Question

Hydrocarbon (P) on reductive ozonolysis gives products which give positive iodoform test and on acidification produces the compound shown. Identify the structure of (P). 

• A. Structure (1)
• B. Structure (2)
• C. Structure (3)
• D. Structure (4)

Answer 1

The Correct Option is A

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. \[ \boxed{\text{Structure (1) is the correct hydrocarbon}} \]