Concept explainers
(a)
Interpretation: Whether the given substituent exhibits an electron-withdrawing or an electron-donating inductive effect is to be identified.
Concept introduction: The atoms which are more electronegative than carbon pull charge density away from carbon and exhibit an electron-withdrawing inductive effect. Alkyl groups donate electron density, thus exhibit an electron donation inductive effect.
(b)
Interpretation: Whether the given substituent exhibits an electron-withdrawing or an electron-donating inductive effect is to be identified.
Concept introduction: The atoms which are more electronegative than carbon pull charge density away from carbon and exhibit an electron-withdrawing inductive effect. Alkyl groups donate electron density, thus exhibit an electron donation inductive effect.
(c)
Interpretation: Whether the given substituent exhibits an electron-withdrawing or an electron-donating inductive effect is to be identified.
Concept introduction: The atoms which are more electronegative than carbon pull charge density away from carbon and exhibit an electron-withdrawing inductive effect. Alkyl groups donate electron density, thus exhibit an electron donation inductive effect.
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Chapter 16 Solutions
ORGANIC CHEMISTRY BOOK& SG/SM
- When the conjugate acid of aniline, C6H5NH3+, reacts with the acetate ion, the following reaction takes place: C6H5NH3+(aq)+CH3COO(aq)C6H5NH2(aq)+CH3COOH(aq) If Kafor C6H5NH3+ is 1.35105 and Kafor CH3COOH is 1.86105 , what is K for the reaction?arrow_forwardComplete the following acid-base reactions. (a) CH3CH2CH2CH2Li + CH3COOH (b) CH3CH2CH2CH2MgBr + CH3CH2OHarrow_forwardThe acid-catalyzed hydrolysis of an ester converts an ester into a carboxylic acid. Although there are two O atoms that can be protonated, the first step in the mechanism is believed to be protonation of the oxygen in the C=0 group. Based on charge stability, why is it favorable to protonate that oxygen? Hint: Draw out the products of each protonation. + Hо HO, НО Carboxylic acid Ester Alcoholarrow_forward
- 4-Methylphenol is more acidic than ethanol (pKa 10.36 vs 16.0) , even though both contain an OH group and a methyl group. Draw the structures of the anions formed from loss of the alcoholic protons from both compounds. Use resonance to explain the difference in their respective acidities.arrow_forwardPQ-27. What is the product of this reaction? (A) 0 OH a (B) (D) H₂ Ptarrow_forwardAn organic acid can be represented by: ‖ CH3 – CH2 – C – H O ‖ CH3 – CH2 – C – OH O ‖ CH3 – C – CH3 CH3 – CH2 - O – CH3arrow_forward
- Account for the fact that one carboxyl group is a considerably stronger acid than the other carboxyl group.arrow_forwarda) If the pH value of an aqueous solution of trimethylamine [(CH3)3N] is 10.75, what should be the molarity of this solution? (CH3)3N + H2O ↔ (CH3)3NH+ + OH-, Kb = 6,3 × 10-5 b) What will be the pH of the solution prepared by dissolving 8.35 g of aniline hydrochloride (C6H5NH3+Cl-) in 750 mL of 0.215 M aniline (C6H5NH2)? Is this solution an effective buffer? Explain (Kb = 7,4 × 10-10 for aniline, C: 12.0 g / mol, H: 1.0 g / mol, N: 14.0 g / mol, Cl: 35.4 g / mol) .arrow_forwarda) If the pH value of an aqueous solution of trimethylamine [(CH3) 3N] is 10.75, what should be the molarity of this solution? (CH3) 3N + H2O ↔ (CH3) 3NH + + OH-, Kb = 6.3 × 10-5 b) What will be the pH of the solution prepared by dissolving 8.35 g of aniline hydrochloride (C6H5NH3 + Cl-) in 750 mL of 0.215 M aniline (C6H5NH2)? Is this solution an effective buffer? Explain (Kb = 7.4 × 10-10 for aniline, C: 12.0 g / mol, H: 1.0 g / mol, N: 14.0 g / mol, Cl: 35.4 g / mol) .arrow_forward
- Explain why phenol (C6H50H) is substantially more acidic than methanol (CH3OH), but benzoic acid (C6H5CO2H) is not much more acidic than acetic acid (CH3CO2H). Нас — ОН ОН H3C- ОН OH pKa = 15.5 pKa = 10.0 pk = 4.75 pk = 4.2arrow_forwardCH₂-CH₂-CH₂-C-OH butanoic acid CH₂ HO–C–CH,CHCH,—CHO 3-methylpentanoic acid Name each carboxylic acid. (a) HO CH₂ CH₂ 99 - 0 B) CHICHICIO CH, Draw a condensed structural diagram for each carboxylic acid (a) Hexanoic acid (b) 3-propyloctanoic acid Iarrow_forwardIn equilibrium, label the stronger acid, the stronger base, the weaker acid, and the weaker base. Also estimate the position of equilibrium.arrow_forward
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