Textbook Question
Using any necessary reagents, show how you would accomplish the following syntheses.
(a)
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Ch. 19 - Amines
Wade9th EditionOrganic ChemistryISBN: 9780135213728
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Table of contents
- Ch.1 - Structure and Bonding107
- Ch. 2 - Acids and Bases; Functional Groups115
- Ch.3 - Structure and Stereochemistry of Alkanes100
- Ch.4 - The Study of Chemical Reactions99
- Ch.5 - Stereochemistry93
- Ch.6 - Alkyl Halides; Nucleophilic Substitution118
- Ch. 7 - Structure and Synthesis of Alkenes; Elimination158
- Ch.8 - Reactions of Alkenes171
- Ch.9 - Alkynes91
- Ch.10 - Structure and Synthesis of Alcohols143
- Ch.11 - Reactions of Alcohols104
- Ch. 12 - Infrared Spectroscopy and Mass Spectrometry22
- Ch. 13 - Nuclear Magnetic Resonance Spectroscopy45
- Ch. 14 - Ethers, Epoxides, and Thioethers72
- Ch. 15 - Conjugated Systems, Orbital Symmetry, and Ultraviolet Spectroscopy89
- Ch. 16 - Aromatic Compounds74
- Ch. 17 - Reactions of Aromatic Compounds126
- Ch. 18 - Ketones and Aldehydes193
- Ch. 19 - Amines129
- Ch. 20 - Carboxylic Acids77
- Ch. 21 - Carboxylic Acid Derivatives141
- Ch. 22 - Condensations and Alpha Substitutions of Carbonyl Compounds175
- Ch. 23 - Carbohydrates and Nucleic Acids93
- Ch. 24 - Amino Acids, Peptides, and Proteins54
Chapter 19, Problem 40e
Show how m-toluidine can be converted to the following compounds, using any necessary reagents.
(e) 
Verified step by step guidance1
Step 1: Begin with m-toluidine (C6H4(CH3)(NH2)) as the starting material. The goal is to introduce a carbonyl group at the methyl position to form m-aminoacetophenone.
Step 2: Oxidize the methyl group (-CH3) on m-toluidine to a carbonyl group (-CO) using an oxidizing agent such as potassium permanganate (KMnO4) or chromic acid (H2CrO4). This step converts the methyl group into a ketone.
Step 3: Ensure that the amino group (-NH2) remains intact during the oxidation process. Use mild conditions or protective groups if necessary to prevent unwanted reactions involving the amino group.
Step 4: After the oxidation, verify the structure of the product to confirm the formation of m-aminoacetophenone (C6H4(COCH3)(NH2)).
Step 5: Purify the product using techniques such as recrystallization or chromatography to isolate the desired compound.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Electrophilic Aromatic Substitution
Electrophilic aromatic substitution (EAS) is a fundamental reaction in organic chemistry where an electrophile replaces a hydrogen atom on an aromatic ring. In the case of m-toluidine, the amino group (-NH2) is a strong activating group that directs electrophiles to the ortho and para positions relative to itself. Understanding EAS is crucial for predicting the products formed when m-toluidine reacts with various reagents.
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EAS Review
Reactivity of Amines
Amines, such as m-toluidine, are nucleophilic and can participate in various reactions, including acylation and alkylation. The amino group can react with carbonyl compounds to form amides, which is a key transformation in the conversion of m-toluidine to the desired product. Recognizing the reactivity of amines helps in selecting appropriate reagents for the synthesis.
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Functional Group Transformation
Functional group transformation involves converting one functional group into another through chemical reactions. In the context of m-toluidine, transforming the amino group into an amide or other derivatives requires specific reagents and conditions. Mastery of functional group transformations is essential for designing synthetic pathways in organic chemistry.
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Related Practice
Textbook Question
Predict the products of the following reactions:
(j)
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Textbook Question
Show how m-toluidine can be converted to the following compounds, using any necessary reagents.
(b)
Textbook Question
Show how m-toluidine can be converted to the following compounds, using any necessary reagents.
(a)
Textbook Question
Show how m-toluidine can be converted to the following compounds, using any necessary reagents.
(f)
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Textbook Question
The mass spectrum of tert-butylamine follows shows an intense base peak at m/z 58, and very little else. Use a diagram to show the cleavage that accounts for the base peak. Suggest why no molecular ion is visible in this spectrum.
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