Animated Solution for Chemistry - Organic Chemistry: The major products of the following reaction are

We are asked to determine the major products after subjecting it to a two-step reaction sequence.

The first step involves potassium tert-butoxide ($K{O}^{t}Bu$) and heat ($\Delta$). The second step involves ozone ($O_3$) followed by hydrogen peroxide ($H_2{O}_2$).

Hofmann Elimination Potassium tert-butoxide is a classic, sterically hindered, bulky base. When faced with an alkyl halide or sulfonate ester, it strongly favors the E2 elimination pathway. However, its massive size prevents it from easily accessing the more sterically hindered internal $\beta$-hydrogens.

Instead of forming the more stable, highly substituted alkene (the Saytzeff product), the bulky base plucks a proton from the more accessible terminal methyl group. This regioselectivity leads to the formation of the Hofmann product, which is the less substituted alkene.

In our molecule, elimination occurs between C2 and C1, yielding 3-methyl-1-butene: $C{H}_3-CH(C{H}_3)-CH(OS{O}_{2}C{H}_3)-C{H}_3\stackrel{K{O}^{t}Bu/\Delta }{\to }C{H}_3-CH(C{H}_3)-CH=C{H}_2$

Oxidative Ozonolysis The second step treats our newly formed alkene with $O_3$ and $H_2{O}_2$. This is an oxidative ozonolysis.

Ozone cleaves the carbon-carbon double bond ($C=C$), initially forming aldehydes or ketones depending on the substitution. However, the presence of the oxidizing agent $H_2{O}_2$ ensures that any resulting aldehydes are further oxidized into carboxylic acids.

Let's cleave 3-methyl-1-butene: - The terminal $=C{H}_2$ group is cleaved and oxidized to formic acid ($HCOOH$). - The remaining $C{H}_3-CH(C{H}_3)-CH=$ fragment is oxidized to isobutyric acid (or 2-methylpropanoic acid, $C{H}_3-CH(C{H}_3)-COOH$).

Comparing these products with the given options, we find a perfect match with Option (d).