Organic Chemistry
6th Edition
ISBN: 9781936221349
Author: Marc Loudon, Jim Parise
Publisher: W. H. Freeman
expand_more
expand_more
format_list_bulleted
Question
Chapter 20, Problem 20.3P
Interpretation Introduction
Interpretation:
Whether the concentration of acetic acid dimer is greater in
Concept introduction:
Carboxylic acid is a class of organic compound that contains a
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
6. (Chapter 15-Q37) Indole is an aromatic heterocyclic that
has a benzene ring fused to a pyrrole ring. Answer the
following questions.
Indole
6(a) What is the hybridization of N in this molecule? =
6(b) How many pi electrons N contributes to the ring? =
6() Which orbitals contribute to form a sigma bond
between N and H in this molecule? =
6(c) What is the electronic relationship of Indole to
naphthalene? Give the answer by comparing number of
rings and number of pi electrons in both compounds, write
x rings, y pi electrons=|
Acetic acid, CH3COOH, is a weak organic acid, pKa 4.76. Write an equation for the equilibrium reaction of acetic acid with each base. Which equilibria lie considerably toward the left? Which lie considerably toward the right?
Q.) H2O
Benzene is one of the compounds used as octane enhancers in unleaded gasoline. It is manufactured by the catalytic
conversion of acetylene to benzene:
3C2H2 (8) C6H6 (8).
Which value of Kwould make this reaction most useful commercially? K 0.01, K1, or K 10. Explain your answer.
Chapter 20 Solutions
Organic Chemistry
Ch. 20 - Prob. 20.1PCh. 20 - Prob. 20.2PCh. 20 - Prob. 20.3PCh. 20 - Prob. 20.4PCh. 20 - Prob. 20.5PCh. 20 - Prob. 20.6PCh. 20 - Prob. 20.7PCh. 20 - Prob. 20.8PCh. 20 - Prob. 20.9PCh. 20 - Prob. 20.10P
Ch. 20 - Prob. 20.11PCh. 20 - Prob. 20.12PCh. 20 - Prob. 20.13PCh. 20 - Prob. 20.14PCh. 20 - Prob. 20.15PCh. 20 - Prob. 20.16PCh. 20 - Prob. 20.17PCh. 20 - Prob. 20.18PCh. 20 - Prob. 20.19PCh. 20 - Prob. 20.20PCh. 20 - Prob. 20.21PCh. 20 - Prob. 20.22PCh. 20 - Prob. 20.23PCh. 20 - Prob. 20.24PCh. 20 - Prob. 20.25PCh. 20 - Prob. 20.26PCh. 20 - Prob. 20.27APCh. 20 - Prob. 20.28APCh. 20 - Prob. 20.29APCh. 20 - Prob. 20.30APCh. 20 - Prob. 20.31APCh. 20 - Prob. 20.32APCh. 20 - Prob. 20.33APCh. 20 - Prob. 20.34APCh. 20 - Prob. 20.35APCh. 20 - Prob. 20.36APCh. 20 - Prob. 20.37APCh. 20 - Prob. 20.38APCh. 20 - Prob. 20.39APCh. 20 - Prob. 20.40APCh. 20 - Prob. 20.41APCh. 20 - Prob. 20.42APCh. 20 - Prob. 20.43APCh. 20 - Prob. 20.44APCh. 20 - Prob. 20.45APCh. 20 - Prob. 20.46APCh. 20 - Prob. 20.47APCh. 20 - Prob. 20.48APCh. 20 - Prob. 20.49APCh. 20 - Prob. 20.50APCh. 20 - Prob. 20.51APCh. 20 - Prob. 20.52APCh. 20 - Prob. 20.53APCh. 20 - Prob. 20.54APCh. 20 - Prob. 20.55APCh. 20 - Prob. 20.56APCh. 20 - Prob. 20.57APCh. 20 - Prob. 20.58APCh. 20 - Prob. 20.59APCh. 20 - Prob. 20.60AP
Knowledge Booster
Similar questions
- a) 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_forwardHow to draw the line bond formula or lewis structure of a methyl ketone with the chemical formula C6H5C3H5O? Based on the results of the solubility tests, the compound can be grouped in VI, which is insoluble in water, 10% NaOH and 10% HCl but soluble in concentrated H2SO4. If the compound is water-insoluble but soluble in concentrated H2SO4, it is most probably an alcohol, aldehyde, anhydride, ester, ether, ketone, or an unsaturated hydrocarbon The functional group/class is identified to be Methyl Ketone, base on the results of the chemical tests on Table 2. CHEMICAL TEST OBSERVATIONS +(compound tested positive for the chemical reaction)/ otherwise (-) Molisch test turbid colorless solution - 2,4-DNP test formation of orange-yellow precipitates + Tollen’s test turbid colorless solution - Ninhydrin test clear pale-yellow solution - iodoform test clear pale-yellow solution +arrow_forwardCa2+(aq)+C2O42 ( aq) CaC204 (s) kc = 2.3 *10-9 1.Explain the components of the equilibrium mixture based on the value of the equilibrium constant . 2. What can be said about the rate of the reaction of ethanol versus the rate of reaction of the oxidation of ethylene glycol ? 3.Large amounts of ethanol are indicated for the treatment of ethylene glycol poison , using the LeChatelier Principle discuss why this treatment is effective.arrow_forward
- At 25°C, for benzoate ion Kp = 1.54 x 10-10 while for ethanoate ion K = 5.5 x 10-10. Choose the correct answer: Select one: Ethanoate ion is a stronger base than benzoate ion but ethanoic acid is a weaker acid than benzoic acid. Ethanoate ion is a stronger base than benzoate ion but ethanoic acid is a stronger acid than benzoic acid. Ethanoate ion is a weaker base than benzoate ion and ethanoic acid is a weaker acid than benzoic acid. Ethanoate ion is a weaker base than benzoate ion but ethanoic acid is a stronger acid than benzoic acid.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_forwardPredict the product of the following organic reaction: CH₂ CH O || C–CH2–CH=CH—(CH2)3 CH3 (CH₂)3 CH=CH- - CH₂ CH3 + 4 H₂ Ni CH,−O -C–CH2–CH=CH–CH=CH–CH2–CH3 Specifically, in the drawing area below, draw the chemical structure of the product P. If there is no product, because this reaction won't happen, check the No reaction box under the drawing area. Click and drag to start drawing a structure. × Ś +arrow_forward
- 4. Ag catalyst ? H2C=CH2 + O2 250 °Carrow_forwardReaction of (CH3)3CH with Cl2 forms two products: (CH3)2CHCH2Cl (63%) and (CH3)3CCl (37%). Why is the major product formed by cleavage of the stronger 1° C–H bond?arrow_forwardDraw the organic product expected from the reaction. Include all hydrogen atoms. Note that K, Cr, 0, is present in excess. H H H, SO, CH, CH,CH, OH + K,Cr,0,(aq) |arrow_forward
- pH= 2.) Calculate the pH in a 0.584 M solution of pentanoic acid. Pentanoic acid is CH3CH2CH2CH2CO2H and it's Ka = 1.44 x 10-5.arrow_forwardWhy is ethanol least acidic?arrow_forwardFor the Reaction: NH3+ H2O = NH4+ OH- When adding NH3+ phenolphthalien+ HCl What stress is applied and what direction does the equilibrium shift? When adding NH3+ phenolphthalien+ NH4Cl What stress is applied and what direction does the equilibrium shift? When adding NH3+ phenolphthalien+ NaOH What stress is applied and what direction does the equilibrium shift? When adding NH3+ phenolphthalien+ HCl What stress is applied and what direction does the equilibrium shift?arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
Chemistry: Principles and ReactionsChemistryISBN:9781305079373Author:William L. Masterton, Cecile N. HurleyPublisher:Cengage Learning
Organic ChemistryChemistryISBN:9781305580350Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. FootePublisher:Cengage Learning
Chemistry: Principles and Reactions
Chemistry
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:Cengage Learning
Organic Chemistry
Chemistry
ISBN:9781305580350
Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. Foote
Publisher:Cengage Learning