Study Notes on Haworth Structures of Monosaccharides

Chapter 15: Carbohydrates

15.4 Haworth Structures of Monosaccharides

  • Learning Goal:
    • To draw and identify the Haworth structures for monosaccharides.
Overview of Monosaccharides
  • The most stable forms of pentose and hexose sugars are five- or six-atom rings.
  • Haworth structures are created from the reaction of a carbonyl group and a hydroxyl group within the same molecule.
Drawing Haworth Structures

  • Step 1: Converting Fischer Projections

    • Rotate the Fischer projection 90° clockwise.
    • The —H and —OH groups on the right of the vertical carbon chain are positioned below the horizontal carbon chain.
    • The groups on the left of the open chain are positioned above the horizontal carbon chain.

  • Step 2: Forming the Ring

    • Fold the horizontal carbon chain into a hexagon.
    • Bond the oxygen atom on carbon 5 to carbon 1.
    • Carbons 2 and 3 will form the base of the hexagon structure, and the remaining carbon atoms will adjust upward.

  • Step 3: Positioning the Hydroxyl Group

    • The —OH group on carbon 1 must be placed:
    • Below the ring for the α isomer.
    • Above the ring for the β isomer.

Mutarotation of D-Glucose

  • When cyclic structures are placed in solution, they undergo interconversion, opening and closing between forms:
    • α-D-glucose can convert to β-D-glucose and vice versa.
    • At any given time, only a small fraction exists in the open-chain form.

Haworth Structures of Specific Monosaccharides

Galactose
  • Classification:
    • Galactose is an aldohexose.
  • Structural Differences:
    • Differs from glucose only in the arrangement of the —OH group on carbon 4.
    • The Haworth structure for galactose resembles that of glucose, with the distinction of the —OH group on carbon 4 being positioned above the ring.
Fructose
  • Classification:
    • Fructose is a ketohexose.
  • Structural Characteristics:
    • Forms a five-atom ring structure, which features carbon 2 at the right corner of the ring.
    • The formation occurs when the —OH group on carbon 5 reacts with carbon 2 in the carbonyl group to create the ring structure.

Study Check: Cyclic Forms of D-Galactose

  • Draw the cyclic form of D-galactose.

Solution Steps for D-Galactose

  • Step 1: Rotate the Fischer projection 90° clockwise.
  • Step 2: Fold the horizontal carbon chain into a hexagon.
    • Rotate groups on carbon 5 and bond the oxygen from carbon 5 to carbon 1.
    • Place the carbon 6 group above the ring.
    • Position the —OH group on carbon 2 below the ring and the —OH groups on carbons 3 and 4 above the ring.
  • Step 3: Position the new —OH group on carbon 1:
    • Below the ring for α-D-galactose.
    • Above the ring for β-D-galactose.
    • Distinctions between α-D-galactose and β-D-galactose are emphasized in drawing their respective structures.