An organic compound "x" where molar ratio of C, O and H are equal, on treatment with 50% KOH under reflux followed by acidification produced "y". The most likely structure of "y" is:
Options
- 1
- 2
- 3
- 4
The Correct Option is B
Solution and Explanation
Step 1: Reactivity with KOH.
KOH will break down a carbonyl compound into simpler products. The molar ratio of C, O, and H indicates the intermediate structure is likely an alkene with a hydroxyl group.
Step 2: Acidification leads to product formation.
After treatment with acid, the likely structure will be an aldehyde or ketone.
Final Answer: \[ \boxed{\text{(B) } CH_3-CH-CH=O} \]
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