The reaction sequence is as follows:
Step 1: Oxidation using Jones Reagent
Jones reagent (\(\text{CrO}_3 + \text{H}_2\text{SO}_4\)) oxidizes primary alcohols (\(\text{CH}_3\text{CH}_2\text{OH}\)) to carboxylic acids. Ethanol (\(\text{CH}_3\text{CH}_2\text{OH}\)) is oxidized to acetic acid (\(\text{CH}_3\text{COOH}\)).
\[\text{CH}_3\text{CH}_2\text{OH} \rightarrow [\text{JonesReagent}] \text{CH}_3\text{COOH}.\]
Step 2: Oxidation using \(\text{KMnO}_4\)
The acetic acid (\(\text{CH}_3\text{COOH}\)) is further oxidized to carbonic acid (\(\text{H}_2\text{CO}_3\)) by \(\text{KMnO}_4\). Carbonic acid is unstable and decomposes to \(\text{CO}_2\) and water.
\[\text{CH}_3\text{COOH} \rightarrow [\text{KMnO}_4]\text{H}_2\text{CO}_3 \rightarrow \text{CO}_2 + \text{H}_2\text{O}.\]
Step 3: Decarboxylation using Soda Lime
The carbon dioxide (\(\text{CO}_2\)) reacts with soda lime (\(\text{NaOH} + \text{CaO}\)) to form methane (\(\text{CH}_4\)) via decarboxylation.
\[\text{CO}_2 + \text{NaOH} \rightarrow \text{CH}_4 + \text{Na}_2\text{CO}_3.\]
Final Product:
The major product \(P\) is methane (\(\text{CH}_4\)).
Final Answer: (1).
Let \( y = f(x) \) be the solution of the differential equation
\[ \frac{dy}{dx} + 3y \tan^2 x + 3y = \sec^2 x \]
such that \( f(0) = \frac{e^3}{3} + 1 \), then \( f\left( \frac{\pi}{4} \right) \) is equal to:
Find the IUPAC name of the compound.
If \( \lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}} = p \), then \( 96 \ln p \) is: 32