Question

An ester (A) with molecular formula \(C_9H_{10}O_2\) was treated with excess \(CH_3MgBr\) and the compound so formed was treated with conc. \(H_2SO_4\) to form olefin (B). Ozonolysis of B gave ketone with formula \(C_8H_8O\) which shows positive iodoform test. The structure of A is :

• A. \(CH_3CH_2COC_6H_5\)
• B. \(C_6H_5COOC_2H_5\)
• C. \(C_6H_5COOC_6H_5\)
• D. \(CH_3COC_6H_4COCH_3\)

Hint:

Iodoform test → always look for \(CH_3CO-\) group (methyl ketone).

Answer 1

The Correct Option is B

Concept:
• Ester + excess Grignard reagent → tertiary alcohol
• Dehydration → alkene
• Ozonolysis → carbonyl compounds
• Iodoform test → presence of \(CH_3CO-\) group

Step 1: Identify final product. Ketone \(C_8H_8O\) gives iodoform test → must be acetophenone (\(C_6H_5COCH_3\)).

Step 2: Work backward (ozonolysis). Ozonolysis of alkene (B) gives acetophenone → alkene must contain \(C_6H_5-C=CH_2\) or \(C_6H_5-C=C-CH_3\).

Step 3: From Grignard reaction. Ester + excess \(CH_3MgBr\) gives tertiary alcohol: \[ C_6H_5COOC_2H_5 \xrightarrow{CH_3MgBr} C_6H_5C(OH)(CH_3)_2 \]

Step 4: Dehydration. \[ C_6H_5C(OH)(CH_3)_2 \xrightarrow{H_2SO_4} C_6H_5C(CH_3)=CH_2 \] This alkene on ozonolysis gives \(C_6H_5COCH_3\) (iodoform positive).

Step 5: Conclusion. Original ester is ethyl benzoate (\(C_6H_5COOC_2H_5\)).