Hydrocarbon (P) on reductive ozonolysis gives products which give positive iodoform test and on acidification produces the compound shown. Identify the structure of (P).
Hydrocarbon (P) on reductive ozonolysis gives products which give positive iodoform test and on acidification produces the compound shown. Identify the structure of (P).
- Structure (1)
- Structure (2)
- Structure (3)
- Structure (4)
The Correct Option is A
Solution and Explanation
Concept:
Ozonolysis cleaves carbon–carbon double bonds to form carbonyl compounds. Reductive ozonolysis (\(O_3/Zn/H_2O\)) converts alkenes into aldehydes and ketones. If one of the resulting products contains a methyl ketone (\(-COCH_3\)), it gives a positive iodoform test. 
Step 1: Ozonolysis reaction. \[ \text{Alkene} \xrightarrow{O_3/Zn,H_2O} \text{Carbonyl compounds} \] The ozonolysis products include a methyl ketone which gives a positive iodoform test. Step 2: Iodoform test condition. Compounds containing the group \[ -COCH_3 \] give yellow precipitate of \(CHI_3\). Step 3: Formation of intermediate products. The ozonolysis of structure (1) produces carbonyl compounds capable of undergoing aldol condensation under basic conditions. \[ \text{Carbonyl compound} \xrightarrow{OH^- / \Delta} \text{Aldol product} \] Step 4: Acidification step. After iodoform reaction: \[ RCOCH_3 \xrightarrow{I_2/NaOH} RCOO^- + CHI_3 \] On acidification: \[ RCOO^- \xrightarrow{H^+} RCOOH \] The final product shown in the question matches the product obtained from alkene structure (1). Step 5: Conclusion. \[ \boxed{\text{Structure (1) is the correct hydrocarbon}} \]
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