Carboxylic acids and related compounds
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When an alcohol and a carboxylic acid form an ester, my teacher says you lose the OH group from the carboxylic acid and the hydrogen from the OH group from the alcohol. The text book says that the reaction attacks the C to O double bond in the carboxylic acid. Which one is it please?
Your two sources of information are talking about different aspects of the reaction. Your teacher is saying that the overall effect of the reaction is that the –OH group of the acid is replaced by the group –OR from an alcohol – in other words that the C–O bond of the acid breaks and not the O–H bond of the acid. The evidence for this is the nice piece of work by Roberts and Urey involving the use of O18 which I expect you know about.
The book is going into the detailed mechanism of the reaction, the first step of which is protonation of the carbonyl oxygen atom of the acid. This is then followed by attack by the lone pair of electrons on the oxygen atom of the alcohol’s –OH group on what was the carbonyl carbon atom of the acid. The resulting complicated ion then loses firstly a water molecule and then a proton giving the ester.
Your textbook will describe the mechanism in the section on esterification – probably in the chapter on carboxylic acids or esters. If you have access to the Nuffield Advanced Chemistry Students’ Book (4th edition) you will find both approaches described on p365.
Ulex
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updated: 26 April 2005
