An Organic compound 'A' \(C_4H_8\) on treatment with \(KMnO_4/H^+\) yields compound 'B' \(C_3H_6O\). Compound 'A' also yields compound 'B' an ozonolysis. Compound 'A' is:
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If an oxidative cleavage of an alkene results in a product with fewer carbon atoms than the starting material, the double bond is likely terminal or the chain is branched. A \(C_4\) alkene yielding a \(C_3\) product indicates the loss of one carbon as \(CO_2\) or \(HCHO\).
Step 1: Understanding the Concept:
Alkenes undergo oxidative cleavage when treated with hot acidic \(KMnO_4\) or ozonolysis (\(O_3/Zn, H_2O\)). Oxidative cleavage of the \(C=C\) bond results in the formation of ketones or carboxylic acids (with \(KMnO_4/H^+\)) and ketones or aldehydes (with reductive ozonolysis). Step 2: Key Formula or Approach:
1. Molecular formula \(C_4H_8\) indicates a degree of unsaturation (DU) of 1. It can be an alkene or a cycloalkane.
2. Since it reacts with \(KMnO_4/H^+\), it must be an alkene.
3. Compound 'B' (\(C_3H_6O\)) has DU = 1, suggesting it is a ketone (acetone) or an aldehyde (propanal). Step 3: Detailed Explanation:
Let's analyze the options:
- But-2-ene (\(CH_3CH=CHCH_3\)): Oxidative cleavage would yield two molecules of ethanoic acid (\(C_2H_4O_2\)) or acetaldehyde (\(C_2H_4O\)). This does not match 'B'.
- 2-Methylpropene (\((CH_3)_2C=CH_2\)):
- Ozonolysis: Reductive ozonolysis cleaves the double bond to give Acetone (\(CH_3COCH_3\), \(C_3H_6O\)) and Formaldehyde (\(HCHO\)).
- Oxidative cleavage (\(KMnO_4/H^+\)): Cleavage gives Acetone (\(C_3H_6O\)) and Formic acid, which further oxidizes to \(CO_2\) and \(H_2O\).
This matches both experimental results described in the question. Step 4: Final Answer:
Compound 'A' is 2-methylpropene because its cleavage consistently yields the \(C_3\) compound acetone (\(C_3H_6O\)).
