(a)
Interpretation:
The products for the given proton transfer reaction are to be drawn and the favored equilibrium side with numerical factor is to be determined.
Concept introduction:
The proton transfer reactions favor the side opposite the stronger acid. Larger the
Answer to Problem 6.41P
The products for the given reaction are:
The equilibrium is favored to product side by a factor of
Explanation of Solution
The given reaction is:
The ion
In the conjugate base formed, the negative charge on nitrogen is delocalized through the electron withdrawing resonance effect of carbonyl group. Thus, amide is a stronger acid than water, and hence, the equilibrium is favored to the product side.
The
The favored equilibrium side with numerical value is determined on the basis of stronger acid and
(b)
Interpretation:
The products for the given proton transfer reaction are to be drawn and the favored equilibrium side with the numerical factor is to be determined.
Concept introduction:
The proton transfer reactions favor the side opposite the stronger acid. Larger the
Answer to Problem 6.41P
The products for the given reaction are:
The equilibrium is favored to the reactant side by a factor of
Explanation of Solution
The given reaction is:
In the given reaction,
The favored equilibrium side with numerical value is determined on the basis of the stronger acid and
(c)
Interpretation:
The products for the given proton transfer reaction are to be drawn and the favored equilibrium side with numerical factor is to be determined.
Concept introduction:
The proton transfer reactions favor the side opposite the stronger acid. Larger the
Answer to Problem 6.41P
The products for the given reaction are:
The equilibrium is favored to product side by a factor of
Explanation of Solution
The given reaction is:
In the given reaction, cyclopentadiene acts as an acid and the negatively charge nitrogen abstracts a proton from diisopropylamine to give the following products:
On the product side, the negative charge on carbon is a resonance stabilized by a conjugated double bond; such stabilization of the negative charge is not possible on the reactant side where the negative charge is on nitrogen bonded to two electron donating isopropyl groups. The acid is stronger when its conjugate base is stable, therefore, cyclopentadiene is a stronger acid than
According to Appendix
The favored equilibrium side with numerical value is determined on the basis of stronger acid and
(d)
Interpretation:
The products for the given proton transfer reaction are to be drawn and the favored equilibrium side with numerical factor is to be determined.
Concept introduction:
The proton transfer reactions favor the side opposite the stronger acid. Larger the
Answer to Problem 6.41P
The products for the given reaction are:
The equilibrium is favored to the product side by a factor of
Explanation of Solution
The given reaction is:
In the given reaction, the hydride ion abstracts the terminal proton of an
As the effective electronegativity of
According to Appendix
The favored equilibrium side with numerical value is determined on the basis of stronger acid and
(e)
Interpretation:
The products for the given proton transfer reaction are to be drawn and the favored equilibrium side with numerical factor is to be determined.
Concept introduction:
The proton transfer reactions favor the side opposite the stronger acid. Larger the
Answer to Problem 6.41P
The products for the given reaction are:
The equilibrium is favored to the product side by a factor of
Explanation of Solution
The given reaction is:
In the given reaction, the propanoate ion abstracts the proton of hydronium ion to give the following products:
According to Appendix
The favored equilibrium side with numerical value is determined on the basis of stronger acid and
(f)
Interpretation:
The products for the given proton transfer reaction are to be drawn and the favored equilibrium side with numerical factor is to be determined.
Concept introduction:
The proton transfer reactions favor the side opposite the stronger acid. Larger the
Answer to Problem 6.41P
The products for the given reaction are:
The equilibrium is favored to the product side. Benzene is the weaker acid by a factor of
Explanation of Solution
The given reaction is:
In the given reaction, the
As the oxygen atom is more electronegative than carbon, the negative charge on oxygen is more stable as compared to carbon. Thus, an anion on the right side, having negative charge on oxygen, is more stable than the anion on the left side where the negative charge is on carbon. Therefore, propanol is more acidic than benzene, and hence, the reaction is favored to the product side.
According to Appendix
The favored equilibrium side with numerical value is determined on the basis of stronger acid and the
(g)
Interpretation:
The products for the given proton transfer reaction are to be drawn and the favored equilibrium side with numerical factor is to be determined.
Concept introduction:
The proton transfer reactions favor the side opposite the stronger acid. Larger the
Answer to Problem 6.41P
The products for the given reaction are:
The equilibrium is favored to the product side. Benzene is the weaker acid by a factor of
Explanation of Solution
The given reaction is:
In the given reaction, the hydride ion abstracts the proton from carboxylic acid and gives the following products:
The conjugate base formed with a negative charge on the oxygen atom is better stabilized by the resonance effect. This makes the carboxylic acid the stronger acid, and the equilibrium is favored to the product side.
According to Appendix
The favored equilibrium side with numerical value is determined on the basis of stronger acid and the
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Chapter 6 Solutions
Organic Chemistry: Principles and Mechanisms (Second Edition)
- (i) Draw the dissociation reaction for a carboxylic acid in water and define the Ka for this reaction. Write the equation that relates pKa to Ka. (ii) A deprotonated carboxylic acid can be drawn in two resonance forms. Draw the two forms and explain what the term “resonance” means. (iii) Draw an energy profile for the above dissociation reaction and describe how the profiles for a strong and a weak acid would differ.arrow_forward6. As the pKa of a proton donor (acid) increases (becomes numerically larger), the nucleophilicity of its conjugate base:arrow_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
- I understand that Ka refers to the rate (constant) that HA acid dissociates into base A- and proton H+. I don't understand the pKa equation. What is the "p"? What does negative logarithm mean? Thank you.arrow_forwardWhat is the conjugate acid of each of the following? What is the conjugate Dase of eacn? (a) NH Parts of the formula in your answer are transposed. Your answer appears to be missing a charge. Conjugate acid (b ) (CH₂)₂NH \table[[chemPad, (9) Help], [xlx =larr|l, Greek -],[(CH),NH,N What is the conjugate acid of each of the following? What is the conjugate base or eacn (UTIL SOLES- (a) NH3 Conjugate acid Conjugate base (b) (CH3)2NH Conjugate acid Conjugate base chemPad XX→→ chemPad XX→ NH₂ NH 2 NH4 NH 4 Parts of the formula in your answer are transposed. Your answer appears to be missing a charge. chemPad XX→ (CH4)2NH2 (CH_4)_2NH_2 chemPad XX→= Greek (CH₂)2N (CH_2) 2N Help Greek Help Greek X Help Greek juur Helparrow_forwardWhich is the conjugate base in each of the pairs below?arrow_forward
- Arrange the following substances in order of increasing stability of conjugate base. 1. H₂Se II. H₂S III. H₂ Te || < ||| < | ||| < | < || | < ||| < || O II < | < || O I < || < ||| ||| < || < |arrow_forwardUsing pKa Values to Determine Relative Acidity and Basicity Rank the following compounds in order of increasing acidity, and then rank their conjugate bases in order of increasing basicity.arrow_forwardWhat is the increasing order of basicity among the following compounds ? I NH OH | < | < III < IV OIV < III < II < 1 I < III < II < IV OII < | < ||| < |V || < ||| < | < IV II NH III NH OH IV OH NHarrow_forward
- Rank the labeled protons (Ha~He) in order of increasing acidityarrow_forwardFor each of the following acid base reactions shown below, draw curved arrows to show proton transfer. Use ARIO to predict if the equilibrium lies to the left or right. Na OH Na но *carbonic acid H,co, has a pk, value of 6.3 Na HOarrow_forwardAde ed arrows to show the Arte attack the proton of the acid (breaking the X-H bond * Identify the nucleophile and electrophile ectants but the charge will be on a different molecule.) Circle the conjugate base. charges of any lons OK a) HO, b) кон SH c) NO. ONa HO + Approximate pK, values of Important organic functional groups pk, Protonated Cacborylic alcohol R. -2 5. 10 Phenol 16 Alcohol 25 Alkyne 38 Amine acid 44 Alkene 50 1. H OH R-OH R -H Alkane R. R. он R H ROH Protonated N amine H. H Thiol R-SHarrow_forward
- Organic ChemistryChemistryISBN:9781305580350Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. FootePublisher:Cengage LearningOrganic Chemistry: A Guided InquiryChemistryISBN:9780618974122Author:Andrei StraumanisPublisher:Cengage Learning