When the (R,R) isomer of the amine shown is treated with an excess of methyl iodide, then silver oxide, then heated, the major product is the Hofman product.

1. Draw the structure of the major (Hofman) product.
2. Some Zaitsev product is also formed. It has the (E) configuration. When the same amine is treated with m-CPBA and heated, the Zaitsev product has the (Z) configuration. Use stereochemical drawings of the transition states to explain these observations.

Hofmann elimination occurs via an E2 mechanism requiring anti-coplanar stereochemistry. Hofmann orientation loses a hydrogen from methyl as well as $\mathrm{N}{\left({\mathrm{CH}}_3\right)}_3$group to make less substituted double bond. The least stable alkene, that is the alkene with least number of substituents is the hofmann alkene.

Hofmann product

Hofmann elimination requires anti-coplanar geometry while cope elimination requires syn-coplanar stereochemistry. Cope elimination takes place in the presence of m-CPBA and heat which forms N-amine oxide and further on heating, highly substituted or Zaitsev alkene forms and Z alkene forms as bond formation and bond-breaking occurs simultaneously and in one orientation, thus rotation of groups do not occur. Hofmann elimination also leads to formation of Zaitsev alkene but it is formed as a minor product.

Formation of products via Hofmann and cope elimination