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: Analyze the Product Structure: Product: 3-Hydroxy-2-methylpentanal. Structure: \( \text{C}_5(\text{CH}_3)-\text{C}_4(\text{H}_2)-\text{C}_3(\text{H})(\text{OH})-\text{C}_2(\text{H})(\text{CH}_3)-\text{C}_1(\text{H})(\text{O}) \) Wait, let's write linear: \( \text{CH}_3-\text{CH}_2-\text{CH}(\text{OH})-\text{CH}(\text{CH}_3)-\text{CHO} \).
Step 2: Retrosynthesis (Breaking the Aldol): The bond formed during Aldol condensation is between the \( \alpha \)-carbon of the nucleophile (which retains the carbonyl) and the \( \beta \)-carbon (which becomes the alcohol). Break the \( \text{C}_2-\text{C}_3 \) bond (bond between alpha and beta carbons). - Part 1 (Carbonyl side): \( -\text{CH}(\text{CH}_3)-\text{CHO} \). Add H to alpha carbon \(\to\) \( \text{CH}_2(\text{CH}_3)-\text{CHO} \) or \( \text{CH}_3\text{CH}_2\text{CHO} \). So, the nucleophile was Propanal. - Part 2 (Alcohol side): \( \text{CH}_3-\text{CH}_2-\text{CH}(\text{OH})- \). Convert C-OH back to C=O. \(\to\) \( \text{CH}_3-\text{CH}_2-\text{CHO} \). So, the electrophile was Propanal.
Step 3: Conclusion: Both reactants A and B are Propanal (\( \text{CH}_3\text{CH}_2\text{CHO} \)). This is a self-aldol condensation of Propanal.