Pictures of the Day CH320M328M

Shown on the left is the BF3 molecule. With three bonds to fluorine (sp2 hybridization), and no lone pairs, there remains one 2p-orbital that is not hybridized and empty. Thus the boron atom needs a lone pair of electrons to give it an octet of electrons. Also, fluorine is highly electronegative, withdrawing electron density from the boron atom. This is represented in the electrostatic potential model at the upper-left, with flourine atoms in yellow-orange (partial negative charge) and boron in blue (partial positive charge). For all these reasons, the molecule BF3 is a good acceptor of electrons and therefore a good Lewis acid. Ammonia (NH3), shown in the middle, has a lone pair of electrons, and since nitrogen is more electronegative than hydrogen, the nitrogen atom has a partial negative charge (red color). In this diagram, NH3 acts as a Lewis base,when it donates its lone pair of electrons to BF3. BF3 acts as a Lewis acid when it accepts the lone pair of electrons that NH3 donates. This reaction fills BF3's empty 2p-orbital, and now boron is sp3 hybridized when previously (as BF3) it was sp2 hybridized. Note the negative charge on the boron half of the molecule (red color) and positive charge on the ammonia half of the molecule (blue color).

The key to understanding alkene chemistry is keeping in mind the special properties of the pi bond (the pi bonding orbital is shown on the left). The electrons in the pi bond are above and below the plane of the bond, so rotating the bond would break it. Thus, as opposed to sigma bonds, pi bonds cannot rotate. This makes possible cis and trans (E and Z) stereoisomers. In addition, the electron density of the pi bond carries with it negative charge, meaning that there is a partial negative charge (red color) above and below the bond. This partial negative charge can interact with molecules having areas of partial or full positive charge. This scenario, in fact, describes almost all of the reactions of pi bonds we will see next chapter.

Most reaction mechanisms we will encounter involve the four common mechanistic elements introduced in class. When drawing mechanisms, always use arrows to indicate the flow of electron density from the electron source, to the electron sink atom. Try to predict which of the mechanistic elements are going to occur under the reaction conditions. In this way you will be learning and understanding the mechanisms (this is good) not simply trying to memorize them (this is bad). If you understand them, you will see how most mechanisms are related, and you will only have to learn a few things before you can successfully write down correct mechanisms (this is good).

For example, the addition of H-Cl can best be understood as "add a proton" mechanistic element followed by "make a bond" between a nucleophile (Cl-) and an electrophile (the carbocation intermediate) . In the first step, electron density of the pi bond attacks the electrophilic hydrogen atom of H-Cl to give a carbocation intermediate and the chloride anion. Note how in the above reaction the more stable of the possible carbocations is produced predominantly, namely the secondary (2°) carbocation (thus explaining Markovinikov’s rule). In the next step, the nucleophilic chloride anion attacks the carbocation electrophile to give the final product. The nucleophile attack can come from the top or the bottom of the trigonal planar (sp2 hybridized) carbocation with equal probability, no matter which face of the alkene the H atom added to, so there is no stereochemical preference of product produced. Click here to see a movie of a related reaction, the addition reaction of H-Br with an alkene.