Organic Chemistry
6th Edition
ISBN: 9781936221349
Author: Marc Loudon, Jim Parise
Publisher: W. H. Freeman
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Chapter 15, Problem 15.74AP
Interpretation Introduction
Interpretation:
The explanation for the fact that the value of
Concept introduction:
A molecule that is cyclic, planar, completely conjugated and follows Hückel’s rule is classified as an aromatic compound.
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Phenol (hydroxybenzene) behaves as a weak acid.
a) Write out the equilibrium equation for its partial dissociation in water.
b) Write out the expression for the acid dissociation constant, Ka.
d) Draw the conjugate base of phenol and show how it is stabilised by resonance.
e) Compare and explain the acidity of phenol (p = 9.9) with that of:
cyclohexanol (pk = 16.0)
3-fluorophenol (pK₁ = 9.3)
4-acetylphenol (pK, = 8.1)
4. The pk, of vanillin is about 9, which is much more acidic than a normal alcohol. Draw a
reaction showing the deprotonation of vanillin with NaOH, and then draw six resonance
structures of the conjugate base. Draw the hybrid structure and clearly indicate how the
negative charge is distributed in the compound.
Chemical Equilibrium
Write the equilibrium-constant expressions and obtain numerical values for each constant in
(a) the basic dissociation of aniline, C6H5NH2.
(b) the acidic dissociation of hypochlorous acid, HClO.
(c) the acidic dissociation of methyl ammonium hydrochloride, CH3NH3Cl.
(d) the basic dissociation of NaNO2.
(e) the dissociation of H3AsO3 to H3O+ and AsO33-
Using step-by-step process
Chapter 15 Solutions
Organic Chemistry
Ch. 15 - Prob. 15.1PCh. 15 - Prob. 15.2PCh. 15 - Prob. 15.3PCh. 15 - Prob. 15.4PCh. 15 - Prob. 15.5PCh. 15 - Prob. 15.6PCh. 15 - Prob. 15.7PCh. 15 - Prob. 15.8PCh. 15 - Prob. 15.9PCh. 15 - Prob. 15.10P
Ch. 15 - Prob. 15.11PCh. 15 - Prob. 15.12PCh. 15 - Prob. 15.13PCh. 15 - Prob. 15.14PCh. 15 - Prob. 15.15PCh. 15 - Prob. 15.16PCh. 15 - Prob. 15.17PCh. 15 - Prob. 15.18PCh. 15 - Prob. 15.19PCh. 15 - Prob. 15.20PCh. 15 - Prob. 15.21PCh. 15 - Prob. 15.22PCh. 15 - Prob. 15.23PCh. 15 - Prob. 15.24PCh. 15 - Prob. 15.25PCh. 15 - Prob. 15.26PCh. 15 - Prob. 15.27PCh. 15 - Prob. 15.28PCh. 15 - Prob. 15.29PCh. 15 - Prob. 15.30PCh. 15 - Prob. 15.31PCh. 15 - Prob. 15.32PCh. 15 - Prob. 15.33PCh. 15 - Prob. 15.34PCh. 15 - Prob. 15.35PCh. 15 - Prob. 15.36PCh. 15 - Prob. 15.37PCh. 15 - Prob. 15.38PCh. 15 - Prob. 15.39PCh. 15 - Prob. 15.40PCh. 15 - Prob. 15.41PCh. 15 - Prob. 15.42APCh. 15 - Prob. 15.43APCh. 15 - Prob. 15.44APCh. 15 - Prob. 15.45APCh. 15 - Prob. 15.46APCh. 15 - Prob. 15.47APCh. 15 - Prob. 15.48APCh. 15 - Prob. 15.49APCh. 15 - Prob. 15.50APCh. 15 - Prob. 15.51APCh. 15 - Prob. 15.52APCh. 15 - Prob. 15.53APCh. 15 - Prob. 15.54APCh. 15 - Prob. 15.55APCh. 15 - Prob. 15.56APCh. 15 - Prob. 15.57APCh. 15 - Prob. 15.58APCh. 15 - Prob. 15.59APCh. 15 - Prob. 15.60APCh. 15 - Prob. 15.61APCh. 15 - Prob. 15.62APCh. 15 - Prob. 15.63APCh. 15 - Prob. 15.64APCh. 15 - Prob. 15.65APCh. 15 - Prob. 15.66APCh. 15 - Prob. 15.67APCh. 15 - Prob. 15.68APCh. 15 - Prob. 15.69APCh. 15 - Prob. 15.70APCh. 15 - Prob. 15.71APCh. 15 - Prob. 15.72APCh. 15 - Prob. 15.73APCh. 15 - Prob. 15.74APCh. 15 - Prob. 15.75APCh. 15 - Prob. 15.76APCh. 15 - Prob. 15.77APCh. 15 - Prob. 15.78APCh. 15 - Prob. 15.79APCh. 15 - Prob. 15.80APCh. 15 - Prob. 15.81APCh. 15 - Prob. 15.82APCh. 15 - Prob. 15.83APCh. 15 - Prob. 15.84APCh. 15 - Prob. 15.85APCh. 15 - Prob. 15.86AP
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- Write the equilibrium-constant expressions and obtain numerical values for each constant in (a) the basic dissociation of aniline, C6H5NH2. (b) the acidic dissociation of hypochlorous acid, HClO. (c) the acidic dissociation of methyl ammonium hydrochloride, CH3NH3Cl. (d) the basic dissociation of NaNO2. (e) the dissociation of H3AsO3 to H3O+ and AsO33- just answer the letters C, D and E.arrow_forwardGive reasons for the following :(i) Phenol is more acidic than methanol.(ii) The C—O—H bond angle in alcohols is slightly less than the tetrahedral angle (190°28′).(iii) (CH3)3C—O—CH3 on reaction with HI gives (CH3)3C—I and CH3—OH as the main products and not (CH3)3C—OH and CH3—I.arrow_forward(a) Explain how NaBH, in CH;OH can reduce hemiacetal A to 1,4-butanediol (HOCH,CH,CH,CH,OH). (b) What product is formed when A is treated with Ph;P=CHCH,CH(CH),? (c) The drug isotretinoin is formed by reaction of X and Y. What is the structure of isotretinoin? Although isotretinoin (trade name Accutane or Roaccutane) is used for the treatment of severe acne, it is dispensed under strict controls because it also causes birth defects. PPha NaOCH,CH3 HO- isotretinoin HO A Br X Yarrow_forward
- (c) Arrange the following compounds in order of increasing acidity, and explain the reasons for your choice of order: phenol, cyclohexanol, 2-fluorocyclohexanol, 2-fluorophenol.arrow_forwardWrite the equilibrium-constant expressions and obtainnumerical values for each constant in. (a) the basic dissociation of aniline, C6H5NH2. (b) the acidic dissociation of hypochlorous acid,HClO. (c) the acidic dissociation of methyl ammoniumhydrochloride, CH3NH3Cl. (d) the basic dissociation of NaNO2. (e) the dissociation of H3AsO3to H3O+and AsO33-. (f) the reaction of C2O42-with H2O to give H2C2O4and OH-. show solutionarrow_forward1. Write the equilibrium-constant expressions and obtain numerical values for each constant in(a) the basic dissociation of aniline, C6H5NH2 .(b) the acidic dissociation of hypochlorous acid, HClO.(c) the acidic dissociation of methyl ammonium hydrochloride, CH3NH3Cl.(d) the basic dissociation of NaNO2 .(e) the dissociation of H3AsO3 to H3O + and AsO33- 2. The chemicals A and B react as follows to produce C and D: A + B ↔ C + DKe = [C] [D] [A] [B] The equilibrium constant Ke has a value of 0.30. Assume 0.20 mol of A and 0.50 mol of B are dissolved in 1.00 L, and the reaction proceeds. Calculate the concentrations of reactants [A], [B] and products [C], [D] at equilibrium. Using step-by-step processarrow_forward
- Arrange the following compounds in the increasing order of their acid strength: p-cresol, p-nitrophenol, phenolarrow_forwardно HO но он The pK, of ascorbic acid (vitamin C) is 4.17, showing that it is slightly more acidic than acetic acid (CH3CO0H, pKa 4.74). (a) Show the fou r different conjugate bases that would be formed by deprotonation of the four different OH groups in ascorbic acid. (b) Compare the stabilities of these four conjugate bases, and predict which OH group of ascorbic acid is the most acidic. (c) Compare the most stable conjugate base of ascorbic acid with the conjugate base of acetic acid, and suggest why these two compounds have similar acidities, even though ascorbic acid lacks the carboxylic acid (COOH) group.arrow_forward(a) Draw the three isomers of benzenedicarboxylic acid.(b) The isomers have melting points of 210 °C, 343 °C, and 427 °C. Nitration of the isomers at all possible positions was once used to determine their structures. The isomer that melts at 210 °C gives two mononitro isomers. The isomer that melts at 343 °C gives three mononitro isomers. The isomer that melts at 427 °C gives only one mononitro isomer. Show which isomer has which melting point.arrow_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_forwardRank the following compounds in order of increasing acidity (1 = least acidic, 3 = most acidic) and in the space provided use resonance (of the conjugate base) to explain why the compound you have labelled “3” is the most acidic.arrow_forwardArrange the following types in order of increasing acidity: carboxylic acids, alcohols, phenols, water. Explain why.arrow_forward
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