1hoffmann 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
plz someone explain the answers of options B and C to me
or in short explain to me :
the cases in which anti-saytzeff elimination takes place
thanks
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3 Answers
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1but 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???
39The 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 SN, 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.