Textbook Question
Long, polarized bonds are also reducible in the same way that the C―C π bond of an alkyne is. Show a mechanism by which a C―Br bond might be reduced by sodium metal.
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Ch. 10 - Alkynes: Electrophilic Addition and Redox Reactions
Mullins1st EditionOrganic Chemistry: A Learner Centered ApproachISBN: 9780137566471
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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
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Mullins 1st EditionCh. 10 - Alkynes: Electrophilic Addition and Redox ReactionsProblem 32
Mullins 1st EditionCh. 10 - Alkynes: Electrophilic Addition and Redox ReactionsProblem 32Chapter 9, Problem 32
The alkyl carbocation is estimated to be 15 kcal/mol more stable than the alkenyl carbocation. If this is also the difference in the energies of the transition state leading to each, what is the expected rate difference?
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Verified step by step guidance1
Step 1: Understand the problem. The question asks for the expected rate difference between two reactions based on the energy difference of their transition states. The energy difference between the transition states is given as 15 kcal/mol.
Step 2: Recall the relationship between activation energy and reaction rate. The rate constant (k) is related to the activation energy (Ea) through the Arrhenius equation: k = A * exp(-Ea / RT), where A is the pre-exponential factor, R is the gas constant, and T is the temperature.
Step 3: Use the ratio of rate constants to compare the rates of the two reactions. The ratio of rate constants (k1/k2) can be expressed as: k1/k2 = exp((Ea2 - Ea1) / RT), where Ea2 and Ea1 are the activation energies for the two reactions.
Step 4: Substitute the given energy difference into the equation. The energy difference between the transition states is 15 kcal/mol, which needs to be converted to cal/mol (1 kcal = 1000 cal). Thus, Ea2 - Ea1 = 15000 cal/mol.
Step 5: Simplify the expression for the rate difference. Plug the values into the equation, ensuring the temperature (T) and gas constant (R) are in consistent units. The final expression will allow you to calculate the expected rate difference between the two reactions.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Carbocation Stability
Carbocations are positively charged carbon species that can vary in stability based on their structure. Alkyl carbocations are generally more stable than alkenyl carbocations due to hyperconjugation and inductive effects from surrounding alkyl groups. This stability influences the energy of the transition states leading to their formation, affecting reaction rates.
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Determining Carbocation Stability
Transition State Theory
Transition state theory posits that during a chemical reaction, reactants pass through a high-energy transition state before forming products. The energy difference between the reactants and the transition state (activation energy, Ea) determines the rate of the reaction. A lower activation energy correlates with a faster reaction rate, making the understanding of transition states crucial for predicting reaction kinetics.
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Arrhenius Equation
The Arrhenius equation relates the rate constant of a reaction to the temperature and activation energy. It states that the rate constant increases exponentially as the activation energy decreases. This relationship allows for the calculation of expected rate differences when comparing reactions with different activation energies, such as the 15 kcal/mol difference noted in the question.
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Related Practice
Textbook Question
We have studied the following reactions in previous chapters. For each, (i) indicate which reaction sheets they should appear on, (ii) show the best structure to use to represent them, and (iii) write the notes you could put in the margin so that the mechanism is implied.
(b) Radical halogenation of an alkane
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Textbook Question
Predict the major products resulting from the addition of one equivalent of HX to the following alkynes.
(b)
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Textbook Question
Predict the alkyne and reactants you might use to make the following haloalkenes. [Providing the reactant and the reagent is how we start thinking about synthesis.]
(a)
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Textbook Question
Predict the alkyne and reactants you might use to make the following haloalkenes. [Providing the reactant and the reagent is how we start thinking about synthesis.]
(b)
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Textbook Question
Sodium amide, the base we use to deprotonate terminal alkynes, is synthesized by reducing ammonia via a mechanism similar to the reduction of alkynes in Figure 10.21. Suggest a mechanism for this reaction.
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