Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 22: Problem 133

Three different organic compounds have the formula $\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}$ . Only two of these isomers react with \(\mathrm{KMn} \mathrm{O}_{4}\) (a strong oxidizing agent). What are the names of the products when these isomers react with excess \(\mathrm{KMnO}_{4} ?\)

Short Answer

Expert verified
The names of the products when the isomers react with excess KMnO4 are Propanoic acid and Propanone (acetone).

Step by step solution

01

Identify the three isomers of C3H8O

Three isomers of C3H8O are: 1. Methoxyethane (\(CH_{3}OCH_{2}CH_{3}\)) 2. Propan-1-ol (1-propanol) (\(CH_{3}CH_{2}CH_{2}OH\)) 3. Propan-2-ol (2-propanol) (\(CH_{3}CH(OH)CH_{3}\)) These isomers have different molecular structures, but the same chemical formula.
02

Determine which two isomers will react with KMnO4

KMnO4, as an oxidizing agent, primarily reacts with alcohols. Thus, the two isomers that will react with KMnO4 are: 1. Propan-1-ol (1-propanol) (\(CH_{3}CH_{2}CH_{2}OH\)) 2. Propan-2-ol (2-propanol) (\(CH_{3}CH(OH)CH_{3}\)) Methoxyethane, which is an ether, is not reactive to strong oxidants like KMnO4, so it's not considered in this analysis.
03

Write balanced chemical equations for the reactions

Balanced chemical equations for the oxidation reactions: 1. Reaction with Propan-1-ol (1-propanol): \[CH_{3}CH_{2}CH_{2}OH + 2 KMnO_{4} + 3 H_{2}SO_{4} \rightarrow CH_{3}CH_{2}COOH + 2 MnSO_{4} + K_{2}SO_{4} + 8 H_{2}O\] 2. Reaction with Propan-2-ol (2-propanol): \[CH_{3}CH(OH)CH_{3} + KMnO_{4} + H_{2}SO_{4} \rightarrow CH_{3}COCH_{3} + MnSO_{4} + K_{2}SO_{4} + 4 H_{2}O\]
04

Identify the products and provide their names

From the balanced chemical equations, we can list the products: 1. Reaction with Propan-1-ol (1-propanol) - The primary alcohol gets converted to a carboxylic acid: - Product: Propanoic acid (\(CH_{3}CH_{2}COOH\)) 2. Reaction with Propan-2-ol (2-propanol) - The secondary alcohol gets converted to a ketone: - Product: Propanone (acetone) (\(CH_{3}COCH_{3}\)) Thus, the names of the products when the isomers react with excess KMnO4 are Propanoic acid and Propanone (acetone).

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

The base sequences in mRNA that code for certain amino acids are Glu: GAA, GAG Val: GUU, GUC, GUA, GUG Met: AUG Trp: UGG Phe: UUU, UUC Asp: GAU, GAC These sequences are complementary to the sequences in DNA. a. Give the corresponding sequences in DNA for the amino acids listed above. b. Give a DNA sequence that would code for the peptide trp–glu–phe–met. c. How many different DNA sequences can code for the tetrapeptide in part b? d. What is the peptide that is produced from the DNA sequence T–A–C–C–T–G–A–A–G? e. What other DNA sequences would yield the same tripeptide as in part d?

There are three isomers of dichlorobenzene, one of which has now replaced naphthalene as the main constituent of mothballs. a. Identify the ortho, the meta, and the para isomers of dichlorobenzene. b. Predict the number of isomers for trichlorobenzene. c. It turns out that the presence of one chlorine atom on a benzene ring will cause the next substituent to add ortho or para to the first chlorine atom on the benzene ring. What does this tell you about the synthesis of m-dichlorobenzene? d. Which of the isomers of trichlorobenzene will be the hardest to prepare?

Give an example of a hydrocarbon that, in theory, exhibits each of the following bond angles: \(60^{\circ}, 90^{\circ}, 109.5^{\circ}, 120^{\circ},\) and \(180^{\circ} .\)

Draw a structural formula for each of the following compounds. $$ \begin{array}{ll}{\text { a. } 2,2-\text { dimethylheptane }} & {\text { c. } 3,3 \text { -dimethylheptane }} \\ {\text { b. } 2,3 \text { -dimethylheptane }} & {\text { d. } 2,4 \text { -dimethylheptane }}\end{array} $$

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
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Analysis

Read Explanation

Physical Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Organic Chemistry

Read Explanation

Chemistry Branches

Read Explanation

The Earths Atmosphere

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