7.5 Bimolecular Elimination (E2) Steps

Each of the elementary steps we have examined so far involve only one or two curved arrows. A bimolecular elimination (E2) step, however, is an example of an elementary step that requires three curved arrows, as shown in Equations 7-14 through 7-16.

Three chemical reactions are shown to represent a bimolecular elimination step (E2), marked by three curved arrows. The first chemical reaction shows a closed six-ring structure, with a hydrogen atom at carbon-2 and bromine carrying three lone pairs of electrons at carbon-3 by a single bond each, labeled as a substrate. A hydroxyl group with an oxygen atom of the hydroxyl group carrying three lone pairs of electrons and a negative charge is labeled strong base. A curved arrow points from an oxygen atom of the hydroxyl group toward the hydrogen atom at carbon-2 while another curved arrow points from a single bond between carbon-2 and hydrogen toward a single bond between carbon 2 and 3. A third curved arrow points from a single bond between carbon-3 and bromine toward bromine. It is followed by a rightward arrow to show the release of a water molecule, with the oxygen of a hydroxyl group in the water molecule carrying two lone pairs of electrons; a closed six-ring structure, with a double bond between carbon-2 and 3, labeled as new bond; and a bromine anion carrying four lone pairs of electrons. The second chemical reaction shows a hydroxyl group with an oxygen atom of the hydroxyl group carrying three lone pairs of electrons and a negative charge labeled strong base. It is shown to react with a condensed structural formula of a compound made up of a chain of five-carbon atoms, with carbon-3 linked to a methyl group by a single bond, carbon-5 linked to a hydrogen atom by a single bond, and carbon-4 linked to a nitrogen atom carrying a positive charge by a single bond. The nitrogen atom is further linked to three methyl groups by a single bond each. A curved arrow points from an oxygen atom of a hydroxyl group toward hydrogen atom at carbon-5. Another curved arrow points from a single bond between carbon-2 and hydrogen toward single bond between a hydrogen atom and carbon-5 toward a single bond between carbon-4 and 5. A third curved arrow points from a single bond between carbon-4 and nitrogen toward nitrogen. The resultants show a release of a water molecule, with an oxygen atom of the hydroxyl group in the water molecule carrying two lone pairs of electrons; a chain of five-carbon atoms, with a double bond between carbon 1 and 2, labeled new bond and a methyl group linked to carbon-3 by a single bond; and a central nitrogen atom carrying a lone pair of electrons surrounded by three methyl groups linked by a single bond each. The third reaction shows the nitrogen atom of an amino group carrying two lone pairs of electrons with a negative charge, labeled strong base. It reacts with a substrate shown as two CH groups linked by a double bond. One of the CH groups connects to a hydrogen atom by a single bond while another connects to a bromine ion carrying two lone pairs of electrons. A curved arrow points from the nitrogen atom of an amino group toward hydrogen atom while another curved arrow points from a single bond between hydrogen and CH group toward a double bond between two CH groups. A third curved arrow points from a single bond between CH and bromine toward bromine. The resultants show a release of three compounds as an addition of a hydrogen atom to the nitrogen of an amino group carrying a lone pair of electrons by a single bond; the formation of triple between two CH groups; and a bromine anion carrying four lone pairs of electrons.

An E2 step can take place when a strong base reacts with a substrate in which a leaving group (L) and a hydrogen atom are on adjacent carbon atoms. That is:

In an E2 step the substrate generally has the form or .

It is called an elimination step because both the H atom and L are eliminated from the substrate. It is a bimolecular step because there are two reactant species in the elementary step—the base and the substrate. Furthermore, by assigning α to the C atom initially bonded to L and β to the adjacent C atom (shown above), we can see why an E2 step is a type of 𝛃 elimination.

Equations 7-14, 7-15, and 7-16 illustrate the diversity of reactants and products that can be involved in E2 steps. The base is HO– in both Equations 7-14 and 7-15, whereas it is H2N– in Equation 7-16. The leaving group can come off as a negatively charged species, as in Equations 7-14 and 7-16, or it can be uncharged, as in Equation 7-15. And the C atoms containing the H atom and the leaving group in the substrate may be joined by either a single bond (Equations 7-14 and 7-15) or a double bond (Equation 7-16).

The primary importance of E2 steps is the incorporation of a carbon–carbon multiple bond into a molecule at a particular site. In Equations 7-14 and 7-15, for example, a C═C double bond is generated. And, in Equation 7-16, a C≡C triple bond is formed. As we discuss later in this chapter, as well as in Chapters 11 and 12, the reactivity of carbon–carbon double and triple bonds is quite important in organic chemistry. We therefore revisit E2 steps in greater detail in Chapters 8, 9, and 10.

The base in an E2 step is the electron-rich species, but the hydrogen atom that the base attacks is not particularly electron poor; instead, the electron-poor atom is the carbon atom bonded to the leaving group. Thus, the movement of electrons from the electron-rich site to the electron-poor site is depicted with two curved arrows (Equation 7-17). One curved arrow is drawn from the negatively charged atom in the base to the hydrogen in the substrate. The second curved arrow is then drawn from the C—H bond to the bonding region between the two C atoms, and the third curved arrow is necessary to depict the departure of L to avoid exceeding the octet on carbon. Overall, the electron-poor C atom gains a share of electrons from the C—H bond. Furthermore, that C atom is no longer electron poor in the products because it is no longer bonded to the electronegative leaving group.

A chemical reaction represents the flow of electrons from an electron-rich to electron-poor site using a curved arrow notation. The chemical reaction shows a boron atom carrying a lone pair of electrons and a negative charge, labeled electron-rich, reacting with a carbon atom linked to a hydrogen atom, a carbon atom, marked delta plus and labeled electron-poor and two vacant single bonds. The second carbon atom is linked to a lithium group and two vacant single bonds. A curved arrow points from a boron atom toward a hydrogen atom while another curved arrow points from a single bond between carbon and hydrogen toward a hydrogen atom. A third curved arrow from a single bond between carbon and lithium toward lithium. The resultant shows addition of hydrogen atom to boron atom; formation of double bond between two carbon atom, with each containing two vacant single bonds; and a lithium carrying a lone pair of electrons and a negative charge.

YOUR TURN 7.9
SHOW ANSWERS

The reactants and products for this E2 step are shown, but the curved arrow notation has been omitted. Supply the missing curved arrow notation and identify the pertinent electron-rich and electron-poor sites.

A chemical reaction represents the bimolecular elimination step, E2 step to check for electron-rich and electron-poor sites. The chemical reaction shows the nitrogen atom of an amino group carrying two lone pairs of electrons and a negative charge reacting with a chain of five carbon atoms carrying a hydrogen atom at carbon-2 and a chlorine atom carrying three lone pairs of electrons at carbon-3. The resultants show addition of hydrogen atom to the nitrogen of the amino group carrying a lone pair of electrons by a single bond; a five-carbon chain with a double bond between carbon 2 and 3; and a chlorine anion with four lone pairs of electrons.

E2 requires three arrows. The curved arrow originates from the electron-rich N of H2N– and points to the H adjacent to the C with the leaving group. The movement of electrons from the electron-rich site to the electron-poor site is depicted using two curved arrows.

A chemical reaction represents the bimolecular elimination step, E2 step to check for electron-rich and electron-poor sites. The chemical reaction shows the nitrogen atom of an amino group which is a electron rich group carrying two lone pairs of electrons which is interacting with hydrogen atom and a negative charge reacting with a chain of five carbon atoms carrying a hydrogen atom at carbon-2, and the single bond is interacting with chlorine atom and the chlorine atom carrying three lone pairs of electrons at carbon-3. The single atom is interacting with other single atom which is an electron poor group. The resultants show addition of hydrogen atom to the nitrogen of the amino group carrying a lone pair of electrons by a single bond; a five-carbon chain with a double bond between carbon 2 and 3; and a chlorine anion with four lone pairs of electrons.

problem 7.15 Supply the appropriate curved arrows and the products for each of the following E2 steps. Hint: Consider simplifying the electron-rich species.

Two chemical reactions to identify the products of the bimolecular elimination step, E2 to mark appropriate curves. The first chemical reaction shows sodium hydroxide reacting with a closed six-ring structure, with a three-carbon side chain attached to its ortho-position. A bromine ion is attached to the second carbon of the side chain by a single bond. It is followed by a rightward arrow to read a question mark. The second reaction shows CH3 Li reacting with a five-carbon zigzag chain having chlorine at carbon-1 and a double bond between carbon-2 and 3. It is followed by a rightward arrow to read a question mark.