Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 15: Problem 317

Synthesize a ketone from each of the following precursors using the reaction of acid chlorides with organocadmium compounds.

Short Answer

Expert verified
To synthesize a ketone from a given precursor using the reaction of acid chlorides with organocadmium compounds, follow these general steps: 1. Identify the precursor. 2. Determine the required acid chloride by finding the same carbon backbone and functional group in the desired ketone, but with a chloride atom replacing the carbonyl oxygen atom. 3. Determine the organocadmium compound by identifying the carbon adjacent to the carbonyl group in the desired ketone structure, along with the attached substituents. 4. Write the reaction equation with the acid chloride and organocadmium compound as reactants, the desired ketone as the product, and cadmium chloride (CdCl2) as a byproduct: \(RCOCl + R'CdCl \rightarrow RCOCH(R') + CdCl_2 \) where R and R' are the organic groups of the acid chloride and organocadmium compound, respectively.

Step by step solution

01

Identify the given precursors

Unfortunately, no precursors have been given in this case. Therefore, we can't proceed with the step-by-step solution. However, we can provide the general steps you can take to solve similar problems when provided with specific precursors.
02

Determine the acid chloride required

Once you have the precursor, you can find the required acid chloride by examining the structure of the desired ketone. The acid chloride should contain the same carbon backbone and functional group, but with a chloride atom replacing the carbonyl oxygen atom.
03

Determine the organocadmium compound required

The organocadmium compound that you need to use in the reaction should have a structure that includes the remaining portion of your target ketone along with the cadmium atom. To find the appropriate organocadmium compound, look at the desired ketone structure and identify the carbon adjacent to the carbonyl group. This carbon atom and its attached substituents would form the organocadmium compound.
04

Write the reaction equation

After identifying the required acid chloride and organocadmium compound, write the reaction equation by placing the acid chloride and organocadmium compound as reactants, and the desired ketone as the product. Additionally, include cadmium chloride (CdCl2) as a byproduct of the reaction. In general, the reaction can be represented as: \(RCOCl + R'CdCl \rightarrow RCOCH(R') + CdCl_2 \) where R and R' are the organic groups of the acid chloride and organocadmium compound, respectively. Keep in mind that the specific reaction and reagents will depend on the given precursors, which have not been provided in this case.

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

Excess methylmagnesium iodide and \(0.1776 \mathrm{~g}\) of compound A of formula \(\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}_{3}\) Sreact to give \(84.1 \mathrm{cc}\) of methane collected over mercury at \(740 \mathrm{~mm}\) and \(25^{\circ} \mathrm{C}\). How many active hydrogens does compound A possess? Suggest a possible structure for the compound given that the infrared spectrum shows no carbonyl absorption and the n.m.r. spectrum shows only three types of hydrogen with areas in the ratio of \(1: 2: 2\).

Give structures of compounds \(\mathrm{A}, \mathrm{B}\) and \(\mathrm{C}\) : ethyl oxalate \(+\) ethyl acetate + sodium ethoxide, then \(\mathrm{H}^{+} \rightarrow \mathrm{A}\left(\mathrm{C}_{8} \mathrm{H}_{20} \mathrm{O}_{5}\right)\) \(\mathrm{A}+\) ethy 1 bromoacetate \(+\mathrm{Zn}\), then \(\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{B}\left(\mathrm{C}_{12} \mathrm{H}_{20} \mathrm{O}_{7}\right)\) \(\mathrm{B}+\mathrm{OH}^{-}+\) heat, then \(\mathrm{H}^{+} \rightarrow \mathrm{C}\left(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{7}\right)\), citric acid.

The following projected synthesis for \(\mathrm{n}\) -butane is not very efficient. Why? \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Br}+\mathrm{CH}_{3} \mathrm{Br}+2 \mathrm{Na} \rightarrow \mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{2} \mathrm{CH}_{3}+2 \mathrm{NaBr}\)

Write balanced equations for the preparation of each of the following organometallic compounds by two different reactions starting from suitable alky1 halides and inorganic reagents. Specify reaction conditions and solvents. In each case, indicate which method of preparation you would prefer from standpoints of yield, convenience, etc. (a) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{Zn}\) (c) \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2}\right)_{4} \mathrm{~Pb}\) (b) \(\mathrm{CH}_{3} \mathrm{MgCl}\) (d) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHLi}\)

An aromatic dibromide \(\mathrm{C}_{7} \mathrm{H}_{6} \mathrm{Br}_{2}\) reacted with aqueous sodium hydroxide. The product of this reaction had lost only one bromo group to give the product \(\mathrm{C}_{7} \mathrm{H}_{7} \mathrm{BrO}\). When the dibromide was converted to a Grignard reagent and then hydrolyzed, the product was toluene. Determine the structure of the dibromide.
See all solutions

Recommended explanations on Chemistry Textbooks

Inorganic Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation

Physical Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Kinetics

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