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-

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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, \ \
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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