ORGANIC CHEMISTRY
5th Edition
ISBN: 9781259977596
Author: SMITH
Publisher: MCG
expand_more
expand_more
format_list_bulleted
Concept explainers
Question
Chapter 15, Problem 15.78P
Interpretation Introduction
Interpretation: A stepwise mechanism for the given reaction is to be drawn.
Concept introduction: The general steps involved in the free-radical reaction are stated below:
1. First step is initiation that involves formation of radical.
2. Second step is propagation.
3. Third step is the termination that involves the formation of stable bond.
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
In the presence of a radical initiator (Z•), tributyltin hydride (R3SnH, R = CH3CH2CH2CH2) reduces alkyl halides to alkanes: R′X + R3SnH → R′H + R3SnX. The mechanism consists of a radical chain process with an intermediate tin radical:
This reaction has been employed in many radical cyclization reactions. Draw a stepwise mechanism for the following reaction.
When exposed to a radical initiator, the C-H bond in HCC13 will cleave homolytically and undergo
radical addition to alkenes. With this in mind, propose a mechanism for the following reaction:
CCB
0:
HCC13
peroxides
Allylic bromination of alkenes is accomplished with the reagent NBS (N-bromosuccinimide).
The reaction proceeds by a radical chain mechanism.
For the following reaction:
NBS
Br
light, CCI4
a Select the species that you would expect to react in the first propagation step and draw
curly arrows to show the mechanism. Use half-headed (fishhook) curved arrows to show
electron reorganization.
When drawing bond formation by pairing of single electrons, terminate the arrows in a
hotspot located to one side of a reacting radical. This hotspot is not associated with any
structural feature.
Arrow-pushing Instructions
将一郎:
Br
H2
Previous
Next
Chapter 15 Solutions
ORGANIC CHEMISTRY
Ch. 15 - Prob. 15.1PCh. 15 - Prob. 15.2PCh. 15 - Prob. 15.3PCh. 15 - Prob. 15.4PCh. 15 - Prob. 15.5PCh. 15 - Problem 15.6 Using mechanism 15.1 as guide, write...Ch. 15 - Prob. 15.7PCh. 15 - Problem 15.8 Which bond in the each compound is...Ch. 15 - Prob. 15.9PCh. 15 - Prob. 15.10P
Ch. 15 - Prob. 15.11PCh. 15 - Prob. 15.12PCh. 15 - Prob. 15.13PCh. 15 - Prob. 15.14PCh. 15 - Prob. 15.15PCh. 15 - Prob. 15.16PCh. 15 - Draw the products of each reaction.
a. b. c.
Ch. 15 - Draw all constitutional isomers formed when each...Ch. 15 - Draw the structure of the four allylic halides...Ch. 15 - Problem 15.20 Which compounds can be prepared in...Ch. 15 - Which CH bond is most readily cleaved in linolenic...Ch. 15 - Prob. 15.22PCh. 15 - Prob. 15.23PCh. 15 - Problem 15.24 When adds to under radical...Ch. 15 - Prob. 15.25PCh. 15 - Prob. 15.26PCh. 15 - Problem 15.27 Draw the steps of the mechanism that...Ch. 15 - Prob. 15.28PCh. 15 - Prob. 15.29PCh. 15 - Prob. 15.30PCh. 15 - Prob. 15.31PCh. 15 - Prob. 15.32PCh. 15 - Prob. 15.33PCh. 15 - Prob. 15.34PCh. 15 - 15.35 What is the major monobromination product...Ch. 15 - Prob. 15.36PCh. 15 - 15.37 What alkane is needed to make each alkyl...Ch. 15 - 15.38 Which alkyl halides can be prepared in good...Ch. 15 - Prob. 15.39PCh. 15 - 15.40 Explain why radical bromination of p-xylene...Ch. 15 - a. What product(s) (excluding stereoisomers) are...Ch. 15 - Prob. 15.42PCh. 15 - 15.43 Draw the products formed when each alkene is...Ch. 15 - 15.44 Draw all constitutional isomers formed when...Ch. 15 - 15.45 Draw the organic products formed in each...Ch. 15 - Prob. 15.46PCh. 15 - 15.47 Treatment of a hydrocarbon A (molecular...Ch. 15 - 15.48 Draw the products formed in each reaction...Ch. 15 - Prob. 15.49PCh. 15 - 15.50 Draw all the monochlorination products that...Ch. 15 - Prob. 15.51PCh. 15 - 15.52 (a) Draw the products (including...Ch. 15 - 15.53 Consider the following bromination: .
a....Ch. 15 - 15.54 Draw a stepwise mechanism for the following...Ch. 15 - Prob. 15.55PCh. 15 - Prob. 15.56PCh. 15 - 15.57 Devise a synthesis of each compound from...Ch. 15 - Prob. 15.58PCh. 15 - Prob. 15.59PCh. 15 - 15.60 Devise a synthesis of each compound using ...Ch. 15 - Prob. 15.61PCh. 15 - Prob. 15.62PCh. 15 - 15.63 As described in Section 9.16, the...Ch. 15 - 15.64 Ethers are oxidized with to form...Ch. 15 - Prob. 15.65PCh. 15 - Prob. 15.66PCh. 15 - 15.67 In cells, vitamin C exists largely as its...Ch. 15 - What monomer is needed to form each...Ch. 15 - Prob. 15.69PCh. 15 - Prob. 15.70PCh. 15 - 15.71 Draw a stepwise mechanism for the following...Ch. 15 - 15.72 As we will learn in Chapter 30, styrene...Ch. 15 - Prob. 15.73PCh. 15 - 15.74 A and B, isomers of molecular formula , are...Ch. 15 - Prob. 15.75PCh. 15 - 15.76 Draw a stepwise mechanism for the...Ch. 15 - Prob. 15.77PCh. 15 - Prob. 15.78PCh. 15 - Prob. 15.79P
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
- Bicyclo-2,5-heptadiene can be prepared in two steps from cyclopentadiene and vinyl chloride. Provide a mechanism for each step.arrow_forwardFollowing 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.arrow_forwardC=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_forward
- Consider the following mechanism of reaction: CH3 CH, CH3 CH;C CH, + H-CI CH;CCH, + C: → CH;CCH, tert-butyl cation CI tert-butyl chloride In this reaction: The alkene and the carbocation intermediate are both nucleophilic HCl and CI" are both nucleophilic The alkene and HCI are both electrophilic The carbocation and CI" are both electrophilic The alkene and CI" are both nucleophilic AMeving to another question will save this responsearrow_forwardBicyclo-2,5-heptadiene can be prepared in two steps from cyclopentadiene and vinyl chloride. Provide a mechanism for each step. heat C,H,ONa + CH,=CHCI C,H,OH CI Bicyclo-2,5-heptadienearrow_forwardDefine the Mechanism of the Radical Addition of HBr to an Alkene ?arrow_forward
- Similar to alkanes, hydrogen gas can undergo radical halogenation according to the following reaction, where X = F, CI, Br, or I. Propose a chain-reaction mechanism for this reaction, including an initiation step, propagation steps, and two plausible termination steps. H-H + X-X 2 H-X hvarrow_forwardWhen the alkene A was treated first with Hg(OAc)2 in MeOH and second with NaBH4 the product was the ether B. Using curved arrows please give the mechanism for the first step of (Hg(OAc)2 in MeOH) this reaction, including any regioselectivity or stereoselectivity. H3C 1. Hg(OAc)2, MeOH H3C O-CH3 =CH2 ✓ -CH3 H3C 2. NaBH H3C A Barrow_forwardWhen 3-methyl-1-butene reacts with HBr, two alkyl halides are formed: 2-bromo-3-methylbutane and 2-bromo-2-methylbutane. Propose a mechanism that explains the formation of these two products.arrow_forward
- H3C CH3 H3C NA C→XT Br Br₂ CH₂Cl₂ H3C Electrophilic addition of bromine, Br₂, to alkenes yields a 1,2-dibromoalkane. The reaction proceeds through a cyclic intermediate known as a bromonium ion. The reaction occurs in an anhydrous solvent such as CH₂Cl₂. CH3 Br In the second step of the reaction, bromide is the nucleophile and attacks at one of the carbons of the bromonium ion to yield the product. Due to steric clashes, the bromide ion always attacks the carbon from the opposite face of the bromonium ion so that a product with anti stereochemistry is formed. Draw curved arrows to show the movement of electrons in this step of the mechanism. Arrow-pushing Instructions Br CH3 H3C CH3arrow_forwardIdentify the following pericyclic reaction; explain the course, product distribution and stereochemistry of the reaction. Where the first product is produced 70% and the second product is produced 30%.arrow_forwardCan someone show the step by step mechanism for this problem. I don't understand where the atoms move and come from.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