(a)
Interpretation:
The more acidic compound from each given pair of compounds should be identified in each of the given anions.
Concept introduction:
Acidic strength in the molecule depends on the stability of the Conjugate anion.
The anion involving in resonance in a molecule will have greater stability when compared to molecules having single anion. In resonance the delocalization of anion would take place.
In a molecule where the anion is carried by a more electronegative atom will show more stability. The greater electronegativity makes anion closer to the nucleus hence more stabilized. If the negative charge located atom is larger enough to hold the negative charge, then anion attains more stability.
The carbanion stability varies with the percentage of ‘s’character in its hybridization state. Since the s-orbital is closer to nucleus than p-orbital, the hybridized orbital having more‘s’ character will have more stability to accommodate anion (due to high nuclear charge). The order is 
.
Inductive effect in a compound will affect the anionic stability. +I groups in the compound stabilizes the negative charge of anion.
To find: the more acidic compound from each given pair of compounds.
(b)
Interpretation:
The more acidic compound from each given pair of compounds should be identified in each of the given anions.
Concept introduction:
Acidic strength in the molecule depends on the stability of the Conjugate anion.
The anion involving in resonance in a molecule will have greater stability when compared to molecules having single anion. In resonance the delocalization of anion would take place.
In a molecule where the anion is carried by a more electronegative atom will show more stability. The greater electronegativity makes anion closer to the nucleus hence more stabilized. If the negative charge located atom is larger enough to hold the negative charge, then anion attains more stability.
The carbanion stability varies with the percentage of ‘s’character in its hybridization state. Since the s-orbital is closer to nucleus than p-orbital, the hybridized orbital having more‘s’ character will have more stability to accommodate anion (due to high nuclear charge). The order is .
Inductive effect in a compound will affect the anionic stability. +I groups in the compound stabilizes the negative charge of anion.
To find: the more acidic compound from each given pair of compounds.
(c)
Interpretation:
The more acidic compound from each given pair of compounds should be identified in each of the given anions.
Concept introduction:
Acidic strength in the molecule depends on the stability of the Conjugate anion.
The anion involving in resonance in a molecule will have greater stability when compared to molecules having single anion. In resonance the delocalization of anion would take place.
In a molecule where the anion is carried by a more electronegative atom will show more stability. The greater electronegativity makes anion closer to the nucleus hence more stabilized. If the negative charge located atom is larger enough to hold the negative charge, then anion attains more stability.
The carbanion stability varies with the percentage of ‘s’character in its hybridization state. Since the s-orbital is closer to nucleus than p-orbital, the hybridized orbital having more‘s’ character will have more stability to accommodate anion (due to high nuclear charge). The order is .
Inductive effect in a compound will affect the anionic stability. +I groups in the compound stabilizes the negative charge of anion.
To find: the more acidic compound from each given pair of compounds.
(d)
Interpretation:
The more acidic compound from each given pair of compounds should be identified in each of the given anions.
Concept introduction:
Acidic strength in the molecule depends on the stability of the Conjugate anion.
The anion involving in resonance in a molecule will have greater stability when compared to molecules having single anion. In resonance the delocalization of anion would take place.
In a molecule where the anion is carried by a more electronegative atom will show more stability. The greater electronegativity makes anion closer to the nucleus hence more stabilized. If the negative charge located atom is larger enough to hold the negative charge, then anion attains more stability.
The carbanion stability varies with the percentage of ‘s’character in its hybridization state. Since the s-orbital is closer to nucleus than p-orbital, the hybridized orbital having more‘s’ character will have more stability to accommodate anion (due to high nuclear charge). The order is .
Inductive effect in a compound will affect the anionic stability. +I groups in the compound stabilizes the negative charge of anion.
To find: the more acidic compound from each given pair of compounds.
(e)
Interpretation:
The more acidic compound from each given pair of compounds should be identified in each of the given anions.
Concept introduction:
Acidic strength in the molecule depends on the stability of the Conjugate anion.
The anion involving in resonance in a molecule will have greater stability when compared to molecules having single anion. In resonance the delocalization of anion would take place.
In a molecule where the anion is carried by a more electronegative atom will show more stability. The greater electronegativity makes anion closer to the nucleus hence more stabilized. If the negative charge located atom is larger enough to hold the negative charge, then anion attains more stability.
The carbanion stability varies with the percentage of ‘s’character in its hybridization state. Since the s-orbital is closer to nucleus than p-orbital, the hybridized orbital having more‘s’ character will have more stability to accommodate anion (due to high nuclear charge). The order is .
Inductive effect in a compound will affect the anionic stability. +I groups in the compound stabilizes the negative charge of anion.
To find: the more acidic compound from each given pair of compounds.
(f)
Interpretation:
The more acidic compound from each given pair of compounds should be identified in each of the given anions.
Concept introduction:
Acidic strength in the molecule depends on the stability of the Conjugate anion.
The anion involving in resonance in a molecule will have greater stability when compared to molecules having single anion. In resonance the delocalization of anion would take place.
In a molecule where the anion is carried by a more electronegative atom will show more stability. The greater electronegativity makes anion closer to the nucleus hence more stabilized. If the negative charge located atom is larger enough to hold the negative charge, then anion attains more stability.
The carbanion stability varies with the percentage of ‘s’character in its hybridization state. Since the s-orbital is closer to nucleus than p-orbital, the hybridized orbital having more‘s’ character will have more stability to accommodate anion (due to high nuclear charge). The order is .
Inductive effect in a compound will affect the anionic stability. +I groups in the compound stabilizes the negative charge of anion.
To find: the more acidic compound from each given pair of compounds.
(g)
Interpretation:
The more acidic compound from each given pair of compounds should be identified in each of the given anions.
Concept introduction:
Acidic strength in the molecule depends on the stability of the Conjugate anion.
The anion involving in resonance in a molecule will have greater stability when compared to molecules having single anion. In resonance the delocalization of anion would take place.
In a molecule where the anion is carried by a more electronegative atom will show more stability. The greater electronegativity makes anion closer to the nucleus hence more stabilized. If the negative charge located atom is larger enough to hold the negative charge, then anion attains more stability.
The carbanion stability varies with the percentage of ‘s’character in its hybridization state. Since the s-orbital is closer to nucleus than p-orbital, the hybridized orbital having more‘s’ character will have more stability to accommodate anion (due to high nuclear charge). The order is .
Inductive effect in a compound will affect the anionic stability. +I groups in the compound stabilizes the negative charge of anion.
To find: the more acidic compound from each given pair of compounds.
(h)
Interpretation:
The more acidic compound from each given pair of compounds should be identified in each of the given anions.
Concept introduction:
Acidic strength in the molecule depends on the stability of the Conjugate anion.
The anion involving in resonance in a molecule will have greater stability when compared to molecules having single anion. In resonance the delocalization of anion would take place.
In a molecule where the anion is carried by a more electronegative atom will show more stability. The greater electronegativity makes anion closer to the nucleus hence more stabilized. If the negative charge located atom is larger enough to hold the negative charge, then anion attains more stability.
The carbanion stability varies with the percentage of ‘s’character in its hybridization state. Since the s-orbital is closer to nucleus than p-orbital, the hybridized orbital having more‘s’ character will have more stability to accommodate anion (due to high nuclear charge). The order is .
Inductive effect in a compound will affect the anionic stability. +I groups in the compound stabilizes the negative charge of anion.
To find: the more acidic compound from each given pair of compounds.
Trending nowThis is a popular solution!
Chapter 3 Solutions
ORGANIC CHEM PRINT STUDY GDE & SSM
- Draw one product of an elimination reaction between the molecules below. Note: There may be several correct answers. You only need to draw one of them. You do not need to draw any of the side products of the reaction. OH + ! : ☐ + Х Click and drag to start drawing a structure.arrow_forwardFind one pertinent analytical procedure for each of following questions relating to food safety analysis. Question 1: The presence of lead, mercury and cadmium in canned tuna Question 2: Correct use of food labellingarrow_forwardFormulate TWO key questions that are are specifically in relation to food safety. In addition to this, convert these questions into a requirement for chemical analysis.arrow_forward
- What are the retrosynthesis and forward synthesis of these reactions?arrow_forwardWhich of the given reactions would form meso product? H₂O, H2SO4 III m CH3 CH₂ONa CH3OH || H₂O, H2SO4 CH3 1. LiAlH4, THF 2. H₂O CH3 IVarrow_forwardWhat is the major product of the following reaction? O IV III HCI D = III ა IVarrow_forward
- The reaction of what nucleophile and substrate is represented by the following transition state? CH3 CH3O -Br อ δ CH3 Methanol with 2-bromopropane Methanol with 1-bromopropane Methoxide with 1-bromopropane Methoxide with 2-bromopropanearrow_forwardWhat is the stepwise mechanism for this reaction?arrow_forward32. Consider a two-state system in which the low energy level is 300 J mol 1 and the higher energy level is 800 J mol 1, and the temperature is 300 K. Find the population of each level. Hint: Pay attention to your units. A. What is the partition function for this system? B. What are the populations of each level? Now instead, consider a system with energy levels of 0 J mol C. Now what is the partition function? D. And what are the populations of the two levels? E. Finally, repeat the second calculation at 500 K. and 500 J mol 1 at 300 K. F. What do you notice about the populations as you increase the temperature? At what temperature would you expect the states to have equal populations?arrow_forward
- 30. We will derive the forms of the molecular partition functions for atoms and molecules shortly in class, but the partition function that describes the translational and rotational motion of a homonuclear diatomic molecule is given by Itrans (V,T) = = 2πmkBT h² V grot (T) 4π²IKBT h² Where h is Planck's constant and I is molecular moment of inertia. The overall partition function is qmolec Qtrans qrot. Find the energy, enthalpy, entropy, and Helmholtz free energy for the translational and rotational modes of 1 mole of oxygen molecules and 1 mole of iodine molecules at 50 K and at 300 K and with a volume of 1 m³. Here is some useful data: Moment of inertia: I2 I 7.46 x 10- 45 kg m² 2 O2 I 1.91 x 101 -46 kg m²arrow_forwardK for each reaction step. Be sure to account for all bond-breaking and bond-making steps. HI HaC Drawing Arrows! H3C OCH3 H 4 59°F Mostly sunny H CH3 HO O CH3 'C' CH3 Select to Add Arrows CH3 1 L H&C. OCH3 H H H H Select to Add Arrows Q Search Problem 30 of 20 H. H3C + :0: H CH3 CH3 20 H2C Undo Reset Done DELLarrow_forwardDraw the principal organic product of the following reaction.arrow_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