Answer to Question #111265 in Organic Chemistry for Priya jaiswal

Question

Explain why Hoffman elimination gives less substituted alkene

Answer 1

Formation of less substituted (less stable) alkene can be explained by two factors involved:

1) Kinetic one: Less substituted $\beta$-atom of hydrogen is more open for base that approaches H-atom and elimiates it. Substitutes are more bulky than hydrogen, so they less promote hydrogen elimination from C-atom they are attached to. That's why strong and at the same time bulky bases (t-BuO^-, i-PrO^-) lead to yield increase of Hoffman product more then Zaytsev product (small amounts of Zaytsev's elimination usualy form too)

2) The higher acidity of H of less sustituted C-atom explained by hyperconjugation effect - primary carbanion is more stable then secondary one. Also reaction involves formation of positively charged quaternary amine, that also promotes more acidity for H-atom of less substituted C atom.

These factors show that in given conditions H-atom of less substituted C-atom is more attractive fo bases, and also more cleavable, so reaction goes in that way.