Question

Draw the Fischer projection for the monosaccharide

Draw the Fischer projection for the monosaccharide drawn as a haworth projection below.

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Answer #1
Concepts and reason

The concept used to solve this problem is converting of Haworth projection to Fischer projection. The cyclic structure is given in the question, so draw in straight chain. Place the –OH group at appropriate positions using the rules of converting cyclic ring structure to straight chain structure.

Fundamentals

The substituents/groups which are facing towards up in Haworth projection, must be placed on left side of the fisher projection. Similarly, the groups which are facing towards down in Haworth projection, must be placed on right side of the fisher projection.

The structure is as follows:

6
CH20H
CH2OH
(up) 5
ОН
н
anomeric carbon
ОН
Н
4
ОН
н (up)
1
ОН
Н
ОН
2
3
он (down)
н
ОН

The Fischer projection is as follows:

н
6
CH-оH
2
5
(up) он
н
anomeric carbon
Н
4
ОН (up)
4
С
Н
ОН
2
3
с
ОН (down)
Н
6 CH-он
со

The structure is as follows:

6
с
CH2OH
2
С
5
(up) он
н
Н
1
4
он (up)
4
ОН
Н
2
3
5
-с
-ОН
он (down)
Н
6 CH-ОН
т
о

6
н
CH-Он
5
(up) он
2
OH
Н
3
Но-
1
4
ОН (up)
4
Но-
-С
ОН
2
3
5
Он (down)
Н
OH
6 CH-ОH
Place Hydrogens
at vacant positions
н
2

Ans:

The Fischer projection is,

Н
Н-
-ОН
—с-
но—с-
—Н
но-
-С—н
-С—он
Н—с-
CH-ОH
I

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