Let's break down this reaction step by step: --- ## **Step 1: Analyzing the Starting Material** - The molecule is enantiomerically pure and contains two **tosylate (OTs)** leaving groups on adjacent stereocenters. - The groups are: - **Left**: OTs (wedge) and (wedge) - **Right**: OTs (dash) and Me (dash) - Both tosylates are on secondary carbons. - The reaction conditions: **KOH** (strong base) and **H₂S** (nucleophile). --- ## **Step 2: Predicting the Reaction Type** - **KOH** is a strong base, can deprotonate H₂S to HS⁻ (a good nucleophile). - **OTs** is an excellent leaving group, so **substitution (S_N2)** is likely. - S_N2 reactions on secondary carbons are possible, especially with strong nucleophiles and good leaving groups. --- ## **Step 3: Mechanism** ### **a. Generation of HS⁻** - KOH deprotonates H₂S: \[ \text{H}_2\text{S} + \text{KOH} \rightarrow \text{HS}^- + \text{H}_2\text{O} + \text{K}^+ \] ### **b. S_N2 Substitution on Both Centers** - Both tosylate groups are replaced by HS⁻ via backside attack, **inverting stereochemistry** at each center. --- ## **Step 4: Drawing the Product** - **Original stereochemistry:** Both OTs are on opposite faces (one wedge, one dash). - **After inversion:** Both HS groups will be on the **opposite faces** compared to OTs. So: - On the left carbon: OTs (wedge) → HS (dash) - On the right carbon: OTs (dash) → HS (wedge) The methyl groups' stereochemistry will remain the same as originally drawn. --- ## **Step 5: Final Product** **The product is:** - 2,4-dimethyl-2,4-dihydrothiopyran (a six-membered ring is NOT formed; chain is kept). - The correct structure: both OTs replaced with SH, with stereochemistry inverted at each center. ### **Drawn Structure:** **(2R, 4S)-2,4-dimethyl-2,4-butanedithiol** (assuming original centers are R and S): ``` Me—CH(SH)—CH₂—CH(SH)—Me (dash) (wedge) ``` ### **Summary Table** | Step | Result | |---------------------------|---------------------------------------| | OTs replaced by HS⁻ | Inversion at both stereocenters | | Stereochemistry | Opposite of starting material | | Product | Enantiomerically pure dithiol | --- ## **Final Answer** The product is **enantiomerically pure 2,4-dimethyl-2,4-butanedithiol** with both stereocenters inverted: **Structure:** ``` Me | HS–C–C–C–C–SH | | Me Me ``` - Left SH (dash), right SH (wedge) **Mechanism:** Two S_N2 substitutions, both invert stereochemistry at each center. The product remains enantiomerically pure but is the enantiomer of the starting material. --- **If you want a drawn structure, let me know!**