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Which of the following order/statement (s) is/are correct? (a) In \(S_{N} 1\) reactions, \(\mathrm{AgNO}_{3}\) increases the rates of solvolysis (b) Soft bases have enhanced nucleophilicities towards \(\mathrm{S}_{\mathrm{N}} 2\) reactions. (c) \(\mathrm{PhSO}_{3}^{-}>\mathrm{Cl}_{3} \mathrm{CCOO}^{-}>\mathrm{CH}_{3} \mathrm{COO}^{-}>\mathrm{PhO}^{-}\)(leav- ing group ability) (d) HCOOH may lead a tertiary halide towards \(\mathrm{S}_{\mathrm{N}}^{2} .\)

Short Answer

Expert verified
Statements (a) and (c) are correct.

Step by step solution

01

Assess statement (a)

For an \(S_N1\) reaction, the rate-determining step is the formation of a carbocation intermediate, which occurs fastest in polar protic solvents. \(\mathrm{AgNO}_3\) can promote the formation of carbocations, as it can precipitate halides, making the leaving group dissociation more favorable. Thus, statement (a) is correct.
02

Analyze statement (b)

Soft bases, according to HSAB (hard and soft acids and bases) theory, prefer soft acids. However, \(S_N2\) reactions often involve hard electrostatic interactions, rather than the polarizable interactions characteristic of soft acids and bases. Hence, this statement tends to oversimplify the role of nucleophiles in \(S_N2\). Therefore, statement (b) is incorrect.
03

Evaluate statement (c)

Leaving group ability is more favorable for groups that can stabilize the negative charge post-dissociation. \(\mathrm{PhSO}_3^-\) is indeed a very good leaving group due to resonance stability, followed by \(\mathrm{Cl}_3\mathrm{CCOO}^-\) as it is stabilized by electron-withdrawing chloro groups. \(\mathrm{CH}_3\mathrm{COO}^-\) is a reasonable leaving group, while \(\mathrm{PhO}^-\) is commonly a poor leaving group due to stability effects. Thus, statement (c) is correct.
04

Consider statement (d)

\(S_N2\) reactions involve a backside attack by the nucleophile, typically leading to inversion of configuration. Tertiary halides are less favorable for \(S_N2\) due to steric hindrance. HCOOH, or formic acid, does not change this outcome as it is not capable of significantly reducing such steric hindrance. Therefore, statement (d) is incorrect.

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Key Concepts

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

Solvolysis
Solvolysis is a unique type of chemical reaction that occurs in a solvent, where the solvent itself acts as the nucleophile, participating in the breaking of bonds within the substrate molecule. In the context of nucleophilic substitution reactions, solvolysis usually refers to the reaction of a substrate with the solvent, often leading to the formation of carbocations, particularly in SN1 reactions.
  • **Polar Protic Solvents:** These solvents, like water and alcohols, provide an environment conducive to the formation of carbocations because they stabilize the positive charges through interactions with solvent molecules.
  • **Common Example:** Water acting as a solvent can react with alkyl halides to substitute the halide ion, forming an alcohol in the process.
AgNO₃ often plays a critical role in solvolysis by precipitating the leaving halide ion as silver halide, therefore removing the leaving group from the reaction mixture, which can facilitate carbocation formation and enhance the rate of solvolysis. Understanding solvolysis is essential for grasping how certain reaction conditions affect the reactivity and outcome of a substitution reaction.
Nucleophilic Substitution
Nucleophilic substitution is at the heart of a wide variety of organic reactions, where a nucleophile, a species rich in electrons, replaces a leaving group in a molecule. There are two primary mechanisms: SN1 and SN2. Understanding each helps in predicting the course and outcome of these reactions.
  • **SN1 Reactions:** These occur in two steps, where the leaving group departs first, forming a carbocation intermediate. It usually happens in polar protic solvents and is typical for tertiary carbon centers where carbocations are more stable.
  • **SN2 Reactions:** These are one-step reactions where the nucleophile attacks the substrate from the opposite side of the existing bond, leading to an inversion of configuration. SN2 reactions are common in primary carbon centers where steric hindrance is minimal.
When considering a nucleophilic substitution, it's crucial to think about the nature of the nucleophile and the substrate since factors like steric hindrance, solvent, and the strength of leaving groups play significant roles.
Leaving Group Ability
A leaving group's ability significantly impacts the efficiency of nucleophilic substitution reactions. It determines how easily the leaving group can depart from the parent molecule. Leaving groups are more effective when they can stabilize the negative charge after separation.
  • **Good Leaving Groups:** Halides like iodide and bromide, and groups like tosylate (PhSO₃⁻), have strong resonance or electronegative components that stabilize post-departure.
  • **Poor Leaving Groups:** Groups like phenoxide (PhO⁻) are less stable once they're detached due to lack of sufficient stabilizing factors compared to halides and other anions.
A good understanding of leaving group ability helps in designing effective nucleophilic substitution reactions where the choice of the leaving group can dictate the speed and success of the reaction.
HSAB Theory
Hard and Soft Acids and Bases (HSAB) theory provides a framework to explain the preferences and reactivity of acids and bases in chemical reactions, including nucleophilic substitutions. It categorizes acids and bases as either "hard" or "soft".
  • **Hard Acids and Bases:** Typically have small, highly charged ions that are not easily polarizable, like fluoride (F⁻) and hydroxide (OH⁻). They prefer interacting with each other due to their charge density, often involved in reactions like SN2.
  • **Soft Acids and Bases:** These are larger, more polarizable ions and molecules, such as thiolates, that engage in softer, more covalent interactions, and they are often involved in reactions where these characteristics align with the transition states.
In nucleophilic substitution reactions, understanding HSAB helps predict reactivity, particularly in tailoring reactions where the nature of the nucleophile and the electrophile aligns with the desired outcome.

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Most popular questions from this chapter

In the following groups: \(-\mathrm{OAc}(\mathrm{I})\) \- OMe (II) \(-\mathrm{OSO}_{2} \mathrm{Me}(\mathrm{III}), \quad-\mathrm{OSO}_{2} \mathrm{CF}_{3}(\mathrm{IV})\) the order of the leaving group ability is (a) \(\mathrm{I}>\mathrm{II}>\mathrm{III}>\mathrm{IV}\) (b) IV > III > I > II (c) \(\mathrm{III}>\mathrm{II}>\mathrm{I}>\mathrm{IV}\) (d) \(\mathrm{II}>\mathrm{III}>\mathrm{IV}>\mathrm{I}\).

The reaction \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{Br} \stackrel{\mathrm{H}_{2} \mathrm{O}}{\longrightarrow}\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{OH}\) is (a) addition reaction (b) substitution reaction (c) free radical reaction (d) elimination reaction

Among the following groups, \(-\mathrm{CH}_{3},-\mathrm{Cl},-\mathrm{OCH}_{3}\), \(-\mathrm{CHO},-\mathrm{CN},-\mathrm{NO}_{2}\) and \(-\mathrm{COOH}\), number of meta directing groups is__ .

The order of stability of the following carbocations: (I) \(\mathrm{CH}_{2}=\mathrm{CH}-\stackrel{\oplus}{\mathrm{C}} \mathrm{H}_{2}\) \(\mathbf{[ 2 0 1 3}]\) (II) \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\stackrel{\oplus}{\mathrm{C}} \mathrm{H}_{2}\) (III) is: [C+]c1ccccc1 (a) \(\mathrm{I}>\mathrm{II}>\mathrm{III}\) (b) \(\mathrm{III}>\mathrm{I}>\mathrm{II}\) (c) \(\mathrm{III}>\mathrm{II}>\mathrm{I}\) (d) \(\mathrm{II}>\mathrm{III}>\mathrm{I}\)

Which of the following compounds possesses the \(\mathrm{C}-\mathrm{H}\) bond with the lowest bond dissociation energy? (a) n-pentane (b) toluene (c) benzene (d) 2,2 -dimethylpropane

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