(a)
Interpretation:
The curved arrow notation for the elimination of
Concept introduction:
Curved arrows are used to represent the movement of electrons in a reaction mechanism. The arrow starts on an electron-rich atom or an electron-rich region such as a pi bond. It ends on an electron poor atom when the movement results in the formation of a new sigma bond. If the result is the formation of a pi bond, the arrow ends in the region between the two atoms that form the bond.
A carbocation is a positively charged carbon atom that is electron-poor, two electrons short of an octet. A nearby bond or a lone pair on a nearby atom acts as an electron-rich region and can transfer the pair of electrons to the electron-poor atom. This can result in the formation of a more stable neutral species, accompanied by the loss of an electrophile. The electrophile may be a proton or another cationic species and is extracted by any base that may be present.
(b)
Interpretation:
The curved arrow notation for the elimination of
Concept introduction:
Curved arrows are used to represent the movement of electrons in a reaction mechanism. The arrow starts on an electron-rich atom or an electron-rich region such as a pi bond. It ends on an electron poor atom when the movement results in the formation of a new sigma bond. If the result is the formation of a pi bond, the arrow ends in the region between the two atoms that form the bond.
A carbocation is a positively charged carbon atom that is electron-poor, two electrons short of an octet. A nearby bond or a lone pair on a nearby atom acts as an electron-rich region and can transfer the pair of electrons to the electron-poor atom. This can result in the formation of a more stable neutral species, accompanied by the loss of an electrophile. The electrophile may be a proton or another cationic species and is extracted by any base that may be present.
Want to see the full answer?
Check out a sample textbook solution
Chapter 7 Solutions
Organic Chemistry: Principles and Mechanisms (Second Edition)
- Draw the structure of each of the following molecules. (a) pentanoic anhydride; (b) hexanoic propanoic anhydride; (c) ethanoic 3-methylpentanoic anhydridearrow_forwardDraw the structure of each of the following molecules. (a) pentanoyl chloride; (b) 4-(2-methylpropyl)heptanedioyl chloride; (c) (S)-5-phenyloctanoyl chloridearrow_forwardClassify the following reactions as addition (A), substitution (S), elimination (E) or rearrangement (R).arrow_forward
- 1) The carbon-oxygen double bond present in aldehydes and ketones is very polar. What does this mean and how does it arise? 2) The carbon-oxygen double bond is readily attacked by nucleophiles like cyanide ions or ammonia. (i) What do you understand by the term nucleophile? (ii) Which part of the carbon-oxygen double bond is attractive to nucleophiles? 3) Why is there a difference between aldehydes and ketones in their response to oxidizing agents such as potassium dichromate(VI) solution acidified with dilute sulfuric acid?arrow_forwardPropylene does not undergo free radical polymerization readily because there are two competing steps after initiation: propagation and hydrogen atom abstraction.(a) Using a generic radical R• as a reactant with propylene, draw the mechanism and products for the two competing steps.(b) Which step produces the more stable product?(c) How do your results explain propylene’s poor reactivity in free radical polymerization?arrow_forwardGiven each of the IUPAC names provided, draw the corresponding structure. (a) 1-ethoxy-3-methoxyhexane;(b) 1,5-dipropoxypentane; (c) 4-butoxy-1,2-dimethoxyheptanearrow_forward
- Identify the major product of this reaction of an alkene. | ||| OH OH 1. Hg(OAc)2, H2O 2. NaBH4, NaOH || IV ㅊ OH OHarrow_forwardDraw out the structure of product (s). If no reaction, write No Reactionarrow_forward(a) Draw the products of the proton transfer reaction shown here. (b) Draw a free energy diagram for this reaction, indicating whether it is endothermic or OH ? exothermic.arrow_forward
- Follow the curved arrows and draw the hydrocarbon product of the following reaction. Include all lone pairs and charges as appropriate. Ignore inorganic byproducts.arrow_forwardConsider the following reactants: CI H₂O Would substitution take place at a significant rate between these reactants? Note for advanced students: you can assume that the reaction mixture is heated mildly, somewhat above room temperature, but strong heat or reflux is not used. If you said substitution would take place, draw all the major products in the upper drawing area below. Be sure you use wedge and dash bonds where necessary, for example to distinguish between major products. If you said substitution would take place, also draw the complete mechanism for one of the major products in the lower drawing area below. If there is more than one major product, you may draw the mechanism for the production of any one of them. Note: be sure you show all the steps in your mechanism. A good check is to make sure the final product of your mechanism is a stable molecule that could be isolated and purified. Oyes Ono 금 Ararrow_forwardHO (c) Alkene + Reagent HO (d) Alkene + Reagentarrow_forward
ChemistryChemistryISBN:9781305957404Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCostePublisher:Cengage Learning
ChemistryChemistryISBN:9781259911156Author:Raymond Chang Dr., Jason Overby ProfessorPublisher:McGraw-Hill Education
Principles of Instrumental AnalysisChemistryISBN:9781305577213Author:Douglas A. Skoog, F. James Holler, Stanley R. CrouchPublisher:Cengage Learning
Organic ChemistryChemistryISBN:9780078021558Author:Janice Gorzynski Smith Dr.Publisher:McGraw-Hill Education
Chemistry: Principles and ReactionsChemistryISBN:9781305079373Author:William L. Masterton, Cecile N. HurleyPublisher:Cengage Learning
Elementary Principles of Chemical Processes, Bind...ChemistryISBN:9781118431221Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. BullardPublisher:WILEY