Electrophilic Addition Reactions II
We have seen that electrophilic addition of HX to an alkene is a 2-step process. The first, rate determining step, is the addition of a proton to one of the doubly bonded carbon atoms. This is followed by addition of the nucleophilic reagent to the resulting carbocation. The regiochemistry of the first step is determined by the relative stabilities of the two carbocations that may be formed by addition of the proton to either carbon of the double bond. In this topic we will consider the stereochemistry of this 2-step process.
The Stereochemistry of Electrophilic Additions
The steroechemical features of electrophilic additon reactions are best studied in cyclic systems where free rotation is either restricted or eliminated. Figure 1 summarizes the stereochemical results of several additions.
Something for Everyone
It should be clear that the stereochemical situation is muddy. In reactions 1 and 3 the elements of HCl add to opposite sides of the double bond. This is called anti addition. The stereochemistry of reaction 4 is also anti. In reaction 5 the elements of deuterated acetic acid add to the same side of the double bond in what is referred to as syn addition. Finally, reaction 2 proceeds with both syn and anti addition of the elements of water. The job of the organic chemist is to rationalize these results.
Perhaps the easiest result to rationalize is the most complicated, i.e. reaction 2. Figure 2 suggests how both syn and anti addition of water might occur.
From Top to Bottom
The first thing to notice is that the starting material is a symmetrical structure, which means that protonation of either carbon is equally likely. Similarly, protonation from the top face or the bottom face of either p orbital is equally likely. Figure 2 illustrates one of those four possibilities. Once the carbocation is formed, nucleophilic addition of a water molecule may occur either syn or anti to the H that added in the electrophilic step.
Exercise 1 Examine a model of the carbocation shown in Figure 2. Do you think addition of water syn to the H or anti to the H is preferred. syn anti
Exercise 2 Select the most stable conformation of the product of syn addition of the elements of water to 1,2-dimethylcylcohexene.
a b c d
Kinetic studies of reaction 3 indicate that it is second order in HCl. In other words, the transition state involves two molecules of HCl in addition to the alkene. Figure 3 outlines one suggestion that is consistent with these results.
Fitting the Facts
Even though only one molecule of HCl is consumed for each molecule of 1,2-dimethylcyclopentene, two molecules are involved in the transition state. Note that this picture specifically avoids the postulate of a carbocationic intermediate. If a carbocation were formed, then you should expect a mixture of syn and anti addition as was the case in reaction 2.
Exercise 3 The addition of HCl to 1,2-dimethylcyclopentene in the manner shown in Figure 3 yields 1-chloro-1,2-dimethylcyclopentane, which contains two chiral centers. What is the configuration of C-1? What is the configuration of C-2?
Exercise 4 Would the 1-chloro-1,2-dimethylcyclopentane obtained in reaction 3 be optically active? Yes No
The rationalization of the stereochemistry of reaction 5 is interesting. The chemists who performed this experiment postulated that the geometry of the alkene is such that the hydrogens shown in red in Figure 4 block access of the deuterated acetic acid to the "bottom" face of the pi bond, thus forcing the addition of the electrophilic deuterium ion and the nucleophilic acetate ion both to occur on the "top" face of the double bond.
One Way Street
Exercise 5 Draw the structure of the regiosiomer that would be formed if the deuteroacetic acid added to the double bond in the reverse sense, i.e. the positions of the D and the acetate are switched. What is the relationship of this product to the one shown in Figure 4? They are the same. They are enantiomers. They are diastereomers.
Finally we will look at a reaction where the electrophilic species is not a hydrogen or deuterium atom but rather a bromine atom. Figure 5 offers the accepted interpretation of the stereochemistry observed in reaction 4.
Yet Another Alternative
This rationale invokes a steric argument as well. In the step labeled 1 the pi electrons of the double bond act as a nucleophile displacing a bromide ion from the dibromine (2). Since a bromine atom is so large, it can simultaneously bond to both p orbitals, forming a bromonium ion. Note that in this ion the bromine is divalent and carries a formal positive charge. In the step labeled 3 the bromide ion effects a nucleophilic substitution on a carbon, displacing the bromine (4) and producing the final product. The bulk of the bromine atom in the bromonium ion forces step 3 to occur anti to step 1.
The fact that reactions 2 and 4 have different stereochemical outcomes implies that they involve different intermediates. If they involved the same mechanism, you would expect the same stereochemical outcome.
Exercise 6 Would the 1,2-dibromo-1,2-dimethylcyclohexane obtained in reaction 4 be optically active? Yes No
Exercise 7 In our discussion of solvent-solute interactions we learned about induced dipole-induced dipole interactions wherein a momentary dipole in one species induced a transient dipole in another. How can you apply this concept to account for the fact that dibromine acts as an electrophilic reagent toward alkenes?
Exercise 8 Reaction 1 in Figure 1 shows both a regiochemical and a stereochemical preference. Select the carbocation that is consitstent with the regiochemical outcome: a b
Is the stereochemical outcome consistent the formation of a free carbocation intermediate? Yes No