### Understanding Reaction Products - Educational Guide#### Given Reaction:The products of the following reaction are:\[ \text{CH}_3\text{CH}_2\text{C} \doteq CO \cdot \text{O-CH}_2\text{CH}_3 + \text{NaOH} \rightarrow \]#### Choices:1. \( \text{CH}_3\text{CH}_2\text{C} \dot{=}{\displaystyle CO^- \cdot \text{Na}^+} + \text{CH}_3\text{CH}_2\text{OH} \)2. \( \text{CH}_3\text{CH}_2\text{C} \dot{=}{\displaystyle COH} + \text{CH}_3\text{CH}_2\text{OH} \)3. \( \text{CH}_3\text{CH}_2\text{C} \dot{=}{\displaystyle COH} + \text{H}_3\text{C-CH}_2\text{O}^-\text{Na}^+ \)4. \( \text{CH}_3\text{CH}_2\text{C} \dot{=}{\displaystyle CO^- \cdot \text{Na}^+} + \text{CH}_3\text{CH}_2\text{O}^- \text{Na}^+ \)#### Detailed Explanation:The reaction in question is an example of a nucleophilic acyl substitution where an ester reacts with a strong base, in this case, NaOH, through a process known as saponification.- **Saponification:** This is the hydrolysis of an ester under basic conditions to produce a carboxylate salt and an alcohol. For the provided reaction: \[ \text{CH}_3\text{CH}_2\text{C} \doteq CO \cdot \text{O-CH}_2\text{CH}_3 + \text{NaOH} \rightarrow \] - The NaOH disassociates into Na+ and OH- ions. - The OH- ion acts as a nucleophile and attacks the carbonyl carbon (C=O) of the ester, breaking the ester bond. - This results in the formation

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Answered: The products of the following reaction…

The products of the following reaction are: CH;CH, C-0-CH,CH; + NAOH CH;CH,C-O Na* + CH;CH,OH CH;CH,C-OH + CH,CH,OH CH;CH, C-OH + CH;CH,O Na* CH,CH, C-O' Na* + CH,CH,0 Na*

Chemistry: The Molecular Science

5th Edition

ISBN:9781285199047

Author:John W. Moore, Conrad L. Stanitski

Publisher:John W. Moore, Conrad L. Stanitski

Chapter10: Fuels, Organic Chemicals, And Polymers

Section: Chapter Questions

Problem 121QRT

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### Understanding Reaction Products - Educational Guide

#### Given Reaction:
The products of the following reaction are:

\[ \text{CH}_3\text{CH}_2\text{C} \doteq CO \cdot \text{O-CH}_2\text{CH}_3 + \text{NaOH} \rightarrow \]

#### Choices:
1. \( \text{CH}_3\text{CH}_2\text{C} \dot{=}{\displaystyle CO^- \cdot \text{Na}^+} + \text{CH}_3\text{CH}_2\text{OH} \)

2. \( \text{CH}_3\text{CH}_2\text{C} \dot{=}{\displaystyle COH} + \text{CH}_3\text{CH}_2\text{OH} \)

3. \( \text{CH}_3\text{CH}_2\text{C} \dot{=}{\displaystyle COH} + \text{H}_3\text{C-CH}_2\text{O}^-\text{Na}^+ \)

4. \( \text{CH}_3\text{CH}_2\text{C} \dot{=}{\displaystyle CO^- \cdot \text{Na}^+} + \text{CH}_3\text{CH}_2\text{O}^- \text{Na}^+ \)

#### Detailed Explanation:
The reaction in question is an example of a nucleophilic acyl substitution where an ester reacts with a strong base, in this case, NaOH, through a process known as saponification.

- **Saponification:** This is the hydrolysis of an ester under basic conditions to produce a carboxylate salt and an alcohol.

For the provided reaction:
\[ \text{CH}_3\text{CH}_2\text{C} \doteq CO \cdot \text{O-CH}_2\text{CH}_3 + \text{NaOH} \rightarrow \]
- The NaOH disassociates into Na+ and OH- ions.
- The OH- ion acts as a nucleophile and attacks the carbonyl carbon (C=O) of the ester, breaking the ester bond.
- This results in the formation

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