What are X and P in the following reaction sequence?
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Remember:
1. Cannizzaro reaction of a dialdehyde yields a hydroxy-acid salt.
2. Treatment of a 1,2-hydroxy-acid with acid ($\text{HCl}$) always drives intramolecular esterification to yield a cyclic ester, known as a lactone.
3. A lactone contains a carbonyl ($C=\text{O}$) and a ring oxygen, immediately eliminating hemiacetal structures like those in (b) and (d).
Step 1: Understanding the Question:
This question asks us to identify the chemical intermediates and products in a two-step reaction sequence starting from an ortho-phthalaldehyde derivative where one of the aldehyde groups is deuterated ($-\text{CDO}$) and the other is a standard aldehyde ($-\text{CHO}$). Step 2: Detailed Explanation:
Let's analyze each reaction step-by-step:
• Step 1: Intramolecular Cannizzaro Reaction:
- The starting material is a phthalaldehyde derivative containing both $-\text{CHO}$ and $-\text{CDO}$ groups.
- Treatment with concentrated sodium hydroxide ($\text{NaOH}$) under heat ($\Delta$) initiates an intramolecular Cannizzaro reaction.
- In this process, one aldehyde group undergoes nucleophilic attack by $\text{OH}^-$, forming a tetrahedral intermediate. This intermediate then transfers a hydride (or deuteride) ion to the adjacent aldehyde carbonyl carbon.
- Because the C-H bond is weaker than the C-D bond, hydride ($\text{H}^-$) transfer from the $-\text{CHO}$ group is kinetically favored over deuteride ($\text{D}^-$) transfer.
- Specifically, $\text{OH}^-$ attacks the $-\text{CHO}$ carbonyl carbon to form a tetrahedral intermediate. This intermediate transfers a hydride ($\text{H}^-$) to the adjacent $-\text{CDO}$ carbonyl carbon.
- This reduces the $-\text{CDO}$ group to a $-\text{CH(O}^-)\text{D}$ group, which protonates to form a deuterated alcohol group: $-\text{CH(OH)D}$.
- The original $-\text{CHO}$ group is oxidized to a carboxylate salt: $-\text{COONa}$.
- Thus, intermediate X consists of a benzene ring with a $-\text{COONa}$ group at the top position and a $-\text{CH(OH)D}$ group at the bottom position. This matches the structure of X in option (a).
• Step 2: Acid-Catalyzed Lactonization:
- Intermediate X is treated with excess hydrochloric acid gas ($\text{HCl}$).
- The acid protonates the sodium carboxylate ($-\text{COONa}$) to form a carboxylic acid ($-\text{COOH}$).
- Because the carboxylic acid group and the alcohol group are located on adjacent positions of the benzene ring, they undergo a rapid, spontaneous intramolecular esterification (cyclization) to form a stable 5-membered lactone ring (phthalide derivative).
- The oxygen of the $-\text{CH(OH)D}$ group attacks the carbonyl carbon of the $-\text{COOH}$ group, eliminating a water molecule.
- This forms product P, which is a lactone containing a carbonyl ($C=\text{O}$) group at the top position and a $-\text{CHD}-$ group at the bottom position of the heterocyclic ring.
- This perfectly matches the structure of P in option (a). Step 3: Final Answer:
The intermediate X and product P are correctly depicted in option (a), so the correct option is (A).
