7.3 Bond-Forming (Coordination) and Bond-Breaking (Heterolysis) Steps

In both the proton transfer and the SN2 steps we have examined so far, a bond is formed and a separate bond is broken simultaneously. It is possible, however, for bond formation and bond breaking to occur as independent steps. In Equations 7-5 and 7-6, for example, only a single covalent bond is formed. These are called coordination steps.

Two chemical reactions are shown to represent examples of a coordination step resulting in the formation of a single covalent bond. The first reaction shows bromine carrying four lone pairs of electron and a negative charge, acting as a Lewis base, nucleophile. It is shown to react with a compound, with central carbon atom carrying a positive charge, single-bonded to three methyl groups, acting as a Lewis acid, electrophile. A curved arrow from bromine is shown to point toward central carbon atom. The resultant labeled, Lewis adduct, shows an addition of a bromine atom to the central carbon atom in a tetrahedral arrangement. The second reaction shows a hypochlorite, with a chlorine atom of hypochlorite carrying three lone pairs of electrons and a negative charge reacting with a compound having a central aluminum atom surrounded by three chlorides by a single bond each. A curved arrow from chlorine is shown to point toward aluminum. The reaction is a reversible reaction leading to an incorporation of hypochlorite, with chlorine linked to the central aluminum atom by a single bond.

You may have learned in general chemistry that coordination steps are also called Lewis acid–base reactions. A Lewis acid is an electron-pair acceptor, having an atom that lacks an octet. A Lewis base, on the other hand, is an electron-pair donor. Thus, in Equation 7-5, (CH3)3C+ is the Lewis acid and Br– is the Lewis base. The product, Br—C(CH3)3, is called the Lewis adduct.

YOUR TURN 7.5
SHOW ANSWERS

Label each species in Equation 7-6 as a Lewis acid, Lewis base, or Lewis adduct.

A chemical reaction is shown to represent example of a coordination step resulting in the formation of a single covalent bond. The reaction shows a hypochlorite, with a chlorine atom of hypochlorite carrying three lone pairs of electrons which is a lewis base and a positive charge reacting with a compound having a central aluminum atom surrounded by three chlorides by a single bond each which is a lewis acid. A curved arrow from chlorine is shown to point toward aluminum. The reaction is a reversible reaction leading to an incorporation of chlorine linked to the central aluminum atom by a single bond which is a lewis adduct.

An elementary step can also occur in which only a single bond is broken and both electrons from that bond end up on one of the atoms initially involved in the bond, as shown in Equations 7-7 and 7-8. These are called heterolytic bond dissociation steps, or heterolysis steps (hetero = different; lysis = break), to emphasize that, once the bond is broken, the two electrons are not distributed equally to the atoms initially involved in the bond. One of those atoms, in particular, will be left without an octet. Viewed in this way, heterolysis steps are the reverse of coordination steps.

Two chemical reactions to represent examples of hydrolysis step resulting in the breakage of a single bond. The first reaction shows a closed ring with iodine carrying three lone pairs of electrons present at its meta-position. A curved arrow drawn from a single bond between two carbon atoms at meta-position point toward iodine. It is followed by a reversible arrow to show a closed ring with a positive charge at carbon at the meta-position and an iodine anion carrying four lone pairs of electrons. The second reaction shows a condensed structural formula of a compound, with a central aluminum atom carrying a negative charge surrounded by four chloride ions by a single bond each. The fourth chloride ion carries two lone pairs of electrons and a positive charge, which is further linked to a carbon atom by a single bond. The carbon atom connects to a methyl group by a single bond and an oxygen atom by a double bond. A curved arrow points from a single bond between a carbon atom and chlorine carrying positive charge toward chlorine. It is followed by a reversible arrow to show two compounds. The first compound consists of a central carbon atom carrying positive charge connected to a methyl group by a single bond and an oxygen atom by a double bond. The second compound consists of a central aluminum atom carrying a negative charge surrounded by four single bonded chloride ions with the fourth chloride ion carrying three lone pairs of electrons.

Unlike proton transfer and SN2 steps, coordination and heterolysis generally do not take place in isolation. This is because one of the reactant or product species lacks an octet and is therefore highly unstable and reactive. Instead, coordination and heterolysis steps usually compose one step of a mechanism involving two or more elementary steps—a so-called multistep mechanism (Chapter 8). We will see, for example, that coordination and heterolysis occur in mechanisms for SN1 and E1 reactions (Chapter 8), electrophilic addition reactions (Chapter 11), and Friedel–Crafts reactions (Chapter 22).

YOUR TURN 7.6
SHOW ANSWERS

In Equations 7-5 through 7-8, identify all atoms lacking an octet.

Two illustrations that show mechanism of a chemical structure that lacks octet. It is shown to react with a compound, with central carbon atom carrying a positive charge, single-bonded to three methyl groups, acting as a Lewis acid, electrophile. The reaction shows a hypochlorite, with a chlorine atom of hypochlorite carrying three lone pairs of electrons and a negative charge reacting with a compound having a central aluminum atom surrounded by three chlorides by a single bond each. It is followed by a reversible arrow to show a closed ring with a positive charge at carbon at the meta-position. It is followed by a reversible arrow to show two compounds. The first compound consists of a central carbon atom carrying positive charge connected to a methyl group by a single bond and an oxygen atom by a double bond. An arrow mark pointing toward the two illustrations reads �Lacks octet.�

In a coordination step, the single curved arrow drawn represents the flow of electrons from an electron-rich site to an electron-poor site. In Equation 7-5, Br– is negatively charged and hence electron rich. (CH3)3C+, on the other hand, has a C atom that is positively charged and has only six shared electrons, so it is electron poor. Thus, as indicated in Equation 7-5, Br– acts as a nucleophile. (CH3)3C+ acts as an electrophile, because it forms a bond with a species bearing a negative charge, the same kind of charge an electron has.

In Equation 7-6, CH3COCl has an electron-rich Cl atom that bears a partial negative charge, whereas the Al atom in AlCl3 is electron poor, largely because it is short of an octet. The Al atom is made even more electron poor by the surrounding electron-withdrawing Cl atoms on AlCl3 (Section 6.6e).

problem 7.12 (a) Draw the appropriate curved arrows for the coordination step between FeCl3 and Cl–. Draw the reaction products. Identify which reactant species is electron rich and which is electron poor. (b) Use curved arrow notation to show the product from part (a) undergoing heterolysis to regenerate FeCl3 and Cl–.