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### Topic: Wanted to teach something new-Tebbe's Reagent  (Read 1825 times)

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#### zsinger

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##### Wanted to teach something new-Tebbe's Reagent
« on: December 03, 2013, 02:39:34 PM »
All,
I am going to teach the Wittig in my review session, and I think it is worth mentioning the Tebbe's reagent, and its synthetic utility when one has a sterically hindered carboxylic acid derivative such an amide or ester groups.  Never run the reaction myself, but the professor of their class said any alternative synthesis with reagents not spoken about in the lecture is grounds for bonus points.  Just so I do not teach it to them wrong, I understand the reaction to proceed in pyridine?  Any other points would be greatly appreciated.  I have already done a literature review, and it seems like a pretty kick-arse little trick for methyleneation of an otherwise "deactivated" carbon.
-Zack
"The answer is of zero significance if one cannot distinctly arrive at said place with an explanation"

#### AlphaScent

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##### Re: Wanted to teach something new-Tebbe's Reagent
« Reply #1 on: December 03, 2013, 03:01:29 PM »
Zach,

After looking at the compound on wikipedia I agree totally that there is some great chemistry going on there.  Quite useful indeed.  I would say keep it simple and explain why you would use that reagent instead of say methyl triphenylphosphonium bromide.  The Wittig reagent used instead of the Tebbe's reagent.

Great idea for sure.

Wonder how expensive the material is..
If you're not part of the solution, then you're part of the precipitate

#### AlphaScent

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##### Re: Wanted to teach something new-Tebbe's Reagent
« Reply #2 on: December 03, 2013, 03:27:47 PM »
Holy expensive!! $620.00 USD from aldrich for 100 mL of a 0.5 M solution. If you're not part of the solution, then you're part of the precipitate #### kriggy • Chemist • Sr. Member • Posts: 1473 • Mole Snacks: +128/-16 ##### Re: Wanted to teach something new-Tebbe's Reagent « Reply #3 on: December 03, 2013, 03:46:25 PM » Holy expensive!!$620.00 USD from aldrich for 100 mL of a 0.5 M solution.
Thats like.. very expensive.
You can synthesise it in the lab if you want, doesnt look "that difficult".