Organic Chemistry

and have to be one of the following pairs: sugars 1 and 2, 3 and 4, 5
and 6, or 7 and 8.

2. and are both oxidized by nitric acid to optically active
   aldaric acids. The aldaric acids of sugars 1 and 7 would not be optically active because
   they have a plane of symmetry. (A compound with a plane of symmetry has a superim-
   posable mirror image; Section 5.10.) Excluding sugars 1 and 7 means that
   and must be sugars 3 and 4 or 5 and 6.

3.Because and are the products obtained when the
Kiliani–Fischer synthesis is carried out on there are only two possibil-
ities for the structure of That is, if and are
sugars 3 and 4, then has the structure shown below on the left; on the
other hand, if and are sugars 5 and 6, then
has the structure shown on the right:

When is oxidized with nitric acid, the aldaric acid that is obtained is
optically active. This means that the aldaric acid does nothave a plane of symmetry.
Therefore, must have the structure shown on the left because the
aldaric acid of the sugar on the right has a plane of symmetry. Thus, and
are represented by sugars 3 and 4.
4.The last step in the Fischer proof was to determine whether is
sugar 3 or sugar 4. To answer this question, Fischer had to develop a chemical method
that would interchange the aldehyde and primary alcohol groups of an aldohexose.
When he chemically interchanged the aldehyde and primary alcohol groups of the
sugar known as he obtained an aldohexose that was different from
When he chemically interchanged the aldehyde and primary alcohol
groups of he still had Therefore, he concluded that
is sugar 3 because reversing the aldehyde and alcohol groups of sugar 3
leads to a different sugar (L-gulose).

If is sugar 3, must be sugar 4. As predicted, when the alde-
hyde and primary alcohol groups of sugar 4 are reversed, the same sugar is obtained.

(+)-glucose (+)-mannose

(+)-glucose

(+)-mannose, (+)-mannose.

(+)-glucose.

(+)-glucose,

(+)-glucose

(+)-mannose

(+)-glucose

(-)-arabinose

(-)-arabinose

HO H

HOH

HOH

CH 2 OH

HC O

the structure of (−)-arabinose if

(+)-glucose and (+)-mannose

are sugars 3 and 4

HOH

HO H

HOH

CH 2 OH

HC O

the structure of (−)-arabinose if

(+)-glucose and (+)-mannose

are sugars 5 and 6

(+)-glucose (+)-mannose (-)-arabinose

(-)-arabinose

(-)-arabinose. (+)-glucose (+)-mannose

(-)-arabinose,

(+)-glucose (+)-mannose

(+)-mannose

(+)-glucose

(+)-Glucose (+)-mannose

(+)-mannose

reverse the aldehyde and primary alcohol groups

HO H

H

HOH

HOH

CH 2 OH

HO

HC O

HO H

H

HOH

HOH

HC O

HO

CH 2 OH

D-mannose D-mannose

drawn upside down

Section 22.9 Stereochemistry of Glucose: The Fischer Proof 933

reverse the aldehyde and primary alcohol groups

HO H

OH

HOH

HOH

CH 2 OH

H

HC O

HO H

OH

HOH

HOH

HC O

H

CH 2 OH

D-glucose L-gulose

drawn upside down