(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.
Want to see the full answer?
Check out a sample textbook solutionChapter 13 Solutions
ORG.CHEM EBOOK W/BBWILEY PLUS>CUSTOM<
- I need help working this problem out step by step, I was trying to use my example from the txt book but all I know how to do is set it up. I need to be shown step by step as I am a visual learner. Please help me.arrow_forwardDon't used hand raitingarrow_forwardDon't used Ai solution and don't used hand raitingarrow_forward
- & Calculate the molar enthalpy of combustion (A combH) of 1.80 g of pyruvic acid (CH3COCOOH; 88.1 g mol-1) at 37 °C when they are combusted in a calorimeter at constant volume with a calorimeter constant = 1.62 kJ °C-1 and the temperature rose by 1.55 °C. Given: R = 8.314 J mol −1 °C-1 and the combustion reaction: AN C3H4O3 + 2.502(g) → 3CO2(g) + 2H2O(l)arrow_forwardAn unknown salt, AB, has the following precipitation reaction:A+(aq) + B-(aq) ⇌ AB(s) the K value for this reaction is 4.50 x10-6. Draw a model that represents what will happen when 1.00 L each of 1.00 M solution of A+(aq) and 1.00M solution of B-(aq) are combined.arrow_forward5. a) Use the rules in Example 4.4 (p. 99) and calculate sizes of octahedral and tetrahedral cavities in titanium and in zirconium. Use values for atomic radii given in Fig. 9.1 (p.291). (3 points) b) Consider the formation of carbides (MC) of these metals. Which metal is able to accommodate carbon atoms better, and which cavities (octahedral or tetrahedral) would be better suited to accommodate C atoms into metal's lattice? (4 points)arrow_forward
- 2. Read paragraph 3.4 in your textbook ("Chiral Molecules"), and explain if Cobalt(ethylenediamine) 33+ shown in previous problem is a chiral species. If yes, draw projections of both enantiomers as mirror images, analogous to mirror projections of hands (below). Mirror (4 points)arrow_forward3. Borane (BH3) belongs to D3h point group. Consider the vibrational (stretching) modes possible for B-H bonds under D3h symmetry. Using the methods we used in class, construct the reducible representation I, and break it down into irreducible representations using the character table provided. Sketch those modes, indicate whether they are IR-active. (6 points) D3h E 2C3 3C2 σh 283 30% A₁' 1 1 1 1 1 1 x² + y², z² 1 -1 1 1 -1 R₂ E' 2 0 2 0 (x, y) (x² - y², xy) " A₁" 1 1 -1 A2" 1 -1 -1 1 Z E" 2 -1 0 -2 1 0 (Ry, Ry) (xz, yz)arrow_forward1. List all the symmetry elements, and assign the compounds to proper point groups: a) HCIBrC-BrCIH Cl Br H (2 points) H Br b) Pentacarbonylmanganese(I)bromide Br OEC-Mn-CEO 00- c) Phenazine (aromatic molecule, with delocalized bonding) 1 d) Cobalt(ethylenediamine)33+ (just the cation) 3+ H₂N H₂ .NH2 (CI)3 NH2 H2 H₂N. (2 points) (2 points) (2 points)arrow_forward
- Hello, I desperately need help figuring out 8-14; I also wanted to see if you would mind letting me know if I picked the right degree as my melting points on the two graphs. Please and thank you in advance! All the information is provided.arrow_forwardThe reaction: A + B ⇌ 2 C, can be represented by the equilibrium expression, KC =[C]2[A][B]=258 at 520K.When 1.00 M of C was allowed to reach equilibrium and 0.055 M of A was formed. If this reaction wasperformed at the same temperature using 0.500 M C, what would the equilibrium concentration of Abe?arrow_forward1. What is the functional group of an alcohol and a phenol? 2. Why are some alcohols soluble in water? 3. Classify each of the following alcohols as primary, secondary or tertiary. a. 3-pentanol b. 2-methyl-2-butanol c. 1-propanolarrow_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