Organic Chemistry

22.14 The Anomeric Effect

We have seen that -D-glucose is more stable than -D-glucose because there is more
room for a substituent in the equatorial position. The relative amounts of -D-glucose and

• D-glucose are only however, so the preference of the OH group for the equatorial
  position is surprisingly small (Section 22.10). Contrast this to the preference of the OH
  group for the equatorial position in cyclohexane, which is (Table 2.10 on p. 99).
  When glucose reacts with an alcohol to form a glucoside, the major product be-
  comes the Since acetal formation is reversible, the must be
  more stable than the The preference of certain substituents bonded to the
  anomeric carbon for the axial position is called the anomeric effect.
  What is responsible for the anomeric effect? One clue is that substituents that prefer
  the axial position have a lone pair on the atom (Z) bonded to the ring. The bond
  has a antibonding orbital. If one of the ring oxygen’s lone pairs is in an orbital that is
  parallel to the antibonding orbital, the molecule can be stabilized by electron density
  from oxygen moving into the orbital. The orbital containing the axial lone pair of the
  ring oxygen can overlap the orbital only if the substituent is axial. If the substituent is
  equatorial, neither of the orbitals that contain a lone pair is aligned correctly for overlap.
  As a result of overlap between the lone pair and the orbital, the bond is longer
  and weaker and the bond within the ring is shorter and stronger than normal.

22.15 Reducing and Nonreducing Sugars

Because glycosides are acetals (or ketals), they are not in equilibrium with the open-
chain aldehyde (or ketone) in neutral or basic aqueous solutions. Because they are not
in equilibrium with a compound with a carbonyl group, they cannot be oxidized by
reagents such as or Glycosides, therefore, are nonreducing sugars—they
cannot reduce or
Hemiacetals (or hemiketals) are in equilibrium with the open-chain sugars in aque-
ous solution. So as long as a sugar has an aldehyde, a ketone, a hemiacetal, or a
hemiketal group, it is able to reduce an oxidizing agent and therefore is classified as a
reducing sugar. Without one of these groups, it is a nonreducing sugar.

PROBLEM 25 SOLVED

Name the following compounds, and indicate whether each is a reducing sugar or a non-

reducing sugar:

a. b.

OH

OCH 3

CH 2 OH

HO

HO

O

OH

OCH 2 CH 2 CH 3

CH 2 OH

HO

HO

O

Ag+ Br 2.

Ag+ Br 2.

Z

Z

OO

equatorial

lone pair

overlapping

orbitals

axial

lone pair

axial

lone pair

C¬O

s* C¬Z

s*

s*

s*

s*

C¬Z

b-glucoside.

a-glucoside. a-glucoside

5.4 : 1

a 2:1,

b

b a

Section 22.15 Reducing and Nonreducing Sugars 941

A sugar with an aldehyde, a ketone, a

hemiacetal, or a hemiketal group is a

reducing sugar. A sugar without one of

these groups is a nonreducing sugar.