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Chapter 15 Carbohydrates

Chapter 15 Carbohydrates. 15.3 Haworth Structures of Monosaccharides. Haworth Structures. Haworth structures are the prevalent form of monosaccharides with 5 or 6 carbon atoms form when the hydroxyl group on carbon 5 (C5) reacts with the aldehyde or ketone group.

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Chapter 15 Carbohydrates

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  1. Chapter 15 Carbohydrates 15.3Haworth Structuresof Monosaccharides

  2. Haworth Structures Haworth structures • are the prevalent form of monosaccharides with 5 or 6 carbon atoms • form when the hydroxyl group on carbon 5 (C5) reacts with the aldehyde or ketone group

  3. Haworth Structures (continued) Haworth structures • are cyclic hemiacetals • form when the C═O group and the —OH are part of the same molecule • of hexoses form when the —OH on C5 reacts with a C═O group • of a D-isomer place the —CH2OH of C6 above the ring

  4. Guide to Drawing Haworth Structures

  5. Drawing the Haworth Structures for Glucose STEP 1Turn the open carbon chain clockwise by 90°.

  6. Drawing the Haworth Structures for Glucose (continued) STEP 2Fold the carbon chain into a hexagon by: • moving C5 (with C6) above C3 • bonding the C5 —OH to C═O of C1 • writing the —OH groups on C2 and C4 below the ring • writing the —OH group on C3 above the ring • writing a new —OH on C1

  7. Drawing the Haworth Structures for Glucose (continued) STEP 2continued

  8. Drawing the Haworth Structures for Glucose (continued) • STEP 3The new —OH on C1: • forms two isomers called anomers • is drawn down for the α anomer • Is drawn up for the β anomer

  9. -D-Glucose and β-D-Glucose in Aqueous Solution When placed in an aqueous solution, • cyclic structures open and close • -D-glucose converts to β-D-glucose and back • there is only a small amount of open chain

  10. Haworth Structures of Fructose Fructose • is a ketohexose • forms Haworth structures when the —OH on C5 bonds to the C═O on C2

  11. Learning Check Draw the Haworth structure of -D-galactose.

  12. Solution STEP 1 Turn the open carbon chain clockwise by 90°. OH H H HO H H

  13. Solution (continued) STEP 2 Fold the carbon chain into a hexagon. HO HO H OH H H

  14. Solution (continued) STEP 3Draw the new —OH on C1 down for the  anomer.

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