Do these reagents work? OC Question.

[dentalmagic]

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Could I use the following reagents to accomplish this reaction?

1) OTs, pyridine
2) NaOEt
3) BH3 . THF
4) H2O2, NaOH

 

[DAT Destroyer]

View attachment 189977

Could I use the following reagents to accomplish this reaction?

1) OTs, pyridine
2) NaOEt
3) BH3 . THF
4) H2O2, NaOH

I like it ! Well done.......The enantiomer would also be produced

Dr. Romano

 

[dentalmagic]

I like it ! Well done.......The enantiomer would also be produced

Dr. Romano

Thank you Dr. Romano.

I had one more question regarding the following:

The answer given is:

I'm not sure if that answer is correct or not. I got this:

 

[DAT Destroyer]

Thank you Dr. Romano.

I had one more question regarding the following:

View attachment 189978

The answer given is:

View attachment 189979

I'm not sure if that answer is correct or not. I got this:

View attachment 189980

Yes.....this is correct......we first form the epoxide,,,,,,then after workup, we get the diol. Since only C2 is chiral, stereochem is shown. Your answer is perfect.

Make sure there are 2 OHs present !!!!

Keep up the good work!

Dr. Romano

 

[w4nt2baDDS]

View attachment 189977

Could I use the following reagents to accomplish this reaction?

1) OTs, pyridine
2) NaOEt
3) BH3 . THF
4) H2O2, NaOH

This reaction confuses me a bit. This is my logic, not sure if I'm in the right ball park out not.

I figure an alkene must be formed which is most likely 2-methyl-2-pentene in order for BH3-THF/H2O2,NaOH to put an anti-markovnikov alcohol on the 3rd carbon atom.

If this is the case, then how is the pentene formed. Is OTs an anion? I found this structure online:

If this is the case, I would imagine it would act as a base and deprotonate the alcohol leaving a unstable oxygen anion behind. Would this unstable anion take a hydrogen away from carbon 3 (reforming the alcohol) and leaving a carbocation at the 3rd carbon? If this does occur, then what happens? How does the -OH group eventually get removed and a double bond form in its place?

 

[dentalmagic]

This reaction confuses me a bit. This is my logic, not sure if I'm in the right ball park out not.

I figure an alkene must be formed which is most likely 2-methyl-2-pentene in order for BH3-THF/H2O2,NaOH to put an anti-markovnikov alcohol on the 3rd carbon atom.

If this is the case, then how is the pentene formed. Is OTs an anion? I found this structure online:
View attachment 190047

If this is the case, I would imagine it would act as a base and deprotonate the alcohol leaving a unstable oxygen anion behind. Would this unstable anion take a hydrogen away from carbon 3 (reforming the alcohol) and leaving a carbocation at the 3rd carbon? If this does occur, then what happens? How does the -OH group eventually get removed and a double bond form in its place?

OH isn't a good leaving group at all so you must convert it to a good leaving group like tosylate (OTs). The tosylate basically takes the place of the OH. An Sn2 reaction takes place when the NaOEt is used. The OTs gets kicked off and a double bond forms (Zaitsev product). BH3-THF/H2O2,NaOH is then used to add OH anti-Mark.

Someone correct me if I'm wrong.

 

[w4nt2baDDS]

Oh, that's interesting. I wonder what the mechanism is which allows OTs to replace the OH group.

I think you meant an E2 reaction takes place when NaOEt is used. I have to try to remember that tertiary leaving groups result in E2 reactions and not SN2.

Thanks for the help dentaalmagic. Maybe I'll be able to research the OH to OTs substitution.

 

[dentalmagic]

Oh, that's interesting. I wonder what the mechanism is which allows OTs to replace the OH group.

I think you meant an E2 reaction takes place when NaOEt is used. I have to try to remember that tertiary leaving groups result in E2 reactions and not SN2.

Thanks for the help dentaalmagic. Maybe I'll be able to research the OH to OTs substitution.

Oops, yep, meant E2. Also the first step should have been TsCl and pyridine. Not OTs and pyridine. For the product, the OH is replaced by OTs.

 

[DAT Destroyer]

Good job guys !

The actual reaction should be treating the alcohol with TsCl. This will split off HCl.....and put the OTs group. This group is a super good leaving group......think of it as a halogen on steroids !!!! This group is well stabilized by resonance and will leave as the OTs anion. On the above tertiary substrate, NaOEt displaces the OTs group in an E2 mechanism.

Hope this helps

Dr. Jim Romano

 

[w4nt2baDDS]

Okay, awesome. I saw some info online, which showed how the Sulfur in the Ts structure becomes an electrophile to O-H groups but the examples always showed Ts initially bound with Cl.

Thanks guys, your explanations helped cleared things up and I learned a new reaction (or at least relearned one from a couple years back haha).