[hoshizora]

Hi guys,

This is a question about organoboron chemistry and involves both the formation of a homopropargyl alcohol and an allenic alcohol. I can rationalize the formation of both products from an allenic borane formed by reagent 1 and 2 (via 1,2 migration of R), but I can't seem to figure out why there will be an equilibrum mixture of A and B at different temperatures. Isn't there only an allenic borane formed (which will then form either 4 or 5)?

[AlphaScent]

This is a great problem to work out.  I myself have also only worked to the Allenic borane formed through the 1,2-shift and the intramolecular attack of the alpha carbon and the chloride leaving.  I am puzzled as to what the rearrangement would be for B. 

My understanding of borates is that the during the shift there is nuclephilic attack of pi electrons on either a center with a good leaving group, tin, selenium or carbonyl compounds (im sure there are more that are good electrophilic centers), which would be intermolecularly.  In this case it is the alpha carbon with a good leaving group for this reaction and is intramolecular. 

As the mixture is heated, would it rearrange to form a conjugated borate system?

How does the nucleophilic attack on the aldehyde then proceed?

[hoshizora]

Is it possible to isolate the intermediate prior to 1,2 migration (at -90 degrees)? But even so, it must still undergo the migration for the propargyl alcohol to form...

[hoshizora]

Guys, please *delete me* >.<

[AlphaScent]

No, you cannot isolate that material.  I believe the rearrangment is to a conjugated borate system followed by nucleophilic addition to the aldehyde.

Have you thought about a mechanism?

[hoshizora]

I can only come up with a mechanism for the route to form 4. I have no idea what B is.

How does your suggested conjugated borate look like?

[hoshizora]

Could there be an attack of the allenic borane on another allenic borane at room temperature (25°C) to form a borate anion? The resulting borate will be highly nucleophilic and should be able to attack the electrophilic carbon of the aldehyde. This route seems more possible. What do you think?

[AlphaScent]

The more I think about this I think I am wrong.  What is your mechanism for forming A and then the homopropargylic alcohol?

[hoshizora]

I'm sure my A and 4 are correct. The base first attacks the borane to form an anionic borate as an intermediate. 1,2-migration of the R group occurs, and from there you can do arrow pushing to form an allene and kick out the chloride group, forming the allenic borane (A).

Following that, a bond forms between the aldehyde O and the boron atom. Push arrows from C-B bond to the C=C bond, and continue doing so from the next C=C bond to a new bond between the terminal C (of the allene) and the electrophilic carbon of the aldehyde. This gives an alkyne. Hydrolysis of the resulting borane gives the homopropargyl alcohol (4).

I even looked up literature for this step, so it's most probably correct. Most literature didn't have any equilibrium occurring at 25°C though, and even if the reaction was brought to room temperature, the same stable allenic borane (in this case A) will form. 

[hoshizora]

Question still unsolved after many, many tries on my side Anyone else wanna give it a shot? >.<

[clarkstill]

The original reference is JACS 1978 5561, and they propose that B is the propargylic borane.  I can't imagine the rearrangement is unimolecular, so maybe this? Not necessarily concerted i guess.

[hoshizora]

OMG you're a genius! It seems to fit in nicely Thanks!!