text{Cl} > (\text{CH}_3)_2\text{CHCl} > (\text{CH}_3)_3\text{CCl}\).

Step by step solution

01

Understand SN2 Mechanism

The SN2 reaction is a type of nucleophilic substitution where the rate-determining step involves a single transition state. It is bimolecular and involves the simultaneous bond making between nucleophile and breaking from the leaving group.
02

Identify Factors Affecting SN2 Reactivity

For SN2 reactions, steric hindrance at the electrophilic carbon significantly impacts reactivity. Less sterically hindered or less bulky substrates are more reactive in SN2 reactions.
03

Analyze Each Compound's Structure

- **\(\text{CH}_3\text{Cl}\)** has no alkyl groups attached, causing minimal steric hindrance. - **\(\text{CH}_3\text{CH}_2\text{Cl}\)** has one methyl group causing minimal steric hindrance.- **\((\text{CH}_3)_2\text{CHCl}\)** has two methyl groups causing moderate steric hindrance.- **\((\text{CH}_3)_3\text{CCl}\)** has three methyl groups, leading to significant steric hindrance.
04

Arrange Based on Steric Hindrance

Based on the steric hindrance:1. \(\text{CH}_3\text{Cl}\) is the least hindered and thus the most reactive.2. \(\text{CH}_3\text{CH}_2\text{Cl}\) comes next due to slight steric hindrance.3. \((\text{CH}_3)_2\text{CHCl}\) is less reactive due to more steric hindrance than \(\text{CH}_3\text{CH}_2\text{Cl}\).4. \((\text{CH}_3)_3\text{CCl}\) is the least reactive due to heavy steric hindrance.
05

Match With Given Options

The correct order of reactivity is: \(\text{CH}_3\text{Cl} > \text{CH}_3\text{CH}_2\text{Cl} > (\text{CH}_3)_2\text{CHCl} > (\text{CH}_3)_3\text{CCl}\).This matches option (d).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Nucleophilic Substitution
Nucleophilic substitution is a fundamental class of reactions in organic chemistry, involving the replacement of a leaving group by a nucleophile. In \(S_N2\) reactions, this substitution process is swift and occurs in a single concerted step, where the nucleophile attacks the substrate while the leaving group exits the molecule. Due to this simultaneous action, \(S_N2\) reactions are considered bimolecular and are highly sensitive to the structure of the substrate. Understanding the balance between the nucleophile’s strength and leaving group’s ability is vital to predicting the reactivity in these reactions. \(S_N2\) reactions commonly occur with primary alcohols and alkyl halides, as these substrates allow sufficient access for the nucleophilic attack.
Steric Hindrance
Steric hindrance refers to the crowding that occurs when atoms or groups within a molecule obstruct a chemical reaction. In \(S_N2\) reactions, steric hindrance plays a crucial role in determining the reactivity of the substrate. Less hindered, or less crowded, substrates react more readily with nucleophiles because the nucleophile can easily approach and attack the electrophilic carbon.
For example, in the case of \((\text{CH}_3)_3\text{CCl}\), the presence of three bulky methyl groups around the central carbon creates significant steric hindrance. This hinders the approach of the nucleophile, making it less reactive. Conversely, a molecule like \(\text{CH}_3\text{Cl}\) has almost no steric hindrance, allowing for a much more efficient reaction. The effect of steric hindrance is a key consideration when predicting the outcome of \(S_N2\) reactions.
Reaction Mechanism
In the \(S_N2\) reaction mechanism, two molecules participate in the rate-determining step, making it "bimolecular." This mechanism features a single transition state: as the nucleophile forms a new bond with the carbon atom, the bond between the carbon atom and the leaving group is simultaneously broken. This simultaneous bond-forming and breaking ensures the process is a concerted mechanism, characterized by a smooth progression without any intermediate stages.
Such reactions typically involve an inversion of configuration at the carbon center, often referred to as a "backside attack." This inversion occurs because the nucleophile attacks from the opposite side to the leaving group, which is a distinct feature of \(S_N2\) reactions. No intermediate forms because the reaction progresses straight to completion from the starting materials to the product through this one step. This provides a clear contrast to \(S_N1\) reactions, which proceed through distinct intermediates.
Reactivity Order
Reactivity order in \(S_N2\) reactions largely depends on the steric accessibility and the nature of the substrate. Primary alkyl halides, like \(\text{CH}_3\text{Cl}\), exhibit the highest reactivity because they have minimal steric hindrance, allowing nucleophiles easy access to the electrophilic center.
Secondary alkyl halides, such as \((\text{CH}_3)_2\text{CHCl}\), show reduced reactivity due to increased steric bulk that partially obstructs the nucleophile's approach. Tertiary alkyl halides, exemplified by \((\text{CH}_3)_3\text{CCl}\), display the lowest reactivity in \(S_N2\) reactions. The significant steric hindrance from adjacent groups forms a substantial barrier for incoming nucleophiles, dramatically decreasing reactivity. Understanding this order helps chemists predict and manipulate reaction conditions to achieve desired transformations efficiently.

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Most popular questions from this chapter

When a primary amine reacts with chloroform in ethanolic KOH, then the product formed is [2002] (a) an aldehyde (b) an isocyanide (c) an cyanide (d) an alcohol

Reaction of ethyl benzene with bromine in presence of \(\mathrm{FeBr}_{3}\) gives (a) 2-bromoethyl benzene (b) 2 -bromoethyl benzene (c) 3-bromoethyl benzene (d) 4 -bromoethyl benzene

In the following chemical reactions: \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NH}_{2}+\mathrm{CHCl}_{3}+3 \mathrm{KOH} \longrightarrow(\mathrm{A})+(\mathrm{B})+3 \mathrm{H}_{2} \mathrm{O}\), the compounds (A) and (B) are respectively [2007] (a) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CONH}_{2}\) and \(3 \mathrm{KCl}\) (b) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NC}\) and \(\mathrm{K}_{2} \mathrm{CO}_{3}\) (c) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NC}\) and \(3 \mathrm{KCl}\) (d) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{CN}\) and \(3 \mathrm{KCl}\)

Identify the final product (C) in the following sequence of reactions. \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{I} \stackrel{\text { Alc. } \mathrm{KOH}, \Delta}{\longrightarrow}(\mathrm{A}) \stackrel{\mathrm{Br}_{2}}{\longrightarrow}(\mathrm{B}) \stackrel{\mathrm{KCN}}{\longrightarrow}(\mathrm{C})\) (a) \(\mathrm{NCCH}_{2} \mathrm{CH}_{2} \mathrm{CN}\) (b) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Br}\) (c) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CN}\) (d) \(\mathrm{BrCH}_{2} \mathrm{CH}_{2} \mathrm{CN}\)

The reaction of the formation of bromobenzyl chloride with \(\mathrm{NaCN}\) in ethanol leads to (a) 4-bromo-2-cyanobenzyl chloride (b) 4 -cyanobenzyl cyanide (c) 4-cyanobenzyl chloride (d) 4-bromobenzyl cyanide

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