Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 22: Problem 71

Using appropriate reactants, alcohols can be oxidized into aldehydes, ketones, and/or carboxylic acids. Primary alcohols can be oxidized into aldehydes, which can then be oxidized into carboxylic acids. Secondary alcohols can be oxidized into ketones, while tertiary alcohols do not undergo this type of oxidation. Give the structure of the product(s) resulting from the oxidation of each of the following alcohols. a. 3-methyl-1-butanol b. 3-methyl-2-butanol c. 2-methyl-2-butanol

Short Answer

Expert verified
The oxidation products for the given alcohols are as follows: a. 3-methyl-1-butanol (primary alcohol) - Aldehyde: 3-methylbutanal and Carboxylic acid: 3-methylbutanoic acid. b. 3-methyl-2-butanol (secondary alcohol) - Ketone: 3-methyl-2-butanone. c. 2-methyl-2-butanol (tertiary alcohol) - No oxidation product as tertiary alcohols do not undergo this type of oxidation.

Step by step solution

01

Identify the type of alcohols

First, we will analyze each alcohol's structure and determine if it's a primary, secondary, or tertiary alcohol. a. 3-methyl-1-butanol The OH group is attached to the first (primary) carbon atom which is bonded to only one other carbon atom. So, 3-methyl-1-butanol is a primary alcohol. b. 3-methyl-2-butanol The OH group is attached to the second (secondary) carbon atom which is bonded to two other carbon atoms. So, 3-methyl-2-butanol is a secondary alcohol. c. 2-methyl-2-butanol The OH group is attached to the second (tertiary) carbon atom, which is bonded to three other carbon atoms. So, 2-methyl-2-butanol is a tertiary alcohol.
02

Determine the oxidation products for each type of alcohol

Now that we have determined the type of alcohol for each compound, we will establish the oxidation products for each alcohol accordingly. a. 3-methyl-1-butanol is a primary alcohol. It can be oxidized into an aldehyde (3-methylbutanal) and further oxidized into a carboxylic acid (3-methylbutanoic acid). b. 3-methyl-2-butanol is a secondary alcohol. It can only be oxidized into a ketone (3-methyl-2-butanone). c. 2-methyl-2-butanol is a tertiary alcohol. Tertiary alcohols do not undergo this type of oxidation, so there is no oxidation product 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

Ethyl caprate is an ester used in the manufacture of wine bouquets. It is sometimes called "cognac essence." Combustion analysis of a sample of ethyl caprate shows it to be 71.89\(\% \mathrm{C}\) , 12.13\(\%\) hydrogen, and 15.98\(\%\) O by mass. Hydrolysis (reaction with water) of the ester produces ethanol and a carboxylic acid. The molar mass of the carboxylic acid is 172 \(\mathrm{g} / \mathrm{mol}\) , and the \(\mathrm{R}\) group attached to the COOH group of the acid is a straight chain alkane. What is the structure of ethyl caprate?

What forces are responsible for the solubility of starch in water?

Draw a structural formula for each of the following compounds. a. 2-methylpropane b. 2-methylbutane c. 2-methylpentane d. 2-methylhexane

If one hydrogen in a hydrocarbon is replaced by a halogen atom, the number of isomers that exist for the substituted compound depends on the number of types of hydrogen in the original hydrocarbon. Thus there is only one form of chloroethane (all hydrogens in ethane are equivalent), but there are two isomers of propane that arise from the substitution of a methyl hydrogen or a methylene hydrogen. How many isomers can be obtained when one hydrogen in each of the compounds named below is replaced by a chlorine atom? $$ \begin{array}{ll}{\text { a. } n \text { -pentane }} & {\text { c. } 2,4 \text { -dimethylpentane }} \\ {\text { b. } 2 \text { -methylbutane }} & {\text { d. methylcyclobutane }}\end{array} $$

What are the structural differences between \(\alpha\) - and \(\beta\) -glucose? These two cyclic forms of glucose are the building blocks to form two different polymers. Explain.
See all solutions

Recommended explanations on Chemistry Textbooks

Chemistry Branches

Read Explanation

Chemical Reactions

Read Explanation

Nuclear Chemistry

Read Explanation

Organic Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation

Ionic and Molecular Compounds

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