Question

The mechanism of $\beta$-keto acid decarboxylation is shown below: Heating of a single enantiomer of a $\beta$-keto acid (1) will afford a decarboxylated product that is:

A. a single monomer
B. diastereomer
C. achiral
D. racemic

Answer 1

Hint: For this problem, we have to study the mechanism of decarboxylation as shown in the question carefully. With the help of this mechanism, we can find the product which will be formed after the decarboxylation of the enantiomer of a $\beta$-keto acid.

Complete step by step answer:
- in the given question, we have to find the correct product which will be formed by the decarboxylation of the enantiomer of a $\beta$-keto acid.
- As we know that decarboxylation is the process of removal of carboxylic acid from the reactant.
- So, as we can see that firstly the double bond of oxygen with the main chain breaks and attaches the hydrogen from the carboxylic group with it to form an alcohol group.
- Meanwhile, the carbon in -COOH also breaks the bond between carbon in the main chain and separates as a carbon dioxide gas.
- Due to which the double bond is formed between 2 and 3 carbon and the product formed is known as but - 2 - ene - 3 - ol.
- Now, in the second reaction, the same mechanism will take place in which the double bond of oxygen with the main chain breaks and attaches the hydrogen from the carboxylic group with it to form an alcohol group.
- And similarly then the carbon in -COOH also breaks the bond between carbon in the main chain and separates as a carbon dioxide gas as shown below:

- So, here the product formed is racemic because it consists of the two enantiomers one in the enol form and one in ketone form.
So, the correct answer is “Option D”.

Note: In the given problem, enantiomers are considered as chiral molecules which have a mirror image of one another. Whereas chiral molecules are those molecules which cannot be superimposed on each other.