Reaction Mechanism of NaOH and MeOH

Question

How exactly would I draw the reaction mechanism of NaOH and MeOH (in regards to transesterification, when preparing biodiesel from cooking oil)?

Answer 1

In a transesterification reaction, one ester is converted to another with the help of an R-OH molecule. NaOH and MeOH can be used in a base-mediated transesterification reaction. NaOH is used as a base and MeOH is used as a nucleophile.

1) To initiate the reaction, the highly-basic hydroxide component (OH^–) of sodium hydroxide (NaOH) is used to remove a proton (H⁺) from MeOH. Now, you'll be left with H₂O (from adding the H⁺ to OH^–) and MeO^–.

2) MeO^– is added to the carbonyl carbon (C=O) of your ester (RCOOR′) in a nucleophilic substitution (S_N2) reaction. The original carbonyl carbon is now single-bonded to 3 different oxygen atoms: (i) the original carbonyl oxygen (C=O), which now bears a negative charge (O^–); (ii) the MeO^– oxygen, which is now neutrally-charged; and (iii) the OR' oxygen from the ester (this bond has remained unchanged thus far).

3) Next, the carbonyl reforms. The pi electrons on the first (i) oxygen atom mentioned in step 2) reform the double bond. This reforms the carbonyl (C=O) on the ester. As a result, the carbonyl carbon has one too many bonds. So, the original OR' group is eliminated from the molecule. What is left is a new ester: RCOOMe. The original OR' group has been replaced with OMe.