Question:medium

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 :

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Iodoform test → always look for \(CH_3CO-\) group (methyl ketone).
Updated On: Apr 18, 2026
  • \(CH_3CH_2COC_6H_5\)
  • \(C_6H_5COOC_2H_5\)
  • \(C_6H_5COOC_6H_5\)
  • \(CH_3COC_6H_4COCH_3\)
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The Correct Option is B

Solution and Explanation

Step 1: Understanding the Concept:
This is a multi-step organic transformation problem involving Grignard reaction on esters, dehydration of alcohols, ozonolysis of alkenes, and the iodoform test.
Step 2: Detailed Explanation:
1. Analysis of the Ketone: The ketone has formula \( \text{C}_{8}\text{H}_{8}\text{O} \) and gives a positive iodoform test. A positive iodoform test implies the presence of a methyl ketone (\( \text{R-CO-CH}_{3} \)) group. The only \( \text{C}_{8}\) methyl ketone is acetophenone (\( \text{C}_{6}\text{H}_{5}\text{COCH}_{3} \)).
2. Retrosynthesis of Olefin (B): Acetophenone is obtained by ozonolysis of B. This suggests B is an alkene where one side of the double bond is \( \text{C}_{6}\text{H}_{5}\text{C(CH}_{3}\text{)=} \).
3. Formation of Alcohol: B was formed by dehydrating a tertiary alcohol produced by reacting ester A with excess \( \text{CH}_{3}\text{MgBr} \).
4. Structure of A: An ester \( \text{C}_{6}\text{H}_{5}\text{COOR} \) reacts with 2 moles of \( \text{CH}_{3}\text{MgBr} \) to give \( \text{C}_{6}\text{H}_{5}\text{C(OH)(CH}_{3}\text{)}_{2} \).
- Molar formula check: \( \text{C}_{6}\text{H}_{5}\text{COOC}_{2}\text{H}_{5} \) is \( \text{C}_{9}\text{H}_{10}\text{O}_{2} \). This matches the given formula.
- Reaction: Ethyl benzoate \( + 2 \text{CH}_{3}\text{MgBr} \to 2\text{-phenylpropan-2-ol} \).
- Dehydration: \( 2\text{-phenylpropan-2-ol} \xrightarrow{\text{H}^{+}} \text{2-phenylpropene} \).
- Ozonolysis: \( \text{2-phenylpropene} \xrightarrow{\text{O}_{3}/\text{Zn,H}_{2}\text{O}} \text{Acetophenone } (\text{C}_{8}\text{H}_{8}\text{O}) + \text{HCHO} \).
- Acetophenone gives a positive iodoform test.
Step 3: Final Answer:
Structure A is Ethyl benzoate (\( \text{C}_{6}\text{H}_{5}\text{COOC}_{2}\text{H}_{5} \)).
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