krock Posted January 23, 2009 Share Posted January 23, 2009 (edited) Hi, hopefully someone can help me with this question: 1. Write an equation to show the reaction between ethanol, C2H5OH and methyllithium, CH3Li. In this reaction, ethanol is a Lewis acid and CH3Li is a Lewis base. Draw all non-bonding electrons and show electron flow with curved arrows. I basically can't figure out the equation: should it produce CH4 and C2H5OLi or LiOH and C5H8? Also then, I need help with the electron flow. Thanks in advance, K. Edited January 23, 2009 by krock Link to comment Share on other sites More sharing options...
UC Posted January 23, 2009 Share Posted January 23, 2009 What is going to be the most acidic part of the ethanol molecule? The alkane-type hydrogens? The methyl carbon? The carbon with the hydroxy group on it? How about the oxygen? But it has those lone pairs on it....so what about the hydrogen attached to the oxygen? The O-H bond is pretty polar right? So, you should know what exact part of your ethanol molecule is the lewis acid (electrophile) now. The methyllithium is practically a lithium salt of a carbanion, so the carbon is going to have electrons it really doesn't want (lewis base, nucleophile). You draw arrows from the nucleophile to the electrophile, indicating that a bond is formed. The electrons in the O-H bond are going to migrate to the oxygen since the hydrogen is recieving enough to make a bond from the methyllithium carbon. So you need to draw an arrow from the O-H bond to the O, indicating that the bond is broken. The net effect is that the Lithium cation has moved to ethoxide and the methyl carbanion has gained a hydrogen forming it's conjugate acid. -1 Link to comment Share on other sites More sharing options...
krock Posted January 23, 2009 Author Share Posted January 23, 2009 Great thanks! Makes more sense now. One question: why does there need to be an arrown from the O-H bond to the O? If you could explain that part in a bit more detail it would be awesome. Link to comment Share on other sites More sharing options...
UC Posted January 23, 2009 Share Posted January 23, 2009 (edited) It's electron pushing. I suppose you should technically draw an arrow from the Li-C bond to the carbon too. There is a bond between the Lithium and carbon, albeit an extremely polar one. In the reaction products, you see that lithium is a +1 ion with ethoxide. It can't become an ion if it's still bonded to carbon. So the bond (which is made up of 1 electron from the lithium and 1 from the carbon) must disconnect from lithium. Hence the arrows showing the migration of both electrons. Now, carbon has two extra electrons, right? Enough to make an entire bond. if it uses them to attach to the hydrogen and the O-H bond isn't broken, the hydrogen is going to have 4 electrons surrounding it. Does that look like a happy compound? IMPORTANT: If you make a bond (arrow from atom to atom) and have the same number of reactants as products, you also have to break a bond somewhere (arrow from bond to atom). Generally you won't draw the ethoxide attacking the lithium cation since they don't really form much of a bond. If you count them as seperate products, you have three products. If you have two reactants and they combine to make three, another bond has to break somewhere to make up for the difference. Edited January 23, 2009 by UC Link to comment Share on other sites More sharing options...
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