Claisen Condensation Mechanism

A Carbon-Carbon bond formation in a reaction in which a single ester and one carbonyl compound or two esters are being used for the organic coupling is known as the Claisen Schmidt's reaction. This reaction can only occur if the base is quite strong from the start of the reaction and the product is beta keto ester or beta-diketone. 

The Claisen reaction was named after the great chemist who first successfully performed this reaction Rainer Ludwig Claisen. Students find this Claisen condensation difficult as a lot is going on with the complex chemical compounds. That is why we planned to break it down into even stages and give proper theoretical knowledge as we move further down the reaction. Below we have provided you with how the reaction takes place along with its mechanism step by step. 

The reaction mechanism starts with the removal of an alpha proton as it reacts with a strong base, which results in the formation of an enolate ion.

Requirements of Claisen Ester Condensation

• For the reaction to occur, you need to have one reagent that can provide an alpha proton and help in the formation of an enolate anion when the deprotonation process occurs. 

• During the reaction, the base has to stay inactive and must not react to nucleophilic substitution reactions.

• When it comes to selecting an ideal base, we suggest you go for Sodium Alkoxide as it is a conjugate base of the alcohol.

• Lastly, the ester's Alkoxy part must act as an excellent leaving group when the reaction takes place to form ethyl and methyl esters. 

Claisen Schmidt Reaction Mechanism

Before we explain the Claisen condensation mechanism, we need students to recognize two units in this process. There are two portions of this reaction: nucleophilic (enolate) and the other is an electrophilic portion that can be found in carbonyl. 

After the reaction is complete nucleophilic enolate will still contain the ester unit, which is -CO2R. Simultaneously, the electrophilic ester will become ketone (C=O) as it loses the (-OR) group during the reaction.

Stage 1 

During the initial stage of the reaction, the protons get removed from a strong base; this causes the generation of an enolate ion. As the enolate ion has a negative charge delocalization, it becomes relatively more stable. In the image given below, you will see how the enolate ion is formed in this reaction. 

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(Formation of enolate ion.)

Stage 2 

Here the enolate ion, formed in the initial stage of the reaction, will start a nucleophilic attack on carbonyl carbon that belongs to the second ester reactant. This attack results in eliminating the alkoxy groups, and the conjugate base of alcohol is regenerated. 

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(Nucleophilic attack.)

In addition to this, the alkoxide ion, which is formed at this stage, will remove double alpha protons and bring in a new enolate anion, which is now being stabilized by resonance.

Stage 3

Now you need to take an aqueous acid. It could be phosphoric acid, or you can also use sulphuric acid. The acid will neutralize the negative charge, which is present in the enolate anion, along with the base, which is still present in the reaction. 

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(Removal of the leaving group.)

Lastly, the remaining group will be removed, and the Claisen Schmidt condensation mechanism gets completed. 

Difference Between Claisen Schmidt Condensation Mechanism & Aldol Condensation

This might be the biggest confusion you have for this reaction as the Claisen reaction is quite identical to the aldol condensation reaction. The main reason behind both of them to be similar is that both of these condensations are organic, and both involve the addition of enolates to organic compounds. 

The difference comes in the addition of enolates, which gets added in ketones or aldehydes, whereas if you look at the Claisen condensation, the enolates get added in esters.