Question

Identify the compounds A and B involved in the formation of given aldol

• A.

  

• B.

  

• C.

  

• D.

  

Hint:

For retro-aldol analysis of an aldol addition product: 1. Locate the carbonyl and the \(\beta\)-hydroxy group. 2. Break the bond between the \(\alpha\) and \(\beta\) carbons. 3. The \(\beta\)-hydroxy fragment becomes one carbonyl reactant (oxidize the -CH(OH)- group to -CHO or -CO-). 4. The \(\alpha\)-carbonyl fragment becomes the other reactant (add a H to the \(\alpha\)-carbon).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The problem shows the product of an aldol addition reaction and asks to identify the reactant aldehydes/ketones (A and B). The aldol addition is a reaction where an enolate ion reacts with a carbonyl compound to form a \(\beta\)-hydroxy aldehyde or \(\beta\)-hydroxy ketone (an aldol). The reaction can be reversed conceptually to find the starting materials. This process is called retro-aldol analysis.
Step 2: Retro-Aldol Analysis:
The product is: \( \text{CH}_3-\text{CH}_2-\overset{3}{\underset{\text{OH}}{\underset{|}{C}}}\text{H}-\overset{2}{\underset{\text{CH}_3}{\underset{|}{C}}}\text{H}-\overset{1}{\text{C}}\text{HO} \)
1. Identify the \(\alpha\) and \(\beta\) carbons relative to the carbonyl group. The CHO is C1, so C2 is the \(\alpha\)-carbon and C3 is the \(\beta\)-carbon.
2. The new C-C bond formed in the aldol reaction is always the bond between the \(\alpha\) and \(\beta\) carbons.
3. To find the reactants, mentally break the \(\alpha-\beta\) bond (the C2-C3 bond).
4. The part of the molecule containing the \(\beta\)-carbon (and the -OH group) was originally the electrophile (the carbonyl acceptor). Add a carbonyl group back to the \(\beta\)-carbon (C3) by removing the -OH group and a hydrogen from the same carbon, forming a C=O double bond.
- Fragment: \( \text{CH}_3-\text{CH}_2-\text{CH(OH)}- \) \(\rightarrow\) \( \text{CH}_3-\text{CH}_2-\text{CHO} \) (Propanal) 5. The part of the molecule containing the \(\alpha\)-carbon was originally the nucleophile (the enolate). Add a hydrogen to the \(\alpha\)-carbon (C2) to make it a neutral carbonyl compound.
- Fragment: \( -\text{CH(CH}_3\text{)-CHO} \) \(\rightarrow\) \( \text{H-CH(CH}_3\text{)-CHO} \) which is \( \text{CH}_3\text{-CH}_2\text{-CHO} \) (Propanal)
Both fragments lead back to the same molecule: propanal (CH₃CH₂CHO).
Step 3: Conclusion:
This means the reaction was a self-condensation of propanal. Both A and B are propanal.
- Molecule A (provides enolate): Propanal
- Molecule B (acts as electrophile): Propanal
This corresponds to option (C).
Step 4: Final Answer:
Both compounds A and B are propanal (CH₃CH₂CHO). Therefore, option (C) is correct.