Destroyer 2012 Road map 2 question

[jules9595]

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Hi,

I have a question regarding the last step in the second row (page 105 in the 2012 destroyer). I- is supposed to attack the less sterically hindered carbon bonded to the O of the ether, and O-R should be part of the leaving group but the answer has C2H5I in as the leaving group- is this a mistake?

Thanks!

 

[sagarkanodiya]

Answer is right but I think leaving group should be C2H5OH. because I- is strong nucleophile and will attack benylic carbon and causing C2H5O- to leave. Since it's in acidic condition, ethoxide turns into Ethanol

 

[jules9595]

right, thanks