ORGANIC CHEMISTRY-OWL V2 ACCESS
8th Edition
ISBN: 9781305582422
Author: Brown
Publisher: CENGAGE L
expand_more
expand_more
format_list_bulleted
Concept explainers
Videos
Textbook Question
Chapter 8, Problem 8.22P
Following is a balanced equation for the allylic bromination of propene.
CH2==CHCH3 + Br2 h CH2==CHCH2Br + HBr
(a) Calculate the heat of reaction, ∆H 0, for this conversion.
(b) Propose a pair of chain propagation steps and show that they add up to the observed stoichiometry.
(c) Calculate the ∆H 0 for each chain propagation step and show that they add up to the observed ∆H 0 for the overall reaction.
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
Draw an approximate reaction-energy diagram showing the curves for the two possible pathways for ionic addition of HBr to 1-methylcyclohexene. (a) Formation of the major product, 1-bromo-1-methylcyclohexane, and (b) formation of the minor product, 1-bromo-2-methylcyclohexane. Point out how these curves show that 1-bromo-1-methylcyclohexane should be formed faster
Q2.
Consider the nucleophilic substitution reaction between 3-chloro-3-methylpentane and the
hydroxide ion:
CI
x
+
OH
(a) Is the alkyl halide 1°, 2° or 3°?
(b) Will the nucleophilic substitution reaction be S₁1 or S№2?
(c) Give the substitution mechanism showing all full and partial positive and negative charges,
electron movements using curved arrows.
Suppose the alkene in the drawing area below is put in strong acid solution, for example a solution of HBr.
Highlight each carbon that might become protonated to form a carbocation intermediate. (The carbocation intermediate might then react with something
else to form a final product.)
سعد
Chapter 8 Solutions
ORGANIC CHEMISTRY-OWL V2 ACCESS
Ch. 8.2 - Prob. 8.1PCh. 8.4 - Name and draw structural formulas for all...Ch. 8.4 - Using the table of bond dissociation enthalpies in...Ch. 8.5 - Prob. 8.4PCh. 8.6 - Given the solution to Example 8.5, predict the...Ch. 8.7 - Prob. 8.6PCh. 8.7 - Linoleic acid is shown below. What makes this...Ch. 8.7 - Prob. BQCh. 8.7 - Prob. CQCh. 8.7 - The strength of the HO bond in vitamin E is weaker...
Ch. 8.7 - Prob. EQCh. 8.8 - Prob. 8.7PCh. 8 - Prob. 8.8PCh. 8 - Prob. 8.9PCh. 8 - Prob. 8.10PCh. 8 - Prob. 8.11PCh. 8 - Account for the fact that among the chlorinated...Ch. 8 - Name and draw structural formulas for all possible...Ch. 8 - Prob. 8.14PCh. 8 - There are three constitutional isomers with the...Ch. 8 - Following is a balanced equation for bromination...Ch. 8 - Prob. 8.17PCh. 8 - Prob. 8.18PCh. 8 - Prob. 8.19PCh. 8 - Cyclobutane reacts with bromine to give...Ch. 8 - Prob. 8.21PCh. 8 - Following is a balanced equation for the allylic...Ch. 8 - Prob. 8.23PCh. 8 - Prob. 8.24PCh. 8 - The major product formed when methylenecyclohexane...Ch. 8 - Prob. 8.26PCh. 8 - Prob. 8.27PCh. 8 - Prob. 8.28PCh. 8 - Write the products of the following sequences of...Ch. 8 - Using your reaction roadmap as a guide, show...Ch. 8 - Prob. 8.31PCh. 8 - Prob. 8.32PCh. 8 - Prob. 8.33PCh. 8 - Prob. 8.34P
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
- A novel heterocyclic compound, M, contains nitrogen and sulfur atoms and exhibits interesting photochemical properties. When exposed to ultraviolet (UV) light at a specific wavelength, M undergoes a photoreaction resulting in two products: N and O. Product N is formed through a [2+2] cycloaddition involving the nitrogen atom in M, while product O results from a homolytic cleavage of a sulfur-sulfur bond in M. Given these reaction pathways, what is the most probable structure of M, and what are the likely structures of N and O? A. M is a thiazole derivative; N is a dimerized product through the nitrogen atoms, and O is a compound with two sulfur-centered radicals. B. M is a diazine derivative; N is a tetra-atomic cyclic compound, and O is a compound with two separate thiol groups. C. M is a thiadiazole derivative; N is a four-membered ring involving the nitrogen atom, and O results in two sulfur-centered radicals. D. M is a dithiolane derivative; N is a dimer involving the nitrogen…arrow_forwardDraw a Lewis structure of sodium ethyl mercaptide (NaSC2H5), showing all relevant lone pairs and formal charges. Based on the structure you drew, do you expect that it would act as a good nucleophile? Do you expect it to act as a strong base?arrow_forwardi cant find the answerarrow_forward
- Please answer both, thank you so much!arrow_forward(1) Predict the outcome of the addition of HBr to (a) trans-2-pentene, (b) 2-methyl-2-butene, and (c) 4-methylcyclohexene. How many isomers can be formed in each case? (2) Addition of HBr to 3,3-dimethyl-1-butene gives a mixture of two isomeric alkyl bromide products. Draw structures for the two products, and give a mechanistic explanation for their formation.arrow_forwardPlease give the main substitution product for each of the following reactions, and indicate the dominant mechanism: (a) 1-bromopropane + NaOCH3 → (b) 3-bromo-3-methylpentane + NaOC2H5 →arrow_forward
- C=CH H20, H2SO4 H9SO4 CH3 Alkynes do not react directly with aqueous acid as do alkenes, but will do so in the presence of mercury(II) sulfate as a Lewis acid catalyst. The reaction occurs with Markovnikov regiochemistry, so the OH group adds to the more highly substituted carbon and the H adds to the less highly substituted carbon. The initial product of the reaction is a vinyl alcohol, also called an enol. The enol immediately rearranges to a more stable ketone via tautomerization. Draw curved arrows to show the movement of electrons in this step of the mechanism. Arrow-pushing Instructions H-OH HO: Hjö: C=CH c=CH Hö Hg Hgarrow_forwardAlkyne anions react with the carbonyl groups of aldehydes and ketones to form alkynyl alcohols, as illustrated by the following sequence. HCI CH,C=C Na* + H-C-H [CH,C=C-CH,O Na*] H,O → CH,C=C-CH,OH Propose a mechanism for the formation of the bracketed compound, using curved arrows to show the flow of electron pairs in the course of the reaction.arrow_forwardReaction of (CH3)3CH with Cl2 forms two products: (CH3)2CHCH2Cl (63%) and (CH3)3CCl (37%). Why is the major product formed by cleavage of the stronger 1° C–H bond?arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
Organic ChemistryChemistryISBN:9781305580350Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. FootePublisher:Cengage Learning
Organic Chemistry
Chemistry
ISBN:9781305580350
Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. Foote
Publisher:Cengage Learning
How to Design a Total Synthesis; Author: Chemistry Unleashed;https://www.youtube.com/watch?v=9jRfAJJO7mM;License: Standard YouTube License, CC-BY