[Heory]

For the prinse reaction, was there any stereochemistry ?
Use Newman projection.

[azmanam]

methyl group wont shift because you'd sacrifice a tertiary carbocation to form a secondary carbocation.  hydride shifts so that you trade a tertiary carbocation for a tertiary carbocation.  in the previous post, the shift was restricted by cyclic constraint (the carbocation and shifting groups were within a 6-membered ring), and you need overlap of the sigma bond with the empty p-orbital to get shift to occur (same with last week's C-Si bond problem).  In this case, free rotation about the C-C single bond allows for both the methyl group or the hydride to have good overlap for shifting.

The reason I didn't have N attack straight away was the disfavored formation of a 4-membered ring, and I liked that I could make a 3- or a 5-membered ring after shifting the carbocation.  There are typically 3(/4) things that happen after a prins.  Attack by water (with elimination to allylic alcohol sometimes), attack by a nucleophile (like N), or coordination of another aldehyde to form a dioxane.

[Heory]

A OH was introduced and the other OH was kicked away by the amine group.
Consider a larger ring that would be formed.

[Heory]

If the hydride shifted, a carbocation would be formed next to the OH group, then what would occur must be pinacol reaction, not the epoxide formation. Am I right?

[azmanam]

piperidine formation?

What text did you see this in?  Did they give a reference?

[philmont702a]

Could you please provide the citation for this reaction?  The reactivity is surprising under the conditions, assuming that the piperidine is formed. 

[Heory]

The product for step 1 is right. There's another named reaction for step 2. Because we haven't finish the whole reaction yet, I think it's not proper to tell where I copied it from.
we shall consider the stereochemistry of step 1 now, or have a discussion on what you think are unreasonalbe.

[philmont702a]

What are you defining as step one above?  Is the piperidine the end of step one, or the prins the end of step one?

[Heory]

philmont702a, are we in the same region
?

[Heory]

step 1 : 1) pH=3
step 2: 2)48%HBr

[philmont702a]

Yes, I understand the steps.  Are you defining the piperidine as the completion of step one, or the prins product as the end of step one? 

No, we are not in the same region unless you're in the US.

[Heory]

the red structure of azmanam's picture is the right product of step 1 (pH=3).
We have the same local time, so I thought we were.
In my text book, only 2 products of each step were given,  full mechanism including stereochemistry was not given. I think the stereoselectivity does exist. I will post my own idea later, we can have a disscussion.

[Heory]

Step 2 includes two major reactions, which have little relationship with each other, and one of which is a named reaction. Have to go to bed, see you tomorrow.

[philmont702a]

From the piperidine structure in red, the HBr is presumed to react with the alcohol and likely to form an electrophilic carbocation or eliminate to form the tetra-substituted alkene.  From there, there are a number of options.  What do you think?

[Heory]

re philmont702a:

path 1: need very high temperature to surmount the high activation barrier, thus not favoured. And the resulted amine will be destroyed by hydolysis, won't it?
path2: the most reasonable of the three. But the much stained 5-memebered ring was formed through an very sterically hindered way, thus overwhelmed by another path.
path3: 7-membered ring's  is not very likely to be formed.

Hint: F-C rxn is right. Under the very acidic condition, oxygen atom of ArOMe was protonated, therefore...Go on