7.2 Bimolecular Nucleophilic Substitution (S_N2) Steps

In a bimolecular nucleophilic substitution (S_N2) step, a molecular species, called a substrate, undergoes substitution in which one atom or group of atoms is replaced by another. Examples are shown in Equations 7-2 and 7-3.

HO– substitutes for Cl– in Equation 7-2, and H₃N substitutes for in Equation 7-3.

Reactions involving S_N2 steps are really important in organic chemistry, especially in organic synthesis (Chapter 13). Sometimes the product of an S_N2 reaction is the compound you want to synthesize. In other cases, an S_N2 reaction, by substituting one atom or group for another, can be used to alter the reactivity of a molecule in ways that make further reactions possible. Because of this central role in synthesis, we revisit S_N2 reactions in Chapters 8–10.

During the course of the S_N2 steps in Equations 7-2 and 7-3, a nucleophile forms a bond to the substrate at the same time a bond to the leaving group—the group that is displaced—is broken. The step is said to be bimolecular because it contains two separate reacting species in an elementary step. In other words, the step’s molecularity is 2. It is called nucleophilic because a nucleophile is the species that reacts with the substrate.

Equations 7-2 and 7-3 also show that a leaving group often comes off in the form of a negatively charged species. Common leaving groups are relatively stable with a negative charge. Applying what we learned in Sections 6.6 and 6.7, we can say that:

Leaving groups thus include Cl–, Br–, and I– (conjugate bases of the strong acids HCl, HBr, and HI, respectively) because, in each case, the negative charge is on a relatively large and/or electronegative atom. They also include anions in which there is substantial resonance and inductive stabilization, such as an alkylsulfonate anion, (the conjugate base of RSO₃H). Water and alcohols (ROH) are also leaving groups because they, too, are the conjugate bases of strong acids (H₃O+ and , respectively).

The term nucleophile literally means “nucleus loving.” It is given this name because the nucleus of an atom bears a positive charge, the type of charge to which a nucleophile is attracted.

Species that act as nucleophiles generally have the following two attributes:

1. A nucleophile has an atom that carries a full negative charge or a partial negative charge. The charge is necessary for it to be attracted to an atom bearing a positive charge.

2. The atom with the negative charge on the nucleophile has a pair of electrons that can be used to form a bond to an atom in the substrate. As we saw in Equations 7-2 and 7-3, those electrons are usually lone pairs.

In the nucleophile in Equation 7-2 (HO–), the O atom possesses a full negative charge and three lone pairs of electrons, as indicated below. In Equation 7-3, on the other hand, the nucleophile (H₃N) has a N atom bearing a partial negative charge and one lone pair of electrons.

Other common, negatively charged nucleophiles include CH₃O–, Cl–, Br–, I–, H₂N–, CH₃NH–, HS–, CH₃S–, N≡C–, and . Other common uncharged nucleophiles include H₂O, CH₃OH, CH₃NH₂, H₂S, and CH₃SH.

YOUR TURN 7.3

SHOW ANSWERS

Draw the complete Lewis structures for three of the negatively charged nucleophiles listed in the previous paragraph (other than HO–) and for three of the uncharged nucleophiles (other than H₃N). Include all lone pairs. For each of the uncharged nucleophiles, write “δ–” next to the atom bearing a partial negative charge.

Solved Problem 7.8

Should CH₄ act as a nucleophile? Why or why not?

The complete Lewis structure for CH₄ is shown here. The C atom has a small partial negative charge (because C is slightly more electronegative than H), but it does not possess a lone pair of electrons that can be used to form a bond with another atom. Thus, CH₄ should not act as a nucleophile.

problem 7.9 Which of the following species can behave as a nucleophile? Explain. (a) SiH₄; (b) NaSCN; (c) ; (d) CH₃Li

Recall from Section 7.1a that the electrons in an elementary step tend to flow from an electron-rich site to an electron-poor site. In the proton transfer in Equation 7-1 (shown again in Equation 7-4a), for example, where HO– acts as a base, a curved arrow is drawn from a lone pair on the O in HO– to the H in HCl. A second curved arrow is drawn to show the initial bond to H being broken, with its electrons becoming a lone pair on Cl–.

A similar thing happens in the S_N2 step shown in Equation 7-4b. In this case, HO– is still relatively electron rich, but the C atom of CH₃Cl is relatively electron poor because it carries a partial positive charge. Thus, HO– acts as a nucleophile, and a curved arrow is drawn from a pair of electrons on O to the C atom to signify bond formation. To avoid five bonds to C (which would exceed the C atom’s octet), a second curved arrow shows that the pair of electrons initially composing the C—Cl bond becomes a lone pair on Cl–.

YOUR TURN 7.4

SHOW ANSWERS

The following S_N2 step is similar to the one in Equation 7-4b:

Label the appropriate reacting species as “electron rich” or “electron poor” and draw in the correct curved arrows.

Solved Problem 7.10

Draw the S_N2 step that would occur between C₆H₅CH₂I and CH₃SNa.

C₆H₅CH₂I will behave as the substrate because it possesses an I, a good leaving group that departs as I–. The conjugate acid of I–, HI, is a very strong acid. CH₃SNa has a metal atom that can be treated as a spectator ion and can thus be ignored. The nucleophile is therefore CH₃S–. In an S_N2 step, a curved arrow is drawn from the lone pair of electrons on the electron-rich S atom to the electron-poor C atom bonded to I. A second curved arrow must be drawn to indicate that the C—I bond is broken (otherwise that C would end up with five bonds).

problem 7.11 Draw the S_N2 step that would occur between the two compounds at the right.