Esters and the Fischer Esterification Rxn: Give the complete specific mechanism

Question

Esters and the Fischer Esterification Rxn:

Give the complete specific mechanism for the formation of one of your esters from the following lab steps:

For the ester made from propionic acid.
Here is how it was prepared in lab:
1. 0.1 mL of Propionic acid was added to test tube
2. 5 drops of H2SO4 was added to test tube
3. Solution heated in warm water bath, removed to cool after 10 min
4. 2mL of water and 1 mL of t-butyl methyl ether was added to test tube
5. 2 layers formed
6. water layer removed with pipet
7. 1 mL 5% aq. sodium bicarbonate added to test tube dropwise (slow)
8. lower aqueous layer removed
9 Small spatula of Na2SO4 added to remaining ester layer

Explanation / Answer

Solution:

First clue is the words ‘formation of ester’ in the question

Esterification is the chemical reaction between carboxylic acid and alcohol resulting in the formation of ester. The general word equation is

carboxylic acid +alcohol ester+ water

Hence for an ester to be formed we need a carboxylic acid which is propionic acid(CH3CH2COOH) in this case.

In the lab experiment no alcohol is taken, but ether is taken. This means this ether is converted into an alcohol. We know that t-Butyl methyl ether (MTBE) is generally considered to be resistant to chemical transformation in aqueous solution. MTBE is sensitive to acid-catalyzed hydrolysis reaction. If it is hydrolyzed there are two possible alcohols which can be formed.

t-Butyl methyl ether(C4H9-O-CH3) C4H9-OH+CH3.HSO4   -------- 1

                                    or

                        (C4H9-O-CH3) C4H9.HSO4-OH+CH3-OH -----------2

In dilute aqueous the sensitive MTBE generates tert-butyl alcohol (TBA) and methanol as products.

Another clue in the question is give- specific mechanism for the formation of one of your esters

Hence it is clear that two esters CH3CH2COO C4H9 and CH3CH2COO CH3 are formed.

Mechanism:

The tertiary carbons are much too hindered for a backside attack. However, tertiary carbocations are relatively stable – and “ionization” (i.e. loss of a leaving group) leaves us with an alcohol (R-OH) and a tertiary carbocation, which can then be attacked by iodide ion to give R-HSO4

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