[youconfusedmetoo]

From structure 9-11, I'm stuck on the part after the grignard reaction. I'm assuming that water will break the bond between the oxygen (ether) and the tertiary Carbon with the allylic double bond, but how exactly. All I've got so far is that the oxygen (ether) would attack one of the protons from water, which would end up with a positive charge. I'm stuck from there.

Also, how exactly does periodate maintain the cyclic hemiacetal structure. I understand it cleaves the 1,2 diol, resulting in an aldehyde still attached and one formaldehyde that left the structure.

[rolnor]

The second step looks strange, if you add BuLi to an aldehyde you get the pentyl alcohol?

[clarkstill]

I think step 1 looks pretty suspect too. Have they replaced the solvent with BuLi in both steps?! I initially thought maybe they were using BuLi to protect the primary alcohol as the lithium alkoxide, but it would be an odd thing to do - most people would just add an extra equivalent of grignard.

Searching scifinder for the first step I only found two hits: Hetereocycles 2002, 421 and Tetrahedron 2004 2569, but neither of these papers has this combination of reagents and steps...

[youconfusedmetoo]

I forgot to mention that this is part of a scheme for "the stereoselective synthesis of the C-ring" according to the paper I'm reading. The first compound was treated with n-BuLi for deprotonation and addition to provide a triol. Maybe that explains the strangeness of it?

@clarkstill The reagents were like so:
For 9-11:
(deprotonation, followed by addition)
1. n-BuLi
2. vinylMgBr, THF
yield: 80%

For 11-13:
(cleavage, then )
1. NaIO4, THF
2. H2O
0°C

[clarkstill]

Can you give a reference for the paper you are looking at?

[youconfusedmetoo]

@clarkstill, I have since figured out/solve the first issue. But I am currently stuck on the NaIO4 part. It should yield a cyclic ether, but I have no idea how H2O can play a role in this.
The paper is: https://pubs.acs.org/doi/abs/10.1021/jacs.7b10177

[clarkstill]

THF/water is just a solvent for the NaIO4 cleavage. If you draw the product of that oxidative cleavage reacton you'll find you have an aldehyde and an unprotected alcohol. Can you see how that alcohol and aldehyde might react to form compound 21 from the paper? Compound 21 is known as a hemi-acetal, so you could have a read about their properties/reactivity.