Organic Chemistry: Principles and Mechanisms (Second Edition)
2nd Edition
ISBN: 9780393663556
Author: Joel Karty
Publisher: W. W. Norton & Company
expand_more
expand_more
format_list_bulleted
Videos
Question
Chapter 18, Problem 18.49P
Interpretation Introduction
Interpretation:
It is to be explained why the given reaction proceeds dramatically more slowly under the neutral conditions than under either acidic or basic conditions.
Concept introduction:
The
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
The decomposition of di-2-methylpropan-2-yl peroxide produces propanone and ethane:
(CH3)3COOC(CH3)3 → 2CH3COCH3 + C₂H6
and is described by the generally accepted mechanism:
(CH3)3COOC(CH3)3 → 2 (CH3)3CO
(CH3)3CO → CH3COCH3 + CH3*
CH3 + CH3 →→ C₂H6
k₁
k₂
k3
1. Identify the intermediates.
2. Explain why this is NOT a chain reaction.
3. Confirm using the steady state approximation that the production of ethane is first order even
though it takes place in a number of steps.
For the reaction shown, the rate of reaction when X=Cl is about the same as that when X=Br. Based on
this information, which statement represents a valid deduction concerning the mechanism of this
reaction?
OCH2CH3
O2N
NO2
NO2
(A) This is a one-step displacement reaction
(B) The reaction proceeds via a benzyne intermediate
(C) The C-X bond is broken in the rate-determining step
(D) The C-X bond is NOT broken in the rate-determining step
O2N
OCH,CH3
NO2
NO2
Identify the overall reaction for the following mechanism.
Step 1: NO + NO ⇌ N2O2
Step 2: N2O2 + O2 2NO2
Chapter 18 Solutions
Organic Chemistry: Principles and Mechanisms (Second Edition)
Ch. 18 - Prob. 18.1PCh. 18 - Prob. 18.2PCh. 18 - Prob. 18.3PCh. 18 - Prob. 18.4PCh. 18 - Prob. 18.5PCh. 18 - Prob. 18.6PCh. 18 - Prob. 18.7PCh. 18 - Prob. 18.8PCh. 18 - Prob. 18.9PCh. 18 - Prob. 18.10P
Ch. 18 - Prob. 18.11PCh. 18 - Prob. 18.12PCh. 18 - Prob. 18.13PCh. 18 - Prob. 18.14PCh. 18 - Prob. 18.15PCh. 18 - Prob. 18.16PCh. 18 - Prob. 18.17PCh. 18 - Prob. 18.18PCh. 18 - Prob. 18.19PCh. 18 - Prob. 18.20PCh. 18 - Prob. 18.21PCh. 18 - Prob. 18.22PCh. 18 - Prob. 18.23PCh. 18 - Prob. 18.24PCh. 18 - Prob. 18.25PCh. 18 - Prob. 18.26PCh. 18 - Prob. 18.27PCh. 18 - Prob. 18.28PCh. 18 - Prob. 18.29PCh. 18 - Prob. 18.30PCh. 18 - Prob. 18.31PCh. 18 - Prob. 18.32PCh. 18 - Prob. 18.33PCh. 18 - Prob. 18.34PCh. 18 - Prob. 18.35PCh. 18 - Prob. 18.36PCh. 18 - Prob. 18.37PCh. 18 - Prob. 18.38PCh. 18 - Prob. 18.39PCh. 18 - Prob. 18.40PCh. 18 - Prob. 18.41PCh. 18 - Prob. 18.42PCh. 18 - Prob. 18.43PCh. 18 - Prob. 18.44PCh. 18 - Prob. 18.45PCh. 18 - Prob. 18.46PCh. 18 - Prob. 18.47PCh. 18 - Prob. 18.48PCh. 18 - Prob. 18.49PCh. 18 - Prob. 18.50PCh. 18 - Prob. 18.51PCh. 18 - Prob. 18.52PCh. 18 - Prob. 18.53PCh. 18 - Prob. 18.54PCh. 18 - Prob. 18.55PCh. 18 - Prob. 18.56PCh. 18 - Prob. 18.57PCh. 18 - Prob. 18.58PCh. 18 - Prob. 18.59PCh. 18 - Prob. 18.60PCh. 18 - Prob. 18.61PCh. 18 - Prob. 18.62PCh. 18 - Prob. 18.63PCh. 18 - Prob. 18.64PCh. 18 - Prob. 18.65PCh. 18 - Prob. 18.66PCh. 18 - Prob. 18.67PCh. 18 - Prob. 18.68PCh. 18 - Prob. 18.69PCh. 18 - Prob. 18.70PCh. 18 - Prob. 18.71PCh. 18 - Prob. 18.72PCh. 18 - Prob. 18.73PCh. 18 - Prob. 18.74PCh. 18 - Prob. 18.75PCh. 18 - Prob. 18.76PCh. 18 - Prob. 18.77PCh. 18 - Prob. 18.78PCh. 18 - Prob. 18.79PCh. 18 - Prob. 18.80PCh. 18 - Prob. 18.81PCh. 18 - Prob. 18.82PCh. 18 - Prob. 18.83PCh. 18 - Prob. 18.84PCh. 18 - Prob. 18.85PCh. 18 - Prob. 18.86PCh. 18 - Prob. 18.87PCh. 18 - Prob. 18.88PCh. 18 - Prob. 18.89PCh. 18 - Prob. 18.90PCh. 18 - Prob. 18.91PCh. 18 - Prob. 18.92PCh. 18 - Prob. 18.93PCh. 18 - Prob. 18.94PCh. 18 - Prob. 18.95PCh. 18 - Prob. 18.96PCh. 18 - Prob. 18.97PCh. 18 - Prob. 18.98PCh. 18 - Prob. 18.99PCh. 18 - Prob. 18.100PCh. 18 - Prob. 18.101PCh. 18 - Prob. 18.102PCh. 18 - Prob. 18.103PCh. 18 - Prob. 18.1YTCh. 18 - Prob. 18.2YTCh. 18 - Prob. 18.3YTCh. 18 - Prob. 18.4YTCh. 18 - Prob. 18.5YTCh. 18 - Prob. 18.6YTCh. 18 - Prob. 18.7YTCh. 18 - Prob. 18.8YTCh. 18 - Prob. 18.9YTCh. 18 - Prob. 18.10YTCh. 18 - Prob. 18.11YTCh. 18 - Prob. 18.12YTCh. 18 - Prob. 18.13YTCh. 18 - Prob. 18.14YTCh. 18 - Prob. 18.15YT
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.Similar questions
- TRUE or FALSE: For a transition state to be considered as the rate- determining step (rds) in a multi-step reaction, it must have the highest change in enthalpy.arrow_forwardAlcohol is removed from the bloodstream by a series of metabolic reactions. The first reaction produces acetaldehyde; then other products are formed. The following data have been determined for the rate at which alcohol is removed from the blood of an average male, although individual rates can vary by 25-30%. Women metabolize alcohol a little more slowly than men: [CH,OH] (M) 4.4 x 10-2 3.3 x 10-2 2.2 x 10-2 Rate (mol/L/h) 2.0 x 10-2 2.0 × 10-2 2.0 x 10-2 Determine the rate equation, the rate constant, and the overall order for this reaction.arrow_forwardThe following reaction proceeds at a rate such that 4 mole of AA is consumed per minute. Given this, how many moles of CC are produced per minute? 2A+2B→4Carrow_forward
- Outline a detailed mechanism for the reaction below. Identify the abbreviation for the reaction mechanism. OH H PO4arrow_forwardThe acid-catalysed conversion of γ-hydroxybutyric acid (GHBA) into its lactone, γ-butyrolactone (GBL) is a reversible reaction. The GHBA to GBL forward reaction is first-order with respect to the GHBA concentration; the GBL to GHBA reverse reaction is first-order with respect to the GBL concentration. An experimental study of the kinetics of this reaction was undertaken in 0.2 mol L-1hydrochloric acid (HCl) at 298 K. The initial concentration of GHBA was 18.23 × 10-3mol L-1 . The concentration of GBL in solution was followed as a function of time (t), as indicated in Table B.3: Time/min 0 21 36 50 65 80 100 ∞ GBL Concentration / 10-3 L-1 0 2.41 3.73 4.96 6.10 7.08 8.11 13.28 Use the data in Table B.3 to determine the equilibrium constant and the first-order rate constants for both forward and reverse reactions.arrow_forwardWhich of the following mechanism types is/are likely in this reaction: CH3 I + KOH → CH3OH +KI i) SN1 ii) SN2 iii) E1 iv) E2arrow_forward
- In aqueous solution buffered at a pH of 6.75 Penicillin G (molecular weight of 334.39 g/mole) undergoes hydrolysis via a pseudo first order reaction that has an activation energy of 77.3 kJ/mole and a pseudo first order rate constant of 1.58 x 10-7 s-1 at 30°C. What would the value of the rate constant be (in s-1) if the temperature was 60°C? a. 9.96 x 10-9 b. 1.59 x 10-7 c. 2.51 x 10-6 d. 7.62 x 10-6 e. None of the abovearrow_forward5. For the chlorination of ethane, represented by the equation: CH3CH3 + Cl2 + hv→ CH3CH2CI + HCI, the following radical-mediated mechanism has been proposed: Cl2 2 Cl· + CI - CH4 + Cl. → CH3· + HCI CH3: + Cl2 → CH3CI + Cl· k1, k.1 k2 k3 CH3* + Cl· 2CH3CI ka, k.4 Additionally, it has been possible to establish, under certain experimental conditions, that the reaction is first order with respect to the methane, and three halves with respect to chlorine. a) Classify the different steps of this mechanism (i.e. initiation, propagation, etc.). b) Derive the rate law for the chlorination of ethane, based on the mechanism proposed. c) Identify which is the rate-determining step. Justify your selection. d) Indicate which are the assumptions and approximations you used so that this model could work (i.e. pre- equilibrium condition, steady state approximation, neglecting the contribution of a step, etc.).arrow_forwardOne possible mechanism for the decomposition of ethane, C2H6, into ethylene, C2H4, and hydrogen, C2H6 - C2H4 + H2 includes the following steps. (1) C2Hg - 2CH3• (2) CH3• + C2H6→ CH4 + C2H5. (3) C2H5• → C2H4 + H. (4) H• + C2Hs -C2H5• + H2 (5) H. + C2H5•→ C2H6 (a) Which step(s) initiate the reaction? O Step (2) O Step (5) O Step (4) O Step (1) O Step (3)arrow_forward
- Suppose the formation of tert-butanol proceeds by the following mechanism: step elementary reaction rate constant 1 CH33CBr (aq) → CH33C+ (aq) + Br− (aq) k1 2 CH33C+ (aq) + OH− (aq) → CH33COH (aq) k2 Suppose also k1 ≫ k2 . That is, the first step is much faster than the second.arrow_forwardIn aqueous solution buffered at a pH of 6.75 Penicillin G (molecular weight of 334.39 g/mole) undergoes hydrolysis via a pseudo first order reaction that has an activation energy of 77.3 kJ/mole and a pseudo first order rate constant of 1.58 x 10-7 s-1 at 30°C. The solubility of the sodium salt of Penicillin G is 25 mg/ml of water. One would expect the solubility of the sodium salt to be ____________ that of the Penicillin G. a. higher than b. the same as c. lower thanarrow_forwardBased on the time and temperature data collected for the reaction of KMnO4 with either malonic acid or oxalic acid, one can conclude that generally as the temperature of a reaction is increased, the rate of the reaction increases. This is because, the activation energy is lowered and the reactant molecules collide with greater energy the activation energy is lowered and the reactant molecules collide more frequently the activation energy is lowered, the reactant molecules collide more frequently and with greater energy per collision O the reactant molecules collide more frequently and with greater energy per collisionarrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
Chemistry: The Molecular ScienceChemistryISBN:9781285199047Author:John W. Moore, Conrad L. StanitskiPublisher:Cengage Learning
Chemistry: Principles and PracticeChemistryISBN:9780534420123Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward MercerPublisher:Cengage Learning
Chemistry: The Molecular Science
Chemistry
ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
Publisher:Cengage Learning
Chemistry: Principles and Practice
Chemistry
ISBN:9780534420123
Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher:Cengage Learning
SAR of Anticancer(Antineoplastic) Drug/ Alkylating agents/ Nitrogen Mustard; Author: Pharmacy Lectures;https://www.youtube.com/watch?v=zrzyK3LhUXs;License: Standard YouTube License, CC-BY