Final answer:
LiAlH4 reduces an ester into two alcohols, as it is a strong reducing agent that breaks ester bonds.
Step-by-step explanation:
When an ester is treated with LiAlH4, a strong reducing agent, the ester undergoes reduction. Unlike alkaline hydrolysis, which would lead to the formation of an alcohol and a carboxylate salt, the reduction by LiAlH4 breaks both ester bonds and fully reduces the ester to two alcohols.
This process is part of a broader category of reactions involving the reduction of carboxylic acid derivatives.
In the presence of LiAlH4, an ester undergoes reduction, resulting in the formation of an alcohol and a corresponding compound. LiAlH4 is a strong reducing agent that donates a hydride ion (H-) to the carbonyl carbon of the ester, leading to the formation of an alkoxide ion.
The alkoxide ion then reacts with a proton source (such as water) to generate the alcohol and the conjugate base of the carboxylic acid.