THE REDUCTION OF ALDEHYDES AND KETONES

THE REDUCTION OF ALDEHYDES AND KETONES This page gives you the facts and mechanisms for the reduction of carbonyl compounds (specifically aldehydes and ketones) using sodium tetrahydridoborate (sodium borohydride) as the reducing agent. Only one UK A level Exam Board (AQA) is likely to ask for these mechanisms, and they are happy with a simplified version of what is quite a complex mechanism. Because of that simplification, these reactions are dealt with entirely on this page - without the "talk through" page that you will find for other mechanisms on this site. The reduction of aldehydes and ketones by sodium tetrahydridoborate The facts Sodium tetrahydridoborate (previously known as sodium borohydride) has the formula NaBH4, and contains the BH4- ion. That ion acts as the reducing agent. There are several quite different ways of carrying out this reaction. Two possible variants (there are several others!) are: The reaction is carried out in solution in water to which some sodium hydroxide has been added to make it alkaline. The reaction produces an intermediate which is converted into the final product by addition of a dilute acid like sulphuric acid. The reaction is carried out in solution in an alcohol like methanol, ethanol or propan-2-ol. This produces an intermediate which can be converted into the final product by boiling it with water. In each case, reduction essentially involves the addition of a hydrogen atom to each end of the carbon-oxygen double bond to form an alcohol. Reduction of aldehydes and ketones lead to two different sorts of alcohol. The reduction of an aldehyde For example, with ethanal you get ethanol: Notice that this is a simplified equation - perfectly acceptable to examiners. The H in square brackets means "hydrogen from a reducing agent". In general terms, reduction of an aldehyde leads to a primary alcohol. A primary alcohol is one which only has one alkyl group attached to the carbon with the -OH group on it. They all contain the grouping -CH2OH.