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Identify the false statement among the following (a) The rate of hydrolysis of tert-butyl chloride does not change by increasing the concentration of \(\mathrm{OH}^{-}\). (b) 1 -phenylethanol reacts with thionyl chloride to give a chloro compound with almost complete retention of configuration. (c) Polar solvents generally increase the rate of \(\mathrm{S}_{\mathrm{N}} 1\) reactions. (d) \(\mathrm{S}_{\mathrm{N}} 2\) reaction of optically active substrate leads to racemization if the leaving group is attached to the chiral carbon.

Short Answer

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a) The rate of hydrolysis of tert-butyl chloride does not change by increasing the concentration of \(\mathrm{OH}^{-}\). b) 1-phenylethanol reacts with thionyl chloride to give a chloro compound with almost complete retention of configuration. c) Polar solvents generally increase the rate of \(\mathrm{S}_{\mathrm{N}} 1\) reactions. d) \(\mathrm{S}_{\mathrm{N}} 2\) reaction of optically active substrate leads to racemization if the leaving group is attached to the chiral carbon. Answer: d) \(\mathrm{S}_{\mathrm{N}} 2\) reaction of optically active substrate leads to racemization if the leaving group is attached to the chiral carbon.

Step by step solution

01

Statement (a) Analysis #

The rate of hydrolysis of tert-butyl chloride does not change by increasing the concentration of \(\mathrm{OH}^{-}\). This statement is true. The hydrolysis of tert-butyl chloride is an \(\mathrm{S}_{\mathrm{N}}1\) reaction, which means it's a unimolecular nucleophilic substitution reaction. In an \(\mathrm{S}_{\mathrm{N}}1\) reaction, the reaction rate depends only on the concentration of the substrate (tert-butyl chloride in this case), not the concentration of the nucleophile (\(\mathrm{OH}^{-}\)). Therefore, increasing the concentration of \(\mathrm{OH}^{-}\) does not affect the rate of hydrolysis of tert-butyl chloride.
02

Statement (b) Analysis #

1-phenylethanol reacts with thionyl chloride to give a chloro compound with almost complete retention of configuration. This statement is true. Reaction of 1-phenylethanol with thionyl chloride is an example of an \(\mathrm{S}_{\mathrm{N}}2\) reaction, in which the nucleophile attacks the carbon attached to the leaving group (in this case the hydroxyl group) from the opposite side of the leaving group. Since the configuration at the chiral center is determined by the three-dimensional arrangement of the groups, an attack from the opposite side will lead to retention of configuration. Therefore, 1-phenylethanol reacts with thionyl chloride to give a chloro compound with almost complete retention of configuration.
03

Statement (c) Analysis #

Polar solvents generally increase the rate of \(\mathrm{S}_{\mathrm{N}} 1\) reactions. This statement is true. In \(\mathrm{S}_{\mathrm{N}}1\) reactions, the reaction rate depends on the formation of a carbocation intermediate. Polar solvents are capable of stabilizing the charged species (the carbocation) and hence lower the activation energy required for the reaction to occur. As a result, polar solvents generally increase the rate of \(\mathrm{S}_{\mathrm{N}}1\) reactions.
04

Statement (d) Analysis #

\(\mathrm{S}_{\mathrm{N}} 2\) reaction of optically active substrate leads to racemization if the leaving group is attached to the chiral carbon. This statement is false. In an \(\mathrm{S}_{\mathrm{N}}2\) reaction, the nucleophile attacks the substrate from the opposite side of the leaving group, leading to inversion of configuration at the chiral center. There is no intermediate formation where two enantiomers can be formed, as seen in the \(\mathrm{S}_{\mathrm{N}}1\) reaction. Hence, there is no racemization in \(\mathrm{S}_{\mathrm{N}}2\) reactions of optically active substrates. In conclusion, statement (d) is the false statement among the given options.

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Key Concepts

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

Nucleophilic substitution reactions
Nucleophilic substitution reactions are fundamental in organic chemistry where a nucleophile attacks an electrophile, replacing another group in the compound. These reactions are categorized into two primary types: \( \mathrm{S}_{\mathrm{N}}1 \) and \( \mathrm{S}_{\mathrm{N}}2 \).
  • The \( \mathrm{S}_{\mathrm{N}}1 \) mechanism involves a two-step process that starts with the formation of a carbocation. This step is dependent on the substrate concentration and the reaction rate is uninfluenced by the nucleophile concentration.
  • The \( \mathrm{S}_{\mathrm{N}}2 \) mechanism is a single-step process where the nucleophile attacks the substrate from the opposite side of the leaving group. The rate of reaction is determined by both the substrate and nucleophile concentrations.
Understanding these mechanisms is crucial for predicting the outcome of organic reactions. Additionally, the choice of solvent and steric factors can significantly affect the reaction pathway and rates.
Tert-butyl chloride
Tert-butyl chloride is a common substrate used in \( \mathrm{S}_{\mathrm{N}}1 \) reactions. Its structure consists of a tertiary carbon bonded to a chlorine atom, making it quite stable as a carbocation intermediate. This stability arises because:
  • The three electron-donating alkyl groups on the tertiary carbon stabilize the carbocation through hyperconjugation and inductive effects.
  • The bulkiness of the tert-butyl group hinders the approach of a nucleophile, favoring the \( \mathrm{S}_{\mathrm{N}}1 \) pathway over \( \mathrm{S}_{\mathrm{N}}2 \).
In reactions, tert-butyl chloride undergoes hydrolysis readily in polar solvents, forming a tert-butyl alcohol. This occurs independently of the concentration of nucleophiles such as \( \mathrm{OH}^{-} \). Therefore, changes in nucleophile concentration do not impact the reaction rate.
Optical activity and racemization
Optical activity refers to a molecule's ability to rotate plane-polarized light due to its chiral nature, having non-superimposable mirror images called enantiomers. In nucleophilic substitution reactions, whether a reaction leads to racemization or retention of configuration depends on the mechanism.
  • \( \mathrm{S}_{\mathrm{N}}1 \) reactions may lead to racemization because of the planar nature of the carbocation intermediate, allowing the nucleophile to attack from either side, forming a mixture of enantiomers.
  • \( \mathrm{S}_{\mathrm{N}}2 \) reactions typically result in inversion of configuration at the chiral center because the nucleophile attacks from the opposite side of the leaving group.
Racemization occurs when an equal amount of both enantiomers is produced, resulting in a racemic mixture which is optically inactive.
Organic reaction mechanisms
Organic reaction mechanisms describe the step-by-step processes by which chemical reactions occur at the molecular level. Understanding these mechanisms helps in predicting and controlling chemical reactivity and outcome.
  • The mechanism of a reaction involves identifying the reactants, products, intermediates, and transition states.
  • It also includes knowledge of the energetic pathways, steric hindrance, and electronic effects that influence reaction rates and selectivities.
In nucleophilic substitution, particularly, the study of mechanisms enables chemists to manipulate reaction conditions to favor one reaction pathway over another, thus controlling product distribution. Familiarity with organic mechanisms is key for success in IIT JEE Chemistry, as these reactions form the basis for many synthetic strategies and problem-solving exercises.

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Most popular questions from this chapter

Which of the following is the correct order of dipole moment of methyl halides? (a) \(\mathrm{CH}_{3} \mathrm{~F}>\mathrm{CH}_{3} \mathrm{Cl}>\mathrm{CH}_{3} \mathrm{Br}>\mathrm{CH}_{3} \mathrm{I}\) (b) \(\mathrm{CH}_{3} \mathrm{Cl}>\mathrm{CH}_{3} \mathrm{~F}>\mathrm{CH}_{3} \mathrm{Br}>\mathrm{CH}_{3} \mathrm{I}\) (c) \(\mathrm{CH}_{3} \mathrm{I}>\mathrm{CH}_{3} \mathrm{Br}>\mathrm{CH}_{3} \mathrm{~F}>\mathrm{CH}_{3} \mathrm{I}\) (d) \(\mathrm{CH}_{3} \mathrm{I}>\mathrm{CH}_{3} \mathrm{Br}>\mathrm{CH}_{3} \mathrm{Cl}>\mathrm{CH}_{3} \mathrm{~F}\)

Directions: Each question contains Statement- 1 and Statement-2 and has the following choices (a), (b), (c) and (d), out of which ONLY ONE is correct. (a) Statement- 1 is True, Statement- 2 is True; Statement- 2 is a correct explanation for Statement-1 (b) Statement- 1 is True, Statement- 2 is True; Statement- 2 is NOT a correct explanation for Statement-1 (c) Statement- 1 is True, Statement- 2 is False (d) Statement- 1 is False, Statement- 2 is True Statement 1 Nucleophilicity does not always match with basicity. and Statement 2 While nucleophilicity is more related to size, polarizability and steric factors, the basicity is the ability to abstract a proton from a substrate.

\(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHBr} \stackrel{\mathrm{Mg} / \mathrm{ether}}{\longrightarrow} \mathrm{A} \stackrel{\mathrm{C}_{6} \mathrm{H}_{5}-\mathrm{CO}-\mathrm{CH}_{3}}{\longrightarrow} \mathrm{B}\) The product \(B\) in the above reaction is (a) 2 -phenyl- 3 -methyl butan- 2 -ol (b) 2 -phenylpentan- 2 -ol (c) 3 -phenyl butan-2-ol (d) propan-2-ol and ethyl benzene

In the preparation of \(\mathrm{CH}_{3}-\mathrm{O}-\mathrm{C}\left(\mathrm{CH}_{3}\right)_{3}\), two methods, namely (i) and (ii) are suggested. Which method is preferable and why? (i) \(\mathrm{CH}_{3} \mathrm{ONa}+\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{Br} \rightarrow \mathrm{CH}_{3}-\mathrm{O}-\mathrm{C}\left(\mathrm{CH}_{3}\right)_{3}\) (ii) \(\mathrm{CH}_{3} \mathrm{Br}+\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{ONa} \rightarrow \mathrm{CH}_{3}-\mathrm{O}-\mathrm{C}\left(\mathrm{CH}_{3}\right)_{3}\)

The correct statements about the following compound are $$ \mathrm{CH}_{3}(\mathrm{OH}) \mathrm{CH}-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}(\mathrm{OH}) \mathrm{CH}_{3} $$ (a) The total number of stereoisomers possible is 6 . (b) The total number of diastereomers possible is \(3 .\) (c) The number of enantiomers possible for the cis isomer is \(2 .\) (d) The number of enantiomers possible for the trans isomer is 4 .

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