Question

What are X and Y respectively, in the following set of reactions ? \[ \text{CH}_3\text{CH}_3 + 3\text{O}_2 \xrightarrow{\text{(CH}_3\text{COO})_2\text{Mn, }\Delta} \text{X} \] \[ \text{CH}_3\text{CH}=\text{CH}_2 \xrightarrow{\text{KMnO}_4/\text{H}^+} \text{Y} \]

• A. \( \text{CH}_3\text{COOH, CH}_3\text{CH(OH)CH(OH)CH}_3 \)
• B. \( \text{CH}_3\text{CH}_2\text{OH, CH}_3\text{CHO} \)
• C. \( \text{CH}_3\text{CHO, CH}_3\text{COOH} \)
• D. \( \text{CH}_3\text{COOH, CH}_3\text{COOH} \)

Hint:

For oxidation reactions: \begin{itemize} \item Controlled oxidation of alkanes (e.g., with Mn(OAc)\(_2\)) often leads to carboxylic acids. \item Hot, acidic KMnO\(_4\) causes oxidative cleavage of C=C bonds. Terminal =CH\(_2\) groups become CO\(_2\), and =CH- groups become -COOH. \end{itemize}

Answer 1

The Correct Option is D

Step 1: Determine Product X from the first reaction.
The first reaction is the oxidation of ethane (CH₃CH₃) with oxygen (O₂) in the presence of manganese acetate ((CH₃COO)₂Mn) and heat ($\Delta$). This is a controlled catalytic oxidation.
Under these conditions, alkanes can be selectively oxidized to carboxylic acids. For ethane, the controlled oxidation leads to the formation of acetic acid. \[ \text{CH}_3\text{CH}_3 + 3\text{O}_2 \xrightarrow{\text{(CH}_3\text{COO})_2\text{Mn, }\Delta} \text{CH}_3\text{COOH} + \text{H}_2\text{O} \] Thus, X = CH₃COOH. Step 2: Determine Product Y from the second reaction.
The second reaction involves the oxidation of propene (CH₃CH=CH₂) with hot, acidic potassium permanganate (KMnO₄/H⁺). Acidic KMnO₄ is a strong oxidizing agent that causes oxidative cleavage of carbon-carbon double bonds. When an alkene is subjected to hot, acidic KMnO₄: \begin{itemize} \item A carbon atom of the double bond that carries at least one hydrogen atom is oxidized to a carboxylic acid. \item A terminal \(=\text{CH}_2\) group is completely oxidized to carbon dioxide (\(\text{CO}_2\)) and water (\(\text{H}_2\text{O}\)). \end{itemize} For propene (CH₃CH=CH₂): \begin{itemize} \item The \(\text{CH}_2=\) group (terminal carbon of the double bond) will be oxidized to \(\text{CO}_2\) and \(\text{H}_2\text{O}\). \item The \(\text{CH}_3\text{CH}=\) group (the other carbon of the double bond with a hydrogen) will be oxidized to a carboxylic acid. The product from this fragment is \(\text{CH}_3\text{COOH}\). \end{itemize} \[ \text{CH}_3\text{CH}=\text{CH}_2 \xrightarrow{\text{KMnO}_4/\text{H}^+} \text{CH}_3\text{COOH} + \text{CO}_2 + \text{H}_2\text{O} \] In organic chemistry problems asking for the major organic product of such cleavages, the organic acid (or ketone) formed is usually the expected answer, rather than \(\text{CO}_2\). Thus, Y = CH₃COOH. Step 3: Match the products with the given options.
Based on our analysis:
X = CH₃COOH
Y = CH₃COOH
Comparing this with the given options:
% Option (1) \(\text{CH}_3\text{COOH, CH}_3\text{CH(OH)CH(OH)CH}_3\)
% Option (2) \(\text{CH}_3\text{CH}_2\text{OH, CH}_3\text{CHO}\)
% Option (3) \(\text{CH}_3\text{CHO, CH}_3\text{COOH}\)
% Option (4) \(\text{CH}_3\text{COOH, CH}_3\text{COOH}\)
Option (4) matches our derived products.