Textbook Question
Identify the bonds that break and form in the following elimination reactions.
(a)
6
views
Ch. 12 - Substitution and Elimination: Reactions of Haloalkanes
Mullins1st EditionOrganic Chemistry: A Learner Centered ApproachISBN: 9780137566471
Not the one you use?Change textbookSelect a different textbook
Table of contents
- Ch. 2 - General Chemistry Translated: Finding the Electrons114
- Ch. 3 - Alkanes and Cycloalkanes: Properties and Conformational Analysis109
- Ch. 4 - Acids and Bases: Electron Flow144
- Ch. 5 - Chemical Reaction Analysis: Thermodynamics and Kinetics118
- Ch. 6 - Stereoisomerism: Arrangement of Atoms in Space112
- Ch. 7 - Principles of Green (Organic) Chemistry3
- Ch. 8 - Alkenes I: Properties and Electrophilic Additions158
- Ch. 9 - Alkenes II: Oxidation and Reduction150
- Ch. 10 - Alkynes: Electrophilic Addition and Redox Reactions101
- Ch. 11 - Properties and Synthesis of Alkyl Halides: Radical Reactions108
- Ch. 12 - Substitution and Elimination: Reactions of Haloalkanes172
- Ch. 13 - Alcohols, Ethers and Related Compounds: Substitution and Elimination239
- Ch. 14 - Structural Identification I: Infrared Spectroscopy and Mass Spectrometry54
- Ch. 15 - Structural Identification II: Nuclear Magnetic Resonance Spectroscopy93
- Ch. 16 - Metals in Organic Chemistry48
- Ch. 17 - Carbonyl Addition Reactions: Aldehydes and Ketones45
- Ch. 18 - Nucleophilic Acyl Substitution I: Carboxylic Acids18
- Ch. 19 - Nucleophilic Acyl Substitution II: Carboxylic Acid Derivatives39
- Ch. 20 - Enolates: Carbonyl Addition and Substitution13
- Ch. 21 - Conjugated Systems I: Stability and Addition Reactions56
- Ch. 22 - Conjugated Systems II: Pericyclic Reactions54
- Ch. 23 - Benzene I: Aromatic Stability and Substitution Reactions64
- Ch. 24 - Benzene II: Reactions Influenced by the Aromatic Ring62
- Ch. 25 - Amines: Structure, Reactions, and Synthesis27
- Ch. 26 - Amino Acids, Proteins, and Peptide Synthesis9
- Ch. 27 - Carbohydrates, Nucleic Acids, and Lipids54
Mullins1st EditionOrganic Chemistry: A Learner Centered ApproachISBN: 9780137566471Not the one you use?Change textbook
All textbooks
Mullins 1st EditionCh. 12 - Substitution and Elimination: Reactions of HaloalkanesProblem 34b
Mullins 1st EditionCh. 12 - Substitution and Elimination: Reactions of HaloalkanesProblem 34bChapter 11, Problem 34b
Identify the bonds that break and form in the following elimination reactions.
(b) 
Verified step by step guidance1
Step 1: Analyze the reaction type. This is an elimination reaction, specifically an E2 mechanism, as the base (NaOEt) is strong and the reaction occurs in a single step without intermediates.
Step 2: Identify the bonds that break. In the substrate, the bond between the carbon atom attached to the chlorine (C-Cl bond) breaks, and the bond between the β-hydrogen (hydrogen on the carbon adjacent to the carbon bonded to chlorine) and its carbon also breaks.
Step 3: Identify the bonds that form. A new π bond forms between the α-carbon (the carbon bonded to chlorine) and the β-carbon (the carbon adjacent to the α-carbon). Additionally, the base (NaOEt) abstracts the β-hydrogen, forming ethanol (EtOH).
Step 4: Understand the stereochemistry. The elimination reaction follows the anti-periplanar geometry, meaning the β-hydrogen and the leaving group (Cl) must be in opposite planes for the reaction to proceed efficiently.
Step 5: Summarize the process. The reaction involves breaking the C-Cl bond and the C-H bond on the β-carbon, forming a double bond between the α- and β-carbons, and producing ethanol as a byproduct.
Verified video answer for a similar problem:
This video solution was recommended by our tutors as helpful for the problem above.
Was this helpful?
Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Elimination Reactions
Elimination reactions involve the removal of atoms or groups from a molecule, resulting in the formation of a double bond. In this specific reaction, a halogen (chlorine) and a hydrogen atom are eliminated from adjacent carbon atoms, leading to the formation of an alkene. Understanding the mechanism of elimination is crucial for predicting the products formed.
Recommended video:
Guided course
00:40
Recognizing Elimination Reactions.
Bonds Breaking and Forming
In the elimination reaction shown, a C-Cl bond breaks as the chlorine atom is removed, while a new C=C double bond forms between the adjacent carbon atoms. Additionally, a C-H bond is also broken to facilitate the formation of the double bond. Recognizing which bonds are involved is essential for understanding the reaction mechanism and the resulting products.
Recommended video:
Guided course
02:31Identifying Bond Breaking
Base Catalysis
The reaction uses sodium ethoxide (NaOEt) as a strong base to facilitate the elimination process. The base abstracts a proton from the carbon adjacent to the carbon bearing the leaving group (Cl), promoting the formation of the double bond. Understanding the role of the base is important for grasping how elimination reactions proceed and the conditions required for them.
Recommended video:
2:50Acid-Base Catalysis Concept 3
Related Practice
Textbook Question
Predict the product of the following substitution reactions, making sure to note whether a rearrangement should occur.
(d)
2
views
Textbook Question
When 2-bromopropane is treated with sodium ethoxide, propene is produced. What molecule is lost from 2-bromopropane in this process?
1
views
Textbook Question
(a) How would you convert propene to 2-bromopropane?
1
views
Textbook Question
Practice your electron-pushing skills by drawing a mechanism for the following E2 reactions.
(a)
3
views
Textbook Question
Predict the product of the following substitution reactions, making sure to note whether a rearrangement should occur.
(c)
1
views