Question

Ch3COOCH3+LiAlH4 then H2O

Answer 1

The products are Ethanol and Methanol.

$\text{CH}_3\text{CH}_2\text{OH} + \text{CH}_3\text{OH}$

Answer 2

• Reaction Type: Reduction of Ester
• Reagent: Lithium Aluminum Hydride (LiAlH4) followed by hydrolysis (H2O)
• Substrate: Methyl Acetate (CH3COOCH3)

Mechanism:

1. Hydride Attack: The hydride ion (H-) from LiAlH4 attacks the electrophilic carbonyl carbon of methyl acetate. This forms a tetrahedral intermediate.

   \begin{blockarray}{ccc} \text{O} \\ \text{||} \\ \text{CH}_3 \text{—C—OCH}_3 & \xrightarrow{\text{LiAlH}_4} & \begin{blockarray}{c} \text{O}^- \\ \text{|} \\ \text{CH}_3 \text{—C—OCH}_3 \\ \text{|} \\ \text{H} \\ \end{blockarray} \end{blockarray}

2. Elimination: The tetrahedral intermediate collapses, expelling the methoxide ion (CH3O-) as a leaving group, and forming an aldehyde (acetaldehyde).

   \begin{blockarray}{c} \text{O}^- \\ \text{|} \\ \text{CH}_3 \text{—C—OCH}_3 \\ \text{|} \\ \text{H} \\ \end{blockarray} \quad \longrightarrow \quad \begin{blockarray}{cc} \text{O} \\ \text{||} \\ \text{CH}_3 \text{—C—H} & + \quad \text{CH}_3 \text{O}^- \\ \end{blockarray}

3. Second Hydride Attack: The newly formed aldehyde (acetaldehyde) is immediately reduced by another hydride ion from LiAlH4, forming an alkoxide.

   \begin{blockarray}{cc} \text{O} \\ \text{||} \\ \text{CH}_3 \text{—C—H} & \xrightarrow{\text{LiAlH}_4} & \begin{blockarray}{c} \text{O}^- \\ \text{|} \\ \text{CH}_3 \text{—C—H} \\ \text{|} \\ \text{H} \\ \end{blockarray} \quad = \quad \text{CH}_3 \text{CH}_2 \text{O}^- \\ \end{blockarray}

4. Hydrolysis (Workup): In the presence of water (H2O), the alkoxide and the methoxide ions are protonated to yield the corresponding alcohols.

   $$\text{CH}_3 \text{CH}_2 \text{O}^- + \text{H}_2 \text{O} \longrightarrow \text{CH}_3 \text{CH}_2 \text{OH} \quad (\text{Ethanol}) + \text{OH}^-$$ $$\text{CH}_3 \text{O}^- + \text{H}_2 \text{O} \longrightarrow \text{CH}_3 \text{OH} \quad (\text{Methanol}) + \text{OH}^-$$

Final Products: The products of the reaction are ethanol ($\text{CH}_3\text{CH}_2\text{OH}$) and methanol ($\text{CH}_3\text{OH}$).

Explanation:

Methyl acetate (an ester) is reduced by lithium aluminum hydride (LiAlH4), a strong reducing agent. Esters are cleaved and reduced to two primary alcohols. The acyl part (CH3CO-) is reduced to ethanol (CH3CH2OH), and the alkoxy part (-OCH3) is reduced to methanol (CH3OH). The reaction proceeds through an aldehyde intermediate which is further reduced. Subsequent hydrolysis with water protonates the alkoxide intermediates to yield the alcohols.