Stereochemical Outcome in Sequential Nucleophilic Substitution Reactions

To determine the correct structure of compound $\mathbf{X}$, let us analyze the stereochemical outcome of each step in the reaction sequence.

1. Analysis of the Reaction Mechanism and Stereochemistry

The reaction sequence consists of three consecutive nucleophilic substitution steps at the stereocenter bearing the hydroxyl group:

1. Step 1: Reaction with $\mathrm{PBr}_3$ in $\mathrm{Et}_2\mathrm{O}$ The reaction of a secondary alcohol with phosphorus tribromide proceeds via the initial formation of a dibromophosphite intermediate ($-\mathrm{OPBr}_2$), followed by backside attack by a bromide ion ($\mathrm{Br}^-$) in an $\mathrm{S_N2}$ fashion: $\text{Alcohol } (\mathrm{R-OH}) \xrightarrow{\mathrm{PBr}_3} \text{Alkyl bromide } (\mathrm{R-Br}) \quad (\text{1st inversion of configuration})$

2. Step 2: Reaction with $\mathrm{NaI}$ in acetone (Finkelstein Reaction) The alkyl bromide reacts with iodide ion ($\mathrm{I}^-$) in acetone via an $\mathrm{S_N2}$ displacement: $\mathrm{R-Br} \xrightarrow{\mathrm{NaI, \, acetone}} \mathrm{R-I} \quad (\text{2nd inversion of configuration})$

3. Step 3: Reaction with $\mathrm{NaN}_3$ in $\mathrm{DMF}$ ($\mathrm{HCONMe}_2$) The alkyl iodide undergoes nucleophilic displacement by azide ion ($\mathrm{N}_3^-$) in a polar aprotic solvent ($\mathrm{DMF}$) via an $\mathrm{S_N2}$ pathway: $\mathrm{R-I} \xrightarrow{\mathrm{NaN_3, \, DMF}} \mathrm{R-N_3} \quad (\text{3rd inversion of configuration})$

Because three sequential $\mathrm{S_N2}$ substitutions occur at the same carbon center, the overall stereochemical outcome is a net inversion of configuration at this stereocenter: $\text{Net stereochemical outcome at the substituted carbon} = \text{Inversion} \times \text{Inversion} \times \text{Inversion} = \mathbf{Inversion}$

The chiral carbon bearing the methyl group ($-\mathrm{Me}$) does not participate in any of the reaction steps, so its absolute configuration remains completely retained.

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2. Determination of Absolute Configuration of the Product

The product shown is an enantiomerically pure 1,3-disubstituted cyclopentane:

• The methyl group ($-\mathrm{Me}$) is attached to $\mathrm{C3}$ with a solid wedge bond.
• The azide group ($-\mathrm{N}_3$) is attached to $\mathrm{C1}$ with a dashed bond.

Assigning the Cahn-Ingold-Prelog $(R/S)$ configurations:

• At $\mathrm{C3}$ (bearing $-\mathrm{Me}$):

  1. Priority 1: $-\mathrm{CH}_2-\mathrm{CH}(\mathrm{N}_3)-$ ($\mathrm{C2}$)
  2. Priority 2: $-\mathrm{CH}_2-\mathrm{CH}_2-$ ($\mathrm{C4}$)
  3. Priority 3: $-\mathrm{CH}_3$
  4. Priority 4: $-\mathrm{H}$ (on a dash, pointing away from the viewer)

  Tracing the path from Priority $1 \to 2 \to 3$ yields a clockwise direction. With the lowest priority group ($-\mathrm{H}$) pointing away, the configuration at $\mathrm{C3}$ is $(3R)$.

• At $\mathrm{C1}$ (bearing $-\mathrm{N}_3$):

  1. Priority 1: $-\mathrm{N}_3$ (on a dash)
  2. Priority 2: $-\mathrm{CH}_2-\mathrm{CH}(\mathrm{Me})-$ ($\mathrm{C2}$)
  3. Priority 3: $-\mathrm{CH}_2-\mathrm{CH}_2-$ ($\mathrm{C5}$)
  4. Priority 4: $-\mathrm{H}$ (on a wedge, pointing towards the viewer)

  Tracing the path from Priority $1 \to 2 \to 3$ on the plane yields a counter-clockwise direction. Since the lowest priority group ($-\mathrm{H}$) is on a wedge, the configuration reverses to $(1R)$.

Thus, the product is $(1R, 3R)$-1-azido-3-methylcyclopentane (trans isomer).

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3. Absolute Configuration Required for Compound $\mathbf{X}$

• At $\mathrm{C3}$: Configuration is retained $\implies \mathbf{(3R)}$
• At $\mathrm{C1}$: Configuration undergoes net inversion $\implies \mathbf{(1S)}$

Therefore, compound $\mathbf{X}$ must have the absolute configuration $(1S, 3R)$, which corresponds to cis-3-methylcyclopentanol.

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4. Evaluation of the Given Structures

• (A): Features a trans relationship (methyl on wedge, hydroxyl on dash), which corresponds to the $(1R, 3R)$ diastereomer. Under net inversion at $\mathrm{C1}$, this would yield a cis product, which does not match the product.

• (B): Features a cis relationship where the five-membered ring has the single bridging methylene carbon ($\mathrm{C2}$) at the top, with both the $-\mathrm{Me}$ group and the $-\mathrm{OH}$ group on solid wedges.

  • At $\mathrm{C3}$: $-\mathrm{Me}$ is on a wedge, $-\mathrm{H}$ is on a dash $\implies (3R)$
  • At $\mathrm{C1}$: $-\mathrm{OH}$ is on a wedge, $-\mathrm{H}$ is on a dash $\implies (1S)$

  This is precisely the required $(1S, 3R)$ enantiomer.

• (C): Features a trans relationship (one wedge, one dash) in an inverted ring orientation, representing the $(1S, 3S)$ isomer.

• (D): Features the ring drawn with the single bridging carbon at the bottom and both substituents on wedges, which represents the enantiomer of (B), having the $(1R, 3S)$ configuration.

Therefore, the only correct structure of $\mathbf{X}$ is (B).