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Ph-underset((P))underset(OH)underset(|)(...

`Ph-underset((P))underset(OH)underset(|)(CH)-overset(O)overset(||)C-Hunderset(H_(2)O)overset(HO^(-))rarrQ.P`and`Q` are isomers. Identify `Q`

A

`Ph-CH_2-overset(O)overset(||)C-CH`

B

`Ph-overset(O)overset(||)C-OCH_3`

C

`Ph-overset(O)overset(||)C-CH_2OH`

D

`H-overset(O)overset(||)C-CH_2-O-Ph`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to identify the isomers Q from the given compound P, which is an aldehyde. The reaction involves the presence of hydroxide ions (OH-) and water (H2O). ### Step-by-Step Solution: 1. **Identify Compound P**: The compound P is given as an aldehyde with the structure: \[ \text{P: } \text{H}_2\text{C=O} \text{ (aldehyde)} \] We can represent it as CH3CHO (acetaldehyde). 2. **Reaction with Hydroxide Ion**: When the aldehyde reacts with hydroxide ions (OH-) in the presence of water, it undergoes a reaction known as an aldol condensation. The hydroxide ion acts as a base and deprotonates the aldehyde, forming an enolate ion. 3. **Formation of Enolate Ion**: The enolate ion is formed by the abstraction of a proton from the alpha carbon of the aldehyde: \[ \text{Enolate: } \text{CH}_2\text{=C(OH)-C(=O)-H} \] 4. **Nucleophilic Attack**: The enolate ion then attacks another molecule of the aldehyde, leading to the formation of a β-hydroxy aldehyde. 5. **Dehydration**: The β-hydroxy aldehyde can then undergo dehydration (loss of water) to form an α,β-unsaturated carbonyl compound: \[ \text{Q: } \text{C=C(OH)-C(=O)-H} \] 6. **Final Structure of Q**: The final structure of Q after the dehydration step is: \[ \text{Q: } \text{CH}_2=CH-C(=O)-H \] This compound is an isomer of P. ### Conclusion: The isomer Q is an α,β-unsaturated aldehyde, specifically: \[ \text{Q: } \text{CH}_2=CH-C(=O)-H \]
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