/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 73 How would you synthesize each of... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 22: Problem 73

How would you synthesize each of the following? a. 1,2-dibromopropane from propene b. acetone (2-propanone) from an alcohol c. tert-butyl alcohol (2-methyl-2-propanol) from an alkene (See Exercise 68.) d. propanoic acid from an alcohol

Short Answer

Expert verified
a. To synthesize 1,2-dibromopropane from propene, perform a halogenation reaction with an excess of Br2 under UV light. b. To synthesize acetone from an alcohol, oxidize 2-propanol using an oxidizing agent such as \(KMnO_4\) or \(H_2CrO_4\). c. (See Exercise 68.) d. To synthesize propanoic acid from an alcohol, oxidize 1-propanol using an oxidizing agent such as \(KMnO_4\) or \(H_2CrO_4\).

Step by step solution

01

Analyzing the reactants and the product

Propene is an alkene with the formula C3H6. The target molecule is 1,2-dibromopropane, a molecule with two bromine atoms attached to adjacent carbon atoms. The required reaction for this transformation is the addition of two bromine atoms to the double bond.
02

Performing the reaction

In order to synthesize 1,2-dibromopropane from propene, we need to perform a halogenation reaction with bromine (Br2). To do this, add an excess of Br2 to propene under UV light. The bromine molecules will be added across the double bond, resulting in 1,2-dibromopropane. b. acetone (2-propanone) from an alcohol
03

Analyzing the reactants and the product

The target molecule is acetone, a ketone with the formula \(C_3H_6O\). The starting material is an alcohol. The required reaction for this transformation is the oxidation of the secondary alcohol to a ketone.
04

Performing the reaction

To synthesize acetone from an alcohol, start with 2-propanol, a secondary alcohol, as the reactant. Oxidize 2-propanol using an oxidizing agent such as potassium permanganate (\(KMnO_4\)) or chromic acid (\(H_2CrO_4\)). This will convert the secondary alcohol to a ketone, forming acetone (2-propanone). c. tert-butyl alcohol (2-methyl-2-propanol) from an alkene (See Exercise 68.) d. propanoic acid from an alcohol
05

Analyzing the reactants and the product

The target molecule is propanoic acid, a carboxylic acid with the formula \(C_2H_5COOH\). The starting material is an alcohol. The required reaction for this transformation is the oxidation of a primary alcohol to a carboxylic acid.
06

Performing the reaction

To synthesize propanoic acid from an alcohol, start with 1-propanol, a primary alcohol, as the reactant. Oxidize 1-propanol using an oxidizing agent such as potassium permanganate (\(KMnO_4\)) or chromic acid (\(H_2CrO_4\)). This will convert the primary alcohol to a carboxylic acid, forming propanoic acid.

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.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Understanding Halogenation
Halogenation is an essential reaction in organic synthesis where halogen atoms are added to a molecule, usually an alkene or alkane. It often involves elements such as chlorine, bromine, or iodine. These reactions can be catalytic or occur in the presence of UV light.

One classic example is the reaction of propene with bromine (Br e_2). In this process, bromine molecules add across the double bond of the alkene. This results in a 1,2-dibromopropane compound. Such transformations are invaluable in the production of many industrially important intermediates.
  • Halogens generally increase the reactivity and solubility of organic compounds.
  • Halogenation often requires specific conditions like UV light to initiate the reaction.
  • Different halogens may work at varying levels of reactivity, with chlorine usually being more reactive than bromine.
Exploring Oxidation Reactions
Oxidation reactions are key processes in organic synthesis where a molecule loses electrons, typically increasing the oxidation state of the molecule. This chemical transformation often involves the conversion of alcohols to ketones or carboxylic acids.

For instance, the oxidation of 2-propanol, a secondary alcohol, can produce acetone. This can be achieved using powerful oxidizing agents such as potassium permanganate ( KMnO4) or chromic acid ( H2CrO4).
  • Oxidation significantly alters the properties and reactivity of the original compound.
  • Choice of oxidizing agent may influence the rate and completeness of the reaction.
  • Secondary alcohols oxidize to ketones, while primary alcohols can oxidize further to form carboxylic acids.
Insights into Alcohol Transformation
Alcohol transformation involves the conversion of an alcohol into another type of compound through various chemical reactions. This transformation is a crucial step in multiple industrial chemical processes.

For example, synthesizing acetone from 2-propanol involves an oxidation reaction, transforming the alcohol's hydroxyl group into a carbonyl group. Similarly, converting 1-propanol into propanoic acid involves the oxidation of the alcohol to a carboxylic acid.
  • Transformations can include oxidation, reduction, substitution, or elimination reactions.
  • Controlling the reaction conditions, such as temperature and the presence of catalysts, is important for a successful transformation.
  • Understanding the nature of the alcohol (primary, secondary, tertiary) is vital in predicting its reactivity and suitable transformations.
Explaining Alkene Reactions
Alkene reactions refer to the diverse chemical processes that involve alkenes, compounds with at least one carbon-carbon double bond. These reactions are fundamental in forming a variety of chemical products.

One such reaction is the halogenation of alkenes, where the double bond in an alkene like propene adds halogen atoms, such as bromine, across its structure. Another common reaction involving alkenes is hydration, where water adds to the double bond, often converting the alkene to an alcohol, like in the synthesis of tert-butyl alcohol from isobutylene.
  • Reactions are often driven by the high reactivity of the carbon-carbon double bond.
  • Common processes include addition reactions, where new atoms are added to the compound.
  • Reactions can be catalytic or non-catalytic, affecting the reaction rate and product formation.

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

In the presence of light, chlorine can substitute for one (or more) of the hydrogens in an alkane. For the following reactions, draw the possible monochlorination products. $$ 2,2 \text { -dimethylpropane }+\mathrm{Cl}_{2} \longrightarrow $$ $$ 1,3 \text { -dimethylcyclobutane }+\mathrm{Cl}_{2} \stackrel{\mathrm{hn}}{\longrightarrow} $$ $$ 2,3 \text { -dimethylbutane }+\mathrm{Cl}_{2} \stackrel{\mathrm{w}}{\longrightarrow} $$

Helicenes are extended fused polyaromatic hydrocarbons that have a helical or screw-shaped structure. a. A \(0.1450-\) g sample of solid helicene is combusted in air to give 0.5063 \(\mathrm{g} \mathrm{CO}_{2}\) . What is the empirical formula of this helicene? b. If a 0.0938 -g sample of this helicene is dissolved in 12.5 g of solvent to give a 0.0175 M solution, what is the molecular formula of this helicene? c. What is the balanced reaction for the combustion of this helicene?

Cis-trans isomerism is also possible in molecules with rings. Draw the cis and trans isomers of \(1,2\) -dimethylcyclohexane. In Exercise \(43,\) you drew all of the noncyclic structural and geometrical isomers of \(\mathrm{C}_{4} \mathrm{H}_{7} \mathrm{F}\) . Now draw the cyclic structural and geometrical isomers of \(\mathrm{C}_{4} \mathrm{H}_{7} \mathrm{F}\) .

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 two examples of saturated hydrocarbons. How many other atoms are bonded to each carbon in a saturated hydrocarbon?
See all solutions

Recommended explanations on Chemistry Textbooks

Organic Chemistry

Read Explanation

Making Measurements

Read Explanation

Nuclear Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Ionic and Molecular Compounds

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.