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) as shown below OH R*+ OH as shown below -R*+ H,0 as shown below B+R-OH RB + OH as shown below [+] as shown below Br -RBr + H0 as shown below ROH + HBr ROH
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) as shown below OH R*+ OH as shown below -R*+ H,0 as shown below B+R-OH RB + OH as shown below [+] as shown below Br -RBr + H0 as shown below ROH + HBr ROH
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.)
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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