1ANS PLZZZZZZZZ
c6h5-ch2-chF-CH3 ON DOING E1CB WILL FORM SAYTZEFF ALKENE OR HOFFMAN
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39in e1cb first electrophile goes then nucleophile goes.
. A strong base abstracts the α proton generating a carbanion. The electron pair then expels the leaving group and the double bond is formed.
sry edited..........
39u mean saytzeff alkene is the answer? (then i got to think a bit){hehehe}
1it will form hoffman whenevr we hav F we get hoffman product in e1cb
19definitely the hoffmann mechanism will predominate because of the presence of strong electron withdrawing group...fluorine !