Animated Solution for Chemistry - Organic Chemistry: The major product in the following reaction is

Welcome to a fascinating journey into the world of strained ring systems! In this problem, we are tasked with predicting the major product when an oxaspiropentane derivative is treated with an aqueous acid ($H_3{O}^{+}$). At first glance, the molecule looks like a ticking time bomb of angle strain, and that is exactly the key to unlocking its reactivity.

Our starting material is 2-methyloxaspiro[2.2]pentane. This molecule features a cyclopropane ring fused to an epoxide ring at a single spiro carbon. Both of these three-membered rings suffer from massive angle strain because their internal bond angles are forced to be $60^{\circ }$ instead of the ideal tetrahedral angle of $109.5^{\circ }$.

When we introduce an acid like $H_3{O}^{+}$, the most basic site on the molecule—the epoxide oxygen—acts as a nucleophile and grabs a proton. This protonation transforms the oxygen into a highly effective leaving group ($O{H}^{+}$), making the adjacent carbons extremely electrophilic.

Normally, breaking a carbon-oxygen bond in an epoxide would leave behind a carbocation. However, breaking the $C-O$ bond to the $C{H}_2$ group would result in a primary carbocation, which is highly unstable.

Instead of forming a discrete primary carbocation, the molecule utilizes its massive ring strain as a driving force for a concerted semi-pinacol rearrangement. As the $C-O$ bond begins to break, one of the $C-C$ bonds from the adjacent cyclopropane ring shifts over to the epoxide carbon. This elegant molecular dance simultaneously breaks the epoxide and expands the three-membered cyclopropane ring into a much more stable four-membered cyclobutane ring.

Which Bond Migrates?

The cyclopropane ring has two bonds connected to the spiro carbon. Which one migrates? The answer lies in the transition state. During the migration, a partial positive charge develops on the carbon that is "losing" the bond.

The bond connected to the carbon bearing the methyl group migrates preferentially. This is because the electron-donating methyl group provides inductive stabilization to the developing positive charge, making this pathway significantly lower in energy compared to the migration of the unsubstituted carbon.

Following the migration of the methyl-substituted carbon, we form a protonated cyclobutanone intermediate. If we carefully trace the atoms, we see that the methyl group ends up exactly at the 3-position relative to the newly formed carbonyl group.

Finally, a water molecule from the solvent acts as a base to deprotonate the oxygen, restoring its neutrality. The final major product is 3-methylcyclobutan-1-one, which perfectly matches option (c).