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During hydrolysis, in which acid derivative the rate determining step is step- II. (A) Ester (B) Acid halide (C) Amide (D) Acid anhydride

Short Answer

Expert verified
The acid derivative where the rate-determining step during hydrolysis is step-II is the acid anhydride. Therefore, the correct answer is (D) Acid anhydride.

Step by step solution

01

General reaction mechanism overview for hydrolysis of acid derivatives

The acid derivatives can be hydrolyzed through either acid-catalyzed or base-catalyzed mechanisms. However, the general pattern of reaction steps remains similar: nucleophilic attack, followed by elimination. We will break down the mechanism into two steps: 1. Nucleophilic attack: A nucleophile (OH- or H2O) attacks the carbonyl carbon, forming a tetrahedral intermediate. 2. Elimination: The electron pair from the tetrahedral intermediate will reform the carbonyl group and expel the leaving group.
02

Ester hydrolysis mechanism overview

Esters undergo hydrolysis to form a carboxylic acid and an alcohol. The rate-determining step is the nucleophilic attack, which is step-I.
03

Acid halide hydrolysis mechanism overview

Acid halides undergo hydrolysis to form a carboxylic acid and a halide ion. The rate-determining step is the nucleophilic attack, which is step-I.
04

Amide hydrolysis mechanism overview

Amides undergo hydrolysis to form a carboxylic acid and an amine or ammonia. The rate-determining step is the nucleophilic attack, which is step-I.
05

Acid anhydride hydrolysis mechanism overview

Acid anhydrides undergo hydrolysis to form two carboxylic acids. The rate-determining step is the elimination, which is step-II.
06

Selecting the answer based on the rate-determining step

Based on the information above, the acid derivative where the rate determining step during hydrolysis is step-II is the acid anhydride. So the correct answer is: (D) Acid anhydride

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Key Concepts

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

Rate-Determining Step
In the context of a chemical reaction mechanism, the term rate-determining step refers to the slowest step in a sequence that defines the time frame in which the reaction proceeds. Understanding this step is crucial for predicting how changes in conditions or reactant concentrations will affect the reaction rate.

When analyzing the hydrolysis of acid derivatives, such as esters, acid halides, amides, and acid anhydrides, it's essential to identify the rate-determining step to understand the kinetics of the reaction. As depicted in the solution overview, esters, acid halides, and amides share a common rate-determining step: the nucleophilic attack. This is because the nucleophile's interaction with the carbonyl carbon is typically slower in comparison to any subsequent steps. However, with acid anhydrides, the elimination stage is slower and thus determines the reaction rate. Understanding this concept is beneficial not only for homework exercises but also for designing experiments and synthesizing organic compounds.
Nucleophilic Attack Mechanism
The mechanism of a nucleophilic attack plays a fundamental role in the field of organic chemistry and is particularly pertinent for students preparing for competitive exams like the JEE (Joint Entrance Examination). A nucleophilic attack involves a nucleophile, a species rich in electrons, approaching and forming a bond with an electrophilic center, often a carbon atom within a carbonyl group.

Using a simple explanation, visualize the carbonyl group as an electron-deficient site due to the oxygen atom's electronegativity. Nucleophiles, with their electron surplus, are attracted to this site. During the hydrolysis of acid derivatives, the nucleophile (such as water or hydroxide ion) attacks the carbonyl carbon, leading to the opening of the carbonyl group and formation of a tetrahedral intermediate. This intermediate is unstable and quickly leads to the regeneration of the carbonyl group and the expulsion of a leaving group, thus proceeding to completion. It's the comparison of various acid derivatives' nucleophilic attack strengths that underpin their differing reactivities and hydrolysis rates.
Organic Chemistry for JEE
For students gearing up for the JEE, understanding organic chemistry's underlying principles is key to solving complex problems. The hydrolysis of acid derivatives, mechanisms like nucleophilic attack, and concepts such as rate-determining step are all topics that routinely appear in the JEE syllabus.

It's important to approach studying organic chemistry methodically by first grasping fundamental mechanisms, which frequently involve electron shifts, bond-making, and bond-breaking processes. Equipped with this knowledge, students can then tackle a wide range of organic transformations. Practice problems, such as the hydrolysis of different acid derivatives, help solidify these concepts and aid in recognizing patterns that will prove invaluable during the actual examination. Understanding the intricacies of these reactions not only assists in solving textbook exercises but also builds a strong foundation necessary for success in competitive chemistry exams like the JEE.

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Most popular questions from this chapter

Which of the following reaction is not possible? (A) \(2 \mathrm{R}-\mathrm{COOH}+2 \mathrm{Na} \longrightarrow 2 \mathrm{R}-\mathrm{COONa}+\mathrm{H}_{2}\) (B) \(\mathrm{R}-\mathrm{COONa}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{R}-\mathrm{COOH}+\mathrm{NaOH}\) (C) \(\mathrm{R}-\mathrm{COOH}+\mathrm{NaHCO}_{3} \longrightarrow \mathrm{R}-\mathrm{COO}^{-} \mathrm{Na}^{\oplus}+\mathrm{H}_{2} \mathrm{O}+\mathrm{CO}_{2}\) (D) \(\mathrm{R}-\mathrm{COOH}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{H}_{3} \mathrm{O}^{\oplus}+\mathrm{R}\)

Correct option(s) regarding reaction, reactant and product respectively given is : (A) Schmidt reaction Acid Primary amine (B) Hoffmann bromamide Primary acid amide Primary amine (C) Schmidt reaction Ketone Secondary acid amide (D) Beckmann rearrangement Ketoxime Secondary acid amide

During reaction of \(\alpha\)-diazo carbonyl compound \([\mathrm{R}\) \(\left(\mathrm{R}-\mathrm{C}-\mathrm{CH}_{2} \mathrm{~N}_{2}\right)\) with \(\mathrm{Ag}_{2} \mathrm{O}\), product obtained and reaction is respectively: (A) \(\mathrm{R}-\mathrm{N}=\mathrm{C}=\mathrm{O}\) Wolf rearrangement (B) \(\mathrm{R}-\mathrm{CH}=\mathrm{C}=\mathrm{O}\) Wolf Kishner reduction (C) \(\mathrm{R}-\mathrm{CH}=\mathrm{C}=\mathrm{O} \quad\) Wolf rearrangement (D) \(\mathrm{R}-\mathrm{CH}=\mathrm{C}=\mathrm{NH} \quad\) Wolf rearrangement

Compound \(\mathrm{R}+\) Compound \(\mathrm{T} \quad \mathrm{NaBH}_{\mathrm{S} \mathrm{CN}}{\longrightarrow}\) Product The product is: (A) \(1^{\circ}\) amine (B) \(2^{\circ}\) amine (C) \(3^{\circ}\) amine (D) Quaternary amine

\((\mathrm{P}) \stackrel{\mathrm{Red} \mathrm{P} / \mathrm{X}_{2}}{\longrightarrow}(\mathrm{Q})\) (P) has M.F. \(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}_{2}\) which gives fruity smell with alcohol and ' \(\mathrm{X}^{\prime}\) is halogen. Choose the incore: option: (A) Product (Q) is \(\beta\)-halocarboxylic acid (B) Reaction is possible when \(\alpha\)-hydrogen is present (C) Reaction is known as HVZ reaction (D) Red \(\mathrm{P}\) is taken in small amount and ' \(\mathrm{X}\) ' can be chlorine or bromine.

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