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

by James

in Alkenes, Alkyl Halides, Organic Chemistry 1, Organic Reagents

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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{ 6 comments… read them below or add one }

Marya November 4, 2015 at 1:35 pm

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.


James November 10, 2015 at 8:02 pm

Yes, so long as they’re not in a ring that would forbid rotation (like a cycloalkane for example)


Lola November 18, 2015 at 10:52 pm

Can t-butanol, not t-butoxide, engage in an E1 reaction?


James November 20, 2015 at 1:22 am

Yes, if the reaction is heated.


J November 22, 2015 at 3:07 pm

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)?


James November 23, 2015 at 4:02 pm

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.


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