Carboxylic acids are reduced to primary alcohols by lithium tetrahydridoaluminate(III), LiAlH4.Conditions: LiAlH4 in dry ether at room temperature, followed by the careful addition of dilute acid. The ether must be dry because LiAlH4 reacts violently with water.The COOH group is reduced to CH2OH. The reaction passes through an aldehyde, but the aldehyde is reduced more easily than the acid, so it cannot be isolated and the product is the primary alcohol. Using [H] for the reducing agent:RCOOH + 4[H] → RCH2OH + H2O Four [H] are needed: two to reduce the acid to the aldehyde (forming water) and two more to reduce the aldehyde to the alcohol.Carboxylic acids are harder to reduce than aldehydes and ketones. Sodium tetrahydridoborate(III), NaBH4, reduces aldehydes and ketones but does not reduce carboxylic acids, so the more powerful LiAlH4 is needed.The number of carbon atoms does not change: propanoic acid gives propan-1-ol, and benzoic acid gives phenylmethanol, C6H5CH2OH.Worked example: ethanoic acid is reduced with LiAlH4:
CH3COOH + 4[H] → CH3CH2OH + H2O
6.00 g of ethanoic acid (Mr = 60.0) is 0.100 mol. The ratio is 1 : 1, so the maximum amount of ethanol is 0.100 mol, which is 0.100 × 46.0 = 4.60 g.
6.00 g of ethanoic acid (Mr = 60.0) is 0.100 mol. The ratio is 1 : 1, so the maximum amount of ethanol is 0.100 mol, which is 0.100 × 46.0 = 4.60 g.