3. Complete the following reactions by giving products in each case. KMnO4 1) LIAIH, 2) H3O* NH2OH "H*" V- H2O (excess) он - H2O PPH3 + Ph3P=O Ph = phenyl = C6H5-
3. Complete the following reactions by giving products in each case. KMnO4 1) LIAIH, 2) H3O* NH2OH "H*" V- H2O (excess) он - H2O PPH3 + Ph3P=O Ph = phenyl = C6H5-
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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![### Reaction Completion
For educational purposes, we will complete the following chemical reactions by identifying the products in each case.
#### Given Reactions:
**1.**
\[ \text{Reactant:} \quad \text{Unknown (Box 1)} \quad \xrightarrow{\text{KMnO}_4} \quad \text{Product:} \quad \text{4-oxopentanal (shown)}
\]
KMnO₄ is an oxidizing agent. To determine the unknown reactant, consider that KMnO₄ oxidation typically converts a primary alcohol to a carboxylic acid or an aldehyde. The shown product is 4-oxopentanal. Thus, the reactant was likely 4-hydroxy-pentanal. Putting this into the reaction:
\[
\text{4-Hydroxy-pentanal} \xrightarrow{\text{KMnO}_4} \text{4-oxopentanal}
\]
**2.**
\[ \text{Reactant:} \quad \text{4-oxopentanal (shown)} \quad \xrightarrow{\text{(excess)} \ \text{NH}_2 \text{OH}, \ \text{H}^+} \quad \text{Product:} \quad \text{Unknown (Box 3)}
\]
With the given reaction conditions, oximes are formed. Thus:
\[
\text{4-oxopentanal} \ + \ \text{NH}_2 \text{OH} \quad \xrightarrow{\text{H}^+} \quad 4-oxo-pentanal \ \text{oxime}
\]
**3.**
\[ \text{Reactant:} \quad \text{4-oxopentanal} \ \text{oxime} \quad \xrightarrow{\text{LiAlH}_4, \ \text{H}_3 \text{O}^+} \quad \text{Product:} \quad \text{Unknown (Box 4)}
\]
Lithium aluminium hydride \( \text{LiAlH}_4 \) reduces oximes to amines. Thus, the product is:
\[
\text{4-oxo-pentanal} \ \text{oxime} \quad \xrightarrow{\text{LiAlH}_4, \ \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F053a7dd4-6035-4ca1-b497-47ec6b123526%2Ffcd148cc-4b0c-43ed-938e-547f535f2389%2Fow9y2hi_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Reaction Completion
For educational purposes, we will complete the following chemical reactions by identifying the products in each case.
#### Given Reactions:
**1.**
\[ \text{Reactant:} \quad \text{Unknown (Box 1)} \quad \xrightarrow{\text{KMnO}_4} \quad \text{Product:} \quad \text{4-oxopentanal (shown)}
\]
KMnO₄ is an oxidizing agent. To determine the unknown reactant, consider that KMnO₄ oxidation typically converts a primary alcohol to a carboxylic acid or an aldehyde. The shown product is 4-oxopentanal. Thus, the reactant was likely 4-hydroxy-pentanal. Putting this into the reaction:
\[
\text{4-Hydroxy-pentanal} \xrightarrow{\text{KMnO}_4} \text{4-oxopentanal}
\]
**2.**
\[ \text{Reactant:} \quad \text{4-oxopentanal (shown)} \quad \xrightarrow{\text{(excess)} \ \text{NH}_2 \text{OH}, \ \text{H}^+} \quad \text{Product:} \quad \text{Unknown (Box 3)}
\]
With the given reaction conditions, oximes are formed. Thus:
\[
\text{4-oxopentanal} \ + \ \text{NH}_2 \text{OH} \quad \xrightarrow{\text{H}^+} \quad 4-oxo-pentanal \ \text{oxime}
\]
**3.**
\[ \text{Reactant:} \quad \text{4-oxopentanal} \ \text{oxime} \quad \xrightarrow{\text{LiAlH}_4, \ \text{H}_3 \text{O}^+} \quad \text{Product:} \quad \text{Unknown (Box 4)}
\]
Lithium aluminium hydride \( \text{LiAlH}_4 \) reduces oximes to amines. Thus, the product is:
\[
\text{4-oxo-pentanal} \ \text{oxime} \quad \xrightarrow{\text{LiAlH}_4, \ \
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