(a)
Interpretation:
The product for the given set of reactions should be identified.
Concept introduction:
Alcohol: It is one type of
Oxidation Reaction: It involves loss of electrons, addition of oxygen atoms or removal of hydrogen atoms. The oxidation of alcohol are achieved by using reagents like
Base: They are substance that gives
Leaving group: it is a fragment that leaves substrate with a pair of electrons via heterolytic bond cleavage.
Nucleophile: donates pair of electrons to positively charged substrate resulting in the formation of
Elimination Reaction: It is just reverse reaction of addition where substituent from the given molecule is removed via E1 (the reaction depends only on the substrate involved in the reaction) or E2 (the reaction depends on both of the substituents in the reaction) mechanism.
Carbocation: carbon atom bears positive charged species with three bonds is called Carbocation and it plays vital intermediate in
Carbocation’s are classified in to three types as primary, secondary and tertiary depending on the number of carbon atoms which is attached to the ionized carbon.
Tertiary carbocation is more stable than secondary carbocation, secondary carbocation is more stable than the primary carbocation as shown below.
(b)
Interpretation:
The product for the given set of reactions should be identified.
Concept introduction:
Alcohol: It is one type of functional group in organic chemistry in which carbon is singly bonded to one
Oxidation Reaction: It involves loss of electrons, addition of oxygen atoms or removal of hydrogen atoms. The oxidation of alcohol are achieved by using reagents like
Base: They are substance that gives
Leaving group: it is a fragment that leaves substrate with a pair of electrons via heterolytic bond cleavage.
Nucleophile: donates pair of electrons to positively charged substrate resulting in the formation of chemical bond.
Elimination Reaction: It is just reverse reaction of addition where substituent from the given molecule is removed via E1 (the reaction depends only on the substrate involved in the reaction) or E2 (the reaction depends on both of the substituents in the reaction) mechanism.
Carbocation: carbon atom bears positive charged species with three bonds is called Carbocation and it plays vital intermediate in organic synthesis. Carbocation migratory aptitude is mainly depends on the stability of the carbocation. This type of carbocation undergoes inter or intra molecule reactions and it form more stable product this type of rearrangement called carbocation rearrangement.
Carbocation’s are classified in to three types as primary, secondary and tertiary depending on the number of carbon atoms which is attached to the ionized carbon.
Tertiary carbocation is more stable than secondary carbocation, secondary carbocation is more stable than the primary carbocation as shown below.
(c)
Interpretation:
The product for the given set of reactions should be identified.
Concept introduction:
Alcohol: It is one type of functional group in organic chemistry in which carbon is singly bonded to one
Oxidation Reaction: It involves loss of electrons, addition of oxygen atoms or removal of hydrogen atoms. The oxidation of alcohol are achieved by using reagents like
Base: They are substance that gives
Leaving group: it is a fragment that leaves substrate with a pair of electrons via heterolytic bond cleavage.
Nucleophile: donates pair of electrons to positively charged substrate resulting in the formation of chemical bond.
Elimination Reaction: It is just reverse reaction of addition where substituent from the given molecule is removed via E1 (the reaction depends only on the substrate involved in the reaction) or E2 (the reaction depends on both of the substituents in the reaction) mechanism.
Carbocation: carbon atom bears positive charged species with three bonds is called Carbocation and it plays vital intermediate in organic synthesis. Carbocation migratory aptitude is mainly depends on the stability of the carbocation. This type of carbocation undergoes inter or intra molecule reactions and it form more stable product this type of rearrangement called carbocation rearrangement.
Carbocation’s are classified in to three types as primary, secondary and tertiary depending on the number of carbon atoms which is attached to the ionized carbon.
Tertiary carbocation is more stable than secondary carbocation, secondary carbocation is more stable than the primary carbocation as shown below.
(d)
Interpretation:
The product for the given set of reactions should be identified.
Concept introduction:
Alcohol: It is one type of functional group in organic chemistry in which carbon is singly bonded to one
Oxidation Reaction: It involves loss of electrons, addition of oxygen atoms or removal of hydrogen atoms. The oxidation of alcohol are achieved by using reagents like
Base: They are substance that gives
Leaving group: it is a fragment that leaves substrate with a pair of electrons via heterolytic bond cleavage.
Nucleophile: donates pair of electrons to positively charged substrate resulting in the formation of chemical bond.
Elimination Reaction: It is just reverse reaction of addition where substituent from the given molecule is removed via E1 (the reaction depends only on the substrate involved in the reaction) or E2 (the reaction depends on both of the substituents in the reaction) mechanism.
Carbocation: carbon atom bears positive charged species with three bonds is called Carbocation and it plays vital intermediate in organic synthesis. Carbocation migratory aptitude is mainly depends on the stability of the carbocation. This type of carbocation undergoes inter or intra molecule reactions and it form more stable product this type of rearrangement called carbocation rearrangement.
Carbocation’s are classified in to three types as primary, secondary and tertiary depending on the number of carbon atoms which is attached to the ionized carbon.
Tertiary carbocation is more stable than secondary carbocation, secondary carbocation is more stable than the primary carbocation as shown below.
(e)
Interpretation:
The product for the given set of reactions should be identified.
Concept introduction:
Alcohol: It is one type of functional group in organic chemistry in which carbon is singly bonded to one
Oxidation Reaction: It involves loss of electrons, addition of oxygen atoms or removal of hydrogen atoms. The oxidation of alcohol are achieved by using reagents like
Base: They are substance that gives
Leaving group: it is a fragment that leaves substrate with a pair of electrons via heterolytic bond cleavage.
Nucleophile: donates pair of electrons to positively charged substrate resulting in the formation of chemical bond.
Elimination Reaction: It is just reverse reaction of addition where substituent from the given molecule is removed via E1 (the reaction depends only on the substrate involved in the reaction) or E2 (the reaction depends on both of the substituents in the reaction) mechanism.
Carbocation: carbon atom bears positive charged species with three bonds is called Carbocation and it plays vital intermediate in organic synthesis. Carbocation migratory aptitude is mainly depends on the stability of the carbocation. This type of carbocation undergoes inter or intra molecule reactions and it form more stable product this type of rearrangement called carbocation rearrangement.
Carbocation’s are classified in to three types as primary, secondary and tertiary depending on the number of carbon atoms which is attached to the ionized carbon.
Tertiary carbocation is more stable than secondary carbocation, secondary carbocation is more stable than the primary carbocation as shown below.
(f)
Interpretation:
The product for the given set of reactions should be identified.
Concept introduction:
Alcohol: It is one type of functional group in organic chemistry in which carbon is singly bonded to one
Oxidation Reaction: It involves loss of electrons, addition of oxygen atoms or removal of hydrogen atoms. The oxidation of alcohol are achieved by using reagents like
Base: They are substance that gives
Leaving group: it is a fragment that leaves substrate with a pair of electrons via heterolytic bond cleavage.
Nucleophile: donates pair of electrons to positively charged substrate resulting in the formation of chemical bond.
Elimination Reaction: It is just reverse reaction of addition where substituent from the given molecule is removed via E1 (the reaction depends only on the substrate involved in the reaction) or E2 (the reaction depends on both of the substituents in the reaction) mechanism.
Carbocation: carbon atom bears positive charged species with three bonds is called Carbocation and it plays vital intermediate in organic synthesis. Carbocation migratory aptitude is mainly depends on the stability of the carbocation. This type of carbocation undergoes inter or intra molecule reactions and it form more stable product this type of rearrangement called carbocation rearrangement.
Carbocation’s are classified in to three types as primary, secondary and tertiary depending on the number of carbon atoms which is attached to the ionized carbon.
Tertiary carbocation is more stable than secondary carbocation, secondary carbocation is more stable than the primary carbocation as shown below.
![Check Mark](/static/check-mark.png)
Want to see the full answer?
Check out a sample textbook solution![Blurred answer](/static/blurred-answer.jpg)
Chapter 12 Solutions
ORGANIC CHEMISTRY-PRINT MULTI TERM
- Identify the molecule as having a(n) E, Z, cis, or trans configuration. H₂C- CH3 О Е ○ cis ○ transarrow_forwardThe decomposition of dinitrogen pentoxide according to the equation: 50°C 2 N2O5(g) 4 NO2(g) + O2(g) follows first-order kinetics with a rate constant of 0.0065 s-1. If the initial concentration of N2O5 is 0.275 M, determine: the final concentration of N2O5 after 180 seconds. ...arrow_forwardDon't used hand raitingarrow_forward
- CS2(g) →CS(g) + S(g) The rate law is Rate = k[CS2] where k = 1.6 × 10−6 s−¹. S What is the concentration of CS2 after 5 hours if the initial concentration is 0.25 M?arrow_forwardCS2(g) → CS(g) + S(g) The rate law is Rate = k [CS2] where k = 1.6 × 10-6 s−1. S Calculate the half-life.arrow_forwardThe following is a first order reaction where the rate constant, k, is 6.29 x 10-3 min-*** What is the half-life? C2H4 C2H2 + H2arrow_forward
- Control Chart Drawing Assignment The table below provides the number of alignment errors observed during the final inspection of a certain model of airplane. Calculate the central, upper, and lower control limits for the c-chart and draw the chart precisely on the graph sheet provided (based on 3-sigma limits). Your chart should include a line for each of the control limits (UCL, CL, and LCL) and the points for each observation. Number the x-axis 1 through 25 and evenly space the numbering for the y-axis. Connect the points by drawing a line as well. Label each line drawn. Airplane Number Number of alignment errors 201 7 202 6 203 6 204 7 205 4 206 7 207 8 208 12 209 9 210 9 211 8 212 5 213 5 214 9 215 8 216 15 217 6 218 4 219 13 220 7 221 8 222 15 223 6 224 6 225 10arrow_forwardCollagen is used to date artifacts. It has a rate constant = 1.20 x 10-4 /years. What is the half life of collagen?arrow_forwardיווי 10 20 30 40 50 60 70 3.5 3 2.5 2 1.5 1 [ppm] 3.5 3 2.5 2 1.5 1 6 [ppm] 1 1.5 -2.5 3.5arrow_forward
- 2H2S(g)+3O2(g)→2SO2(g)+2H2O(g) A 1.2mol sample of H2S(g) is combined with excess O2(g), and the reaction goes to completion. Question Which of the following predicts the theoretical yield of SO2(g) from the reaction? Responses 1.2 g Answer A: 1.2 grams A 41 g Answer B: 41 grams B 77 g Answer C: 77 grams C 154 g Answer D: 154 grams Darrow_forwardPart VII. Below are the 'HNMR, 13 C-NMR, COSY 2D- NMR, and HSQC 2D-NMR (similar with HETCOR but axes are reversed) spectra of an organic compound with molecular formula C6H1003 - Assign chemical shift values to the H and c atoms of the compound. Find the structure. Show complete solutions. Predicted 1H NMR Spectrum 4.7 4.6 4.5 4.4 4.3 4.2 4.1 4.0 3.9 3.8 3.7 3.6 3.5 3.4 3.3 3.2 3.1 3.0 2.9 2.8 2.7 2.6 2.5 2.4 2.3 2.2 2.1 2.0 1.9 1.8 1.7 1.6 1.5 1.4 1.3 1.2 1.1 f1 (ppm) Predicted 13C NMR Spectrum 100 f1 (ppm) 30 220 210 200 190 180 170 160 150 140 130 120 110 90 80 70 -26 60 50 40 46 30 20 115 10 1.0 0.9 0.8 0 -10arrow_forwardQ: Arrange BCC and Fec metals, in sequence from the Fable (Dr. R's slides) and Calculate Volume and Density. Aa BCC V 52 5 SFCCarrow_forward
- ChemistryChemistryISBN:9781305957404Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCostePublisher:Cengage LearningChemistryChemistryISBN:9781259911156Author:Raymond Chang Dr., Jason Overby ProfessorPublisher:McGraw-Hill EducationPrinciples 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 EducationChemistry: Principles and ReactionsChemistryISBN:9781305079373Author:William L. Masterton, Cecile N. HurleyPublisher:Cengage LearningElementary Principles of Chemical Processes, Bind...ChemistryISBN:9781118431221Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. BullardPublisher:WILEY
![Text book image](https://www.bartleby.com/isbn_cover_images/9781305957404/9781305957404_smallCoverImage.gif)
![Text book image](https://www.bartleby.com/isbn_cover_images/9781259911156/9781259911156_smallCoverImage.gif)
![Text book image](https://www.bartleby.com/isbn_cover_images/9781305577213/9781305577213_smallCoverImage.gif)
![Text book image](https://www.bartleby.com/isbn_cover_images/9780078021558/9780078021558_smallCoverImage.gif)
![Text book image](https://www.bartleby.com/isbn_cover_images/9781305079373/9781305079373_smallCoverImage.gif)
![Text book image](https://www.bartleby.com/isbn_cover_images/9781118431221/9781118431221_smallCoverImage.gif)