Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 21: Q29E (page 726)

The following structure represents a tetrahedral alkoxide-ion intermediate formed by addition of a nucleophile to a carboxylic acid derivative. Identify the nucleophile, the leaving group, the starting acid derivative, and the ultimate product (green = Cl).

Short Answer

Expert verified

Intermediate can be prepared acid chloride and nucleophile

The final product amide can be prepared as from the intermediate as

Step by step solution

01

The Nucleophile and the Starting Acid Derivative

Acid chloride reacts with NH3 nucleophile to give the intermediate:

02

Leaving Group and Final Product

Intermediate leaves the Cl- ion and produces the amide.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

When ethyl benzoate is heated in methanol containing a small amount of HCL, Methyl benzoate is formed. Propose a mechanism for the reaction.

Treatment of 5- aminopentanoic acid DCC(dicyclohexylcarbodiimide) yields a lactam. Show the structure of the product and the mechanism of the reaction.

21-75 One frequently used method for preparing methyl esters is by reactionof carboxylic acids with diazomethane, CH2N2

The reaction occurs in two steps: (1) protonation of diazomethane bythe carboxylic acid to yield methyldiazonium ion,, CH3N2+plus a carboxylate ion; and (2) reaction of the carboxylate ion with CH3N2+.
(a) Draw two resonance structures of diazomethane, and account forstep 1.
(b) What kind of reaction occurs in step 2?

What product would you expect from reaction of one equivalent of methanol with a cyclic anhydride, such as phthalic anhydride (1,2-benzenedicarboxylic anhydride)? What is the fate of the second “half” of the anhydride?

Bacteria typically develop a resistance to penicillins and other β-lactam antibiotics (see Something Extra in this chapter) due to bacterial synthesis of β -lactamase enzymes. Tazobactam, however, is able to inhibit the activity of the β -lactamase by trapping it, thereby preventing a resistance from developing.

(a) The first step in trapping is the reaction of a hydroxyl group on the β -lactamase to open the β -lactam ring of tazobactam. Show the mechanism.
(b) The second step is opening the sulfur-containing ring in tazobactam to give an acyclic imine intermediate. Show the mechanism.
(c) Cyclization of the imine intermediate gives the trapped β-lactamase product. Show the mechanism.
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Analysis

Read Explanation

Physical Chemistry

Read Explanation

Making Measurements

Read Explanation

The Earths Atmosphere

Read Explanation

Kinetics

Read Explanation

Inorganic Chemistry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free