Animated Solution for Chemistry - Organic Chemistry: The number of hydroxyl group(s) in Q is

Let's embark on a fascinating journey through a classic organic reaction sequence. We begin with a complex, substituted decalin system. If you look closely at the reactant, you'll notice it's packed with functional groups: a hydroxyl group ($-OH$), a gem-dimethyl group, and an isolated alkene in the adjacent ring.

The problem asks us to subject this molecule to two distinct chemical environments sequentially and determine the final number of hydroxyl groups in the ultimate product, Q.

Dehydration and Rearrangement

The first step involves treating our reactant with an acid ($H^{+}$) and heat. This is the classic recipe for the dehydration of an alcohol. The most basic site on our molecule is the oxygen atom of the hydroxyl group. It rapidly picks up a proton from the acidic medium to form an alkyloxonium ion ($-O{H}_2^{+}$).

Once protonated, the hydroxyl group transforms from a poor leaving group into an excellent one—water! As the water molecule departs, it takes the bonding electrons with it, leaving behind a positively charged carbon atom. We have now generated a secondary carbocation at carbon 3.

Seeking Stability

Nature always seeks the lowest energy state. Secondary carbocations are relatively unstable, especially when a more stable configuration is just a bond shift away. Look at the adjacent carbon 4—it is a quaternary carbon bonded to two methyl groups.

This is a perfect setup for a 1,2-methyl shift. One of the methyl groups, along with its bonding electrons, migrates to the adjacent positively charged carbon. Consequently, the positive charge shifts to carbon 4. Why does this happen? Because carbon 4 is now a tertiary carbocation, which is significantly more stable than the secondary one due to enhanced hyperconjugation and inductive effects.

Since the reaction mixture is being heated, the system favors an $E1$ elimination pathway over substitution. To neutralize the positive charge and form a stable neutral molecule, a proton is lost from the adjacent carbon 5.

The electrons from the breaking $C-H$ bond swing down to form a new $\pi$ bond between carbon 4 and carbon 5. This yields our major intermediate product, P. Notice that product P is a diene; it contains the newly formed double bond in the left ring and the original, untouched double bond in the right ring.

In the final phase, we treat product P with aqueous dilute potassium permanganate ($KMn{O}_4$) at $0^{\circ }C$. This specific mixture is famously known as Baeyer's reagent, a classic chemical test for unsaturation.

Baeyer's reagent acts as a mild oxidizing agent that performs syn-dihydroxylation on alkenes. It will attack both of the double bonds present in molecule P. The reaction proceeds via a cyclic manganate ester intermediate, which upon hydrolysis, adds two hydroxyl groups across each $\pi$ bond.

The question simply asks for the total number of hydroxyl groups in the final product, Q.

Since product P had exactly two double bonds, and each double bond undergoes dihydroxylation to yield two $-OH$ groups, the math is straightforward:

Therefore, the final structure Q contains exactly 4 hydroxyl groups.