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What are the likely products from the following chemical reaction? OH Xs H-I a) b) HO -ОН HO OH OH c) d) OH HO

What are the likely products from the following chemical reaction? OH Xs H-I a) b) HO -ОН HO OH OH c) d) OH HO

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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### Understanding the Likely Products of a Chemical Reaction #### Reaction Analysis The question at hand asks us to identify the likely products from the following chemical reaction: **Chemical Equation:** - Reactant: Benzyl alcohol with 2-hydroxyethyl group - Reagent: Excess Hydrogen Iodide (H-I) In this reaction, benzyl alcohol (C₆H₅CH₂OCH₂CH₂OH) reacts with excess hydrogen iodide (H-I). The products of this reaction can be deduced based on the reactivity of the -OH groups present in the molecule. #### Options **a)** - **Product 1:** Benzyl iodide (C₆H₅CH₂I) - **Product 2:** Ethylene glycol (HOCH₂CH₂OH) **b)** - **Product 1:** Benzyl alcohol (C₆H₅CH₂OH) - **Product 2:** Ethylene glycol (HOCH₂CH₂OH) **c)** - **Product 1:** Benzyl iodide (C₆H₅CH₂I) - **Product 2:** Ethylene diiodide (CH₂I-CH₂I) **d)** - **Product 1:** Benzyl alcohol (C₆H₅CH₂OH) - **Product 2:** Ethylene diiodide (CH₂I-CH₂I) #### Explanation of Each Option 1. **Option (a):** Suggests that benzyl alcohol will convert to benzyl iodide, and the ethylene glycol will remain unchanged. 2. **Option (b):** Proposes that both groups remain unchanged as their original alcohol forms. 3. **Option (c):** Indicates that both the benzyl alcohol converts to benzyl iodide, and the ethylene glycol fully converts to ethylene diiodide. 4. **Option (d):** This option suggests that benzyl alcohol remains unchanged while ethylene glycol converts to ethylene diiodide. For synthesizing and understanding which products are more likely, benzyl alcohol and primary alcohol are subjected to substitution reactions with HI, where -OH groups are usually replaced by -I groups. **Conclusion:** - Excess HI would likely convert both -OH groups in the compound to -I. - Therefore, the most accurate product formation would include benzyl iodide and ethylene diiodide
Transcribed Image Text:### Understanding the Likely Products of a Chemical Reaction #### Reaction Analysis The question at hand asks us to identify the likely products from the following chemical reaction: **Chemical Equation:** - Reactant: Benzyl alcohol with 2-hydroxyethyl group - Reagent: Excess Hydrogen Iodide (H-I) In this reaction, benzyl alcohol (C₆H₅CH₂OCH₂CH₂OH) reacts with excess hydrogen iodide (H-I). The products of this reaction can be deduced based on the reactivity of the -OH groups present in the molecule. #### Options **a)** - **Product 1:** Benzyl iodide (C₆H₅CH₂I) - **Product 2:** Ethylene glycol (HOCH₂CH₂OH) **b)** - **Product 1:** Benzyl alcohol (C₆H₅CH₂OH) - **Product 2:** Ethylene glycol (HOCH₂CH₂OH) **c)** - **Product 1:** Benzyl iodide (C₆H₅CH₂I) - **Product 2:** Ethylene diiodide (CH₂I-CH₂I) **d)** - **Product 1:** Benzyl alcohol (C₆H₅CH₂OH) - **Product 2:** Ethylene diiodide (CH₂I-CH₂I) #### Explanation of Each Option 1. **Option (a):** Suggests that benzyl alcohol will convert to benzyl iodide, and the ethylene glycol will remain unchanged. 2. **Option (b):** Proposes that both groups remain unchanged as their original alcohol forms. 3. **Option (c):** Indicates that both the benzyl alcohol converts to benzyl iodide, and the ethylene glycol fully converts to ethylene diiodide. 4. **Option (d):** This option suggests that benzyl alcohol remains unchanged while ethylene glycol converts to ethylene diiodide. For synthesizing and understanding which products are more likely, benzyl alcohol and primary alcohol are subjected to substitution reactions with HI, where -OH groups are usually replaced by -I groups. **Conclusion:** - Excess HI would likely convert both -OH groups in the compound to -I. - Therefore, the most accurate product formation would include benzyl iodide and ethylene diiodide
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