Organic Reagents
Reagent Friday: Potassium tert-butoxide [KOC(CH3)3]
Last updated: January 29th, 2020 |
Potassium tert-Butoxide (KOt-Bu) Is A Bulky Base
In a blatant plug for the Reagent Guide and the Reagents App for iPhone, each Friday I profile a different reagent that is commonly encountered in Org 1/ Org 2.
Sometime back in general chemistry you (hopefully) learned that hydroxide ion (HO-) is a strong base. It’s the conjugate base of water – that is to say, that’s what’s left behind once we’ve ripped a proton (H+) off of it. Likewise, alcohols (ROH) are strong bases too – once you remove the proton to get the conjugate base (RO-). Similar in strength to the hydroxide ion, these are called alkoxides.
Potassium tert-Butoxide (KOt-Bu) Is A Bulky Base
Today’s reagent, potassium tert-butoxide (KOt-Bu), is a strong base just like all alkoxides, but there’s something about it that makes it special. If you’ve come across the tert-butyl group before, you should be able to remember one main thing: it’s really darn fat (although in these more sensitive times, “bulky” is the preferred nomenclature). And as the conjugate base of t-butanol, that makes t-butoxide a bulky base just like our old friend LDA. (Note that the “potassium” isn’t so crucial here and we can leave it out or replace it with sodium or lithium- it’s really the “t-butoxide” part we really care about.)
Potassium t-butoxide is like a really angry Sumo wrestler. It attacks things that are out in the open with fierce and sturdy determination. However, anything that requires the least bit of navigation through a narrow opening (like a doorway) is going to be difficult. In the chemical sense, this means that tert-butoxide is very sensitive to steric interactions. (What are “steric interactions” ? Think about 4 hungry Sumos trying to fit themselves around your tiny dinner table). More specifically, steric interactions are the repulsive interactions between electron clouds that happens when atoms “bump” into each other.
Potassium tert-Butoxide Is A Poorer Nucleophile Than Other Alkoxides Due To Steric Hindrance
So what does this mean for the chemical reactions of t-butoxide? Two things.
First, tert-butoxide is a poorer nucleophile than smaller alkoxides (like ethoxide, methoxide and so on) in nucleophilic substitution reactions (like the SN2). Why? Because the SN2 is very sensitive to steric interactions, and tert-butoxide is bulky.
As A Base, tert-Butoxide Tends To Favor The “Non-Zaitsev” of “Hofmann” Product In Elimination Reactions
- tert-butoxide can be used to form the “less substituted” alkenes in elimination reactions (the E2, specifically). Most of the time, elimination reactions favor the “more substituted” alkene – that is, the Zaitsev product. However, when tert-butoxide is used, it will preferentially remove the proton from the smaller group. This produces the so-called “Hoffmann” product. Let’s have a look.
Mechanism: Formation Of Alkenes Using tert-Butoxide As A Base
P.S. You can read about the chemistry of KOtBu and more than 80 other reagents in undergraduate organic chemistry in the “Organic Chemistry Reagent Guide”, available here as a downloadable PDF. The Reagents App is also available for iPhone, click on the icon below!
(Advanced) References And Further Reading
- Einwirkung der Wärme auf die Ammoniumbasen
W. Hofmann
Chem. Ber. 1881, 14 (1), 659-669
DOI: 10.1002/cber.188101401148
The original paper by W. Hofmann on a new method for olefin synthesis. He was a very productive organic chemist in the 19th century and his name has been attached to a variety of transformations, including amide degradation, isonitrile synthesis, and a few others.
- Olefins from Amines: The Hofmann Elimination Reaction and Amine Oxide Pyrolysis
Cope, Arthur C.; Trumbull, Elmer R.
React. 1960, 11, 317-493
DOI: 10.1002/0471264180.or011.05
Organic Reactions, published and maintained by the ACS division of Organic Chemistry, is a source of comprehensive reviews on various transformations in organic chemistry. This particular review is written by Prof. Cope (MIT, of the Cope rearrangement). Detailed experimental procedures are provided towards the end. - Hofmann-Type Elimination in the Efficient N-Alkylation of Azoles: Imidazole and Benzimidazole
András Horváth
Synthesis 1994; 1994 (1): 102-106
DOI: 1055/s-1994-25414
Alkylation of cyanoethyl-substituted azoles followed by heating with a strong base yields acrylonitrile via a Hofmann elimination. - Cyclization in the Course of Clarke—Eschweiler Methylation
Arthur C. Cope and W. Dickinson Burrows
The Journal of Organic Chemistry 1965 30 (7), 2163-2165
DOI:1021/jo01018a011
Two Hofmann eliminations are indicated in this paper, with compounds 5 and 7.
I’m curious to know: if the leaving group and the hydrogen are not antiperiplanar, can the molecule be rotated so that they are in order to do an E2 reaction.
Yes, so long as they’re not in a ring that would forbid rotation (like a cycloalkane for example)
Can t-butanol, not t-butoxide, engage in an E1 reaction?
Yes, if the reaction is heated.
When you say ”it will preferentially remove the proton from the smaller group”, do you mean the least substituted group?? Also the Zaitsev rule state that ”the poorer get poorer” (in Hydrogen), is that why it favours the ”more substituted alkene”? (less hydrogens= more functional groups= more substituted)?
Yes, the hydrogen should be removed from the least substituted group (e.g. CH3 vs. CH2CH3) in the case of KOtBu.
Zaitsev favors the “most substituted alkene” because these tend to be more thermodynamically stable. The reason takes a while to explain – it involves hyper conjugation. I don’t go into that here.
Even if tert-butoxide is sterically large, can it engage in Sn2 reactions if the leaving group is primary? If so, would that mean tert-butoxide was in fact a strong nucleophile, just that it cannot be a nucleophile if the substrate is secondary.
Yes, it can engage in SN2 reactions if the leaving group is primary. It’s not as good a nucleophile as less bulky alkoxides, but still OK
Can we use tert-Butoxide for the removal of hydrogen from -OH group.
Yes, absolutely.
After the removal of the hydrogen on pentanol, would a double bond form or would oxygen carry -1 charge?
Oxygen would carry a -1 charge. Ain’t nuthin’ gonna make the O- leave to form a double bond after that!
Can use tert-butoxide for the removal of proton from secondary alcohol
Yeah, but you’ll set up an equilibrium between your secondary alkoxide and t-butoxide. So make sure you add it in excess.
Is secondary butoxide should also consider as bulky base?
Not really.
You said the cation doesn’t make a difference (Li, Na, K), why isn’t sodium t-butoxide mentioned more in literature preps if sodium is so much cheaper than potassium?
*for our purposes* the cation does not make a difference. For introductory organic purposes. For *practical* work, KOtBu has a weaker O-K ionic bond, is more dissociated, better solubility in organic solvents, and tends to be the most popular choice at least for small-scale lab work.
I would like to use this reagent for an elimination reaction of a geminal dichloride to form an alkenyl halide and I am not sure what temperature to use. Is the addition performed at very low temperatures and then the temperature of the reaction is increase to favour elimination? If so, up to which point should I increase it? Would r.t. be a good idea?
Here is an experimental procedure.
Henry N. C. Wong, Peter J. Garratt, and Franz Sondheimer
Journal of the American Chemical Society 1974 96 (17), 5604-5605
DOI: 10.1021/ja00824a066
What will be the good solvent to dissolve potassium tert-butoxide except water?
The best solvent for potassium t-butoxide is tert-butanol. All other alcohols will result in mixtures of alcohols / alkoxides.
can we use tert-Butoxide to remove sulfur from ring compounds?
does the mechanism stay the same?
You’d have to attach a picture of what you’re trying to do.
how can i get potassium butoxide from 1-butanol
Dissolve in a solvent like diethyl ether or THF, then add potassium hydride.
Best website on the internet for organic chemistry concepts . Thank you so much !
Glad you find it useful Vivek!
can potassium t-butoxide be used to synthesis propyne from 1,2-dibromopropane? like double elimination to remove the bromin atoms and form propyne?
Yes, there is precedent for KOtBu to form acetylenes. I don’t have a reference for propypne itself but you could check section 3.3 in this paper: 2. Elimination Strategy for Aromatic Acetylenes
Orita, H.; Otera, J. Chem. Rev. 2006, 106, 5387
DOI: 10.1021/cr050560m
Section 3.3 in this review covers the synthesis of alkynes by double dehydrobromination reactions from vic-dibromoalkanes
I have a question.
i saw a reaction with indanone, NaH, Diethylcarbonate in THF solution and it needs reflux condition.
Besides i found some reaction can be done in room temperature , adding KOt-Bu catalytic amount.
I’m curious what might be the role of KOt-Bu in this reaction. What makes this reaction work in room temperature.
I am trying to find some information but not easy for me to find the exact reason.
Thanks
What happens here is that the first equiv of KOTbu deprotonates the indanone hydrogen, and then when the ester is formed (from reaction with carbonate) EtO(-) is expelled. This acts as the base in the next reaction. In other words, they chose a reagent that generates its own base. This would not work if one used, say, ethyl chloroformate, because it kicks off Cl(-) and Cl(-) is not a strong base.
I want to selectively alkylate (Ethylation [Mono ethylation]) on alpha position of Ethyle Aceto Acetate by using Et-Br/ KOC(CH3)3. does it work?. Actually I want to avoid dialkylation. is it suitable base?
Not sure what you mean by “alpha position” but if you mean that you want to alkylate the CH3 group, you need to add two equivalents of base. 1) KOtbu 2) n-BuLi. Then it should alkylate on the end.
when t butoixde is used which reaction is preferred E1 or E2 ?
In the presence of a strong base such as t-butoxide, the reaction will almost certainly be E2.
Dear Dr James
You gave an example with 2-bromo-3- methyl Butane. what will happen if we use less branched alkyl halide 2 – bromo – Butne?. I ask you because a student in Tel – Aviv University read your openion and argued with his lecturer. The lecturer said that it works only if the Alkyl halide is also brunched threfore with 2 – bromo – Butane’ it will be Zeitsev product
What happens if I use 1-propanol with t-butoxide followed by CH3I?
I’m assuming you mean 1 equivalent. t-butoxide is not a great choice of base here. Since propanol and t-butanol are very close in acidity, there will be an equilibrium between the conjugate base of propanol and the t-buOH formed by deprotonation of propanol. You risk forming t-butyl methyl ether as a byproduct, which I assume is not what you’re going for.
what are the expression rate for t-butoxide with 1-choloro-2-methylpropane? is it rate= k[alkylhalide][bulkybase]?
Yes, that is correct.
What would be the product if potassium tert-butoxide was used on 3-bromopentane?
Would 2-pentene be the Zaitsev product?
In that case there is only one possible product, and there is no “Zaitsev” or “Hoffmann” product.
So you need a Leaving group for this reaction to proceed correct?
You most certainly do!
What will be the product when 3-Bromocyclohexene is treated with
pottasium tert-butoxide ?
1,3-hexadiene.
Is it possible for tert-butoxide to favor a zaitsev product over hoffman?