6.5: An Example of a Polar Reaction - Addition of HBr to Ethylene

Let’s look at a typical polar process—the addition reaction of an alkene, such as ethylene, with hydrogen bromide. When ethylene is treated with HBr at room temperature, bromoethane is produced. Overall, the reaction can be formulated as

The reaction is an example of a polar reaction type known as an electrophilic addition reaction and can be understood using the general ideas discussed in the previous section. Let’s begin by looking at the two reactants.

What do we know about ethylene? We know from Section 1.8 that a carbon–carbon double bond results from the orbital overlap of two sp²-hybridized carbon atoms. The σ part of the double bond results from sp²–sp² overlap, and the π part results from p–p overlap.

What kind of chemical reactivity might we expect from a $\text{C═}$

Figure 6.3: A comparison of carbon–carbon single and double bonds. A double bond is both more accessible to approaching reactants than a single bond and more electron-rich (more nucleophilic). An electrostatic potential map of ethylene indicates that the double bond is the region of highest negative charge

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What about the second reactant, HBr? As a strong acid, HBr is a powerful proton (H⁺) donor and electrophile. Thus, the reaction between HBr and ethylene is a typical electrophile–nucleophile combination, characteristic of all polar reactions.

We’ll see more details about alkene electrophilic addition reactions shortly, but for the present we can imagine the reaction as taking place by the pathway shown in Figure 6.4. The reaction begins when the alkene nucleophile donates a pair of electrons from its $\text{C═C}$ bond to HBr to form a new C−H bond plus Br⁻, as indicated by the path of the curved arrows in the first step of Figure 6.4. One curved arrow begins at the middle of the double bond (the source of the electron pair) and points to the hydrogen atom in HBr (the atom to which a bond will form). This arrow indicates that a new C−H bond forms using electrons from the former $\text{C═C}$ bond. Simultaneously, a second curved arrow begins in the middle of the H−Br bond and points to the Br, indicating that the H−Br bond breaks and the electrons remain with the Br atom, giving Br⁻.

Figure 6.4 MECHANISM The electrophilic addition reaction of ethylene and HBr. The reaction takes place in two steps, both of which involve electrophile–nucleophile interactions.

When one of the alkene carbon atoms bonds to the incoming hydrogen, the other carbon atom, having lost its share of the double-bond electrons, now has only six valence electrons and is left with a positive charge. This positively charged species—a carbon-cation, or carbocation—is itself an electrophile that can accept an electron pair from nucleophilic Br⁻ anion in a second step, forming a C−Br bond and yielding the observed addition product. Once again, a curved arrow in Figure 6.4 shows the electron-pair movement from Br⁻ to the positively charged carbon.

The electrophilic addition of HBr to ethylene is only one example of a polar process; there are many others that we’ll study in depth in later chapters. But regardless of the details of individual reactions, all polar reactions take place between an electron-poor site and an electron-rich site and involve the donation of an electron pair from a nucleophile to an electrophile.

What product would you expect from reaction of cyclohexene with HBr? With HCl?

Reaction of HBr with 2-methylpropene yields 2-bromo-2-methylpropane. What is the structure of the carbocation formed during the reaction? Show the mechanism of the reaction.