Chemistry
Chemistry
10th Edition
ISBN: 9781305957404
Author: Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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Question
In the reaction between 1-butanol and HBr (forming butyl bromide and water as the products), which of the following elementary steps is the correct rate-limiting step? (R represents the alkyl group of the alcohol) 1. \( \text{R-OH} \rightarrow \text{R}^+ + \text{OH}^- \) - as shown below - (Diagram shows the breakup of an alcohol into an alkyl cation and a hydroxide ion.) 2. \( \text{R-OH} + \text{H}^+ \rightarrow \text{R-OH}_2^+ \) - as shown below - (Diagram indicates the protonation of the alcohol forming an oxonium ion.) 3. \( \text{R-OH}_2^+ + \text{Br}^- \rightarrow \text{RBr} + \text{H}_2\text{O} \) - as shown below - (Diagram illustrates the reaction of the protonated alcohol with bromide to form butyl bromide and water.) 4. \( \text{R-OH} + \text{HBr} \rightarrow \text{R-Br} + \text{H}_2\text{O} \) - as shown below - (Diagram shows the direct reaction between alcohol and hydrobromic acid forming the alkyl bromide and water.) 5. \( \text{ROH} + \text{HBr} \rightarrow \text{R}^+ + \text{H}_2\text{O} + \text{Br}^- \) - as shown below - (Diagram depicts the alcohol reacting with hydrobromic acid to form an alkyl cation, water, and a bromide ion.)
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Transcribed Image Text:In the reaction between 1-butanol and HBr (forming butyl bromide and water as the products), which of the following elementary steps is the correct rate-limiting step? (R represents the alkyl group of the alcohol) 1. \( \text{R-OH} \rightarrow \text{R}^+ + \text{OH}^- \) - as shown below - (Diagram shows the breakup of an alcohol into an alkyl cation and a hydroxide ion.) 2. \( \text{R-OH} + \text{H}^+ \rightarrow \text{R-OH}_2^+ \) - as shown below - (Diagram indicates the protonation of the alcohol forming an oxonium ion.) 3. \( \text{R-OH}_2^+ + \text{Br}^- \rightarrow \text{RBr} + \text{H}_2\text{O} \) - as shown below - (Diagram illustrates the reaction of the protonated alcohol with bromide to form butyl bromide and water.) 4. \( \text{R-OH} + \text{HBr} \rightarrow \text{R-Br} + \text{H}_2\text{O} \) - as shown below - (Diagram shows the direct reaction between alcohol and hydrobromic acid forming the alkyl bromide and water.) 5. \( \text{ROH} + \text{HBr} \rightarrow \text{R}^+ + \text{H}_2\text{O} + \text{Br}^- \) - as shown below - (Diagram depicts the alcohol reacting with hydrobromic acid to form an alkyl cation, water, and a bromide ion.)
**Question:** In the reaction between 2-methyl-2-butanol and HBr (forming butyl bromide and water as the products), which of the following elementary steps is the correct rate-limiting step? (R represents the alkyl group of the alcohol) **Options:** 1. **Option A:** \( \text{R-OH} \rightarrow \text{R}^+ + \text{H}_2\text{O} \) (as shown below) 2. **Option B:** \( \text{R-OH} + \text{H}^+ \rightarrow \text{R-OH}_2^+ \) \( \text{R-OH}_2^+ \rightarrow \text{R}^+ + \text{H}_2\text{O} \) (as shown below) 3. **Option C:** \( \text{R}^+ + \text{Br}^- \rightarrow \text{R-Br} \) (as shown below) 4. **Option D:** \( \text{Br}^- + \text{R-OH} \rightarrow \text{R-Br} + \text{OH}^- \) (as shown below) 5. **Option E:** \( \text{R-OH} \rightarrow \text{R}^+ + \text{OH}^- \) (as shown below) **Diagram Explanation:** - In **Option A**, the hydroxyl group (OH) leaves, forming a carbocation (\( \text{R}^+ \)) and water (\( \text{H}_2\text{O} \)). - **Option B** involves two steps: first, the alcohol \( (\text{R-OH}) \) gains a proton to form a protonated alcohol (\( \text{R-OH}_2^+ \)), which then loses a water molecule, forming a carbocation (\( \text{R}^+ \)). - In **Option C**, the formed carbocation (\( \text{R}^+ \)) reacts with the bromide ion (\( \text{Br}^- \)) to form butyl bromide (\( \text{R-Br} \)). - **Option D**
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Transcribed Image Text:**Question:** In the reaction between 2-methyl-2-butanol and HBr (forming butyl bromide and water as the products), which of the following elementary steps is the correct rate-limiting step? (R represents the alkyl group of the alcohol) **Options:** 1. **Option A:** \( \text{R-OH} \rightarrow \text{R}^+ + \text{H}_2\text{O} \) (as shown below) 2. **Option B:** \( \text{R-OH} + \text{H}^+ \rightarrow \text{R-OH}_2^+ \) \( \text{R-OH}_2^+ \rightarrow \text{R}^+ + \text{H}_2\text{O} \) (as shown below) 3. **Option C:** \( \text{R}^+ + \text{Br}^- \rightarrow \text{R-Br} \) (as shown below) 4. **Option D:** \( \text{Br}^- + \text{R-OH} \rightarrow \text{R-Br} + \text{OH}^- \) (as shown below) 5. **Option E:** \( \text{R-OH} \rightarrow \text{R}^+ + \text{OH}^- \) (as shown below) **Diagram Explanation:** - In **Option A**, the hydroxyl group (OH) leaves, forming a carbocation (\( \text{R}^+ \)) and water (\( \text{H}_2\text{O} \)). - **Option B** involves two steps: first, the alcohol \( (\text{R-OH}) \) gains a proton to form a protonated alcohol (\( \text{R-OH}_2^+ \)), which then loses a water molecule, forming a carbocation (\( \text{R}^+ \)). - In **Option C**, the formed carbocation (\( \text{R}^+ \)) reacts with the bromide ion (\( \text{Br}^- \)) to form butyl bromide (\( \text{R-Br} \)). - **Option D**
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