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The Hofmann Elimination

The Hofmann Elimination is an elimination reaction of alkylammonium salts that forms C-C double bonds [pi bonds]. [note] It proceeds through a concerted E2 mechanism. In contrast with most elimination reactions that yield alkenes, which follow the Zaitsev (Saytzeff) rule, the Hofmann elimination tends to provide the less substituted alkene. In this post we go through the difference between Hofmann elimination and Zaitsev elimination and explain the key features in the Hofmann degradation mechanism that result in its preference for the “less substituted” alkene.

Quick Review: Zaitsev’s Rule

Conventional elimination reactions that occur via the E2 mechanism follow  Zaitsev’s rule. The major product will be the more substituted alkene (that is, the alkene with the most carbons directly attached to the alkene).

For instance in the first example below, the “trisubstituted” alkene is favoured over the “mono-substituted” alkene.

Why? The thermodynamic stability of alkenes increases in the order

mono-substituted < disubstituted < trisubstituted <tetrasubstituted.

The energy differences are quite small – about 2 kcal/mol, but that’s enough to deliver an 80:20 ratio of products! [How do we know this? It can be obtained by plugging 2 kcal/mol into the equation ln K = –ΔG/RT]

“Non-Zaitsev” Products Can Dominate When Sufficient Steric Hindrance Is Present 

Sometimes  “non-Zaitsev” products can be obtained through the use of a bulky base for the elimination reaction. A classic example is to use sodium or potassium t-butoxide (KOt-Bu); another is to use lithium di-isopropyl amide (LDA).   The idea here is that the bulky base will react more quickly with the least sterically hindered proton on a beta-carbon, which results in formation of the least substituted alkene. [for more, see: Bulky Bases In Elimination Reactions].

You sometimes might see these “non-Zaitsev” products be referred to as “Hofmann products”. Why?

Back to amines.

The “Hofmann Degradation”

Back in 1851, not many techniques for analyzing complex molecules were available. One method for determining the structure of an unknown compound was to break it down into simpler pieces and look for clues in the fragments, a process called degradation. August Wilhelm von Hofmann developed a two-step degradation method for amines that was later to bear his name.

The first step is to treat an amine with a vast excess of methyl iodide [CH3I], which results in an ammonium salt [we saw this reaction, “exhaustive methylation”,  in a previous post].

The second step is to distill the ammonium salt under low pressure in the presence of a strong base. Silver oxide (Ag2O) is often used.

It might not immediately look like it, but tertiary amines ( :NR3  ) are relatively weak bases (pKaH = 10) and therefore decent leaving groups. [recall that good leaving groups = weak bases]

Heating with strong base results in an elimination reaction:  NR3 is lost and and a new alkene is formed.

The interesting observation here  is that the alkene product from this process tends to be the least substituted alkene (“Hofmann product”)  not the Zaitsev product.

What’s going on?

The Hofmann Elimination Has An Extremely Bulky Leaving Group

It’s not that there’s something about the product alkene that makes it more stable than the Zaitsev product (it isn’t).

The answer lies in the relative energies of the transition states leading to the two products.

It might help to look at the mechanism for the reaction again. Recall that the E2 mechanism demands an antiperiplanar (180°) arrangement of the C-H and C-LG bonds.

It really helps to visualize this by drawing out Newman projections. When we do that, what do you notice?

For most elimination reactions, the steric hindrance of the leaving group isn’t a factor we need to consider. Even though leaving groups like I and Br have a large Van Der Waals radius, their bonds to carbon are long, and being single atoms they don’t interfere with adjacent groups.

Contrast that to the NR3 group, which is like a big-ass ceiling fan spinning around its three alkyl groups – and each of the alkyl groups themselves is like a mini-ceiling fan spinning around three hydrogen atoms. It takes up a lot of space!

The conformation that leads to the “Zaitsev” product has a lot more steric hindrance (two gauche interactions!) than the conformation that leads to the “Hofmann” product, because of the extremely bulky N(CH3)3 leaving group!

These extra steric interactions are enough to disfavour the Zaitsev transition state relative to the Hofmann transition state, and lead to the Hofmann product as the major product.

The Hofmann elimination is just another example of how tweaking a single variable in a chemical reaction can flip the outcome. We saw earlier how increasing the steric hindrance of the base can lead to the non-Zaitsev product. Here, we’re increasing the steric hindrance of the leaving group.

A few kcal/mol difference in a transition state might not sound like a lot, but it’s more than enough to change the identity of the major product.  This is what makes organic chemistry so frustrating to beginners… yet also so deeply interesting!

P.S. Not strictly limited to ammonium salts. There are also examples of “Hofmann-type” eliminations with phosphonium (PR3+) leaving groups as well.



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