
(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, Third Edition Binder Ready Version
- Select the stronger base from each pair of compounds. (a) H₂CNH₂ or EtzN (b) CI or NH2 NH2 (c) .Q or EtzN (d) or (e) N or (f) H or Harrow_forward4. Provide a clear arrow-pushing mechanism for each of the following reactions. Do not skip proton transfers, do not combine steps, and make sure your arrows are clear enough to be interpreted without ambiguity. a. 2. 1. LDA 3. H3O+ HOarrow_forwardb. H3C CH3 H3O+ ✓ H OHarrow_forward
- 2. Provide reagents/conditions to accomplish the following syntheses. More than one step is required in some cases. a. CH3arrow_forwardIdentify and provide an explanation that distinguishes a qualitative and quantitative chemical analysis. Provide examples.arrow_forwardIdentify and provide an explanation of the operational principles behind a Atomic Absorption Spectrometer (AAS). List the steps involved.arrow_forward
- Instructions: Complete the questions in the space provided. Show all your work 1. You are trying to determine the rate law expression for a reaction that you are completing at 25°C. You measure the initial reaction rate and the starting concentrations of the reactions for 4 trials. BrO³¯ (aq) + 5Br¯ (aq) + 6H* (aq) → 3Br₂ (l) + 3H2O (l) Initial rate Trial [BrO3] [H*] [Br] (mol/L) (mol/L) | (mol/L) (mol/L.s) 1 0.10 0.10 0.10 8.0 2 0.20 0.10 0.10 16 3 0.10 0.20 0.10 16 4 0.10 0.10 0.20 32 a. Based on the above data what is the rate law expression? b. Solve for the value of k (make sure to include proper units) 2. The proposed reaction mechanism is as follows: i. ii. BrО¸¯ (aq) + H+ (aq) → HBrO3 (aq) HBrO³ (aq) + H* (aq) → H₂BrO3* (aq) iii. H₂BrO³* (aq) + Br¯ (aq) → Br₂O₂ (aq) + H2O (l) [Fast] [Medium] [Slow] iv. Br₂O₂ (aq) + 4H*(aq) + 4Br(aq) → 3Br₂ (l) + H2O (l) [Fast] Evaluate the validity of this proposed reaction. Justify your answer.arrow_forwardе. Д CH3 D*, D20arrow_forwardC. NaOMe, Br Brarrow_forward
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