Animated Solution for Chemistry - Organic Chemistry: In the following reaction sequence, the correct structure(s) of X is (are)

Imagine you are trying to solve a puzzle, but instead of looking at the final picture, you have to work backward from the last piece to the very first. That is exactly what we are doing in this problem. We are given an enantiomerically pure final product and a sequence of three chemical reactions. Our mission is to reverse-engineer this sequence to uncover the exact 3D structure of our starting material, X.

The key to unlocking this puzzle lies in understanding the mechanism of the reactions involved. All three steps in this sequence share a common, beautiful trait: they proceed via the $S_{N}2$ mechanism.

Let's start by closely examining our final product. It is a cyclopentane ring with two substituents. The stereochemistry is explicitly given: - The methyl group ($-C{H}_3$) is on a solid wedge, meaning it is pointing out of the plane towards us. - The azide group ($-N_3$) is on a dashed line, meaning it is pointing into the plane away from us.

Because the methyl group is never involved in any of the reactions, it acts as a spectator. It will remain on a solid wedge throughout our entire backward journey. The action happens entirely at the other carbon atom.

The Azide Substitution (Inversion 1)

The final reaction uses sodium azide (${\text{NaN}}_3$) in dimethylformamide (${\text{HCONMe}}_2$ or DMF). DMF is a classic polar aprotic solvent. It solvates cations perfectly but leaves anions "naked" and highly reactive. This environment strongly favors the $S_{N}2$ mechanism.

The hallmark of an $S_{N}2$ reaction is the inversion of configuration. The nucleophile attacks from the back, flipping the stereocenter like an umbrella in a strong wind.

Since the incoming azide group ended up on a dash, the leaving group it replaced must have been on a wedge. Looking at the previous step, the leaving group was an iodine atom. Therefore, in Intermediate 2, the iodine atom must be on a solid wedge.

The Finkelstein Reaction (Inversion 2)

Moving one step backward, we encounter the reaction of sodium iodide ($\text{NaI}$) in acetone (${\text{Me}}_2\text{CO}$). This is the famous Finkelstein reaction, used to convert alkyl chlorides or bromides into alkyl iodides.

Acetone is another polar aprotic solvent, and this reaction also proceeds via an $S_{N}2$ mechanism. This means we have a second inversion of configuration!

Since the Finkelstein reaction produced an iodine atom on a wedge (as we just deduced), the bromine atom it replaced must have been pointing in the opposite direction. Thus, in Intermediate 1, the bromine atom must be on a dashed line.

Bromination of the Alcohol (Inversion 3)

Finally, we reach the very first step. The starting material X reacts with phosphorus tribromide (${\text{PBr}}_3$) in ether (${\text{Et}}_2\text{O}$). ${\text{PBr}}_3$ is a standard reagent for converting primary and secondary alcohols into alkyl bromides.

Crucially, this transformation also occurs via an $S_{N}2$ mechanism. The oxygen attacks the phosphorus, creating a good leaving group, and then the bromide ion attacks from the back. This gives us our third and final inversion of configuration.

Let's summarize the chain of events: 1. The ${\text{PBr}}_3$ reaction inverted the stereocenter to give a dashed bromine. 2. This means the original hydroxyl group ($-OH$) in reactant X must have been on a solid wedge.

So, our starting material X must have the methyl group on a wedge (since it never changed) and the hydroxyl group on a wedge (due to the triple inversion).

Looking at our options, the structure with both the $-Me$ and $-OH$ groups on solid wedges perfectly matches Option (B).

By carefully tracking the stereochemistry through three consecutive $S_{N}2$ inversions, we have successfully reverse-engineered the entire sequence!