Organic Chemistry (6th Edition)
6th Edition
ISBN: 9781260119107
Author: Janice Gorzynski Smith
Publisher: McGraw Hill Education
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Chapter 11.1, Problem 1P
Neopheliosyne B is a novel acetylenic fatty acid isolated from a New Caledonian marine sponge. (a) Label the most acidic H atom. (b) Which carbon-carbon
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Nepheliosyne B is a novel acetylenic fatty acid isolated from a New Caledonian marine sponge. (a) Label the most acidic H atom. (b) Which carbon–carbon σ bond is shortest? (c) How many degrees of unsaturation does nepheliosyne B contain? (d) How many bonds are formed from Csp–Csp3? (e) Label each triple bond as internal or terminal.
5.
(a) HF is an acid. What is it's conjugate base ?
(b) CIO₂ is a base. What is it's conjugate acid ?
(c) What is the conjugate base of H₂PO4¹ ?
(d) What is the conjugate acid of H₂PO4-¹ ?
Using the data in Appendix C, determine which of the following bases is strong enough to deprotonate acetonitrile (CH3CN), so that equilibrium favors the products: (a) NaH; (b) Na2CO3; (c) NaOH; (d) NaNH2; (e) NaHCO3.
Chapter 11 Solutions
Organic Chemistry (6th Edition)
Ch. 11.1 - Problem 11.1 Neopheliosyne B is a novel acetylenic...Ch. 11.2 - Give the IUPAC name for each compound.Ch. 11.2 - Give the structures corresponding to each of the...Ch. 11.3 - Prob. 4PCh. 11.5 - Prob. 5PCh. 11.6 - Which bases can deprotonate acetylene? The pKa...Ch. 11.7 - Draw the organic products formed when each alkyne...Ch. 11.7 - Draw additional resonance structures for each...Ch. 11.8 - Problem 11.9 Draw the products formed when is...Ch. 11.8 - Explain the following result. Although alkenes...
Ch. 11.9 - Problem 11.11 Draw the keto tautomer of each...Ch. 11.9 - Prob. 12PCh. 11.9 - a Draw two different enol tautomers of...Ch. 11.10 - Prob. 14PCh. 11.10 - Problem 11.15 Draw the organic products formed in...Ch. 11.11 - Problem 11.16 What acetylide anion and alkyl...Ch. 11.11 - Problem. 11.17 Show how , and can be used to...Ch. 11.11 - Prob. 18PCh. 11.11 - Draw the products of each reaction. a. b.Ch. 11.11 - Prob. 20PCh. 11 - Prob. 25PCh. 11 - 11.25 Answer the following questions about...Ch. 11 - 11.26 Give the IUPAC name for each alkyne.
a. ...Ch. 11 - Prob. 28PCh. 11 - Which of the following pairs of compounds...Ch. 11 - Prob. 30PCh. 11 - 11.30 How is each compound related to A? Choose...Ch. 11 - Prob. 32PCh. 11 - 11.33 Draw the products formed when is treated...Ch. 11 - What reagents are needed to convert (CH3CH2)3CCCH...Ch. 11 - 11.36 What alkynes give each of the following...Ch. 11 - 11.37 What alkyne gives each compound as the only...Ch. 11 - 11.38 Draw the organic products formed in each...Ch. 11 - 11.42 What reactions are needed to convert alcohol...Ch. 11 - 11.50 What acetylide anion and alkyl halide are...Ch. 11 - 11.52 Devise a synthesis of each compound using ...Ch. 11 - Prob. 58PCh. 11 - 11.59 N-Chlorosuccinimide (NCS) serves as a source...Ch. 11 - 11.60 Draw a stepwise mechanism for the following...Ch. 11 - 11.61 Draw a stepwise mechanism for the following...
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- 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_forwardPQ-16. What is the major product of this reaction? OH (A) (B) 2) H3O+ H (C) (D) Harrow_forwardRank the following compounds in order of increasing basicity. CH3 CH3 O₂N ||| fo H3C CH3 CH3 IV A) || < | < V < III < IV B) || < V < | < ||| < IV C) IV < I< III < V < || D) || < V < ||| < | < |V H3Carrow_forward
- PQ-27. What is the product of this reaction? (A) 0 OH a (B) (D) H₂ Ptarrow_forwardThe pKa of ascorbic acid (vitamin C, page 55) is 4.17, showing that it is slightly more acidic than acetic acid (CH3COOH, pKa 4.74). 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_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
- но 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_forwardNepheliosyne B is a novel acetylenic fatty acid isolated from a New Caledonian marine sponge. (a) Label the most acidic H atom. (b) Which carbon–carbon σ bond is shortest? (c) How many degrees of unsaturation does nepheliosyne B contain? (d) How many bonds are formed from Csp–Csp3? (e) Label each triple bond as internal or terminal.arrow_forwardCarboxylic Acid Derivatives Resolve by synthesis, step by step.arrow_forward
- Briefly, but clearly, explain why the –OH hydrogen in acetic acid (CH3CO2H) is more acidic than in ethanol (C2H5OH).arrow_forward78) Why are primary alcohols more acidic than tertiary alcohols, in general? Because tertiary alcohols have less acidic hydrogens. Because primary alcohols have less electron-donating groups which create more effective charge separation between oxygen and hydrogen atom, making it more acidic. Because primary alcohols are more polar than tertiary alcohols. Because primary alcohols have less electron-donating groups which decrease electron density on oxygen, making it more susceptible for H+ to depart from the structure.arrow_forwardDraw the products formed when propan-2-ol [(CH3)2CHOH], the main ingredient in rubbing alcohol, is treated with each acid or base: (a) NaH (b) H2SO4 (c) Li+−N[CH(CH3)2]2 (d) CH3CO2H.arrow_forward
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