Ochem rxn

[Dencology]

hey guys:
what do you think the ans. is?
see the attachment.

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[tsav23]

The answer should be A. Because for the first reaction the ethoxy group will direct the sulfination para to it. Then for the second reaction, the ethoxy group will direct the Cl to its other ortho position. Then for the last reaction, the ethoxy will get protonated from HI, and then I- will do an SN2 reaction and from CH3I while leaving the OH as part of the benzene ring.

 

[Dencology]

would the third reaction be the reverse of Williamson ether synthesis which is SN2?

The answer should be A. Because for the first reaction the ethoxy group will direct the sulfination para to it. Then for the second reaction, the ethoxy group will direct the Cl to its other ortho position. Then for the last reaction, the ethoxy will get protonated from HI, and then I- will do an SN2 reaction and from CH3I while leaving the OH as part of the benzene ring.

 

[arpitpatel86]

im looking at the road map 2 on destroyer the one where it adds excess HI/heat.....al it does is takes RCH2-O-C2H5....and cleaves at O and replaces with I....so it ended up being RCH2I + IC2H5 soo for this reaction i agree its A but why does the reaction proceed differently with the HI unlike what they show on destroyer? and if u are right then on the destroyer shouldnt it be RCH2OH + C2H5I

 

[tsav23]

im looking at the road map 2 on destroyer the one where it adds excess HI/heat.....al it does is takes RCH2-O-C2H5....and cleaves at O and replaces with I....so it ended up being RCH2I + IC2H5 soo for this reaction i agree its A but why does the reaction proceed differently with the HI unlike what they show on destroyer? and if u are right then on the destroyer shouldnt it be RCH2OH + C2H5I

Oh man I don't have destroyer, but I am a TA for organic chemistry so I might be able to help if I can interpret your post.

When HI and heat in excess react with an ether (such as RCH2-O-CH2R). The oxygen on the ether takes a hydrogen from HI, causing it to be protonated. Then I- comes in and reacts with one side CH2R, leaving one mole of RCH2OH and one mole of RCH2I.

Now because it is in excess HI, RCH2OH, will get protonated again, causing it to fall off as water, as this happens, I- attacks the other carbon side of the ether forming another mole of RCH2I.

Therefore you overall reaction would be 2 RCH2I and 1 mole of H2O.

Now in reference to the original post. The reason why it doesn't work on a benzene ring when HI is in excess is because of the stability of the benzene ring. OH on the benzene ring help activate the ring, while Iodine would deactivate the ring. There is not enough strength to break the bond between OH and the benzene ring. Therefore the excess HI doesn't matter for that specific reaction.

Hopefully that helps.