I would love some help with these, thank you so much!Screenshot 2026-01-24 at 8.59.39 PM.png
For this sequence, the blank box should show the deprotected alcohol plus a deuterohydrin (OH + D) added across the original double bond.
Step 1: TMSCl, pyridine
Protect the existing alcohol: ROH → ROTMS (configuration at that carbon is unchanged).
Step 2: peracid (mCPBA-type)
Epoxidation of the C=C happens from the less hindered face, i.e. opposite the bulky OTMS substituent.
Since the starting OH is drawn behind (dash), the protected OTMS is also "behind", so the epoxide oxygen is formed mainly on the front face.
Step 3–4: LAD, then H₂O
LiAlD₄ opens the epoxide by backside attack (anti opening).
Attack occurs at the less hindered epoxide carbon (the one farther from the OTMS side), so:
D ends up on that attacked carbon (back face)
The epoxide oxygen becomes OH on the other carbon (front face after workup)
So D and the new OH are trans (anti).
Step 5: TBAF
Removes the TMS group → restores the original OH (still drawn as dash, same stereochemistry as the start).
Cyclic Synthesis using epoxides .jpg
sequence converts the epoxide into an allylic bromide (alkene).jpg
This sequence converts the epoxide into an allylic bromide (alkene) by:
(acidic epoxide opening → make a good leaving group → E2 elimination).
Step 1: HBr (acidic epoxide opening)
The epoxide oxygen is protonated first.
Br⁻ attacks the more substituted epoxide carbon (the carbon bearing the Me), because the opening has carbocation-like character under acidic conditions.
Ring opens anti, giving a bromohydrin:
Br ends up on the Me-bearing carbon
OH ends up on the adjacent carbon
Step 2: TsCl, pyridine
Converts the alcohol into a tosylate:
–OH → –OTs (excellent leaving group)
Now you have a bromo-tosylate on adjacent carbons.
Step 3: tBuOK / heat (E2)
E2 elimination occurs, and it will preferentially expel OTs (better leaving group than Br).
The tosylate-bearing carbon has β-carbons on both sides, but:
The Me/Br-bearing carbon has no β-H (it is fully substituted: ring bonds + Me + Br)
Therefore elimination cannot form the C=C back to that carbon.
So the base must remove the β-H on the other adjacent carbon, giving the alkene one position away.
✅ Final product: 1-bromo-1-methylcyclohex-2-ene
(Br and Me on the same carbon; double bond between the next two carbons).