Sciencegeeknm Posted December 8, 2016 Share Posted December 8, 2016 Hi, The following reaction(see photo attached) was talked about on a lecture I watched on YouTube. There was no mechanism mentioned so I tried to work it out myself. Can you confirm if I'm on the right track? Firstly the Na H (in excess) would de protonate all of the OH groups on the glucose molecule. Then the now O- groups would react by SN2 pathway on the benzyl chloride molecules. Now all the former O- groups would be joined to benzyl CH2 units? Thanks. Nick. Link to comment Share on other sites More sharing options...
hypervalent_iodine Posted December 9, 2016 Share Posted December 9, 2016 Your picture hasn't attached. Could you try again? Link to comment Share on other sites More sharing options...
Sciencegeeknm Posted December 9, 2016 Author Share Posted December 9, 2016 Sorry, here it is. Here it is. Link to comment Share on other sites More sharing options...
hypervalent_iodine Posted December 9, 2016 Share Posted December 9, 2016 Okay, then I would agree with your mechanism except that it would be SN1, not 2. Link to comment Share on other sites More sharing options...
Sciencegeeknm Posted December 10, 2016 Author Share Posted December 10, 2016 Interesting that you say SN1 as the teacher says this would go by SN2 alkylation. If it went by SN1 then you get a primary carbocation? Link to comment Share on other sites More sharing options...
hypervalent_iodine Posted December 10, 2016 Share Posted December 10, 2016 Yes, but it's benzylic. They happen to be quite stable due to the aromatic ring. Link to comment Share on other sites More sharing options...
Sciencegeeknm Posted December 10, 2016 Author Share Posted December 10, 2016 Ah ok. Stabilisation through resonance. So would it ever be SN2? Link to comment Share on other sites More sharing options...
hypervalent_iodine Posted December 10, 2016 Share Posted December 10, 2016 I believe so, it's just more often SN1. Link to comment Share on other sites More sharing options...
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