Organic Chemistry

Section 10.10 The Role of the Solvent in SN 2 and SN 1 Reactions 389

What percentage of the reaction takes place by the mechanism when these conditions

are met?

a. b.

SOLUTION TO 26a

10.10 The Role of the Solvent in

and Reactions

The solvent in which a nucleophilic substitution reaction is carried out also influences
whether an or an reaction will predominate. Before we can understand how a
particular solvent favors one reaction over another, however, we must understand how
solvents stabilize organic molecules.
The dielectric constantof a solvent is a measure of how well the solvent can insu-
late opposite charges from one another. Solvent molecules insulate charges by cluster-
ing around a charge, so that the positive poles of the solvent molecules surround
negative charges while the negative poles of the solvent molecules surround positive
charges. Recall that the interaction between a solvent and an ion or a molecule dis-
solved in that solvent is called solvation(Section 2.9). When an ion interacts with a
polar solvent, the charge is no longer localized solely on the ion, but is spread out to the
surrounding solvent molecules. Spreading out the charge stabilizes the charged species.

Polar solvents have high dielectric constants and thus are very good at insulating
(solvating) charges. Nonpolar solvents have low dielectric constants and are poor
insulators. The dielectric constants of some common solvents are listed in Table 10.7.
In this table, solvents are divided into two groups: protic solvents and aprotic
solvents. Recall that protic solventscontain a hydrogen bonded to an oxygen or a
nitrogen, so protic solvents are hydrogen bond donors. Aprotic solvents, on the other
hand, do not have a hydrogen bonded to an oxygen or a nitrogen, so they are not
hydrogen bond donors.

δδ−−

δ−

δ−

δ+

δ+

δ+

δ+

δ−

δ+

δ+

δ+

δ+ H

Y− HO Y+

H

H

O

δ−

δ−

δ−

δ+

δ+ δ+

δ+

H

H

H H

δ+H Hδ+

O

O

O

H

H

O

δ−

δ+

δ+

H

H

O

ion–dipole interactions

between a negatively

charged species and water

ion–dipole interactions

between a positively

charged species and water

H

O H

SN 2 SN 1

SN 1

SN 2

=96%

=

3.20* 10 -^5

3.20* 10 -^5 +0.15* 10 -^5

* 100 =

3.20* 10 -^5

3.35* 10 -^5

* 100

=

3.20* 10 -^5 [2-bromobutane] 11 # 002 * 100

3.20* 10 -^5 [2-bromobutane] 11 # 002 +1.5* 10 -^6 [2-bromobutane]

percentage by SN 2 =

SN 2

SN 2 +SN 1

* 100

[HO-]=1.00 M [HO-]=0.001 M

SN 2

AU: OK as

changed?