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Organic Reagents

By James Ashenhurst

Reagent Friday: Potassium t-butoxide [KOC(CH3)3]

Last updated: February 10th, 2016 |

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.

Today’s reagent, potassium t-butoxide, 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 t-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.

So what does this mean for the chemical reactions of t-butoxide? Two things.

  1. t-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 t-butoxide is bulky.
  2. t-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 t-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.
So how does it work? Let’s have a closer look.
If you’ve read parts of this blog before, you might recognize this effect.  That’s because t-butoxide’s tendency to break Zaitsev’s rule is on of the most annoying exceptions in organic chemistry 1. But when you want to use a strong, bulky, poorly-nucleophilic base, potassium t-butoxide is a good choice.

P.S. You can read about the chemistry of Grignard reagents 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!

 

 

 

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37 thoughts on “Reagent Friday: Potassium t-butoxide [KOC(CH3)3]

  1. 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.

  2. 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)?

    1. 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.

  3. 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.

  4. 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?

    1. *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.

  5. 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?

    1. 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

  6. can potassium t-butoxide be used to synthesis propyne from 1,2-dibromopropane? like double elimination to remove the bromin atoms and form propyne?

    1. 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

  7. 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

    1. 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.

  8. 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?

    1. 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.

  9. 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

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