OpenStax Organic Chemistry: A Tenth Edition Student Solutions Manual
Chapter 25 – Biomolecules: Carbohydrates
Solutions to Problems
25.1
(a) (b) (c) (d)
25.2 Horizontal bonds of Fischer projections point out of the page, and vertical bonds point
into the page.
(a)
(b)
(c)
25.3 To decide if two Fischer projections are identical, use the two allowable rotations to
superimpose two groups of each projection. If the remaining groups are also
superimposed after rotation, the projections represent the same enantiomer.
(a) Since –H is in the same position in both A and B, keep it steady, and rotate the
other three groups. If, after rotation, all groups are superimposed, the two
projections are identical. If only two groups are superimposed, the projections are
enantiomers. Thus, A is identical to B.
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, OpenStax Organic Chemistry: A Tenth Edition Student Solutions Manual
Projections A, B and C are identical, and D is their enantiomer.
25.4 Rotate the structure 180° around the horizontal axis to arrive at a drawing having the
hydrogen at the rear. Assign the R,S configuration as usual, and draw the Fischer
projection
25.5 Draw the skeleton of the Fischer projection and add the –CHO and –CH2OH groups to
the top and bottom, respectively. Look at each carbon from the direction in which the –H
and –OH point out of the page, and draw what you see on the Fischer projection.
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, OpenStax Organic Chemistry: A Tenth Edition Student Solutions Manual
25.6 The hydroxyl group bonded to the chiral carbon farthest from the carbonyl group points
to the right in a D sugar, and points to the left in an L sugar.
(a) (b) (c)
25.7
25.8
(a) (b) (c)
25.9 An aldoheptose has 5 chirality centers. Thus, there are 25 = 32 aldoheptoses – 16 D
aldoheptoses and 16 L aldoheptoses.
25.10 See Problem 25.5 for the method of solution.
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Chapter 25 – Biomolecules: Carbohydrates
Solutions to Problems
25.1
(a) (b) (c) (d)
25.2 Horizontal bonds of Fischer projections point out of the page, and vertical bonds point
into the page.
(a)
(b)
(c)
25.3 To decide if two Fischer projections are identical, use the two allowable rotations to
superimpose two groups of each projection. If the remaining groups are also
superimposed after rotation, the projections represent the same enantiomer.
(a) Since –H is in the same position in both A and B, keep it steady, and rotate the
other three groups. If, after rotation, all groups are superimposed, the two
projections are identical. If only two groups are superimposed, the projections are
enantiomers. Thus, A is identical to B.
1 10/4/2023
, OpenStax Organic Chemistry: A Tenth Edition Student Solutions Manual
Projections A, B and C are identical, and D is their enantiomer.
25.4 Rotate the structure 180° around the horizontal axis to arrive at a drawing having the
hydrogen at the rear. Assign the R,S configuration as usual, and draw the Fischer
projection
25.5 Draw the skeleton of the Fischer projection and add the –CHO and –CH2OH groups to
the top and bottom, respectively. Look at each carbon from the direction in which the –H
and –OH point out of the page, and draw what you see on the Fischer projection.
11/6/2023 2
, OpenStax Organic Chemistry: A Tenth Edition Student Solutions Manual
25.6 The hydroxyl group bonded to the chiral carbon farthest from the carbonyl group points
to the right in a D sugar, and points to the left in an L sugar.
(a) (b) (c)
25.7
25.8
(a) (b) (c)
25.9 An aldoheptose has 5 chirality centers. Thus, there are 25 = 32 aldoheptoses – 16 D
aldoheptoses and 16 L aldoheptoses.
25.10 See Problem 25.5 for the method of solution.
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