g. Write a balanced equation for the reaction that takes place between the organic product from part f and NaOH(aq). h. Draw the name of the products that form when the acid from part a is heated in the presence of (CH3CH2)2NH. i. Write a balanced equation for the reaction that takes place when the organic product from part h is heated in the presence of H2O and H+.
g. Write a balanced equation for the reaction that takes place between the organic product from part f and NaOH(aq). h. Draw the name of the products that form when the acid from part a is heated in the presence of (CH3CH2)2NH. i. Write a balanced equation for the reaction that takes place when the organic product from part h is heated in the presence of H2O and H+.
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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g. Write a balanced equation for the reaction that takes place between the organic product from part f and NaOH(aq).
h. Draw the name of the products that form when the acid from part a is heated in the presence of (CH3CH2)2NH.
i. Write a balanced equation for the reaction that takes place when the organic product from part h is heated in the presence of H2O and H+.
![### Chemistry Problem 8.182
a) **Which acid in Table 8.6 is the strongest?**
- Definitions:
- \( K_a \) = acid dissociation constant
- \( pK_a \) = \(-\log K_a\)
- Given:
- Glycolic acid \( pK_a = 3.83 \)
- Lactic acid \( pK_a = 3.08 \)
- Conclusion:
- **Lactic acid is the strongest.**
b) **Name the conjugate base of this acid.**
- Reaction:
\[
\text{CH}_3\text{-CH-COOH} \xrightarrow{H^+} \text{CH}_3\text{-CH-}\dfrac{O^-}{\|}\text{C} \text{O} \]
- **Conjugate base of lactic acid:**
c) **Write the equation for the reaction that takes place between this acid and water.**
- Equation:
\[
\text{CH}_3\text{-CH-COOH} + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{-CH-}\dfrac{O^-}{\|}\text{C} + \text{H}_3\text{O}^+ \]
d) **Which is there more of at equilibrium, the acid in your answer or its conjugate base?**
- Explanation:
- Lactic acid is stronger than water.
- Water can easily accept \( H^+ \) from lactic acid, forming \(\text{CH}_3\text{-CH-}\dfrac{O^-}{\|}\text{C}\).
- **At equilibrium: Conjugate base is present in more amount.**
e) **Draw and name the salt that forms when the acid from part a reacts with KOH.**
- Reaction:
\[
\text{CH}_3\text{-CH-COOH} + \text{KOH} \rightarrow \text{CH}_3\text{-CH-}\dfrac{O^-}{\|}\text{C}\text{O}^-\text{K}^+ + \text{H}_2\text{O}
\]
- **](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F11097a2d-0fa1-4296-84d1-b894fe5eb4da%2Fb699afed-c059-4022-a081-3b3b84ac2f81%2Frsikr0d_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Chemistry Problem 8.182
a) **Which acid in Table 8.6 is the strongest?**
- Definitions:
- \( K_a \) = acid dissociation constant
- \( pK_a \) = \(-\log K_a\)
- Given:
- Glycolic acid \( pK_a = 3.83 \)
- Lactic acid \( pK_a = 3.08 \)
- Conclusion:
- **Lactic acid is the strongest.**
b) **Name the conjugate base of this acid.**
- Reaction:
\[
\text{CH}_3\text{-CH-COOH} \xrightarrow{H^+} \text{CH}_3\text{-CH-}\dfrac{O^-}{\|}\text{C} \text{O} \]
- **Conjugate base of lactic acid:**
c) **Write the equation for the reaction that takes place between this acid and water.**
- Equation:
\[
\text{CH}_3\text{-CH-COOH} + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{-CH-}\dfrac{O^-}{\|}\text{C} + \text{H}_3\text{O}^+ \]
d) **Which is there more of at equilibrium, the acid in your answer or its conjugate base?**
- Explanation:
- Lactic acid is stronger than water.
- Water can easily accept \( H^+ \) from lactic acid, forming \(\text{CH}_3\text{-CH-}\dfrac{O^-}{\|}\text{C}\).
- **At equilibrium: Conjugate base is present in more amount.**
e) **Draw and name the salt that forms when the acid from part a reacts with KOH.**
- Reaction:
\[
\text{CH}_3\text{-CH-COOH} + \text{KOH} \rightarrow \text{CH}_3\text{-CH-}\dfrac{O^-}{\|}\text{C}\text{O}^-\text{K}^+ + \text{H}_2\text{O}
\]
- **
Transcribed Image Text:**Exercise: Drawing and Naming Esterification Products**
**Task:**
- Draw and name the products that form when the acid from part a reacts with HOCH₂CH(CH₃)₂ and H⁺.
**Chemical Structures Provided:**
1. Reactants:
- \( \text{CH}_3 - \text{CH} - \text{C} - \text{OH} \)
- \( \left|\right. \)
- \( \text{OH} \)
- \( \text{HO-CH}_2-\text{CH(CH}_3)_2 \)
- \( \text{H}^+ \)
2. Intermediate:
- Two reactants shown interacting or combining under acidic conditions.
3. Product:
- \( \text{CH}_3 - \text{CH} - \text{C} - \text{O-CH}_2 - \text{CH(CH}_3)_2 \)
- \( \left|\right. \)
- \( \text{OH} \)
- \( \text{CH}_3 \)
**Name of the Ester:**
- 2-Methylpropyl 2-hydroxypropanoate
**Explanation of the Reaction:**
The task involves an esterification reaction. In this reaction, an acid and an alcohol react under acidic conditions to form an ester. The given reactants include a hydroxyl acid derivative and an alcohol, which leads to the formation of 2-Methylpropyl 2-hydroxypropanoate as the ester product.
**Diagrams and Explanation:**
- **Reactants:**
- The chemical structures indicate a secondary alcohol and an acid functionality.
- The acid's hydroxyl group (\( \text{OH} \)) is positioned on the central carbon.
- **Reaction Mechanism:**
- The reaction occurs in the presence of \( \text{H}^+ \), indicating an acid-catalyzed esterification.
- The hydroxyl group (\( \text{OH} \)) from the acid and the hydrogen from the alcohol form water, facilitating the ester linkage.
- **Product:**
- The ester product is formed by the joining of the acid and alcohol through the elimination of water (condensation).
- The naming correctly reflects the alcohol and acid parts contributing to the final ester name.
This reaction is a classic example of
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