(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.
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Chapter 3 Solutions
ORG.CHEM EBOOK W/BBWILEY PLUS>CUSTOM<
- Nonearrow_forwardRedraw the molecule below as a skeletal ("line") structure. Be sure to use wedge and dash bonds if necessary to accurately represent the direction of the bonds to ring substituents. Cl. Br Click and drag to start drawing a structure. : ☐ ☑ Parrow_forwardK m Choose the best reagents to complete the following reaction. L ZI 0 Problem 4 of 11 A 1. NaOH 2. CH3CH2CH2NH2 1. HCI B OH 2. CH3CH2CH2NH2 DII F1 F2 F3 F4 F5 A F6 C CH3CH2CH2NH2 1. SOCl2 D 2. CH3CH2CH2NH2 1. CH3CH2CH2NH2 E 2. SOCl2 Done PrtScn Home End FA FQ 510 * PgUp M Submit PgDn F11arrow_forward
- given cler asnwerarrow_forwardAdd curved arrows to the reactants in this reaction. A double-barbed curved arrow is used to represent the movement of a pair of electrons. Draw curved arrows. : 0: si H : OH :: H―0: Harrow_forwardConsider this step in a radical reaction: Br N O hv What type of step is this? Check all that apply. Draw the products of the step on the right-hand side of the drawing area below. If more than one set of products is possible, draw any set. Also, draw the mechanism arrows on the left-hand side of the drawing area to show how this happens. O primary Otermination O initialization O electrophilic O none of the above × ☑arrow_forward
- Nonearrow_forwardCan I get a drawing of what is happening with the orbitals (particularly the p orbital) on the O in the OH group? Is the p orbital on the O involved in the ring resonance? Why or why not?arrow_forward1) How many monochlorination products-including stereochemistry- are there for the molecule below:arrow_forward
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