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(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
ORGANIC CHEMISTRY,SOLN.MAN.+...-ACCESS
- Page of 3 4. Calculate AG for the following reaction at 25°C. Will the reaction occur (be spontaneous)? How do you know? NH3(g) + HCl(g) → NH4Cl(s) AH=-176.0 kJ AS-284.8 J-K-1arrow_forwardtrue or false The equilibrium constant for this reaction is 0.20. N2O4(g) ⇔ 2NO2(g) Based on the above, the equilibrium constant for the following reaction is 5. 4NO2(g) ⇔ 2N2O4(g)arrow_forwardtrue or false The equilibrium constant for this reaction is 0.20. N2O4(g) ⇔ 2NO2(g) Based on the above, the equilibrium constant for the following reaction is 0.4. 2N2O4(g) ⇔ 4NO2(g)arrow_forward
- true or false Using the following equilibrium, if heat is added the equilibrium will shift toward the reactants. N2(g) + 3H2(g) ⇔ 2NH3(g) + heatarrow_forwardTrue or False Using the following equilibrium, if heat is added the equilibrium will shift toward the products. N2O4(g) + heat ⇔ 2NO2(g)arrow_forwardtrue or false Using the following equilibrium, if solid carbon is added the equilibrium will shift toward the products. C(s) + CO2(g) ⇔ 2CO(g)arrow_forward
- Provide the complete mechanism for the reaction below. You must include appropriate arrows,intermediates, and formal charges. Please also provide a reason to explain why the 1,4-adduct is preferred over the 1,3-adduct.arrow_forwardWhich of the following pairs are resonance structures of one another? I. III. || III IV + II. :0: n P !༠ IV. EN: Narrow_forwardPredict the major organic product(s) and byproducts (either organic or inorganic) for thefollowing reactions.arrow_forward
- A 8.25 g sample of aluminum at 55°C released 2500 J of heat. The specific heat of aluminum is 0.900 J/g°C. The density of aluminum is 2.70 g/mL. Calculate the final temperature of the aluminum sample in °C.arrow_forwardPredict the major organic product(s) and byproducts (either organic or inorganic) for thefollowing reactions.arrow_forwardPredict the major organic product(s) and byproducts (either organic or inorganic) for thefollowing reaction.arrow_forward
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