FailureInLife Posted October 13, 2013 Report Share Posted October 13, 2013 effect of electron withdrawing group WITH resonance on carbocations? Link to comment Share on other sites More sharing options...
Savac Posted October 13, 2013 Report Share Posted October 13, 2013 I'm a little rusty on organic chemistry, but the answer looks correct to me. Halogens definitely destablize a carbocation due to inductive electron withdrawal. All of the sources that you found were completely correct. The problem is, halogens can also donate their lone pair of electrons by resonance. The problem becomes determining which effect is stronger: destabilization due to induction, and stabilization due to resonance. Even in an aromatic system, when resonance is of critical importance, halogens are said to be weakly deactivating groups. This means that the destabilization from induction overrides the stabilization from resonance. Even though all halogens are deactivating groups, I'll give a little bit of information about Bromine. It's less electronegative than fluorine for example, so it's inductive electron withdrawal is less pronounced. That being said, its lone pairs lie in a 4p orbital, and are therefore much less effective at donating electron density to the carbocation (the 4p and 2p orbitals don't overlap very well). I hope this makes sense. Edit: I was about to close this tab, and then I noticed that I wanted to add something that I think is important. Another reason that III must be the slowest has to do with the stability of the transition state (ie. right before the carbocation is formed). The transition state is "similar" to the carbocation (since they're close in energy), and is thus destabilized by the bromine atom. A less stable transition state increases the activation energy for the reaction, which unquestionably decreases the rate. Link to comment Share on other sites More sharing options...
apache Posted October 14, 2013 Report Share Posted October 14, 2013 depends how many electrons they can donate halogens are donating 1, which is in actuality nil, oxygen can sapre 1... like to you see in tautomerization in step 9 of glycolysis to create a double bond and more free energy states... bromine is hardly electron withdrawing, to much inter-electron repulsion from being in such a high orbital group... i'd assume 1 is more stable than 2 since there'd be an easier time shuffling the positive charge (or using the extra electron from teh benzene on a nearer carbocation... (this isn't an mcat question at all, but it's an mcat topic and i couldn't find any other section to post this in..) hi, just have a quick question about carbocation stability for this question for comparing III and IV, for simplicity let's imagine the rings are gone. so we are comparing a 2-bromo butane carbocation (with positive charge on the 2nd carbon) with a pentane carbocation (with positive charge on the 2nd carbon) (for this one put the positive charge and take out the double bond) so what i'm confused about is why does it say bromine destabilizes the carbocation? i searched this online and almost every source said electron withdrawing groups stabilize carbocations by donating a lone pair of electrons to create resonance.. online sources: http://forums.studentdoctor.net/showthread.php?t=938858 http://www.masterorganicchemistry.com/2011/03/21/three-factors-that-destabilize-carbocations/ http://forums.studentdoctor.net/archive/index.php/t-411177.html http://www.chem.ucla.edu/harding/tutorials/cc.pdf and then some sources say electron withdrawing groups destabilize carbocations by taking away electrons making it more positive (less sources say this, which is probably wrong). getting some contradicting information.. so shouldn't the answer be E? (since III is more stable carbocation than IV, making the order IV > III > II > I from slowest to fastest meaning unstable to stable carbocation) would appreciate a clarification, thanks! Link to comment Share on other sites More sharing options...
Guest LeCreuset Posted October 17, 2013 Report Share Posted October 17, 2013 anyone know why the first carbocation is LESS stable than the third even though the first is a tertiary? do resonance structures take priority over # of alkyl groups or something? OCH3 is an electron donating group which stabilizes the carbocation. I miss Felix Lee's orgo, I feel old. Link to comment Share on other sites More sharing options...
silenceofthelambs Posted October 17, 2013 Report Share Posted October 17, 2013 so electron donating groups have precedence over # of alkyl groups? what if we were comparing a tertiary carbocation to a secondary carbocation with O (no OCH3... in other words what exactly is the point of CH3 on the O? the CH3 is partial positive so if anything doesn't it take electron density from the oxygen) and what if we were comapring a tertiary carbocation to a primary carbocation with OCH3? and lastly, is it safe to assume that only oxygen are good electron donating groups, because obviously in the first example (first post) halogens are MORE electron withdrawing than donating thanks for the help No. Anything not considered to be electronegative, or less electronegative than C can be considered donating. E.g N-H. You can consider OCH3 to be another carbon group, think of that example as still tertiary. The reason tertiary > secondary > primary is because the CH3 group is Electron Donating. OCH3 is just a better donator than CH3. Resonance > Number of donating groups always (in the context of your course.) It could just as easily be OH, or OCH2CH2CH2CH3. -O is stabilizing via resonance. The CH3 is not really relavent for your course, it could be O - anything. O-X is EWG by induction, as you may note it is a withdrawing group by electronegativity. However because of the lone pair it is donating by resonance, it can double bond with the C moving the positive charge onto the oxygen. Link to comment Share on other sites More sharing options...
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