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RANGUEIL 2015 - QCM 20 E


Go to solution Solved by Chat_du_Cheshire,

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Posted

re dinedine,

 

du coup on a une SN1 ici avec 2R 3S au départ. Le Cl du 2R devient un OH, et SN1 nous donne 50%R 50%S. En position 2 on aura donc 50%R 50%S et en position on garde notre 3S.

Soit au final 2R 3S et 2S 3S !

Posted (edited)

waoooouw j'ai bugué pendant 10 min @Porthos merci de m'avoir débloqué en fait c'est le 3 qui est S à la bases qui va tout simplement rester S

@chatduCheChiant tu vas trop vite ...

 

Edited by DrWho

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