saytzeff anti saytzeff

hoffmann elimination occurs wen...
1) teh base is bulky
2) steric hindrace at γ-C
3)wen leaving grp is poor

there must b more conditions...i can recall dese 3 at da moment

but how does the base being bulky affect saytezff/hoffmann?? I can understand E1 and E2 differences but Saytzeff and Hoffmann due to bulky base ...how??

Plus in the eg. C doesnt have a bulky base , nor is Br a very bad leaving group. Then why Hoffmann???

The base being bulky cannot abstract a proton as easily from a hindered carbon in the substrate. Hence it chooses the proton of that carbon which is least hindered, beta to the leaving group. Steric hindrance affects almost everything.
C₂H₅O^- is a little bulkier compared to CH₃O^- but not as much as to facilitate a Hoffman elimination I think...strange. When only the beta carbon is hindered, it ruins chances of S_N, not of Zaitsev elimination.

Note the difference where Zaitsev elimination normally prefers that beta carbon which has more methyl groups or has more hyperconjugative support in the transition state.