# Why is the less substituted alkene favoured in Hofmann elimination?

In the reaction, $\ce{OH-}$ group attacks a hydrogen ion in an E2 reaction, forming a transition state in which simultaneously $\ce{NCH3}$ is removed. In this reaction, the $\ce{OH-}$ should attack, like in other case of E2 so as to form the Zaitsev product, but due to "steric hindrance" the less substituted alkene is formed. But surely the $\ce{OH-}$'s attack is not affected by $\ce{CH3}$ groups? It can be affected by the $\ce{NCH3}$ group - but whichever way it attacks, it's going at the beta carbon - so is at the same distance from that group in all cases. How can steric hindrance affect its attack? Why is the Hofmann product (less substituted product) favoured?

• This site supports $\LaTeX$ formatting and it is generally preferred to HTML. For more information on how to do this yourself please see here. – bon Jun 9 '15 at 13:16
• Also can you give the actual reaction that is taking place here. I'm not quite sure what the substrate is. – bon Jun 9 '15 at 13:17
• @bon I imagine it is something like this – orthocresol Jun 9 '15 at 14:22
• Yes, it is that reaction. – Charles Jun 11 '15 at 10:00

There are 3 main types of elimination reactions, the E1, E2 and E1cb. Here is a generalized picture of the transition state for the E2 elimination reaction.

Breaking of the $\ce{C-H}$ bond and the $\ce{C-LeavingGroup}$ bond are roughly equally advanced, as is formation of the double bond. Because the 2 bonds are broken to roughly the same degree, it can be said that the E2 reaction is in the "middle" of the possible elimination mechanisms.

If we change this pattern and make the $\ce{C-H}$ bond breaking more advanced than the $\ce{C-LeavingGroup}$ bond breaking (we can do this by making the leaving group a poor leaving group [i.e. groups like ammonium, $\ce{-NR3^+}$, sulfonium, $\ce{-SR2^+}$, etc.]), then we shift the mechanism towards the E1cb side of the elimination spectrum.

In the E1cb case, because breaking of the $\ce{C-H}$ bond is advanced, partial negative charge develops on the carbon where the proton is leaving from. Now primary carbanions are more stable than secondary carbanions, which, in turn, are more stable than tertiary carbanions (electron releasing alkyl groups will destabilize a negative charge). Therefore, when we have a choice of which proton to remove, we remove the proton on the least substituted carbon as this carbon will better stabilize the developing negative charge. As the LeavingGroup departs and the reaction completes, this will produce the olefin with the least substituted double bond, the Hofmann elimination product.

Sometimes it is argued that base size is responsible for elimination patterns and that when a large base is used the proton on the least hindered carbon is abstracted. This is true but the effect is relatively small. Changing the base from methoxide to t-butoxide only increases the amount of Hoffmann product by a modest 10-20% in simple haloalkanes.

So while sterics play a role, it is usually a small role with the electronic factors discussed above being more important.

For completeness, let me briefly mention the E1 elimination which is on the "other side" of the elimination spectrum of mechanisms. In this case, breaking of the $\ce{C-LeavingGroup}$ bond is advanced and partial positive charge develops on the carbon where the LeavingGroup is departing from.

Elimination of a proton from this intermediate can often proceed in two directions. Since more highly substituted double bonds are more stable than less substituted double bonds and since the transition state for proton elimination involves partial double bond character, the transition state leading to the more stable double bond (e.g. the more highly substituted double bond) will be lower in energy. Hence, the product with the more highly substituted double bond (Zaitsev product) will form in E1 reactions.

Edit: Response to OP's comment

I meant the pyrolytic elimination of quaternary ammonium hydroxide, which follows E2 mechanism

I wasn't saying that this reaction (the thermal Hofmann elimination) goes through an E1cb mechanism, rather I said "we shift the mechanism towards the E1cb side of the elimination spectrum." In other words, it is an E2 reaction but $\ce{C-H}$ bond breaking is somewhat more advanced than $\ce{C-LeavingGroup}$ bond breaking. Therefore the primary $\ce{C-H}$ bond is more likely to be involved than the secondary or tertiary $\ce{C-H}$ bond, for the reasons outlined above.

Here's another approach. Instead of electronic factors, let's look at steric factors - in this reaction they both lead to the same conclusion.

E2 eliminations prefer a trans or antiperiplanar geometry. Below are Newman projections of the 2 conformations with this geometrical arrangement of the substituents.

Conformation A is set up to eliminate towards the least substituted carbon, the methyl group (Hofmann), while conformation B depicts elimination towards the more highly substituted carbon (Zaitsev). The trialkyammonium group is extremely large and generates an unfavorcbale steric interaction in conformation B. This makes conformation B higher in energy, hence elimination from conformation A to produce the less-substituted double bond will be preferred.

• What about the stochastical argument? Creating a terminal double bond one has the choice of three protons which one as a base can abstract, while (assuming something like a dimethylcyclohexane derivate) the other direction can have as little as one hydrogen left, meaning a $3:1$ ratio assuming ideal equivalence. – Jan Jun 9 '15 at 17:28
• Yes, statistical factors play a role in all reactions where a choice of pathways is available.. However, in mechanistically switching from Zaitsev to Hofmann type elimination they are not the controlling factor. – ron Jun 9 '15 at 20:19
• I did not mean the E1cb mechanism. I meant the pyrolytic elimination of quaternary ammonium hydroxide, which follows E2 mechanism. E2 mechanism forms Zaitsev product normally. In this reaction, the Hofmann product is formed instead. – Charles Jun 11 '15 at 9:59
• If one tries to look at the molecule entirely in one go - instead of as two different projections - then both the alpha H in the antiperiplanar configuration seem to be at the same distance from the bulky N group. Then the approaching base should suffer similar repulsion if it attacks either H? – Charles Jul 21 '15 at 15:02
• If I understand your question correctly, then the answer is "yes". – ron Jul 21 '15 at 15:35