Textbook Question
What products are formed when each of the following reacts with HO−?
e. +CH3
f. FeBr3
3
views
Ch. 2 - Acids and Bases: Central to Understanding Organic Chemistry
Bruice8th EditionOrganic ChemistryISBN: 9780135213711
Not the one you use?Change textbookSelect a different textbook
Table of contents
- Ch. 1 - Remembering General Chemistry: Electronic Structure and Bonding (Part 1)31
- Ch. 1 - Remembering General Chemistry: Electronic Structure and Bonding (Part 2)65
- Ch. 2 - Acids and Bases: Central to Understanding Organic Chemistry84
- Ch. 3 - An Introduction to Organic Compounds:Nomenclature, Physical Properties, and Structure183
- Ch. 4 - Isomers: The Arrangement of Atoms in Space121
- Ch. 5 - Alkenes: Structure, Nomenclature, and an Introduction to Reactivity • Thermodynamics and Kinetics83
- Ch. 6 - The Reactions of Alkenes • The Stereochemistry of Addition Reactions175
- Ch. 7 - The Reactions of Alkynes • An Introduction to Multistep Synthesis119
- Ch. 8 - Delocalized Electrons: Their Effect on Stability, pKa, and the Products of a Reaction • Aromaticity and Electronic Effects: An Introduction to the Reactions of Benzene148
- Ch. 9 - Substitution and Elimination Reactions of Alkyl Halides212
- Ch. 10 - Reactions of Alcohols, Ethers, Epoxides, Amines, and Sulfur-Containing Compounds131
- Ch. 11 - Organometallic Compounds60
- Ch. 12 - Radicals90
- Ch. 13 - Mass Spectrometry; Infrared Spectroscopy; UV/Vis Spectroscopy62
- Ch. 14 - NMR Spectroscopy95
- Ch. 15 - Reactions of Carboxylic Acids and Carboxylic Acid Derivatives123
- Ch. 16 - Reactions of Aldehydes and Ketones • More Reactions of Carboxylic Acid Derivatives141
- Ch. 17 - Reactions at the Alpha-Carbon101
- Ch. 18 - Reactions of Benzene and Substituted Benzenes144
- Ch. 19 - More About Amines • Reactions of Heterocyclic Compounds49
- Ch. 20 - The Organic Chemistry of Carbohydrates80
- Ch. 21 - Amino Acids, Peptides, and Proteins57
- Ch. 22 - Catalysis in Organic Reactions and in Enzymatic Reactions35
- Ch. 23 - The Organic Chemistry of the Coenzymes—Compounds Derived from Vitamins1
- Ch. 28 - Pericyclic Reactions58
All textbooks
Bruice 8th EditionCh. 2 - Acids and Bases: Central to Understanding Organic ChemistryProblem 52c,d
Bruice 8th EditionCh. 2 - Acids and Bases: Central to Understanding Organic ChemistryProblem 52c,dChapter 3, Problem 52c,d
What products are formed when each of the following reacts with HO−?
c. CH3N+H3
d. BF3
Verified step by step guidance1
Step 1: Analyze the reaction of CH3N+H3 with HO−. CH3N+H3 is a positively charged ammonium ion, and HO− is a hydroxide ion, which acts as a base. The hydroxide ion will deprotonate the ammonium ion by removing a proton (H+). This is an acid-base reaction.
Step 2: Write the chemical equation for the reaction of CH3N+H3 with HO−. The hydroxide ion will abstract a proton from CH3N+H3, resulting in the formation of CH3NH2 (methylamine) and H2O (water). The equation can be expressed as:
Step 3: Analyze the reaction of BF3 with HO−. BF3 is a Lewis acid because boron has an incomplete octet and can accept electron pairs. HO− is a Lewis base because it has a lone pair of electrons. The hydroxide ion will donate its lone pair to boron, forming a coordinate covalent bond.
Step 4: Write the chemical equation for the reaction of BF3 with HO−. The hydroxide ion will bond with BF3, resulting in the formation of a tetrahedral complex, BF3OH−. The equation can be expressed as:
Step 5: Summarize the products formed. For CH3N+H3 reacting with HO−, the products are CH3NH2 and H2O. For BF3 reacting with HO−, the product is BF3OH−. These reactions demonstrate acid-base and Lewis acid-base interactions, respectively.
Verified video answer for a similar problem:
This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
5mWas this helpful?
Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Nucleophilicity
Nucleophilicity refers to the ability of a species to donate an electron pair to form a chemical bond. In the context of the question, HO− (hydroxide ion) acts as a nucleophile, attacking electrophilic centers in the reactants. Understanding which species can act as nucleophiles and how they interact with electrophiles is crucial for predicting the products of the reactions.
Recommended video:
Guided course
08:27
Nucleophilic Addition
Electrophiles
Electrophiles are species that accept electron pairs from nucleophiles during a chemical reaction. In this question, CH3N+H3 (methylammonium ion) and BF3 (boron trifluoride) are both electrophiles. Recognizing the nature of these electrophiles helps in understanding how they will react with the hydroxide ion, leading to the formation of specific products.
Recommended video:
Guided course
03:Nucleophile or Electrophile
Acid-Base Reactions
Acid-base reactions involve the transfer of protons (H+) between species. In the case of CH3N+H3, it can act as a Brønsted acid, while HO− acts as a base, leading to the formation of products through proton transfer. For BF3, which is a Lewis acid, it can accept a pair of electrons from HO−, illustrating the broader concept of acid-base interactions in organic chemistry.
Recommended video:
Guided course
02:49The Lewis definition of acids and bases.
Related Practice
Textbook Question
b. How does the presence of an electronegative substituent such as Cl affect the acidity of a carboxylic acid?
c. How does the location of the substituent affect the acidity of the carboxylic acid?
Textbook Question
Explain the relative acidities.
3
views
Textbook Question
Draw the products of the following reactions. Use curved arrows to show where the pair of electrons starts and where it ends up.
a.
b.
5
views
Textbook Question
What products are formed when each of the following reacts with HO−?
a. CH3OH
b. +NH4
3
views
Textbook Question
Write the equation that shows how a buffer made by dissolving CH3COOH and CH3COO−Na+ in water prevents the pH of a solution from changing appreciably when
a. a small amount of H+ is added to the solution.
b. a small amount of HO− is added to the solution.
1
views