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**

Since you have asked multiple question, we will solve the first question for you. If youwant any…

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

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...

See similar textbooks

Similar questions

Please solve this problem
(Incorrect) A reaction occurs via the following sequence of elementary steps. What is the rate law based on this reaction mechanism? A+B 2C + D (slow) С-Е D→2F step 1: step 2: step 3: (fast) (fast) rate = k[C]²[D] (Your answer) rate = K[E][F]² rate = k[E] rate = k[C] rate= K[A][B] (Correct answer)
Answer Q
The rxn I- + OCl- ® IO- + Cl- is first order with respect to I- and first order with respect to OCl-. The rate constant is 8.0 x 10-2 L/mol×s. What is the rate of rxn when [I-] = 0.10 M and [OCl-] = 0.20 M? a) 4.4 x 10-4 M/s b) 1.6 x 10-3 M/s c) 3.2 x 10-3 M/s d) 2.2 x 10-4 M/s e) 2.4 x 10-5 M/s
Based on the provide elementary steps below, determine the net reaction. NO2Cl(g) ----> NO2(g) + Cl (g) elementary step 1 NO2Cl(g) + Cl (g) ----> NO2(g) + Cl (g) elementary step 2
16. The reaction between A and B is %23 as follows: * A + 2B 2C The rate of reaction with variation of concentration of A and B at the same temperature is shown in Table 2. Table 2 Experiment [A]/mol dm [B/mol dm 2.00 1.50 1.00 0.50 1.00 1.00 1.50 2.00 Rate of reaction Imol dm s 1.20 x 10 9.00 x 10 1.35 x 10 II III IV What is the value of x? 3.00x 10 B..00 x 103 C. 1.20x 102 D. 2.40 x 102 A.3.00X 10-7 B.6.00 x10 C.1-20 X10-3 b- 240 x 10-3 you 20
Name: Lyna Ammi Date: 1/26/2022 Section: 02L Data Report Sheet: The Reaction of Methyl Violet with NaOH: A Kinetic Study 4. What is the value of x, the order of the reaction with respect to [OH]? k'/k" = (1/0.5)*
Please describe what would happen to the rate if given changes were applied to following reactions: Ex.: Doubling the concentration of both KOH and starting material in the reaction below Br OH KOH It is a 2nd order reaction. Doubling the concentration of both will result in 4x the speed of the reaction. DMF 1. Doubling the concentration of starting material and alcohol in the reaction below Br Br CH3OH 2. Using half of the alcohol in the reaction shown below CH3OK CH3OH
Please help me
Given the following reaction and table of data, determine the rate law for this reaction: CH3X + 3X2 --> CX4 + 3 HX [CH3X] (M) [X2] (M) Initial Rate (M/s) 0.050 0.050 0.014 0.100 0.050 0.028 0.100 0.100 0.056 0.100 0.200 0.112 Group of answer choices rate = k[CH3X][X2]2 rate = k[CH3X]2[X2] rate = k[CH3X] rate = k[CH3X][X2] rate = k[X2]
both question for upvote
Explain your reasoning, please. Thank you.