![]() ![]() This page is the property of William Reusch. An alternative Type III ring closure to 1-methyl-3-cyclohexenol does not occur. The first is a Type I retro ene reaction, facilitated by relief of ring strain. There are two ene transformations in equation 4. Reaction 3 has two equivalent alkyne chains suitably oriented for a Type I reaction, and both engage sequentially to yield a novel tricyclic "propellane" compound. The last two examples show an interesting variant of the ene reaction in which the enol tautomer of a carbonyl function serves as the "ene" component, and a carbon-carbon double or triple bond is the "enophile". There are actually two different ene reactions that take place, and these may be distinguished by the location of the product double bond and the origin of the transferred hydrogen atom (colored red and blue in the above illustration). is a rare example of a Type III ene reaction. The alkene moiety is colored green, and the new sigma-bond is blue.Įquation 2. demonstrates a Type II ene reaction in which the enophile is a carbonyl group (colored red). Most intramolecular ene reactions, including earlier examples, are Type I.Įquation 1. These different arrangements are defined in the following diagram, where the X,Y & Z atoms are colored blue, and the transferred hydrogen is green. Type II reactions have the enophile joined to the alkene carbon bearing the Z–H group, and in Type III reactions the enophile is linked directly to the Z atom. The most common relationship is Type I, in which the enophile is joined to the alkene carbon farthest from the Z–H group. The "ene" (C=C–Z–H) and "enophilic" (X=Y) moieties of an intramolecular ene reaction may assume different relative orientations that depend on the nature of the linking structure. it is not attracted by an externally applied magnetic field.Chemical Reactivity Classes of Intramolecular Ene Reactions On the other hand, the neutral #"C"_2# molecule has no unpaired electrons, so it is diamagnetic, i.e. it is attracted by an externally applied magnetic field. I won't show the calculation here because I'm not sure you're familiar with bond orders yetĪlso, an unpaired electron will make the #"C"_2^(-)# ion paramagnetic, i.e. Notice that because the extra electron is added to a bonding MO, the bond order of the #"C"_2^(-)# will actually be higher than the bond order of the neutral #"C"_2# molecule. The electron configuration of the #"C"_2^(-)# ion will be The electron configuration of the neutral #"C"_2# molecule is - I'll use the notation given to you in the diagram The lowest energy unoccupied molecular orbital is #2p_(sigma)#, so that is where the extra electron will be added. It will be added to the lowest energy unoccupied molecular orbital, or lowest unoccupied molecular orbital, #"LUMO"#, that follows that highest energy occupied molecular orbital, or highest occupied molecular orbital, #"HOMO"#.Īs you can see in the diagram, the two #2p_(pi)# orbitals, let's say #2p_(pix)# and #2p_(piy)#, are the highest energy occupied molecular orbitals. So, where would this extra electron go in terms of molecular orbitals? It thus follows that the #"C"_2^(-)# species will have This, of course, implies that a #"C"_2# molecule has a total of As you know, a neutral carbon atom has a total of #6# electrons. You need to add an electron and not remove one because of the overall negative charge that exists on the molecule. The problem provides you with the MO diagram for the #"C"_2# molecule, so all you really have to do here is add an electron to that diagram.
0 Comments
Leave a Reply. |