Hi all,
I was curious if anyone else was having difficulty in grasping why particular reaction routes were taken in some of our reactions in class and on Sapling. There sometimes seems to be multiple routes to get to the same outcome, and I am finding it difficult to always find the "correct" one. By correct, I mean the resultant product would minimize getting mixture of stereoisomers, achieving the shortest/least intensive route and preventing competing reactions with some byproducts. I seem to waste an inordinate amount of time thinking about these factors, and assume that this will pass with more practice/time.
One such case came up in class concerning the possibility of chloride ion interfering with our SN2 reaction with methoxide when converting an alcohol to an ether. In that case, we had converted the secondary alcohol to a sulfonate ester using pyridine and TSCl. Then we used methoxide in the next step, in a SN2 reaction to convert the sulfonate ester to an ether. I think it was explained well why the protonated pyridine would "trap" the chloride ion, but wouldn't that also "trap" the methoxide ion to a lesser extent? Would there be a mixture as a result?
Brian
I'm actually wondering the same thing now, thanks for pointing this out! I can think of two things, but I am unsure about both of them.
ReplyDelete-Purification techniques to remove the protonated pyridine and chlorine from the solution before adding the methoxide.
-The methoxide is actually MeOCl, and it just pushes more of the pyridine to deprotonate the alcohol to "claim" (so to speak) the excess chlorine (Le Châtlier's Principle). This might produce a really fun mixture, though.
Or maybe the methoxide is just that strong of a nucleophile to where the reaction happens without any of the methoxide getting trapped by the pyridine? I guess I have three thoughts.
Anybody else have better insight?
The chloride ion will not interfere with the second reaction, because the reactions should be worked up between the two steps.
ReplyDeleteWhenever you draw reaction going to a product then draw a second reaction arrow for a second reaction, it is assumed that you have cleaned and purified your product from the first reaction. So disregard the reagents from the previous reaction.
If this was a two-step(one reaction arrow) reaction where the TsCl and pyridine were used in the first step(labeled 1.) and the methoxide was added as the second step(labeled 2.), the pyridine and chloride ion would more than likely have an affect your reaction.
Also one minor correction to the response above me, the methoxide ion(MeO-) would have a positively counter ion such as sodium or potassium not a negative ion such as a chlorine ion.
Oops! Thanks for that! Also thanks for the clarification!
ReplyDelete