Step 1: Ozonolysis (formation of compound A):} The double bond in the cycloalkene undergoes ozonolysis in the presence of ozone (\( \text{O}_3 \)) followed by reduction with Zn/\( \text{H}_2\text{O} \). This cleaves the double bond, producing two aldehyde groups on adjacent carbons. 2.
Step 2: Haloform reaction (formation of compound B): The aldehyde (or ketone) group in compound A reacts with \( \text{NaOH}_{(\text{alc})} \) and \( \text{I}_2 \) (haloform reaction). This cleaves the terminal methyl ketone or aldehyde group to produce sodium formate (\( \text{HCOONa} \)) and iodoform (\( \text{CHI}_3 \)), leaving a carboxylic acid group. The final product, compound B, contains a carboxylate ion (\( \text{COO}^- \)) and a secondary alcohol group. The complete reaction mechanism ensures the correct conversion of "A" to "B."
The best reagent for converting propanamide into propanamine is:
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