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Is this reaction mechanism correct?

Is this reaction mechanism correct?

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Is this reaction mechanism correct?

The image depicts a series of chemical reactions involving chloroalkenes with sodium hydroxide (NaOH). 1. **First Reaction**: - A chlorinated alkene (R^1-CH=CH-Cl) is shown reacting with NaOH. - The chlorine atom is leaving, forming a hydroxyl group (OH) attached to the alkene chain. - Products: An alcohol (HO-R^1-CH=CH) and sodium chloride (NaCl). 2. **Second Reaction**: - The same type of chlorinated alkene is depicted. - NaOH reacts, leading to the formation of a dehydrohalogenation product. - Intermediate steps show electron movement: The base abstracts a proton (H^+), leading to the formation of an alkene with a double bond. - Products: An internal alkene (R^1-CH=CH) and byproducts such as Na^+ and H2O. 3. **Third Reaction**: - Continuation of the same series of reactions. - Further deprotonation occurs, stabilizing the alkene. - Products: An alkene (R^1-CH=CH) and hydroxide ion (OH^-). The diagram illustrates the mechanism of a base-induced elimination reaction, highlighting the nucleophilic substitution and elimination pathways. Curved arrows indicate electron movement, demonstrating the formation and breaking of bonds through the reaction sequence.
Transcribed Image Text:The image depicts a series of chemical reactions involving chloroalkenes with sodium hydroxide (NaOH). 1. **First Reaction**: - A chlorinated alkene (R^1-CH=CH-Cl) is shown reacting with NaOH. - The chlorine atom is leaving, forming a hydroxyl group (OH) attached to the alkene chain. - Products: An alcohol (HO-R^1-CH=CH) and sodium chloride (NaCl). 2. **Second Reaction**: - The same type of chlorinated alkene is depicted. - NaOH reacts, leading to the formation of a dehydrohalogenation product. - Intermediate steps show electron movement: The base abstracts a proton (H^+), leading to the formation of an alkene with a double bond. - Products: An internal alkene (R^1-CH=CH) and byproducts such as Na^+ and H2O. 3. **Third Reaction**: - Continuation of the same series of reactions. - Further deprotonation occurs, stabilizing the alkene. - Products: An alkene (R^1-CH=CH) and hydroxide ion (OH^-). The diagram illustrates the mechanism of a base-induced elimination reaction, highlighting the nucleophilic substitution and elimination pathways. Curved arrows indicate electron movement, demonstrating the formation and breaking of bonds through the reaction sequence.
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