5+ Year Member
May 10, 2015
Look at the reactant. Notice that the nitrogen has been fully alkylated. This is step 1 of Hoffman elimination. Normally you then react this molecule with silver oxide (or silver iodide i forget which one. Out right now. Check my notes for correct workup) and then dehydrate and heat it but it gies directly to step 3. Notice how the molecule reacts with oh- and heat. This signifies you are doing Hoffmann elimination and thus will generate the less substituted double bond.

If you download my notes and go to the last chapter (amines), I go over the Hoffman elimination. You should check it out


Lifetime Donor
15+ Year Member
Mar 12, 2005
New York City
Why does the given reaction go Hoffman?
This reaction is simply the Hofmann elimination, where we form the LESS substituted alkene. The reaction involves forming a quaternary ammonium salt which is set up for the leaving group..... The argument is STERIC. We can discuss transition states and various interactions, but simply put.....it is EASIER to remove the hydrogen on the outside since the path is least blocked. The less sterically hindered the path, the LOWER will be the energy of activation, hence the faster the reaction goes. Removal of an inner hydrogen would be more difficult since it raises the energy of the transition state, and requires a greater energy of activation to achieve. Bottom Line......If you see this leaving group......always form the LESS substituted alkene. I have done this reaction for over 25 years, and it works great !

Hope this helps.

Dr. Romano
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