Essential Organic Chemistry, Global Edition
3rd Edition
ISBN: 9781292089034
Author: Paula Yurkanis Bruice
Publisher: PEARSON
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Chapter 13, Problem 27P
Explain why the pKa of a hydrogen bonded to the sp3 carbon of propene is greater (pKa = 42) than that of any of the carbon acids listed in Table 13.1, but is less than the pKa of an
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True or False
1. With alkynes having 2 pi bonds, they have more electron-rich region than alkenes and, thus, are more reactive than alkenes.
2. Hydrobromination of an alkene follows anti-Markovnikov’s rule in the presence of a peroxide.
3. The reaction of BH3 with 2-methylbut-2-ene results in the attachment of a hydroxyl group, OH, on the third carbon since this carbon is less substituted.
4. The reduction of an alkyne using a strong base such as LiNH2 gives an alkane when 2 equivalents of hydrogen gas is added.
Chapter 13 Solutions
Essential Organic Chemistry, Global Edition
Ch. 13.1 - Identify the most acidic hydrogen in each...Ch. 13.1 - Prob. 2PCh. 13.1 - Prob. 3PCh. 13.1 - Prob. 4PCh. 13.1 - Explain why HO cannot remove a proton from the...Ch. 13.2 - Prob. 6PCh. 13.2 - Prob. 7PCh. 13.3 - Prob. 8PCh. 13.3 - Prob. 9PCh. 13.3 - Prob. 10P
Ch. 13.4 - Prob. 11PCh. 13.5 - Prob. 12PCh. 13.5 - Prob. 13PCh. 13.6 - Prob. 14PCh. 13.7 - Prob. 16PCh. 13.8 - Prob. 17PCh. 13.8 - Prob. 18PCh. 13.8 - Prob. 19PCh. 13.9 - Prob. 20PCh. 13.10 - Propose a mechanism for the formation of...Ch. 13.10 - Prob. 22PCh. 13.10 - a. If the biosynthesis of palmitic acid were...Ch. 13 - Draw the enol tautomers for each of the following...Ch. 13 - Number the following compounds in order from...Ch. 13 - Prob. 26PCh. 13 - Explain why the pKa of a hydrogen bonded to the...Ch. 13 - Prob. 28PCh. 13 - Prob. 29PCh. 13 - Prob. 30PCh. 13 - Prob. 31PCh. 13 - Prob. 32PCh. 13 - Prob. 33PCh. 13 - Using cyclopentanone as the reactant, show the...Ch. 13 - Prob. 35PCh. 13 - Prob. 36PCh. 13 - Prob. 37PCh. 13 - Prob. 38PCh. 13 - Prob. 39PCh. 13 - Prob. 40PCh. 13 - Prob. 41PCh. 13 - Prob. 42PCh. 13 - Prob. 43PCh. 13 - Prob. 44PCh. 13 - Describe how the following compounds can be...Ch. 13 - Prob. 46PCh. 13 - Which would require a higher temperature:...Ch. 13 - Prob. 48PCh. 13 - Propose a mechanism for the following reaction:Ch. 13 - Show how the following compounds could be...Ch. 13 - Prob. 51PCh. 13 - Prob. 52P
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- Give the organic reaction products for the following reaction a. b. a. b. CH3CH2-C. ง ว NaOH ? + ? N(CH3)2 H₂O H*/H₂O CH3-C-OCH2CH3 ? + ?arrow_forwardBut-2-yne has a pK.~ 38. Evaluate the ability of bases to deprotonate the provided acid.arrow_forwardSpecialized reagents and their acronyms: MCPBA= meta-chloroperbenzoic acid; HOAC = acetic acid, NaOAc = Sodium Acetate; Ph= phenyl = C6H5-; P(Ph)3 = triphenyl phosphine; NBS = N- bromosuccinimide, PBR3, SOCI2, AICI3, FeBr3, H2SO4, Li, Na, Mg, Br2, CrO3, LDA = lithium diisopropylamide, PCC = pyridinium chlorochromate, KO-C(CH3)3, LAH = LIAIH4, (sia)2BH = disecondary isobutyl borane, KMNO4, HIO4 = periodic acid %3D write the missing reagents and solvents over the arrows. Some transformation require multiple steps. In some cases there will be multiple arrows. Ph -Ph Ph Ph Brarrow_forward
- Specialized reagents and their acronyms: MCPBA= meta-chloroperbenzoic acid; HOAC = acetic acid, NaOAc = Sodium Acetate; Ph= phenyl = C6H5-; P(Ph)3 = triphenyl phosphine; NBS = N- bromosuccinimide, PBr3, SOCI2, AICI3, FeBr3, H2SO4, Li, Na, Mg, Br2, CrO3, LDA = lithium diisopropylamide, PCC = pyridinium chlorochromate, KO-C(CH3)3, LAH = LIAIH4, (sia)2BH = disecondary isobutyl borane, KMNO4, HIO4 = periodic acid %3D %3D %3D %3D %3D write the missing reagents and solvents over the arrows. Some transformation require multiple steps. In some cases there will be multiple arrows. Ph CH CH o 0om ilubongo CEC-Ph In questions 14-24, write the products of the following transformations. If there is no reaction write NR. Indicate R or S or racemic or indicate cis or trans where appropriate OH + LDA THF solvent (S) Pharrow_forwardSpecialized reagents and their acronyms: MCPBA= meta-chloroperbenzoic acid; HOAC = acetic acid, NaOAc = Sodium Acetate; Ph= phenyl = C6H5-; P(Ph)3 = triphenyl phosphine; NBS = N- bromosuccinimide, PBR3, SOCI2, AICI3, FeBr3, H2SO4, Li, Na, Mg, Br2, CrO3, LDA = lithium diisopropylamide, PCC = pyridinium chlorochromate, KO-C(CH3)3, LAH = LIAIH4, (sia)2BH = disecondary isobutyl borane, KMNO4, HIO4 = periodic acid %3D %3D %3D %3D %3D %3D %3D write the missing reagents and solvents over the arrows. Some transformation require multiple steps. In some cases there will be multiple arrows. ОН H. (R) Br (S) HO OHarrow_forward2. An attempted dehydration of propan-1-ol (CH3CH₂CH₂OH) with H₂SO4 at 135°C gave a product (b.p. 89°C, C6H140) that when tested gave the following results: Baeyer Test: Br₂/CC14 Test: Chromic Acid Test: Negative Negative Negative Propose a structure for the odd product and explain its formation.arrow_forward
- Specialized reagents and their acronyms: MCPBA= meta-chloroperbenzoic acid; HOAC = acetic acid, NaOAc = Sodium Acetate; Ph= phenyl = C6H5-; P(Ph)3 = triphenyl phosphine; NBS = N- bromosuccinimide, PBR3, SOCI2, AlCl3, FeBr3, H2SO4, Li, Na, Mg, Br2, CrO3, LDA = lithium diisopropylamide, PCC = pyridinium chlorochromate, KO-C(CH3)3, LAH = LIAIH4, (sia)2BH = disecondary isobutyl borane, KMNO4, HIO4 = periodic acid %3D %3D %3D %3D %3D write the missing reagents and solvents over the arrows. Some transformation require multiple steps. In some cases there will be multiple arrows. Br Brarrow_forwardAccount for the fact that treating propenoic acid (acrylic acid) with HCl gives only 3-chloropropanoic acid.arrow_forwardEarly structural studies on benzene had to explain the following experimental evidence. When benzene was treated with Br2 (plus a Lewis acid), a single substitution product of molecular formula C6H5Br was formed. When this product was treated with another equivalent of Br2, three different compounds of molecular formula C6H4Br2 were formed.a. Explain why a single Kekulé structure is consistent with the first result, but does not explain the second result.b. Then explain why a resonance description of benzene is consistent with the results of both reactions.arrow_forward
- What accounts for the geometry (pyramidalization) of the NH2 group in aniline? 1. Resonance between the NH2 group and the benzene ring. 2. The electronic withdrawing nature of the sp2 carbons in the phenyl group. 3. Participation of the nitrogen lone pair to make the system aromatic. 4. Both 1 and 3.arrow_forwardThe synthesis of carbohydrates can be particularly difficult because of the large number of chiral centers and OH functional groups present. Epoxides can be useful synthetic intermediates in carbohydrate synthesis. Draw the product of the reaction of a phenyl Gilman reagent with this epoxide. Aa • Use the wedge/hash bond tools to indicate stereochemistry where it exists. Explicitly define the cis/trans stereochemistry of fused rings using wedge or hash bonds. If enantiomers are formed, draw both. Separate multiple products using the sign from the drop-down menu. .arrow_forwardAcid (phosphiruc acid) is catalyzing this dehydration reaction(cyclphexanol to cyclohexene). What does that mean? How many equivalents of acid were needed for the reaction to proceed?arrow_forward
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