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Ethanedioic acid, HO2C2O2H, is reduced using an excess of lithium aluminium hydride, LiAlH4 - CIE - A-Level Chemistry - Question 28 - 2018 - Paper 1
Question 28
Ethanedioic acid, HO2C2O2H, is reduced using an excess of lithium aluminium hydride, LiAlH4. What is the organic product of the reaction? A. ethanol B. ethane-1,2-d... show full transcript
Worked Solution & Example Answer:Ethanedioic acid, HO2C2O2H, is reduced using an excess of lithium aluminium hydride, LiAlH4 - CIE - A-Level Chemistry - Question 28 - 2018 - Paper 1
Step 1
What is the organic product of the reaction?
Answer
When ethanedioic acid (also known as oxalic acid) undergoes reduction using lithium aluminium hydride (LiAlH4), it typically reacts to form ethane-1,2-diol. This is because LiAlH4 is a strong reducing agent that can reduce carboxylic acids to alcohols.
To summarize the reaction pathway:
- Initial reaction with LiAlH4 reduces the carboxylic acid groups ( -COOH) to alcohol groups (-OH).
- In the case of ethanedioic acid, both carboxylic groups are reduced, leading to the formation of ethane-1,2-diol.
Thus, the correct answer is B. ethane-1,2-diol.