Organic Chemistry: Principles and Mechanisms (Second Edition)
2nd Edition
ISBN: 9780393663556
Author: Joel Karty
Publisher: W. W. Norton & Company
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Chapter 6, Problem 6.8P
Interpretation Introduction
Interpretation:
For the given proton transfer reaction, which side of the reaction is favored, and by what numerical factor, is to be predicted.
Concept introduction:
A proton transfer reaction is the one in which a Bronsted Lowry base reacts with a Bronsted Lowry acid to form a conjugate base and a conjugate acid respectively.
A reaction’s tendency to form products is described by its equilibrium constant Keq. A very large Keq favors the products side, while a very small Keq value heavily favors the reactants.
A proton transfer reaction favors the side opposite the stronger acid (i.e., opposite to the lower pKa value). Each difference of one in pKa values between the two acids corresponds to a power of ten by which that side of the reaction is favored.
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3. Use curved arrows to show electron movement in the reactant side and draw the product/s of the Lewis
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1. Rank the following species in order of increasing acidity. Explain your reasons for ordering them as you do. HF NH3 H2SO4 CH3OH CH3COOH H3O+ H2O2.
Consider the following compounds that vary from nearly nonacidic to strongly acidic. Draw the conjugate bases of these compounds, and explain why the acidity increases so dramatically with substitution by nitro groups.
CH4 CH3NO2 CH2(NO2)2 CH(NO2)3
Select the stronger base and then draw its conjugate acid below.
CICH₂CH₂O
or
CH3CH₂O-
• You do not have to consider stereochemistry.
• You do not have to explicitly draw H atoms.
• You do not have to include lone pairs in your answer.
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Chapter 6 Solutions
Organic Chemistry: Principles and Mechanisms (Second Edition)
Ch. 6 - Prob. 6.1PCh. 6 - Prob. 6.2PCh. 6 - Prob. 6.3PCh. 6 - Prob. 6.4PCh. 6 - Prob. 6.5PCh. 6 - Prob. 6.6PCh. 6 - Prob. 6.7PCh. 6 - Prob. 6.8PCh. 6 - Prob. 6.9PCh. 6 - Prob. 6.10P
Ch. 6 - Prob. 6.11PCh. 6 - Prob. 6.12PCh. 6 - Prob. 6.13PCh. 6 - Prob. 6.14PCh. 6 - Prob. 6.15PCh. 6 - Prob. 6.16PCh. 6 - Prob. 6.17PCh. 6 - Prob. 6.18PCh. 6 - Prob. 6.19PCh. 6 - Prob. 6.20PCh. 6 - Prob. 6.21PCh. 6 - Prob. 6.22PCh. 6 - Prob. 6.23PCh. 6 - Prob. 6.24PCh. 6 - Prob. 6.25PCh. 6 - Prob. 6.26PCh. 6 - Prob. 6.27PCh. 6 - Prob. 6.28PCh. 6 - Prob. 6.29PCh. 6 - Prob. 6.30PCh. 6 - Prob. 6.31PCh. 6 - Prob. 6.32PCh. 6 - Prob. 6.33PCh. 6 - Prob. 6.34PCh. 6 - Prob. 6.35PCh. 6 - Prob. 6.36PCh. 6 - Prob. 6.37PCh. 6 - Prob. 6.38PCh. 6 - Prob. 6.39PCh. 6 - Prob. 6.40PCh. 6 - Prob. 6.41PCh. 6 - Prob. 6.42PCh. 6 - Prob. 6.43PCh. 6 - Prob. 6.44PCh. 6 - Prob. 6.45PCh. 6 - Prob. 6.46PCh. 6 - Prob. 6.47PCh. 6 - Prob. 6.48PCh. 6 - Prob. 6.49PCh. 6 - Prob. 6.50PCh. 6 - Prob. 6.51PCh. 6 - Prob. 6.52PCh. 6 - Prob. 6.53PCh. 6 - Prob. 6.54PCh. 6 - Prob. 6.55PCh. 6 - Prob. 6.56PCh. 6 - Prob. 6.57PCh. 6 - Prob. 6.58PCh. 6 - Prob. 6.59PCh. 6 - Prob. 6.60PCh. 6 - Prob. 6.61PCh. 6 - Prob. 6.62PCh. 6 - Prob. 6.63PCh. 6 - Prob. 6.64PCh. 6 - Prob. 6.65PCh. 6 - Prob. 6.66PCh. 6 - Prob. 6.67PCh. 6 - Prob. 6.68PCh. 6 - Prob. 6.69PCh. 6 - Prob. 6.70PCh. 6 - Prob. 6.71PCh. 6 - Prob. 6.72PCh. 6 - Prob. 6.73PCh. 6 - Prob. 6.74PCh. 6 - Prob. 6.75PCh. 6 - Prob. 6.76PCh. 6 - Prob. 6.77PCh. 6 - Prob. 6.78PCh. 6 - Prob. 6.79PCh. 6 - Prob. 6.80PCh. 6 - Prob. 6.81PCh. 6 - Prob. 6.82PCh. 6 - Prob. 6.83PCh. 6 - Prob. 6.84PCh. 6 - Prob. 6.85PCh. 6 - Prob. 6.86PCh. 6 - Prob. 6.87PCh. 6 - Prob. 6.88PCh. 6 - Prob. 6.1YTCh. 6 - Prob. 6.2YTCh. 6 - Prob. 6.3YTCh. 6 - Prob. 6.4YTCh. 6 - Prob. 6.5YTCh. 6 - Prob. 6.6YTCh. 6 - Prob. 6.7YTCh. 6 - Prob. 6.8YTCh. 6 - Prob. 6.9YTCh. 6 - Prob. 6.10YTCh. 6 - Prob. 6.11YTCh. 6 - Prob. 6.12YTCh. 6 - Prob. 6.13YTCh. 6 - Prob. 6.14YTCh. 6 - Prob. 6.15YTCh. 6 - Prob. 6.16YTCh. 6 - Prob. 6.17YTCh. 6 - Prob. 6.18YTCh. 6 - Prob. 6.19YTCh. 6 - Prob. 6.20YT
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- Parts A and B A. Circle the molecule with rhe lowest bp. HO- HO, B. Rank from most acidic (1) to least acidic (4) phenol p-nitrophenol p-methylphenol tert-butanolarrow_forwardWhy is compound la stonger base than compound I? H - H N. II O 1. Because compound II is more sterically hindered. O2. Because compound I is more stable than compound II. O 3. Because the lone pair on compound Iis part of a delocalized pi-system. O4. Because the lone pair on compound II is part of a delocalized pl-system.arrow_forwardFrom each pair, select the stronger base. Then draw the conjugate acid of the strong base below. CH3CH20 or CH3CEC • You do not have to consider stereochemistry. You do not have to explicitly draw H atoms. • You do not have to include lone pairs in your answer. C opy astearrow_forward
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- Indicate which is the weaker base and explain why. Be specific! NH3 or NF3arrow_forwardDraw the products of each acid-base reaction. EXAMPLE: a. b. CH3CH₂ propanoic acid OH || + CH3CH₂CH₂ OH + K+ OH This proton is transferred from the acid to the base. NaOH + base Na₂CO3 + H-O-H CH₂CH₂ O K+ The carboxylate anion is formed as a potassium salt.arrow_forwardHighlight in red each acidic location on the organic molecule at left. Highlight in blue each basic location on the organic molecule at right. Note for advanced students: we mean acidic or basic in the Brønsted-Lowry sense only. H,N ОН OH ни Он OH x сarrow_forward
- Curved arrows are used to illustrate the flow of electrons. In this reaction, a pi bond acts as a Bronsted base. Using the provided starting structures, draw the curved electron- pushing arrows for the acid-base reaction. Be sure to account for all bond-breaking and bond-making steps. Then draw the organic product of the reaction. + Include all lone pairs in the structures. Ignore inorganic byproducts and counterions. I I HH :CI. co THE I Select to Edit Arrowsarrow_forwardAn acid-base reaction is shown below. ONa + Base Acid a. Draw the arrow-pushing mechanism for this reaction. b. Draw the products for the reaction and label them as "conjugate base" and "conjugate acid." Be sure to assign formal charges. c. Consider the base and conjugate base. Using your four tools, identify the weakest conjugate base and put a box around it. Which tool did you use to determine your answer? d. Using the pKa chart in you lecture notes, label the acid and conjugate acid with their pKa values. Which functional group is the stronger acid? e. Equilibrium moves away from the stronger acid (it wants to dissociate) and toward the more stable conjugate base (lower energy side). Based on your answers to the previous questions, which side of the reaction is favored by equilibrium, left or right?arrow_forward7. Draw the conjugate base for each. CH;CH2OH → CH:CH;CH3 → CH:CH2NH2 – CH:CH2SH –arrow_forward
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