IIT JEE , AIEEE Forum - Chemistry , Organic Chemistry

punnima

punnima

well the ans 2
C6H5- CH2-O-C6H5 +HI --------> C6H5CH2-I +C6H5-OH
is because I- being a gud nucleophile doent get attached 2 benzene ring(u all knoe prep of iodo benzene is possible only by addition of KI 2 its diazonium salt )
hence all phenolic rings attached 2 an ether -O- directly form only phenol and not iodobenzene ( simply remem as a rule)

ramkumar_november

In the first reaction iodide ion attacks less hindered carbon atom and thus we get
CH₃I   and (CH₃)₃C-OH

but in this reaction
C₆H₅CH₂OC₆H₅   +   HI --------> C₆H₅CH₂I   + PHENOL

WE GET   C₆H₅CH₂I   because the benzylic carbocation C₆H₅CH₂+
is stabilised by resonance ........... so iodide ion attacks benzylic carbon atom and thus we get C₆H₅CH₂I

another example of this kind is.................

1. C₆H₅OCH₃ + HI   ----------->    CH₃I    +   C₆H₅OH

to explain this reaction ...... first in C₆H₅OCH₃the oxygen gets protonated to give

C₆H₅-O⁺-CH₃
           |
          H

in this intermediate the   phenyl oxygen bond is stronger than methyl oxygen bond (due to resonance ) so iodide ion attacks the methyl group to give CH₃I
and phenol

thus we have to see either carbocation stability or steric hindrance to predict the product............

mohit_jitani

in a rxn we shud only concentrate on the RDS n not on the steps that r very fast like the nucleophilic or electrophilic attack...in this rxn the RDS is formation of carbocation.....so more is the stability of carbocation more will b the amt of product obtained....the 1st rxn is absolutely correct...m sure abt this and ne doubts r ever welcome......

yours truly

varshavallig

1) in the first one, tertiary carbocation is formed which is more stabler than primary. so v get tertiary iodide and not primary.

2)in the second one,it is due to partial double bond character of C-O bond that v get C6H5CH2-I and not C6H5-I.

!!!!!!!!!!!!!!!

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