I have posted all of the slides from today’s lecture (see Syllabus or here). I did not finish my description of electron-donating groups (EDG), but I had saved the easiest material for last so I would like you to look over slides 15-18 on your own. The next exam covers all of the material through this class. The exam will not cover material from Friday’s upcoming class.
Here is a verbal description of what you should get from slides 15-18: Slide 15 – Alkyl halides that can generate conjugated carbocations react readily by the SN1 pathway (exception alert: if a really strong Nu is available and the conjugated system is unhindered, the SN2 pathway may be preferred). We see that just one conjugating group (vinyl, aryl) has the same energetic effect as two alkyl groups so every allyl halide and benzyl halide can be viewed as a potential source of secondary carbocations, tertiary carbocations, or even more stable carbocations. SN1 is always a viable pathway.
We also saw on slide 10 that adding alkyl groups (or other EDG) to an allyl cation provides additional stabilization. Because the charge is spread over multiple C, it doesn’t matter which of these C the EDG is attached to. For example, the data on slide 10 show that positioning two methyl groups on one charged C has the same effect as positioning the methyl groups on different charged C.
Coming back to slide 15, you also see reactions that show how you should think about the outcome of solvolysis reactions involving allyl and benzyl halides. An allyl halide can produce TWO products. The pathway that leads to a product with the double bond in a new position is called allyl inversion. A benzyl halide produces only ONE product. Although the charge in the intermediate benzyl cation is delocalized over three ring atoms, the Lewis base/Nu does not attack any of these atoms. Attack always occurs at the benzylic carbon.
Slide 16 – We have seen that hyperconjugated carbocations “steal” sigma bonding electrons to form a partial pi bond to C+. Conjugated carbocations “steal” pi bonding electrons to form a partial pi bond to C+ and this works even better. What about attaching N, O, S, or F to C+? Is it possible to “steal” nonbonding electrons to form a partial pi bond to C+?
The data (again, these are “delta E aqueous” for RCl going to R+ and Cl-) show that it works very poorly with F. This is not surprising. F is extremely electronegative and doesn’t want to give up its lone pair electrons. I would bet that carbocations do not form when F is attached to the alpha C.
On the other hand, O, S, and N are very effective at stabilizing a carbocation and they are even better at it than a vinyl group. Electron-donating groups form a spectrum that extends from very very strong (N) to very strong (S > O) to strong (aryl > vinyl) to weak (alkyl).
Slide 17 – Models confirm that the heteroatom transfers electron density to C+. This can be seen in the potential maps (not very positive/blue near C+; compare to tertiary cation in upper corner). This can also be seen in the bond distances. The C-EDG distance (“b”) is like a double bond distance (short); contrast with the “normal” C-EDG distance (“a”).
Another important point concerns the “planar” geometry of these carbocations. The atoms attached to the heteroatom (N, S, O) lie in the same plane as the atoms attached to C+. This, too, suggests the formation of a double bond between the EDG heteroatom and C+.
BTW – VSEPR might not predict a planar trigonal geometry around the heteroatoms and this points out a grave danger in relying on VSEPR to predict the geometries of resonance hybrids. If you were to draw these carbocations with single bonds between C+ and N (S,O) as shown on slide 16, VSEPR would lead you to predict a non-planar tetrahedral N with 109 degree bond angles, but the model shows that N (S, O) is a planar trigonal atom. Take-home lesson: recognize resonance hybrids before you apply VSEPR and don’t count delocalized lone pairs as electron domains.
Slide 18 – Very simple ideas explain what is going on in these heteroatom-stabilized carbocations. Lewis’ bonding theory says that atoms share electrons wherever they can in order to achieve octets. This should lead you right past the “minor” carbocation resonance structure (6 electron C+, 8 electron Z) to the “major” double bond resonance structure (8 electron C, 8 electron Z+). Since “Z+” is a very electronegative atom, the double bond is very polar so it is best to see this ion as a resonance hybrid.
This is also a very simple situation from the MO perspective. Z has two electrons in a lone pair orbital. C+ has an empty 2p orbital. If we adjust Z’s geometry properly, we can get the lone pair orbital to be a 2p (or 3p for S) orbital and have it overlap with the 2p on C+. The result: a pi BMO that is strongly polarized towards Z (the lower energy AO dominates the BMO) and a pi ABMO that is strongly polarized towards C.