Fisher Esterification

From: ::Rose::
Date: 11 Jun 2008, 21:46

There are several ways of forming an ester. One very easy way is called the Fischer Esterification

Reactants:
1)Carboxylic acid (not normal acid like HCl, bcos we need the carbonyl group for this reaction)
2)EXCESS amount of alcohol (acts as nucleophile}
3)A few drops of strong acid e.g. sulphuric acid as a catalyst.

The by product are always always an ester and water.

Carboxylic acid + alcohol --> ester + water
R1-COOH + R2-OH --> R1-COOR2 + H2O

whereby R1 and R2 = alkyls, aryls, etc

The bonds which are broken are indicated by /

R1-CO/OH + R2-O/H --> products

KEYPOINTS:
1)The carbonyl group of the carboxylic acid always always remain intact, it never gets broken. Because double bond = very very strong.

2)For the alcohol, RO always always remain intact because the R group helps stabilise the lone pair on the oxygen. H in the otherhand is rubbish at stabilising, so it is always the one to go.


OOC: Do ask if you are unclear. Gosh I am revising! It's good though otherwise I woudnt remember this things!

>>Brilliant brilliant brilliant!