Textbook Question
Each of these compounds can react as a nucleophile. In each case, use curved arrows to show how the nucleophile would react with the strong electrophile BF3.
(e) CH3CH2OH
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Ch. 2 - Acids and Bases; Functional Groups
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 2, Problem 53f
Each of these compounds can react as a nucleophile. In each case, use curved arrows to show how the nucleophile would react with the strong electrophile BF3.
(f) (CH3)2S
Verified step by step guidance1
Identify the nucleophile and electrophile in the reaction. In this case, the nucleophile is dimethyl sulfide ((CH3)2S) and the electrophile is boron trifluoride (BF3).
Understand the nature of the nucleophile and electrophile. Dimethyl sulfide ((CH3)2S) has a lone pair of electrons on the sulfur atom, making it a good nucleophile. Boron trifluoride (BF3) is an electron-deficient molecule, making it a strong electrophile.
Use curved arrows to represent the movement of electrons. Draw a curved arrow from the lone pair on the sulfur atom of (CH3)2S to the boron atom in BF3. This indicates the formation of a coordinate covalent bond between sulfur and boron.
Consider the geometry and hybridization changes. The boron atom in BF3 is sp2 hybridized and trigonal planar. Upon accepting the lone pair from sulfur, it becomes sp3 hybridized and adopts a tetrahedral geometry.
Verify the stability of the resulting adduct. The formation of the S-B bond results in a stable adduct due to the complete octet around boron, and the positive charge on sulfur is stabilized by the electron-donating methyl groups.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Nucleophiles
Nucleophiles are species that donate an electron pair to form a chemical bond in reaction mechanisms. They are typically rich in electrons and can be negatively charged ions or neutral molecules with lone pairs. Understanding nucleophiles is crucial for predicting how they will interact with electrophiles in organic reactions.
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Nucleophilic Addition
Electrophiles
Electrophiles are species that accept an electron pair, often characterized by a positive charge or an electron-deficient atom. In the context of the question, BF3 acts as a strong electrophile due to its electron-deficient boron atom, which makes it highly reactive towards nucleophiles like (CH3)2S.
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Curved Arrow Notation
Curved arrow notation is a fundamental tool in organic chemistry used to depict the movement of electrons during chemical reactions. It helps visualize how nucleophiles donate electron pairs to electrophiles, indicating the formation or breaking of bonds. Mastery of this notation is essential for accurately representing reaction mechanisms.
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Related Practice
Textbook Question
The pKa of ascorbic acid (vitamin C, page 55) is 4.17, showing that it is slightly more acidic than acetic acid (CH3COOH, pKa 4.74).
a. Show the four different conjugate bases that would be formed by deprotonation of the four different OH groups in ascorbic acid.
b. Compare the stabilities of these four conjugate bases, and predict which OH group of ascorbic acid is the most acidic.
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Textbook Question
The pKa of ascorbic acid (vitamin C, page 55) is 4.17, showing that it is slightly more acidic than acetic acid (CH3COOH, pKa 4.74).
c. Compare the most stable conjugate base of ascorbic acid with the conjugate base of acetic acid, and suggest why these two compounds have similar acidities, even though ascorbic acid lacks the carboxylic acid (COOH) group.
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Textbook Question
Each of these compounds can react as an electrophile. In each case, use curved arrows to show how the electrophile would react with the strong nucleophile sodium ethoxide, Na+ −OCH2CH3.
(c) CH3CH2Br
(d) BH3
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Textbook Question
Many naturally occurring compounds contain more than one functional group. Identify the functional groups in the following compounds:
a. Penicillin G is a naturally occurring antibiotic.
b. Dopamine is the neurotransmitter that is deficient in Parkinson’s disease.
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Textbook Question
Each of these compounds can react as a nucleophile. In each case, use curved arrows to show how the nucleophile would react with the strong electrophile BF3.
(a)
(b)
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